Theorem (1 + 1.75) on the Goldbach conjecture

Abstract.

Every sufficiently large even integer $N$ has a representation $N=p+rq$, where $p,q$ are prime, $r=1$ or prime, and $r\leq q^{a-1}=q^{3/4}$. In the terminology of Li and Liu, this proves Proposition $(1+a)$ for $a=1.75=7/4$. A weighted sieve combines thirteen signed row estimates with a switched upper bound. The directed comparison gives $D_{B}>1.967088903489691335582703173494$ and $c_{K2,G0}I(b)<1.966859949178424046530804965344$. The exact margin enclosure is $D_{B}-c_{K2,G0}I(b)>0.000228954311267289051898208151$.

1. Introduction

We represent large even integers by two primes and one factor. Write $\mathbb{P}$ for the primes. Chen’s theorem represents every sufficiently large even integer as a prime plus an integer with at most two prime factors [5]. Paper II supplies the quantitative smaller-factor starting point and the signed sieve sums used here. We change the exterior and switched coefficients and add retained and sixth-sum gains. Table 1.1 lists their separate costs. We use the terminology Proposition $(1+a)$ introduced by Li and Liu [19, p. 2, (1.6)]. Runbo Li proved lower bounds for $D_{1,2}(N)$ [29]; we do not use that result.

Cai established a small-prime variant of Chen’s theorem [4]. Wu developed Chen’s double sieve for Goldbach and twin-prime problems [30]. His later paper continues this double-sieve program [31]. Lichtman modified the linear sieve and studied bounds for the twin-prime counting function [20]. He also proved results on large-modulus distribution and Goldbach beyond the square-root barrier [21]. Maynard’s first paper proves large-modulus estimates for fixed residue classes [22]. His second gives well-factorable estimates in that setting [23]. His third treats estimates uniform over residue classes [24]. Pascadi studies exponents of distribution for primes and smooth numbers [28]. Li and Liu proved Proposition $(1+1.9)$, with a lower bound for $D_{1,1.9}(N)$ [19, Theorem 1.1]. The present proof does not use their estimates.

Theorem 1 (Representation theorem).

Let $\sigma_{*}$ be a rational number satisfying (9.14), and let $N_{0}^{*}(\sigma_{*})$ be the threshold (2.5). Then every even $N\geq N_{0}^{*}(\sigma_{*})$ admits

(1.1) $$N=p+rq,\qquad p,q\in\mathbb{P},\qquad r\in\{1\}\cup\mathbb{P},\qquad r\leq q^{a-1}=q^{3/4}.$$

Section 5 proves the retained comparison (A1) of Table 1.2. Section 6 proves the sixth-sum estimate (A2). Section 7 proves the exterior comparison (A3). Section 8 proves the switched comparison (A4). The constant bounds (A5) follow from the cited Paper I and Paper II estimates, including Paper I, Theorem 2.6, (2.31) [26, Theorem 2.6, (2.31)]. Hence a rational $\sigma_{*}$ as in the theorem exists; $N_{0}^{*}(\sigma_{*})$ may be ineffective.

The proof uses Paper I, [26, Theorems 2.6, 2.8, 3.2 and 4.1], and Paper II, [27, (1.7)–(1.12), (6.4)–(6.4d)]. We restate each input below with its precise scope.

1.1. A positive count

Put

(1.2) $$\varepsilon=\frac{1}{10000},\qquad a=\frac{7}{4},\qquad b=\frac{3}{7}-\frac{1}{10^{6}}=\frac{2999993}{7000000}.$$

Define

(1.3) $$\mathcal{A}_{N} =\{N-p:p\in\mathbb{P},\ p<(1-\varepsilon)N\},$$

$$\mathcal{P}_{b} =\mathbb{P}\cup\{rq:r,q\in\mathbb{P},\ r\leq q,\ r<N^{b}\},$$

$$G_{b}(N) =\#(\mathcal{A}_{N}\cap\mathcal{P}_{b}).$$

For a counted semiprime,

(1.4) $$\begin{gathered}q>\varepsilon N^{1-b},\\ q^{3/4}>\varepsilon^{3/4}N^{b+\kappa_{\mathrm{tr}}}\geq N^{b}>r\qquad(N\geq N_{\mathrm{tr}}).\end{gathered}$$

The transfer parameters are defined in (2.3), and Lemma 9.2 proves (1.4). A counted prime gives $r=1$. Hence $G_{b}(N)>0$ implies (1.1).

Use Paper II’s normalization in (2.2) [27, (1.2), §6.1]. The convergent product gives $C_{2}>0$. Thus $W(N)>0$.

1.2. The comparisons

Use the sums of Paper II [27, (1.7)–(1.11)]. Put

(1.5) $$R={} 3S_{1}-2S_{3}-S_{4}-S_{5}-S_{7}+S_{8}+S_{9}+S_{10}-2S_{11}-S_{12}-S_{13},$$

$$\Sigma_{W}={} R+S_{2}-S_{6}-S_{\mathrm{sw}},$$

$$B_{b}(N)={} \sum_{\begin{subarray}{c}r,q\in\mathbb{P}\\ N^{b}\leq r\leq\sqrt{N}\\ \varepsilon N<rq<N\end{subarray}}1_{N-rq\in\mathbb{P}}.$$

Here $S_{2}$ is the complete Paper II sum. All original coprimality conditions remain imposed. The last sum counts each eligible ordered prime pair; both orientations are counted only when both satisfy the displayed cutoff, and a square is counted once. Here $R=R_{\mathrm{fix}}$ and $\Sigma_{W}=\mathcal{W}$ in Section 4.

Write $\overline{J}_{6}$ for Paper II’s sixth upper endpoint [27, (6.1), (6.3a)]. For the optional sharper arithmetic comparison, record

(1.6) $$\begin{gathered}I(b)=\log\frac{1-b}{b},\\ c=\frac{6834282367132846348358930380956825846442683629}{10^{45}}.\end{gathered}$$

Both the exterior and switched comparisons use the common coefficient

(1.7) $$c_{K2,G0}=\frac{6836825163875517756286846321650754372931094689}{10^{45}}.$$

Its definition and source comparison are (7.36) and (7.67). Put

(1.8) $$D_{\mathrm{II}} =\frac{45849916338009170593}{25000000000000000000},$$

$$\overline{H} =\frac{389253}{5000000},$$

$$M_{5,\mathrm{internal}} =-K_{\mathrm{hi}}\frac{6073714261}{10^{9}}-L_{G1},$$

$$g_{\mathrm{RA,lo}} =\frac{989598353}{10^{10}},$$

$$\Delta D_{L3} =\left(2-\frac{c_{K2,G0}}{4}\right)\overline{H}.$$

The two mass constants are

(1.9) $$K_{\mathrm{hi}} =\frac{3989480791070578466390443009830933753748442947478110112223}{10^{58}},$$

$$L_{G1} =-\frac{243950428265194656254206921939947762868871361874409552095097}{100433627766186892221372630771322662657637687111424552206336}.$$

These are Paper II, (A.10d) and (4.84), respectively [27, (A.10d), (4.84)]. The constant $\overline{H}$ bounds the switched integral. It differs from the packing ceiling $153/256$. The exact rational $c_{\mathrm{O1}}$ is (8.60), with Table 8.1. It and $c$ belong to the optional certificate comparison only; neither is used to price the main count. The sixth-sum comparison uses the certified floor $g_{\mathrm{RA,lo}}$. The integral gain is denoted $\mathfrak{g}_{\mathrm{RA}}$, with certified enclosure $[0.0989598353353,0.0989598353355]$ in Section 6. Thus $g_{\mathrm{RA,lo}}$ is rational and $g_{\mathrm{RA,lo}}\leq\mathfrak{g}_{\mathrm{RA}}$. It is neither the integral nor its central estimate.

The required comparisons are

(1.10) $$4G_{b}(N) \geq\Sigma_{W}-4B_{b}(N)-W(N)E_{\mathrm{exc}}(N),$$

$$\frac{R+S_{2}}{W(N)} \geq D_{\mathrm{II}}+\overline{J}_{6}+2\overline{H}+\frac{5719}{10^{6}}$$

$$\qquad+M_{5,\mathrm{internal}}-E^{\prime}(N;\sigma),$$

$$\frac{S_{6}}{W(N)} \leq\overline{J}_{6}-g_{\mathrm{RA,lo}}+E_{\mathrm{RA}}(N;\sigma),$$

$$\frac{S_{\mathrm{sw}}}{W(N)} \leq\frac{c_{K2,G0}}{4}\overline{H}+E_{\mathrm{sw}}(N;\sigma).$$

These errors are nonnegative by (5.2), (6.72) and (8.71); the first includes the inherited envelope $E_{\mathrm{van}}$. Use the following error identifications:

(1.11) $$E_{\mathrm{exc}}(N) =\frac{4E_{\mathrm{pt}}(N)}{W(N)},$$

$$E_{\mathrm{sw}}(N;\sigma) =C_{\mathrm{sw}}\sigma+P_{\mathrm{sw}}\sigma^{9}+V_{\mathrm{sw}}(N;\sigma).$$

Here $E_{\mathrm{pt}}$ is (4.12). The switched quantities are (8.61) and (8.71). Their estimates form input (A4). The retained error includes the conversion errors:

(1.12) $$E^{\prime}(N;\sigma)={} E_{\mathrm{van}}(N;\sigma)+\sum_{j\in\mathcal{H}}\bigl{(}|\Lambda^{0}_{j,N}|+|\Lambda_{j,N}|\bigr{)}$$

$$+|\Lambda_{G,N}|.$$

The row set $\mathcal{H}$ is defined in (5.2). Its two qualified labels denote $S_{5}$’s lower part and retained $S_{10}$. The inherited error $E_{\mathrm{van}}$ is (5.11); the saved-array contribution $\Lambda^{0}_{j,N}$ is included once, as in (5.2). The differences $\Delta_{j}$ are (5.22). The conversion arrays are (5.35) and (5.56); the equivalent notation $\Lambda_{j,N}=\Lambda_{N}(\Delta_{j})$ and $\Lambda_{G,N}=\Lambda_{N}(\Delta_{G_{1}})$ is fixed in (5.3). The saved-array identifications and conversion bounds are proved in Section 5, as part of (A1). The two finite integral bounds are Propositions B.1 and B.2.

The exterior estimate required is

(1.13) $$-\frac{4B_{b}(N)}{W(N)}\geq-c_{K2,G0}I(b)-E_{K}(N;b,\sigma),\qquad E_{K}(N;b,\sigma)\geq 0.$$

Its convolution input is Paper II, [27, Lemma 5.1, (5.44)–(5.52)].

Define the constant by

(1.14) $$\boxed{\begin{aligned} D_{B}={}&D_{\mathrm{II}}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}\\ &+g_{\mathrm{RA,lo}}-\frac{1}{10000}+\Delta D_{L3}.\end{aligned}}$$

Table 1.1 records each summand. Its decimals are rounded; the defining formulas are exact.

Table 1.1. Components of $D_{B}$

Component	Value	Definition
$D_{\mathrm{II}}$	$1.833996653520367$	(1.8)
$5719/10^{6}$	$0.005719$	(5.25)
$M_{5,\mathrm{internal}}$	$0.005874949945026$	(1.8), (1.9)
$g_{\mathrm{RA,lo}}$	$0.0989598353$	(1.8)
$-1/10000$	$-0.0001$	(1.14)
$\Delta D_{L3}$	$0.022638464724298$	(1.8)
$D_{B}$	$1.967088903489691$	(1.14)
$1.967088903489691335582703173494<D_{B}$
$D_{B}<1.967088903489691335582703173495$

Suppose the optional six-endpoint estimate is available. Then Proposition 5.10 would add $g_{6}=0.0020549454896034927680\ldots$ of (5.18) to the retained comparison. The assembled retained-count lower bound below would then increase by $g_{6}$. Theorem 1 does not use this. Let

(1.15) $$\mathcal{E}_{B}=E^{\prime}+E_{\mathrm{RA}}+E_{\mathrm{sw}}+E_{K}+E_{\mathrm{exc}},\qquad\delta_{\mathrm{err}}=\frac{1}{10000}.$$

Adding the established comparisons gives

(1.16) $$\frac{4G_{b}(N)}{W(N)}\geq D_{B}+\delta_{\mathrm{err}}-c_{K2,G0}I(b)-\mathcal{E}_{B}(N;b,\sigma).$$

By (A1)–(A5), Proposition 9.1 proves

(1.17) $$\mathcal{E}_{B}(N;b,\sigma_{*})\leq\delta_{\mathrm{err}}\qquad(N\geq N_{0}^{*}(\sigma_{*}),\ N\in 2\mathbb{Z}).$$

Table 1.2 lists the five inputs, their exact statements and their contributions to the final constant. The first-line source count in (1.10) is (4.13), valid for $N\geq N_{\mathrm{pt}}=10^{12}+1$. Lemma 9.2 gives $N_{\mathrm{pt}}<N_{\mathrm{tr}}\leq N_{0}^{*}(\sigma_{*})$.

Table 1.2. Analytic inputs and their contributions

Input	Exact statement and use	Contribution and error
A1	Retained comparison (5.2) with $g_{6}$ replaced by $0$, Proposition 5.10, at $\sigma=\sigma_{*}$; used in line 2 of (1.10); proved in Section 5.	$+5719/10^{6}+M_{5,\mathrm{internal}}$; error $E^{\prime}$.
A2	Proposition 6.12 at $\sigma=\sigma_{*}$, giving the sixth-sum estimate (6.72) of Proposition 6.13; used in line 3 of (1.10).	$+g_{\mathrm{RA,lo}}$; error $E_{\mathrm{RA}}$.
A3	Proposition 7.4 at $\sigma=\sigma_{*}$, giving the exterior estimate (7.67) of Proposition 7.5; used in (1.13).	$-c_{K2,G0}I(b)$; error $A_{K}C_{\mathrm{AA}}\sigma+P_{K}\sigma^{9}+r_{K}$.
A4	The sieve data and the estimates (8.8), (8.53) and (8.62) of Section 8 for the source (8.3), at $\sigma=\sigma_{*}$, giving the switched estimate (8.72) and the limit (8.73); used in line 4 of (1.10). This application uses Paper I, Theorem 2.8, at its stated inner spectral stage.	$+\Delta D_{L3}$; error $C_{\mathrm{sw}}\sigma+P_{\mathrm{sw}}\sigma^{9}+V_{\mathrm{sw}}$.
A5	The constants $C_{E}$ in (8.25) and $C_{\mathrm{AA}}$ in (7.39) are finite, positive and independent of $\sigma$ and $N$, and $C_{E}\geq C_{E}^{\mathrm{II}}$.	Fixes the caps in (9.14).

The constants $C_{E}$ and $C_{\mathrm{AA}}$ are distinct. Here $C_{E}$ is the common maximum of the switched and Paper II sieve constants. The four-row K2 error constant is $C_{\mathrm{AA}}=4C_{\mathrm{AA37}}$.

The exact constants satisfy

(1.18) $$0.000228954311267289051898208151 <D_{B}-c_{K2,G0}I(b),$$

$$D_{B}-c_{K2,G0}I(b) <0.000228954311267289051898208152,$$

$$1.749974884137892433433207484622 <a_{\mathrm{buf}}^{*},$$

$$a_{\mathrm{buf}}^{*} <1.749974884137892433433207484623<\frac{7}{4}.$$

The last constant is defined in (9.26). The rational bounds are shown in Section 9. They need the stated analytic comparisons; the arithmetic alone gives no positive-count estimate. The value $a^{*}_{\mathrm{buf}}$ is a scalar diagnostic for this fixed $b$. It does not prove the theorem for smaller $a$; that would require comparisons at a different cutoff. The optional sharper scalar comparison (9.29) uses a separate constant $D_{B,c}$; it does not enter the main count.

1.3. Order of the parameters

Fix $b$ and all absolute error constants first. Choose one sufficiently small rational $\sigma_{*}$. Then put $\tau_{*}=(\sigma_{*})^{9}$. Freeze the finite partitions and all coefficient bounds next. Fix every logarithmic precision before the common threshold. Thus the order is

(1.19) $$\begin{gathered}\text{absolute constants}\ \longrightarrow\ \sigma_{*}\in\mathbb{Q}\cap(0,10^{-10}]\ \longrightarrow\ \tau_{*}=(\sigma_{*})^{9},\\ \longrightarrow\ \text{partitions, coefficient bounds and precisions}\ \longrightarrow\ N_{0}^{*}(\sigma_{*})\ \longrightarrow\ N.\end{gathered}$$

The common threshold is defined in (2.5). It includes every length, structural and error threshold. Every limit is taken at fixed $\sigma_{*}$. These thresholds exist by (A1)–(A5) and Proposition 9.1. The final section combines the comparisons and the error bounds. For the main construction, the comparison estimates used are:

1. (A1): the actual retained-source comparison in Section 5, including all saved-array conversions. It supplies $5719/10^{6}+M_{5,\mathrm{internal}}$ beyond the original retained bound.

2. (A2): the sixth-sum distribution and coefficient estimates in Section 6, supplying $g_{\mathrm{RA,lo}}$.

3. (A3): the actual semiprime-source distribution, principal-term and peel estimates in Section 7, giving the exterior cost $c_{K2,G0}I(b)$.

4. (A4): the switched-source comparison, packing, distribution and peel estimates in Section 8, giving $\Delta D_{L3}=(2-c_{K2,G0}/4)\overline{H}$.

Each requirement includes its exact nonnegative error estimates and the fixed-$\sigma_{*}$ limits allocated in Section 9. The source-counting identity and pointwise bound are supplied in Section 4. The list only records dependencies. The cited sections prove the estimates, with precise scopes and error bounds; the scalar arithmetic here combines them.

The use of AI tools is described at the end.

2. Notation and parameters

All logarithms are natural. Put $L=\log N$. The integer $N$ is even throughout. Table 2.1 fixes the common notation.

Table 2.1. Common notation

Symbol	Meaning
$\mathbb{P}$	The set of primes.
$\varphi(n),d(n)$	Euler’s totient and divisor function. The inherited notation $\tau(n)=d(n)$ has an integer argument; the scalar $\tau=\sigma^{9}$ has none.
$\mu_{\mathrm{Mob}}(n),\mu(n)$	The Möbius function; use $\mu_{*}$ for the sieve cutoff.
$\Omega(n)$	Prime factors counted with multiplicity.
$P^{-}(n)$	Least prime factor; $P^{-}(1)=\infty$.
$\mathbf{1}_{E}$	Indicator of $E$.
$\gamma,\gamma_{E}$	Euler’s constant, with $\gamma_{E}=\gamma$.
$a,b$	Representation exponent and smaller-factor cutoff.
$\varepsilon$	Fixed truncation parameter.
$\alpha,\beta,\mu_{*},\lambda,t_{0},u$	Fixed sieve exponents in (2.1); cutoff $\mu=\mu_{*}$.
$C(N),U_{C}(N),W(N)$	Normalizations in (2.2).
$\sigma,\tau,\rho$	Partition parameters: $\tau=\sigma^{9}$, $\rho=1+\tau$.
$\sigma_{*},\tau_{*},\rho_{*}$	Their fixed choices for the final argument.
$\kappa_{\mathrm{tr}},N_{\mathrm{tr}}$	Transfer parameters in (2.3).
$\delta_{\mathrm{err}}$	Fixed total error bound, $10^{-4}$.
$\delta_{\mathrm{pk}}$	The parameter in the factor-partition theorem.
$h_{G},h,\delta_{h}$	Profile displacement $h_{G}=9/5000$; $h=h_{G}$ in (5.1), $\delta_{h}=h_{G}/\alpha$.
$\epsilon_{s},\delta_{s}$	Local sieve parameters, $\delta_{s}=\epsilon_{s}^{9}$; applications use $\epsilon_{s}=\sigma$.
$\eta_{L}$	The level margin, $10^{-4}$; separate from $\varepsilon$ and $\delta_{\mathrm{err}}$.
$r,q$	In (1.1), $q$ is prime and $r$ is either $1$ or prime; other uses are locally bound variables.
$c,I(b)$	Optional sharper coefficient and the logarithm in (1.6).
$c_{K2,G0}$	Common K2/L3 coefficient in (1.7).
$C_{E},C_{\mathrm{AA}}$	Error constants in (8.25) and (7.38).
$K_{\mathrm{SW}}(A),C_{\mathrm{K2},\sigma},C_{\mathrm{L3},\sigma}$	The logarithmic Siegel–Walfisz level and the K2 and L3 coefficient bounds; each is local to its stated application.
$R,R_{\mathrm{fix}}$	The signed sum in (1.5); elsewhere $R$ is a local level or remainder.
$\Sigma_{W},\mathcal{W}$	Equal signed source sums; see (4.5).
$W_{13},W_{rs}$	Arithmetic source weight and spectral operator, respectively. Neither denotes $W(N)$.
$\overline{H},\overline{H}_{L}$	Equal switched-integral bounds, $389253/5000000$.
$H_{\mathrm{pk,max}},H_{\max}$	Equal packing ceilings, $153/256$.
$D_{\mathrm{II}},\overline{J}_{6}$	Original Paper II lower constant and sixth upper endpoint; (1.8), (6.59).
$g_{\mathrm{RA,lo}}$	Certified floor $0.0989598353$, used in (1.14).
$E^{\prime},E_{\mathrm{van}}$	Complete retained error and inherited envelope; (1.12), (5.11).
$\begin{gathered}\Lambda_{j,N},\\ \Lambda_{G,N}\end{gathered}$	Retained conversion differences, defined in (5.35) and (5.56).
$\Delta_{G},\Lambda_{G,N}$	Joint-profile gain and conversion difference in (5.56). The spelling $G1$ in subscripts denotes the same group.
$\begin{gathered}E_{\mathrm{RA}},E_{\mathrm{sw}},\\ E_{K},E_{\mathrm{exc}}\end{gathered}$	Nonnegative complete errors in (1.10)–(1.11) and (1.13).
$\mathcal{E}_{B}$	Their total, including $E^{\prime}$, in (1.15).
$D_{B}$	The constant in (1.14); evaluated in (9.1).
$\mathcal{A}_{N},\mathcal{P}_{b},G_{b},B_{b}$	Shifted-prime source, admissible factors, positive count and exterior count in (1.3), (1.5).
$S_{1},\ldots,S_{13},S_{\mathrm{sw}}$	Paper II arithmetic sums retained in (1.5); definitions in Section 4.
$\mathfrak{g}_{\mathrm{RA}},g_{\mathrm{RA,lo}}$	Integral gain and its smaller rational floor; only the floor enters $D_{B}$.
$\begin{gathered}g_{6},M_{5,\mathrm{internal}},\\ K_{\mathrm{hi}},L_{G1}\end{gathered}$	Optional six-endpoint gain, shifted-profile gain and mass constants; the main proof sets $g_{6}=0$, see (5.18), (1.8), (1.9).
$\Delta D_{L3}$	Switched gain $(2-c_{K2,G0}/4)\overline{H}$.
$G_{*},G_{0},U_{0},c_{\mathrm{O1}}$	Common unshifted K2/L3 functional; coefficient cap $G_{0}=6088440363/10^{9}$; optional sharper upper certificate $U_{0}$; its exact coefficient in (8.60).
$\mathcal{H},\Lambda^{0}_{j,N}$	Retained row set and saved-array contribution in (5.2).
$\mathcal{E},E_{\mathrm{pt}}$	Pointwise exceptional set and error in (4.9), (4.12).
$N_{0}^{*},N_{\mathrm{floor}},T_{\bullet}$	Common threshold, numerical floor and named threshold requirements in (2.5).
$\zeta,a_{\mathrm{buf}}^{*},\mathfrak{E}_{*}$	Buffer $10^{-6}$, scalar endpoint and its exponential in (9.26).
$W_{\mathrm{AA},K},W_{*}$	K2 source-weight bound in (7.38) and Paper II bound $10^{14}$; these are distinct.
$A_{K},P_{K},r_{K}$	K2 error multiplier, peel multiplier and vanishing remainder in (9.8).
$C_{\mathrm{sw}},P_{\mathrm{sw}},V_{\mathrm{sw}}$	Switched error multipliers and positive remainder in (8.61), (8.71).
$L,g_{\mathrm{AA}}$	$\log N$ and Paper II’s fixed error allowance in (9.15).

Cusps carry the index $\mathfrak{a}$ in spectral sums. The switched blocks carry the index $\mathfrak{b}$. Partition cells are $c$ or $\mathfrak{c}$. Coefficient sequences are $a_{n},b_{m}$, with dummy indices interchanged when needed. Spectral moduli and cusp numerators are locally bound variables. They do not redefine the constants $c$ and $u$. Packing coordinates also have local scope. Bare $T$ and $w$ are local endpoints or weights; the named saved and root-A scales are $T_{\mathrm{save}},T_{\mathrm{int}}$, and $w_{\mathrm{RA}}$. The exponent $A_{0}>0$ in distribution estimates is a logarithmic precision. The normalization factor is written $2e^{-\gamma}$.

2.1. Fixed parameters

Keep $a,b,\varepsilon$ from (1.2). The sieve parameters are

(2.1) $$\alpha =\frac{100}{1149}, \beta =\frac{50}{309}, \mu_{*} =\frac{25}{128},$$

$$\lambda =\frac{93}{400}, t_{0} =\frac{57}{224}, u =\frac{1}{500}.$$

These follow Paper II’s notation [27, (1.5)]; $\varepsilon$ is $\varepsilon_{A}$ in [27, (1.3)]. Put

(2.2) $$C_{2} =\prod_{\begin{subarray}{c}\ell\in\mathbb{P}\\ \ell>2\end{subarray}}\left(1-\frac{1}{(\ell-1)^{2}}\right),$$

$$C(N) =C_{2}\prod_{\begin{subarray}{c}\ell\in\mathbb{P},\ \ell\mid N\\ \ell>2\end{subarray}}\frac{\ell-1}{\ell-2},$$

$$U_{C}(N) =\frac{C(N)N}{L^{2}},\qquad W(N)=4(1-\varepsilon)U_{C}(N).$$

We often suppress the argument $N$.

Define the transfer parameters by

(2.3) $$\kappa_{\mathrm{tr}} =a(1-b)-1=\frac{7}{4000000},$$

$$N_{\mathrm{tr}} =\varepsilon^{-(a-1)/\kappa_{\mathrm{tr}}}=\varepsilon^{-3000000/7}.$$

Here $a-1=3/4$ is the exponent on the larger prime in the bound for the smaller factor. The transfer argument appears in the final section.

2.2. Order of choice

First fix all $\sigma$-independent constants in the estimates. These include $C_{E}$ in (8.25) and $C_{\mathrm{AA}}$ in (7.38). Then choose one rational $\sigma_{*}$, with

(2.4) $$0<\sigma_{*}\leq 10^{-10},\qquad\tau_{*}=(\sigma_{*})^{9},\qquad\rho_{*}=1+\tau_{*}.$$

The final section gives the remaining upper restrictions. Apply every partition estimate with $\sigma=\sigma_{*}$. Freeze its finite partitions and coefficient bounds next. Freeze the resulting finite-family bounds and logarithmic precisions too. These choices precede the common threshold.

Use the following named thresholds. Each requires its listed statement for every even $N$ above it, at the fixed $\sigma_{*}$. The cited comparison propositions establish these statements. The numerical floor supplies none of them. The resulting threshold exists but may be ineffective.

1. $T_{\mathrm{src}}$: the actual-source identity and first line of (1.10), including the pointwise hypotheses in (4.11) and (4.13). Include the explicit pointwise threshold $N_{\mathrm{pt}}=1+\lceil\max\{2^{1/\alpha},\varepsilon^{-3},\varepsilon^{-1/(1-\lambda-\beta)}\}\rceil$ from Section 4.

2. $T_{\mathrm{ret}}$: the retained comparison, saved-array identifications and nonnegativity of $E_{\mathrm{van}}$ in Section 5.

3. $T_{\mathrm{RA}}$: the sixth-sum comparison (6.72), with every distribution and coefficient bound specified in Section 6. It also covers the finite source threshold $N_{6}$, chosen so that every even $N\geq N_{6}$ satisfies

   $$N^{1/6}>L^{\max(B_{L},B_{\mathrm{tail}})},\quad N^{v_{\mathrm{RA}}-1/3}>L^{B},\quad\varepsilon N^{1/2}>N^{\mu_{*}},\quad NL^{-333}\geq N^{\mu_{*}}.$$

   Thus the operative $T_{\mathrm{RA}}$ is the maximum of the sixth-sum comparison threshold and this $N_{6}$.

4. $T_{K}$: the actual-source exterior comparison (7.67) and its nonnegative error decomposition, the $r_{K}$ formula displayed after (7.67).

5. $T_{\mathrm{sw}}$: the switched comparison (8.72) and the nonnegative decomposition (8.71).

6. $T_{\mathrm{len}}$: all scale-conversion, zero-mode, factorization, packing, interval-length and logarithmic-precision conditions of the preceding applications, with the fixed finite partitions. In particular, require $N^{\sigma_{*}}\geq 2$, $N^{60\sigma_{*}2^{-126}}\geq 12L^{334}$, (7.61) and (8.46), and include $N_{\mathrm{dom}},N_{z,0},N_{\mathrm{LEN,app}}$ from (8.75). The remaining application conditions are exactly those of the cited comparison propositions. The floor implies none of them.

7. $T_{E^{\prime}},T_{V},T_{E_{\mathrm{RA}}},T_{r},T_{E_{\mathrm{exc}}}$: respectively the eventual bounds

   $$\begin{gathered}E^{\prime}\leq\frac{\delta_{\mathrm{err}}}{20},\qquad V_{\mathrm{sw}}\leq\frac{\delta_{\mathrm{err}}}{20},\qquad E_{\mathrm{RA}}\leq\frac{\delta_{\mathrm{err}}}{10},\\ r_{K}\leq\frac{\delta_{\mathrm{err}}}{10},\qquad E_{\mathrm{exc}}\leq\frac{\delta_{\mathrm{err}}}{10}.\end{gathered}$$

   All arguments are $(N;\sigma_{*})$, or $(N;b,\sigma_{*})$ for $r_{K}$.

These are finitely many requirements after all finite choices. Define the common threshold once by

(2.5) $$\mathcal{T}_{*}={} \{T_{\mathrm{src}},T_{\mathrm{ret}},T_{\mathrm{RA}},T_{K},T_{\mathrm{sw}},T_{\mathrm{len}},$$

$$\hskip 34.1433ptT_{E^{\prime}},T_{V},T_{E_{\mathrm{RA}}},T_{r},T_{E_{\mathrm{exc}}}\},$$

$$N_{0}^{*}(\sigma_{*}) =2\left\lceil\frac{1}{2}\max\bigl{(}\{N_{\mathrm{tr}},N_{\mathrm{floor}}\}\cup\mathcal{T}_{*}\bigr{)}\right\rceil,$$

$$N_{\mathrm{floor}} =2\left\lceil e^{2^{150}}/2\right\rceil.$$

The floor is a convention added to Paper I’s selected-interval length hypothesis [26, §3.2, length hypothesis preceding (3.12)]. It proves none of the listed estimates. The named application estimates establish this threshold. The resulting $N_{0}^{*}$ may be ineffective.

This implements the order in (1.19). Afterwards, $N$ is even and $N\geq N_{0}^{*}(\sigma_{*})$.

Let $\epsilon_{\mathrm{err}}(N;\sigma_{*})$ be a real- or complex-valued error function. Write $\epsilon_{\mathrm{err}}(N;\sigma_{*})=o_{\sigma_{*}}(1)$ if and only if

(2.6) $$\begin{gathered}\forall\eta>0\ \exists N_{\eta}(\sigma_{*})\ \forall N\in 2\mathbb{N},\\ N\geq N_{\eta}(\sigma_{*})\ \Longrightarrow\ |\epsilon_{\mathrm{err}}(N;\sigma_{*})|\leq\eta.\end{gathered}$$

All preceding fixed data are suppressed here. The parameter $\sigma_{*}$ remains fixed in this limit. Uniformity in other variables is stated with each estimate.

3. Classical inputs

We use counting-measure $\ell^{2}$ norms. Write $n\sim Y$ for $Y<n\leq 2Y$. The symbols $\mu_{\rm Mob},\Lambda,\varphi$ have their usual meanings. Write $d(n)$ for the divisor function. The parameters in each fact are local. Put $P^{-}(1)=\infty$ and $e(t)=\exp(2\pi it)$. Statements (P1.x) and (P2.x) restate results proved in Papers I and II, respectively, with their original scope. The structured sieve result is Paper I, Theorem 2.6. Its classical source is Iwaniec’s Theorem 4 [13, Theorem 4]. The $S$-form displayed below specializes the $W_{R}^{\pm}$ estimates of Paper I, Theorem 2.6. It follows from them by $S\leq W_{R}^{+}$ and $S\geq W_{R}^{-}$. Theorem 4 does not state it verbatim. Each new application needs its own coefficient, support and uniformity checks.

Fact 3.1 ((F1) Prime estimates).

For every $A>0$, one may take $B_{\psi}(A)=A+60$ in

(3.1) $$\begin{gathered}\sum_{q\leq Q}\max_{(a,q)=1}\left|\sum_{\begin{subarray}{c}n\leq y\\ n\equiv a\ (q)\end{subarray}}\Lambda(n)-\frac{y}{\varphi(q)}\right|\ll_{A}y(\log y)^{-A},\\ y\geq 3,\quad Q\leq y^{1/2}(\log y)^{-B_{\psi}(A)}.\end{gathered}$$

For fixed $A,C>0$, uniformly for $q\leq(\log y)^{C}$,

(3.2) $$\pi(y;q,a)=\frac{\operatorname{Li}(y)}{\varphi(q)}+O_{A,C}(y(\log y)^{-A}),\quad(a,q)=1,\quad y\geq 3.$$

These are Paper I, (2.20)–(2.21) [26, (2.20)–(2.21), p. 9]. The endpoint $y$ is fixed. The second constant is ineffective.

Fact 3.2 ((F1^′) Maximal-endpoint prime estimates).

Use the prime Siegel–Walfisz estimate (3.2). For every fixed $S>0$ and $X\geq 3$, put $Q_{X}=X^{1/2}(\log X)^{-(S+60)}$. Then

(3.3) $$\sum_{q\leq Q_{X}}\max_{(a,q)=1}\sup_{0\leq y\leq X}\left|\sum_{\begin{subarray}{c}n\leq y\\ n\equiv a\ (q)\end{subarray}}\Lambda(n)-\frac{y}{\varphi(q)}\right| \ll_{S}X(\log X)^{-S},$$

$$\sum_{q\leq Q_{X}}\max_{(a,q)=1}\sup_{0\leq y\leq X}\left|\pi(y;q,a)-\frac{\operatorname{Li}(y)}{\varphi(q)}\right| \ll_{S}X(\log X)^{-S}.$$

Here $\operatorname{Li}(y)=0$ for $0\leq y<2$. The level is imposed at $X$; the endpoints may be chosen separately for each modulus. An interval difference costs at most twice the bound. This is the classical maximal form of Bombieri–Vinogradov [6, Chapter 28]. Removing prime powers and partial summation pass from the von Mangoldt form to the prime-counting form. The constants may be ineffective through (3.2).

Fact 3.3 ((F2) Full-support bilinear distribution).

Let $a_{m},b_{n}$ have supports $m\sim M,n\sim Y$. Assume $x=MY\geq 3$, set $L_{x}=\log x$, and fix $\epsilon,B>0$. Assume $\epsilon\leq\log Y/\log x\leq 1-\epsilon$. For every $A_{0}>0$, assume uniformly in $h,k\geq 1$ and integers $\ell\neq 0$ with $(\ell,k)=1$ that

(3.4) $$\left|\sum_{\begin{subarray}{c}n\equiv\ell\ (k)\\ (n,h)=1\end{subarray}}b_{n}-\frac{1}{\varphi(k)}\sum_{(n,hk)=1}b_{n}\right|\leq C_{A_{0}}\|b\|_{2}Y^{1/2}d(h)^{B}(\log(2Y))^{-A_{0}}.$$

Define

$$\Delta_{\rm fs}(q,a)=\sum_{mn\equiv a\ (q)}a_{m}b_{n}-\frac{1}{\varphi(q)}\sum_{(mn,q)=1}a_{m}b_{n}.$$

For every $A>0$, some $B_{\rm BV}=B_{\rm BV}(A,B,\epsilon)>0$ gives

(3.5) $$\sum_{q\leq Q}\max_{(a,q)=1}|\Delta_{\rm fs}(q,a)|\ll\|a\|_{2}\|b\|_{2}x^{1/2}L_{x}^{-A},\qquad Q\leq x^{1/2}L_{x}^{-B_{\rm BV}}.$$

Both sums use the full supports. The constants depend on the family constants $C_{A_{0}}$. This follows from the proof of BFI Theorem 0(b) [3, (A1)–(A2), p. 206; Theorem 0(b), (1.5), p. 212; proof, pp. 213–214].

Proof.

Character orthogonality and primitive induction give

left side of (3.5) $$\leq\sum_{e\leq Q}\frac{1}{\varphi(e)}\sum_{2\leq f\leq Q/e}\frac{1}{\varphi(f)}$$

$$\qquad\cdot\sum_{\chi\ (f)}^{*}|A_{e}(\chi)B_{e}(\chi)|,$$

$$A_{e}(\chi) =\sum_{(m,e)=1}a_{m}\chi(m).$$

Define $B_{e}$ similarly. Split at $F=L_{x}^{K}$. Since $M\geq x^{\epsilon}$, large $x$ gives $F^{2}\leq M$. This permits the small-conductor estimate. The hypothesis handles $f\leq F$. The multiplicative large sieve handles $f>F$. Before summing $e$, the bounds are

$$\ll\|a\|_{2}\|b\|_{2}x^{1/2}d(e)^{B}F^{3/2}L_{x}^{-A_{0}},\qquad\ll L_{x}^{2}\|a\|_{2}\|b\|_{2}(Q+M^{1/2}+Y^{1/2}+x^{1/2}/F).$$

Choose $B_{2}$ with $\sum_{e\leq Q}d(e)^{B}/\varphi(e)\ll L_{x}^{B_{2}}$. In the large-conductor bound, summing $1/\varphi(e)$ costs one more logarithm, since $\sum_{e\leq Q}1/\varphi(e)\ll L_{x}$. Take $K=A+B_{2}+4$, $A_{0}=A+B_{2}+3K/2+1$, and $B_{\rm BV}=K$. Since $M,Y\geq x^{\epsilon}$, these bounds prove (3.5). The large sieve is BFI (1.6), p. 207, restated in Paper I, (2.19) [3, (1.6), p. 207][26, (2.19), p. 9]. ∎

We do not use BFI86, Theorem 0(a).

Fact (F2) is the standalone full-support bound. Section 8 checks the coefficients, relative levels and rough factors of the first three switched rows on the actual source.

Fact 3.4 ((F3) The spectral gap).

For every congruence subgroup $\Gamma\subset\mathrm{SL}_{2}(\mathbb{Z})$,

(3.6) $$\lambda_{1}(\Gamma)\geq\frac{975}{4096},\qquad 0<\lambda=\frac{1}{4}-t^{2}<\frac{1}{4},\quad t\geq 0\Longrightarrow 0\leq t\leq\frac{7}{64},\quad\vartheta=2t\leq\frac{7}{32}.$$

Here $\lambda_{1}(\Gamma)$ is the smallest nonzero Laplace eigenvalue. The implication applies to every eigenvalue $\lambda$ of $\Delta$ on $L^{2}(\Gamma\backslash\mathbb{H})$ in the indicated range. The first bound is Kim–Sarnak [17, Appendix 2]. The implications are elementary.

Fact 3.5 ((F4) Deligne’s bound).

Let $f=\sum_{n\geq 1}\psi_{f}(n)e(nz)$ be a primitive holomorphic cuspidal newform. Assume $\psi_{f}(1)=1$, integer weight $k\geq 2$, level $M$, and character $\chi$, with $f((az+b)/(cz+d))=\chi(d)(cz+d)^{k}f(z)$ on $\Gamma_{0}(M)$. This agrees with Deligne’s convention $\chi(a)^{-1}$. For $p\nmid M$, Deligne proved

(3.7) $$|\psi_{f}(p)|\leq 2p^{(k-1)/2}$$

[8, Théorème (8.2), p. 302]. The same source states that the roots $\alpha_{p},\beta_{p}$ of $T^{2}-\psi_{f}(p)T+\chi(p)p^{k-1}$ both have modulus $p^{(k-1)/2}$. Since $|\alpha_{p}|^{2}=|\beta_{p}|^{2}=p^{k-1}$, we have $\overline{\alpha_{p}}=p^{k-1}/\alpha_{p}$ and likewise for $\beta_{p}$. With $\alpha_{p}\beta_{p}=\chi(p)p^{k-1}$, this gives

(3.8) $$\chi(p)\overline{\psi_{f}(p)} =\frac{\alpha_{p}\beta_{p}}{p^{k-1}}\left(\frac{p^{k-1}}{\alpha_{p}}+\frac{p^{k-1}}{\beta_{p}}\right)=\psi_{f}(p).$$

Choose $\eta_{p}^{2}=\chi(p)$, $|\eta_{p}|=1$. Then $b_{p}=\eta_{p}^{-1}p^{-(k-1)/2}\psi_{f}(p)$ is real and $|b_{p}|\leq 2$. With $T=p^{(k-1)/2}\eta_{p}Z$, the polynomial becomes

$$T^{2}-\psi_{f}(p)T+\chi(p)p^{k-1}=p^{k-1}\chi(p)(Z^{2}-b_{p}Z+1).$$

Write $b_{p}=2\cos\omega_{p}$, $0\leq\omega_{p}\leq\pi$. The normalized roots are $e^{\pm i\omega_{p}}$, also in the repeated case $b_{p}=\pm 2$. An earlier exposition of the modular-form setting is [7].

Fact 3.6 ((F5) Averaged Kloosterman sums).

For $C,M,Y\geq 1$ and $\epsilon>0$,

(3.9) $$\sum_{c\leq C}\left|\sum_{m\leq M}\sum_{n\leq Y}S(m,n;c)\right|\ll_{\epsilon}(CMY)^{\epsilon}C(C+MY)$$

[9, Theorem 14, equation (8.13), p. 275]. The sums run over integers $1\leq c\leq C$, $1\leq m\leq M$, and $1\leq n\leq Y$. Here $S(m,n;c)=\sum_{u\ (c)}^{*}e((mu+n\bar{u})/c)$.

Use the trivial character on $\Gamma_{0}(q)\backslash\mathbb{H}$. The sum over $f$ runs through a Petersson-orthonormal basis of $S_{k}(\Gamma_{0}(q))$. The sum over $j$ runs through an orthonormal basis of Maass cusp forms, including exceptional forms. The sum over $\mathfrak{a}$ runs through inequivalent cusps with width-one scaling matrices. The measure is $dx\,dy/y^{2}$. At infinity, the DI coefficients satisfy

(3.10) $$u_{j}(z) =\sum_{n\neq 0}\rho_{j}(n)\sqrt{y}K_{i\kappa_{j}}(2\pi|n|y)e(nx),\quad\lambda_{j}=\tfrac{1}{4}+\kappa_{j}^{2},$$

$$f(z) =\sum_{n\geq 1}\psi_{f}(n)e(nz),$$

$$E_{\mathfrak{a}}(z,s) =\text{constant terms}+\sqrt{y}\sum_{n\neq 0}\frac{2\pi^{s}|n|^{s-1/2}}{\Gamma(s)}\varphi_{\mathfrak{a}\infty n}(s)K_{s-1/2}(2\pi|n|y)e(nx).$$

For $0<\lambda_{j}<1/4$, set $\vartheta_{j}=2|\Im\kappa_{j}|=2\sqrt{1/4-\lambda_{j}}$. See Paper I, (2.2)–(2.4) [26, (2.2)–(2.4), p. 5].

Fact 3.7 ((F6) Spectral large sieves).

For $q,K\geq 1$, $Y\geq 1/2$, and $a_{n}$ supported on $n\sim Y$, put

$$\Sigma_{\rm hol}(K,a) =\sum_{\begin{subarray}{c}2\leq k\leq K\\ k\ {\rm even}\end{subarray}}\frac{(k-1)!}{(4\pi)^{k-1}}\sum_{f}\left|\sum_{n\sim Y}a_{n}n^{-(k-1)/2}\psi_{f}(n)\right|^{2},$$

$$\Sigma_{\rm Maa}(K,a) =\sum_{|\kappa_{j}|\leq K}\frac{1}{\cosh(\pi\kappa_{j})}\left|\sum_{n\sim Y}a_{n}\rho_{j}(n)\right|^{2},$$

$$\Sigma_{\rm Eis}(K,a) =\sum_{\mathfrak{a}}\int_{-K}^{K}\left|\sum_{n\sim Y}a_{n}n^{it}\varphi_{\mathfrak{a}\infty n}(1/2+it)\right|^{2}dt.$$

The large-sieve estimates are

(3.11) $$\begin{gathered}\Sigma_{\rm hol}(K,a),\Sigma_{\rm Maa}(K,a),\Sigma_{\rm Eis}(K,a)\ll_{\epsilon}(K^{2}+Y^{1+\epsilon}/q)\|a\|_{2}^{2},\\ \sum_{j\ {\rm exceptional}}V^{\vartheta_{j}}\left|\sum_{n\sim Y}a_{n}\rho_{j}(n)\right|^{2}\ll_{\epsilon}(1+\sqrt{YV/q})(1+\sqrt{Y^{1+\epsilon}/q})\|a\|_{2}^{2},\quad V\geq 1.\end{gathered}$$

These are the specialized forms in Paper I, (2.5)–(2.8) [9, Theorem 2, (1.28)–(1.30), p. 230; Theorem 5, (1.38), p. 232] [26, (2.5)–(2.8), pp. 6–7].

For $\phi\in C_{c}^{\infty}((0,\infty))$, define

(3.12) $$\widehat{\phi}(t) =\frac{\pi}{\sinh\pi t}\int_{0}^{\infty}\frac{J_{2it}(x)-J_{-2it}(x)}{2i}\phi(x)\frac{dx}{x},$$

$$\check{\phi}(t) =\frac{4}{\pi}\cosh\pi t\int_{0}^{\infty}K_{2it}(x)\phi(x)\frac{dx}{x},\qquad\widetilde{\phi}(l)=\int_{0}^{\infty}J_{l}(x)\phi(x)\frac{dx}{x}.$$

The value at $t=0$ is the removable limit.

Fact 3.8 ((F7) Corrected same-cusp trace formula).

For $q,m,n\geq 1$ and the test class above,

(3.13) $$\sum_{h\geq 1}\frac{S(m,n;qh)}{qh}\phi\left(\frac{4\pi\sqrt{mn}}{qh}\right)=\frac{H_{\phi}(m,n)}{\pi}-M_{\phi}(m,n)-\frac{E_{\phi}(m,n)}{\pi},$$

where

$$H_{\phi}(m,n) =\sum_{\begin{subarray}{c}k\geq 2\\ k\ {\rm even}\end{subarray}}\frac{i^{k}(k-1)!\widetilde{\phi}(k-1)}{(4\pi\sqrt{mn})^{k-1}}\sum_{f}\overline{\psi_{f}(m)}\psi_{f}(n),$$

$$M_{\phi}(m,n) =\sum_{j}\frac{\widehat{\phi}(\kappa_{j})}{\cosh\pi\kappa_{j}}\overline{\rho_{j}(m)}\rho_{j}(n),$$

$$E_{\phi}(m,n) =\sum_{\mathfrak{a}}\int_{\mathbb{R}}(n/m)^{it}\overline{\varphi_{\mathfrak{a}\infty m}(1/2+it)}\varphi_{\mathfrak{a}\infty n}(1/2+it)\widehat{\phi}(t)\,dt.$$

No extra diagonal term occurs, also for $m=n$. This is the corrected same-cusp specialization proved in Paper I, (2.7.T15)–(2.7.T18) [26, (2.7.T15)–(2.7.T18)]. Compare [10, Lemma 4.5, (4.8)–(4.17)], in the DI convention [9, (1.15)–(1.17)].

Fact 3.9 ((F8) Mixed-cusp formulas).

For $q=rs$, $(r,s)=1$, use the trivial character and

$$\sigma_{1/s}=\begin{pmatrix}\sqrt{r}&0\\ s\sqrt{r}&1/\sqrt{r}\end{pmatrix},\quad\mathcal{C}(\infty,1/s)=\{cs\sqrt{r}:c\geq 1,\ (c,r)=1\}.$$

DI82 (1.6), p. 224, gives the arithmetic identity [9, (1.6), p. 224]

$$S_{\infty,1/s}(m,n;cs\sqrt{r})=e(n\bar{s}/r)S(m\bar{r},n;cs),\qquad m,n\in\mathbb{N}.$$

Here $s\bar{s}\equiv 1\pmod{r}$ and $r\bar{r}\equiv 1\pmod{cs}$. The accompanying trace formulas, proved in Paper I, are

$$\mathcal{K}_{+}=H_{\infty,1/s}/\pi-M_{\infty,1/s}-E_{\infty,1/s}/\pi,\qquad\mathcal{K}_{-}=(\pi/2)M^{-}_{\infty,1/s}+(1/2)E^{-}_{\infty,1/s}.$$

Here $\mathcal{K}_{\pm}=\sum_{\gamma\in\mathcal{C}}\gamma^{-1}S_{\infty,1/s}(m,\pm n;\gamma)\phi(4\pi\sqrt{mn}/\gamma)$. The plus terms have the definitions in (F7), with their second coefficient at $1/s$. For $M^{-}$ and $E^{-}$, replace $\widehat{\phi}$ by $\check{\phi}$ and $n$ by $-n$ only in the second Fourier coefficient. The cusps are singular, and the scaling matrices have width one. The classical width of $1/s$ is $r$. For $r>1$ they are inequivalent. The normalization is that of Paper I, (2.13)–(2.14). Compare [10, Lemmas 4.3, 4.5, (4.7)–(4.19)] [26, (2.13)–(2.14), p. 8].

Fact 3.10 ((F9) Bessel representations).

For $x>0$, $n\in\mathbb{Z}$, and $|\Re\nu|<1$,

$$J_{n}(x) =\pi^{-1}\int_{0}^{\pi}\cos(x\sin u-nu)\,du$$

$$\quad=i^{-n}\pi^{-1}\int_{0}^{\pi}e^{ix\cos u}\cos(nu)\,du,$$

$$J_{\nu}(x) =(2/\pi)\int_{0}^{\infty}\sin(x\cosh u-\pi\nu/2)\cosh(\nu u)\,du,$$

$$Y_{\nu}(x) =-(2/\pi)\int_{0}^{\infty}\cos(x\cosh u-\pi\nu/2)\cosh(\nu u)\,du,$$

$$K_{\nu}(x) =\sec(\pi\nu/2)\int_{0}^{\infty}\cos(x\sinh u)\cosh(\nu u)\,du.$$

The infinite integrals are improper. These are DLMF (10.9.2), (10.9.8), and (10.32.7) [25, (10.9.2), (10.9.8), (10.32.7)]. Paper I uses them on printed p. 20, (2.7.T31)–(2.7.T32).

Fact 3.11 ((F10) Kontorovich–Lebedev inversion).

For $g$ on $(0,\infty)$, the inversion pair is

$$f(y)=\frac{2y\sinh\pi y}{\pi^{2}}\int_{0}^{\infty}g(x)K_{iy}(x)\frac{dx}{x},\qquad g(x)=\int_{0}^{\infty}f(y)K_{iy}(x)\,dy.$$

It holds under either of two conditions. (a) $g/x\in C^{1}$, with $xg$ and $x(g/x)^{\prime}$ absolutely integrable. (b) Local piecewise continuity and bounded variation, with

$$\int_{0}^{1/2}\frac{|g(x)|}{x}\log(1/x)\,dx<\infty,\qquad\int_{1/2}^{\infty}\frac{|g(x)|}{\sqrt{x}}\,dx<\infty.$$

Under condition (b), the inverse equality is asserted at continuity points of $g$. In our applications, the inverse equality is used only at continuity points of $g$. See DLMF (10.43.30)–(10.43.32) [25, (10.43.30)–(10.43.32)]. These hypotheses hold for $g=\phi/x$, $\phi\in C_{c}^{\infty}((0,\infty))$. They also hold for $g=xK_{2it}(x)$ when $|\Im t|\leq 7/64$. Near zero, $K_{2it}(x)\ll_{t}x^{-2|\Im t|}(1+|\log x|)$, so the first integral in condition (b) converges absolutely. At infinity, $K_{2it}(x)\ll_{t}x^{-1/2}e^{-x}$, so the second converges too. Both functions are smooth on compact subintervals.

Fact 3.12 ((F11) Hecke operators on the full space).

For trivial character and $(k,q)=1$, Paper I proves

$$\|T_{k}\|_{\mathrm{Maass},q}\leq d(k)k^{7/64},\qquad\|T_{k}\|_{\mathrm{hol},q}\leq d(k).$$

Their proof uses the orthogonal cuspidal representation decomposition of Jacquet–Langlands, Chapter III, §16 [16, Chapter III, §16]. It also uses the holomorphic oldform classes of Atkin–Lehner [1, Lemmas 6, 13–15; Theorems 3(i), 5]. Atkin–Lehner’s weight $k$ means classical weight $2k$. For an unramified normalized newform, the algebra needed is

$$\lambda_{f}(p^{r})=\sum_{j=0}^{r}\alpha_{p}^{r-j}\beta_{p}^{j},\qquad\lambda_{f}(mn)=\lambda_{f}(m)\lambda_{f}(n)\quad((m,n)=1).$$

The Maass Satake bound is $|\alpha_{p}|,|\beta_{p}|\leq p^{7/64}$ [17, Appendix 2]. The holomorphic bound is (F4). Full-space diagonalization and unitary transport give the displayed norms, as in Paper I, (2.7.T37)–(2.7.T40) [26, (2.7.T37)–(2.7.T40), pp. 21–22].

Fact 3.13 ((F12) Eisenstein inputs).

Paper I uses three classical Eisenstein properties. The Eisenstein family has a meromorphic continuation, which allows compact Fourier-coefficient extraction. It has an arithmetic Fourier normalization. There is a complete irredundant cusp set. For the cusp $u/w$, let $d_{\rm den}$ be the lower-row denominator variable in $cz+d_{\rm den}$ and write the printed variable as $d=-d_{\rm den}$. The printed denominator is therefore $cz-d$, and the Fourier phase changes with this sign choice. Thus replacing $d$ by $-d$ changes the congruence and the phase together, not one sign in isolation. The normalization includes the orbit conditions

$$(c,d)=1,\quad(c,q)=w,\quad cd\equiv uw\pmod{(w^{2},q)}.$$

With $c=w\gamma$, the last condition is $\gamma d\equiv u\pmod{(w,q/w)}$. The cusp set has the form

$$u/w,\quad w\mid q,\quad u\pmod{(w,q/w)},\quad(u,(w,q/w))=1,$$

with admissible lifts chosen coprime to $q$. The sources are Iwaniec, Chapter 6 and Theorem 3.4; DI82 (1.17), Lemmas 2.3 and 3.6; and Kiral–Young, Proposition 3.1 and Corollary 3.2 [15, Chapter 6; Theorem 3.4] [9, (1.17), p. 227; Lemma 2.3, p. 240; Lemma 3.6, p. 247] [18, Proposition 3.1, Corollary 3.2]. These are the conventions used in Paper I, (2.7.T20)–(2.7.T24) and (2.7.T41)–(2.7.T43) [26, (2.7.T20)–(2.7.T24), (2.7.T41)–(2.7.T43), pp. 18–19, 22–23].

Fact 3.14 ((F13) Further spectral estimates).

For exceptional eigenfunctions, the first line below follows from (F3), since $975/4096>768/4096=3/16$ and $7/32<1/2$. Paper I derives the interval estimate from (F5), (F6), (F7), (P1.1) and this first line. The last line is its supported imaginary-transform estimate:

(3.14) $$\begin{gathered}\lambda_{j}\geq 3/16,\qquad 0<\vartheta_{j}\leq 1/2,\\ \sum_{2\leq q\leq Q}\sum_{j\ {\rm exceptional}}^{(q)}V^{\vartheta_{j}}\left|\sum_{n\in I}\rho_{j}(n)\right|^{2}\ll_{\epsilon}(QY)^{\epsilon}(Q+Y+\sqrt{YV})Y,\\ Q,Y,V\geq 1,\qquad I\subset(Y,2Y]\cap\mathbb{Z}\quad\text{an integer interval},\\ |\widehat{\phi}(i\nu)|+|\check{\phi}(i\nu)|\ll\mathfrak{B}(1+|\log Z|)(1+Z^{-2\nu})/(1+Z),\quad 0<\nu\leq 1/4.\end{gathered}$$

The last line uses the supported test class of (P1.1). The interval argument uses DI82’s proof of Theorem 7, (8.17), with the corrected same-cusp normalization (F7) [9, proof of Theorem 7, (8.17)] [26, (2.9), (2.7.T32), pp. 6, 20]. The weaker gap $3/16$ is also DI82, Theorem 4, p. 232, which we do not use separately. The reconstruction of the last estimate uses (F9).

Fact 3.15 ((F14) Transform identities and interchanges).

Paper I proves the following identities and accompanying interchange bounds in (2.7.T26)–(2.7.T32). Use the transforms (3.12) without rescaling. For $\phi\in C_{c}^{\infty}((0,\infty))$, put

(3.15) $$\begin{gathered}\mathcal{R}_{\phi}(v)=\int_{0}^{\infty}K_{2iv}(x)\phi(x)x^{-2}\,dx,\qquad k_{\phi}(v)=\frac{4}{\pi^{2}}v\sinh(2\pi v)\mathcal{R}_{\phi}(v),\\ \mathcal{H}(t,v)=\frac{\cosh\pi t}{\cosh\pi(t-v)\cosh\pi(t+v)},\\ \int_{\mathbb{R}}k_{\phi}(v)K_{2iv}(x)\,dv=\phi(x)/x,\\ \int_{\mathbb{R}}v\sinh(2\pi v)\mathcal{H}(t,v)K_{2iv}(x)\,dv=\cosh(\pi t)xK_{2it}(x).\end{gathered}$$

By (F10), the KL hypotheses hold for both applications, to $\phi(x)/x$ and to $xK_{2it}(x)$. For the latter, the product-integral condition is $|\Re(2it)|+|\Re(2iv)|<1$, with $v\in\mathbb{R}$. The permitted $t$ are real or exceptional, $|\Im t|\leq 7/64$. The differentiations and contour, limit, sum, and integral interchanges in (2.7.T26)–(2.7.T32) hold in these ranges. This includes absolute summability after the spectral estimates. For each fixed smooth test family, the decay bound is

$$\mathcal{R}_{\phi}(v)\ll_{\phi,j}e^{-\pi|v|}(1+|v|)^{1-2j}\qquad(j\geq 0).$$

Let $\operatorname{supp}\phi\subset[Z/21,9Z]$. The integrations by parts and differentiated estimates retain the five-derivative seminorm $\mathfrak{B}$ of (P1.1), uniformly in every auxiliary parameter. For the following bounds $U>0$, and the oscillatory kernels are the normalized combinations in (3.12); these estimates are not assertions about an unnormalized $J_{2it}$ tail. The hyperbolic factor in the raw Bessel representation is cancelled only after this normalization. For truncated normalized oscillatory representations, the tails are bounded by

$$O\!\left(\frac{1+|t|}{x\sinh U}\right)\quad(t\in\mathbb{R}),\qquad O\!\left(\frac{(1+\nu)e^{2\nu U}}{x\sinh U}\right)\quad(t=i\nu,\ 0<\nu\leq 1/4).$$

These tails permit the stated limiting interchanges. The derivative and decay constants are uniform in the families consumed by Theorem 2.8, with the losses prescribed there. The first inversion identity follows from (F10). The product integral, second kernel identity, and uniform differentiation and interchange steps are proved in the same passage [26, (2.7.T26)–(2.7.T32), pp. 19–20]. A reconstruction from the individual Bessel representations uses (F9).

Fact 3.16 ((P1.1) Supported real transforms).

For $Z>0$, let $\operatorname{supp}\phi\subset[Z/21,9Z]$ and set $\mathfrak{B}=\max_{0\leq j\leq 5}\sup_{x>0}x^{j}|\phi^{(j)}(x)|$. Then

(3.16) $$|\widehat{\phi}(t)|+|\check{\phi}(t)| \ll\frac{\mathfrak{B}(1+|\log Z|)}{1+Z}\min\{1,((1+Z)/|t|)^{3}\},\quad t\in\mathbb{R},$$

$$|\widetilde{\phi}(l)| \ll\frac{\mathfrak{B}(1+|\log Z|)}{1+Z}\min\{1,((1+Z)/l)^{3}\},\quad l\in\mathbb{N}.$$

The constants are absolute. The minimum at $t=0$ is $1$. This is the real-parameter part of Paper I, (2.7.T31) [26, (2.12), (2.7.T31), pp. 8, 20]. The real-transform bound uses exactly the Bessel representations in (F9). It does not cover the imaginary-parameter or interchange steps.

Write $\gamma_{\!E}$ for Euler’s constant. The continuous linear-sieve functions are normalized by

(3.17) $$\begin{gathered}sF(s)=2e^{\gamma_{\!E}}\quad(0<s\leq 3),\qquad sf(s)=0\quad(0<s\leq 2),\\ (sF(s))^{\prime}=f(s-1)\quad(s>3),\qquad(sf(s))^{\prime}=F(s-1)\quad(s>2).\end{gathered}$$

Thus $F(1)=2e^{\gamma_{\!E}}$ and $f(s)=0$ for $0<s\leq 2$. This is Paper I, (2.23), p. 10 [26, (2.23), p. 10].

Fact 3.17 ((P1.2) Structured linear sieve).

Let $\mathcal{A}$ be a finite integer sequence with nonnegative multiplicities, let $\mathcal{P}$ be a set of primes, and let $\mathcal{X}>0$. For a multiplicative function $\omega$, set

(3.18) $$\begin{gathered}P(z)=\prod_{\begin{subarray}{c}p<z\\ p\in\mathcal{P}\end{subarray}}p,\qquad S(\mathcal{A},\mathcal{P},z)=\sum_{a\in\mathcal{A}}1_{(a,P(z))=1},\\ |\mathcal{A}_{d}|=\frac{\omega(d)}{d}\mathcal{X}+r(\mathcal{A},d),\qquad V(z)=\prod_{\begin{subarray}{c}p<z\\ p\in\mathcal{P}\end{subarray}}\left(1-\frac{\omega(p)}{p}\right).\end{gathered}$$

Assume $0<\omega(p)<p$ on $\mathcal{P}$, $\omega(p)=0$ off $\mathcal{P}$, and $\omega(p^{a})\geq 0$. Suppose fixed $K\geq 2$ and $L_{\omega}>1$ satisfy, for every $z>w\geq 2$,

(3.19) $$\begin{gathered}\prod_{\begin{subarray}{c}w\leq p<z\\ p\in\mathcal{P}\end{subarray}}\left(1-\frac{\omega(p)}{p}\right)^{-1}\leq\frac{\log z}{\log w}\left(1+\frac{K}{\log w}\right),\\ \sum_{\begin{subarray}{c}w\leq p<z\\ p\in\mathcal{P}\end{subarray}}\sum_{a\geq 2}\frac{\omega(p^{a})}{p^{a}}\leq\frac{L_{\omega}}{\log(3w)}.\end{gathered}$$

Fix $D\geq 2$, $0<\epsilon_{s}<1/3$ and $\delta_{s}=\epsilon_{s}^{9}$. Define

(3.20) $$\mathcal{G}(D)=\{D^{\epsilon_{s}^{2}(1+\delta_{s})^{n}}:n\geq 0\}.$$

Let $\mathcal{H}(D)$ consist of the finite tuples $\mathbf{D}=(D_{1},\ldots,D_{k})$ from this grid such that $D_{k}\leq\cdots\leq D_{1}<D^{1/2}$. Set

(3.21) $$\mathcal{D}^{+}(D) =\{\mathbf{D}\in\mathcal{H}(D):D_{1}\cdots D_{2l}D_{2l+1}^{3}<D\quad(0\leq l\leq(k-1)/2)\},$$

$$\mathcal{D}^{-}(D) =\{\mathbf{D}\in\mathcal{H}(D):D_{1}\cdots D_{2l-1}D_{2l}^{3}<D\quad(1\leq l\leq k/2)\}.$$

Each of these sets has fewer than $\exp(8\epsilon_{s}^{-3})$ elements. There are coefficients $\phi_{d}^{\nu}$ and $\lambda_{d}^{\nu}(\mathbf{D})$, $\nu\in\{+,-\}$, with $|\phi_{d}^{\nu}|,|\lambda_{d}^{\nu}(\mathbf{D})|\leq 1$, independent of $\mathcal{A},\mathcal{X},\omega$, for which

(3.22) $$S(\mathcal{A},\mathcal{P},z) \leq\mathcal{X}V(z)\{F(s)+E_{s}\}+R_{1}^{+}+R^{+} (2\leq z\leq D),$$

$$S(\mathcal{A},\mathcal{P},z) \geq\mathcal{X}V(z)\{f(s)-E_{s}\}-R_{1}^{-}-R^{-} (2\leq z\leq D^{1/2}),$$

$$s =\frac{\log D}{\log z},\qquad E_{s}\ll\epsilon_{s}+\epsilon_{s}^{-8}e^{K+L_{\omega}}(\log D)^{-1/3}.$$

The implied constant in the last line is absolute. The signed remainders are

(3.23) $$R_{1}^{\nu}=\sum_{\begin{subarray}{c}d<D^{\epsilon_{s}}\\ d\mid P(D^{\epsilon_{s}^{2}})\end{subarray}}\phi_{d}^{\nu}r(\mathcal{A},d)$$

and

(3.24) $$R^{\nu}=\sum_{\mathbf{D}\in\mathcal{D}^{\nu}(D)}\sum_{\begin{subarray}{c}d<D^{\epsilon_{s}}\\ d\mid P(D^{\epsilon_{s}^{2}})\end{subarray}}\lambda_{d}^{\nu}(\mathbf{D})\sum_{\begin{subarray}{c}D_{i}\leq p_{i}<D_{i}^{1+\delta_{s}}\\ p_{i}\mid P(z)\\ p_{i}\neq p_{j}\ ((i,j)\in I(\mathbf{D}))\end{subarray}}r(\mathcal{A},dp_{1}\cdots p_{k}).$$

Here $I(\mathbf{D})$ imposes distinctness between the equal-label positions in Iwaniec’s construction. Unequal grid intervals are disjoint. The coefficients may depend on $\mathcal{P},D,z,\epsilon_{s},\mathbf{D}$ and these distinctness conditions. We do not claim that coefficients at different cutoffs agree. See Paper I, Theorem 2.6, (2.22), (2.24), (2.27)–(2.33), pp. 10–12 [26, Theorem 2.6, (2.22), (2.24), (2.27)–(2.33), pp. 10–12]. Its classical source is Iwaniec, Theorem 4 [13, Theorem 4].

Below, the prime set is $\mathcal{P}_{N}=\{p:p\nmid 2N\}$. Take the nominal density $\omega(p)=1$ on $\mathcal{P}_{N}$, $\omega(p)=0$ elsewhere and $\omega(p^{a})=0$ for $a\geq 2$. Then the second sum in (3.19) is zero. Mertens’ product estimate gives one fixed absolute $K_{\mathrm{M}}\geq 2$ with

(3.25) $$\prod_{w\leq p<z}(1-1/p)^{-1}\leq\frac{\log z}{\log w}\left(1+\frac{K_{\mathrm{M}}}{\log w}\right)\qquad(z>w\geq 2).$$

The first product in (3.19) is no larger than this product. Hence $K=K_{\mathrm{M}}$, $L_{\omega}=2$ are valid for this nominal density. The factor in (3.22) is then $e^{K_{\mathrm{M}}+2}$. See [26, pp. 10–11]. Each actual sequence still needs its own decomposition (3.18) and error estimate.

For every relative level $L_{c}\geq 2$ we use, exactly as in Paper I, (2.33a)–(2.33b), p. 12 [26, (2.33a)–(2.33b), p. 12],

(3.26) $$\begin{gathered}0<\sigma\leq 10^{-10},\qquad\tau=\sigma^{9},\qquad\epsilon_{s}=\sigma,\quad\delta_{s}=\tau,\quad D=L_{c},\\ \mathcal{G}(L_{c})=\{L_{c}^{\sigma^{2}(1+\tau)^{n}}:n\geq 0\},\qquad d<L_{c}^{\sigma},\quad d\mid P(L_{c}^{\sigma^{2}}).\end{gathered}$$

Thus the prime intervals, the smooth-divisor support and the level all refer to the same $L_{c}$. First fix $\sigma>0$. Then let $N$ tend to infinity.

Fact 3.18 ((P1.3) Finite factorization).

Let $X\geq 2$, $K\in\mathbb{N}$, and $z\geq 1$. Assume $z^{K}>X$. For every integer $1\leq n\leq X$,

(3.27) $$\Lambda(n)=\sum_{j=1}^{K}(-1)^{j-1}\binom{K}{j}\sum_{\begin{subarray}{c}m_{1}\cdots m_{j}n_{1}\cdots n_{j}=n\\ m_{i}\leq z\end{subarray}}\mu_{\rm Mob}(m_{1})\cdots\mu_{\rm Mob}(m_{j})\log n_{1}.$$

This is Paper I, Lemma 2.7, (2.36), p. 13 [26, Lemma 2.7, (2.36), p. 13].

Proof.

For $\Re s>1$, set $M_{z}(s)=\sum_{m\leq z}\mu_{\rm Mob}(m)m^{-s}$ and $Y_{z}(s)=\zeta(s)M_{z}(s)$. The polynomial identity $1-(1-Y_{z})^{K}=\sum_{j=1}^{K}(-1)^{j-1}\binom{K}{j}Y_{z}^{j}$ gives

$$-\frac{\zeta^{\prime}}{\zeta}=\sum_{j=1}^{K}(-1)^{j-1}\binom{K}{j}(-\zeta^{\prime})\zeta^{j-1}M_{z}^{j}+\left(-\frac{\zeta^{\prime}}{\zeta}\right)(1-Y_{z})^{K}.$$

Every nonzero coefficient of $Y_{z}-1$ has index greater than $z$. The last term therefore has no coefficient at $n\leq X<z^{K}$; coefficient extraction proves (3.27). ∎

For $n\leq 2X_{m}$, choose $K=3$, $z=2X_{m}^{1/3}$. Then $z^{3}=8X_{m}>2X_{m}$. Any interval partition is imposed after (3.27). The product condition and all representation multiplicities remain in the resulting coefficients. See [26, (2.36)–(2.37), p. 13].

For the packing theorem, put $\mu_{*}=25/128$ and define the packing ceiling by

$$H_{\rm pk,max}=\frac{153}{256}.$$

Use this constant in the packing hypotheses below. Put

(3.28) $$\begin{gathered}v_{\rm pk}(x)=\min\left\{\frac{1+x}{2},\frac{32-34x}{39},\frac{57}{64}-\frac{3}{2}x\right\}\quad(x\geq 0),\\ \psi_{\rm pk}(x,y)=x1_{x\geq y}\quad(x\in\mathbb{R},\ y\geq 0),\\ w_{\rm pk}(x_{1},x_{2},x_{3})=\psi_{\rm pk}\left(\min\left\{\frac{132-68x_{3}}{214},\frac{153-96x_{3}}{256},1-2x_{2}\right\},\frac{1+x_{1}}{2}\right).\end{gathered}$$

For $x_{1}\geq x_{2}\geq x_{3}\geq 0$, define

(3.29) $$H(x_{1},x_{2},x_{3})=\max\{ v_{\rm pk}(x_{1}),v_{\rm pk}(x_{2}),v_{\rm pk}(x_{1}+x_{2}),v_{\rm pk}(x_{1}+x_{2}+x_{3}),$$

$$w_{\rm pk}(x_{1},x_{2},x_{3}),w_{\rm pk}(x_{1},x_{3},x_{2}),$$

$$\psi_{\rm pk}(v_{\rm pk}(x_{1}+x_{3}),x_{1}+2x_{2}+x_{3}),$$

$$\psi_{\rm pk}(v_{\rm pk}(x_{2}+x_{3}),2x_{1}+x_{2}+x_{3})\},$$

$$\widetilde{H}(x_{1},x_{2},x_{3}) =\begin{cases}H(x_{1},x_{2},x_{3}),&x_{1}\leq\mu_{*},\\ v_{\rm pk}(x_{1}),&x_{1}>\mu_{*}.\end{cases}$$

These are the definitions preceding Paper I, Theorem 4.1, p. 53 [26, p. 53].

Fact 3.19 ((P1.4) Three-factor packing).

Let $0<\delta_{\rm pk}\leq 3\cdot 10^{-9}$, $1/2\leq h\leq H_{\rm pk,max}$, $d_{*}=h-2\delta_{\rm pk}$, and $\delta_{\rm pk}\leq g\leq 1/3-\delta_{\rm pk}/2$. Let $e_{1}\geq\cdots\geq e_{m}\geq 0$ be a finite labelled list. Equal values keep distinct labels. The empty list is allowed. Extend the list by zeros, writing $\bar{e}_{i}=0$ for $i>m$. Suppose

(3.30) $$\sum_{i<j}e_{i}+2e_{j}\leq h-2\delta_{\rm pk}\quad(1\leq j\leq m),\qquad h\leq\widetilde{H}(\bar{e}_{1},\bar{e}_{2},\bar{e}_{3}).$$

The rule of Paper I, Section 4.1, partitions the labels as $J_{1}\sqcup J_{2}\sqcup J_{3}$. With $\Sigma_{s}=\sum_{i\in J_{s}}e_{i}$ and $\theta_{\rm sp}=7/32$,

(3.31) $$\Sigma_{1} \leq g,$$

$$2g+\Sigma_{2}+2\Sigma_{3} \leq 1-4\delta_{\rm pk},$$

$$\theta_{\rm sp}+g+\Sigma_{1}+5\Sigma_{2}+2\Sigma_{3} \leq 2-4\delta_{\rm pk},$$

$$(1+2\theta_{\rm sp})g+\Sigma_{1}+(5-2\theta_{\rm sp})\Sigma_{2}+(2+2\theta_{\rm sp})\Sigma_{3} \leq 2-4(1+\theta_{\rm sp})\delta_{\rm pk}.$$

The partition depends only on these data. This is Paper I, Theorem 4.1, (4.1)–(4.3), pp. 53–54 [26, Theorem 4.1, (4.1)–(4.3), pp. 53–54].

Fact 3.20 ((P1.5) Continuous Hecke comparison).

Assume the three Eisenstein statements in (F12). Let $w\mid q$, put $w^{\prime}=(w,q/w)$, and let $\chi$ be a Dirichlet character modulo $w^{\prime}$. Then Paper I, (2.7.T41)–(2.7.T43), gives

(3.32) $$\varphi_{u/w,\infty,n}(z) =(w^{\prime}/(wq))^{z}\sum_{\begin{subarray}{c}\gamma\geq 1\\ (\gamma,q/w)=1\end{subarray}}\gamma^{-2z}\sum_{\begin{subarray}{c}d\ (\gamma w)\\ (d,\gamma w)=1\\ \gamma d\equiv u\ (w^{\prime})\end{subarray}}e(-nd/(\gamma w)),$$

$$\mathcal{V}_{\chi}(n;z) =\sum_{u\ (w^{\prime})}^{*}\chi(u)\varphi_{u/w,\infty,n}(z),$$

$$\mathcal{V}_{\chi}(kn^{\prime};z) =\overline{\chi(k)}\left(\sum_{d\mid k}\chi(d)^{2}d^{1-2z}\right)\mathcal{V}_{\chi}(n^{\prime};z),\quad(k,q)=(k,n^{\prime})=1.$$

The first identity initially holds for $\Re z>1$. The last extends meromorphically. For finitely supported $v_{n^{\prime}}$ in this coprime range,

(3.33) $$\sum_{\mathfrak{a}}\left|\sum_{n^{\prime}}v_{n^{\prime}}|kn^{\prime}|^{it}\varphi_{\mathfrak{a}\infty,kn^{\prime}}(1/2+it)\right|^{2}$$

$$\qquad\leq d(k)^{2}\sum_{\mathfrak{a}}\left|\sum_{n^{\prime}}v_{n^{\prime}}|n^{\prime}|^{it}\varphi_{\mathfrak{a}\infty,n^{\prime}}(1/2+it)\right|^{2}.$$

It holds almost everywhere in real $t$, for either index sign [26, (2.7.T41)–(2.7.T43), pp. 22–23].

Proof.

In (3.32), write the modulus $c_{1}=w\gamma=c_{1}^{\prime}c_{1}^{\prime\prime}$, where $(c_{1}^{\prime},q)=1$ and every prime of $c_{1}^{\prime\prime}$ divides $q$. CRT factors the inner character sum as $\chi(c_{1}^{\prime})G_{\chi}(-n;c_{1}^{\prime\prime})c_{c_{1}^{\prime}}(n)$. Here $G_{\chi}$ is the corresponding finite Gauss sum and $c_{c_{1}^{\prime}}$ is the Ramanujan sum. The outer character supplies another $\chi(c_{1}^{\prime})$. The coprime factor is

$$L^{(q)}(2z,\chi^{2})^{-1}\sum_{\begin{subarray}{c}d\mid|n|\\ (d,q)=1\end{subarray}}\chi^{2}(d)d^{1-2z},$$

where $L^{(q)}$ omits Euler factors at primes dividing $q$. For $(k,q)=(k,n^{\prime})=1$, its divisor sum factors at $n=kn^{\prime}$. Changing the Gauss-sum variable supplies $\overline{\chi(k)}$. Absolute convergence for $\Re z>1$, followed by meromorphic continuation, proves the last identity in (3.32). At $z=1/2+it$, the character multiplier is

$$\overline{\chi(k)}\sum_{d\mid k}\chi(d)^{2}(k/d^{2})^{it}.$$

Its modulus is at most $d(k)$. Orthogonality over the complete cusp set proves the inequality. ∎

Fact 3.21 ((P1.6) Exceptional parameters on the full space).

Apply (F3) to $\Gamma_{0}(q)$. Every exceptional eigenfunction in the full $\Gamma_{0}(q)$ space then has

$$\vartheta_{j}=2\sqrt{1/4-\lambda_{j}}\leq 7/32.$$

The bound depends only on the eigenvalue, so it covers oldforms. These are the parameters in Paper I, (2.7.T32), (2.7.T52), and (2.7.T56) [26, (2.7.T32), (2.7.T52), (2.7.T56), pp. 20, 24–25].

Fact 3.22 ((P1.7) The spectral estimate).

Paper I, Theorem 2.8, proves the estimate below. Its proof uses the large sieves (F6) and the trace formulas (F7)–(F8). It also uses the transform bounds (P1.1), (F13)–(F14), the full-space Hecke bounds (F11), and the continuous and exceptional comparisons (P1.5)–(P1.6). The normalized unitary cusp transport, proved in Paper I, (2.7.T33)–(2.7.T40), is as follows. Choose $rw-sy=1$ and use

(3.34) $$\begin{gathered}W_{rs}=r^{-1/2}\begin{pmatrix}r&y\\ rs&rw\end{pmatrix},\qquad W_{rs}\Gamma_{0}(rs)W_{rs}^{-1}=\Gamma_{0}(rs),\\ \rho_{F,1/s}(n)=e(n\bar{s}/r)\rho_{F\circ W_{rs},\infty}(n),\qquad\psi_{F,1/s}(n)=e(n\bar{s}/r)\psi_{F|W_{rs},\infty}(n),\\ \varphi_{\mathfrak{a},1/s,n}(z)=e(n\bar{s}/r)\varphi_{W_{rs}^{-1}\mathfrak{a},\infty,n}(z),\\ T_{k}F(z)=k^{-1/2}\sum_{hd=k}\sum_{j\ (d)}F((hz+j)/d),\qquad c_{T_{k}F}(n)=\sum_{h\mid(k,n)}c_{F}(kn/h^{2}).\end{gathered}$$

The last line is the Maass normalization, with $(k,rs)=1$. For holomorphic forms use the normalized coefficients of (F11). These formulas fix the normalization of the following theorem.

Let $C,D,Y,R,S,A_{T}\geq 1$ and $0<\epsilon_{0}<1/100$. Let $a_{T}\in\mathbb{Z}$ satisfy $0<|a_{T}|\leq A_{T}$. Let $b_{n,r,s}$ be supported on

(3.35) $$0<n\leq Y,\qquad r\sim R,\qquad s\sim S,\qquad(a_{T},rs)=1.$$

Let $g$ be smooth, supported in $(C,2C]\times(D,2D]\times\mathbb{R}_{>0}^{3}$, with

(3.36) $$|\partial^{\nu}g(c,d,n,r,s)|\leq G_{\nu}(c^{-\nu_{1}}d^{-\nu_{2}}n^{-\nu_{3}}r^{-\nu_{4}}s^{-\nu_{5}})^{1-\epsilon_{0}}\quad(\nu\in\mathbb{Z}_{\geq 0}^{5}).$$

Set $\mathcal{X}=2+A_{T}CDYRS$ and $\theta_{\rm sp}=7/32$. For every $\epsilon>0$,

(3.37) $$\left|\sum_{(r,s)=1}\sum_{0<n\leq Y}b_{n,r,s}\sum_{\begin{subarray}{c}c,d\geq 1\\ (rd,sc)=1\end{subarray}}g(c,d,n,r,s)e\left(\frac{a_{T}n\overline{rd}}{sc}\right)\right|\ll_{\epsilon,\epsilon_{0},\{G_{\nu}\}}\mathcal{X}^{\epsilon+18\epsilon_{0}}J_{*}(a_{T}),$$

where the inverse is modulo $sc$, and

(3.38) $$J_{*}(a_{T})^{2}={} \sum_{\begin{subarray}{c}n^{\prime\prime}\mid a_{T}^{\infty}\\ n^{\prime\prime}\leq 2Y\end{subarray}}A_{T}^{\theta_{\rm sp}}(n^{\prime\prime})^{\theta_{\rm sp}}$$

$$\quad\times\left[CS\left(\frac{Y}{n^{\prime\prime}}+RS\right)(C+DR)+|a_{T}|YRS\right]\|\widetilde{b}(n^{\prime\prime})\|_{2}^{2}$$

$$+\left[CS(CD\sqrt{R})^{2\theta_{\rm sp}}(Y+RS)^{1-\theta_{\rm sp}}\right.$$

$$\qquad\left.\times(C+DR)^{1-2\theta_{\rm sp}}+D^{2}YR\right]\|b\|_{2}^{2},$$

$$\|b\|_{2}^{2}={} \sum_{n,r,s}|b_{n,r,s}|^{2},\qquad\|\widetilde{b}(n^{\prime\prime})\|_{2}^{2}=\sum_{\begin{subarray}{c}n^{\prime},r,s\\ (n^{\prime},a_{T})=1\end{subarray}}|b_{n^{\prime}n^{\prime\prime},r,s}|^{2}.$$

Here $n^{\prime\prime}\mid a_{T}^{\infty}$ means that every prime factor of $n^{\prime\prime}$ divides $a_{T}$. It includes $n^{\prime\prime}=1$; when $a_{T}=\pm 1$, only $n^{\prime\prime}=1$ occurs. The bound is uniform in $(a_{T},A_{T},C,D,Y,R,S,b)$. These are Paper I, Theorem 2.8, (2.38)–(2.43) [26, Theorem 2.8, (2.38)–(2.43), pp. 14–15].

Section 8 checks the masked coefficients, intervals and support of the fourth switched row.

4. Finite source identities and decomposition

Throughout this section, $N$ is an even integer with $N\geq 2$. Further size hypotheses appear at each display or just before it. Use $L=\log N$ from Section 2. We prove the finite identities of this section below. They count each occurrence with its multiplicity. The later analytic estimates are separate hypotheses. Use $\mathcal{A}_{N},G_{b}(N),\varepsilon,a,b$ from (1.2)–(1.3) and $\alpha,\beta,\mu_{*},\lambda,t_{0},u$ from (2.1). Take the roughness cutoff $z_{s}=N^{\alpha}$ from the Root–A scale definition in (6.1). Throughout, $n$ is a positive source integer. When $n\in\mathcal{A}_{N}$, $n=N-p$ for a prime $p$. In a source sum, $e$ is the integer being counted. The letters $p,p_{i},a_{i},r$ denote prime variables, $q$ is a positive modulus, and $m$ is the factor or cofactor specified by its sum. Paper-II counts use $n=dm$; the marked count of Section 6 uses $m=(N-e)/p$. These roles are local to the displayed source. Here $\Omega(n)$ counts prime factors with multiplicity. For squarefree $d$ with $(d,N)=1$, define

(4.1) $$S^{\mathrm{II}}(d;y)=\#\{n=dm\in\mathcal{A}_{N}:P^{-}(m)\geq y,\ (m,Nd)=1\}.\qquad(2\mid N,\ N\geq 2).$$

Set this count to zero for other $d$. For admissible $d$, it is Paper II’s (1.7) count in our notation [27, (1.7)]. The zero value for other $d$ is our convention. Put $x_{i}=\log p_{i}/\log N$. The tenth row uses

(4.2) $$\begin{aligned} C_{10}&=\frac{57}{64}-\frac{1}{500}-\frac{11}{5}\alpha,\\ B_{10}(x)&=\begin{cases}\beta,&\beta\leq x<\mu_{*},\\ \min\{\beta,\max\{\alpha,C_{10}-5x/2\}\},&\mu_{*}\leq x<\lambda.\end{cases}\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

Table 4.1 lists all thirteen sums with every divisor and cutoff of [27, (1.6)–(1.7), row table]. All $p_{i}$ in the table are prime. Each admissible divisor contributes separately.

Table 4.1. The Paper II source sums. Each row sums $S^{\mathrm{II}}(d;y)$.

Row	Coefficient	Divisor and domain	Cutoff $y$
$S_{1}$	$3$	$d=1$	$N^{\alpha}$
$S_{2}$	$1$	$d=1$	$N^{\beta}$
$S_{3}$	$-2$	$d=p_{1},\ \alpha\leq x_{1}<\mu_{*}$	$N^{\alpha}$
$S_{4}$	$-1$	$d=p_{1},\ \mu_{*}\leq x_{1}<1/4$	$N^{\alpha}$
$S_{5}$	$-1$	$d=p_{1},\ 1/4\leq x_{1}<t_{0}$	$N^{\alpha}$
$S_{6}$	$-1$	$d=p_{1},\ t_{0}\leq x_{1}<1/3$	$N^{\alpha}$
$S_{7}$	$-1$	$d=p_{1},\ \mu_{*}\leq x_{1}<\lambda$	$N^{\alpha}$
$S_{8}$	$1$	$d=p_{1}p_{2},$ $\alpha\leq x_{2}<x_{1}<\beta$	$N^{\alpha}$
$S_{9}$	$1$	$d=p_{1}p_{2},$ $\alpha\leq x_{2}<\beta\leq x_{1}<\mu_{*}$	$N^{\alpha}$
$S_{10}$	$1$	$d=p_{1}p_{2},\ \mu_{*}\leq x_{1}<\lambda,$ $\alpha\leq x_{2}<B_{10}(x_{1})$	$N^{\alpha}$
$S_{11}$	$-2$	$d=p_{1}p_{2},\ \lambda\leq x_{1}<1/3,$ $x_{1}<x_{2}<(1-x_{1})/2$	$p_{2}$
$S_{12}$	$-1$	$d=p_{1}p_{2},$ $\alpha\leq x_{1}<1/3\leq x_{2}<(1-x_{1})/2$	$p_{2}$
$S_{13}$	$-1$	$d=p_{1}p_{2},$ $\beta\leq x_{1}<\lambda\leq x_{2}<(1-x_{1})/2$	$\sqrt{N/(p_{1}p_{2})}$

Let $\mathcal{D}_{i}$ be the displayed divisor domain and $y_{i}(d)$ its cutoff. Define the row occurrences by

(4.3) $$\vartheta_{i}(n)=\sum_{\begin{subarray}{c}d\in\mathcal{D}_{i}\\ d\mid n\\ d\ \mathrm{squarefree},\ (d,N)=1\end{subarray}}1_{P^{-}(n/d)\geq y_{i}(d)}1_{(n/d,Nd)=1}.\qquad(2\mid N,\ N\geq 2).$$

Thus $S_{i}=\sum_{d\in\mathcal{D}_{i}}S^{\mathrm{II}}(d;y_{i}(d))$. Keep every cofactor and endpoint restriction. Define

(4.4) $$\begin{aligned} S_{i}&=\sum_{n\in\mathcal{A}_{N}}\vartheta_{i}(n)\qquad(1\leq i\leq 13),\\ W_{13}(n)&=3\vartheta_{1}(n)+\vartheta_{2}(n)-2\vartheta_{3}(n)-\vartheta_{4}(n)-\vartheta_{5}(n)-\vartheta_{6}(n)-\vartheta_{7}(n)\\ &{}+\vartheta_{8}(n)+\vartheta_{9}(n)+\vartheta_{10}(n)-2\vartheta_{11}(n)-\vartheta_{12}(n)-\vartheta_{13}(n),\\ s_{\mathrm{sw}}(n)&=1_{\Omega(n)\geq 3}\max\{W_{13}(n),0\},\\ S_{\mathrm{sw}}&=\sum_{n\in\mathcal{A}_{N}}s_{\mathrm{sw}}(n),\qquad w(n)=W_{13}(n)-s_{\mathrm{sw}}(n),\qquad\mathcal{W}=\sum_{n\in\mathcal{A}_{N}}w(n).\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

This is the complete switched source of Paper II, (1.9) [27, (1.9)]. Every divisor prime and cofactor cutoff is at least $N^{\alpha}$. For $S_{13}$ the cutoff exceeds $N^{(1-\lambda)/4}>N^{\alpha}$. Thus every $\vartheta_{i}(n)$ vanishes when $P^{-}(n)<N^{\alpha}$, so the extra roughness indicator of Paper II is redundant. Use $P^{-}(1)=\infty$ as in Table 2.1. Finite linearity gives

(4.5) $$\begin{aligned} \mathcal{W}={}&3S_{1}+S_{2}-2S_{3}-S_{4}-S_{5}-S_{6}-S_{7}\\ &+S_{8}+S_{9}+S_{10}-2S_{11}-S_{12}-S_{13}-S_{\mathrm{sw}}.\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

4.1. The representation count

Put

(4.6) $$\begin{aligned} g_{b}(e)&=\#\{(p_{1},p_{2})\in\mathbb{P}^{2}:N^{b}\leq p_{1}\leq\sqrt{N},\ \varepsilon N<p_{1}p_{2}=e<N\},\\ B_{b}&=\sum_{e}g_{b}(e)1_{N-e\in\mathbb{P}},\qquad P_{2}=\#\{n\in\mathcal{A}_{N}:\Omega(n)\leq 2\}.\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

When $e$ has distinct prime factors and both choices satisfy the first-factor cutoff, both orientations are counted. For $e=p^{2}$, there is only the ordered pair $(p,p)$. Hence

(4.7) $$0\leq g_{b}(e)\leq 2,\qquad P_{2}-G_{b}(N)\leq B_{b}\qquad(\varepsilon N>1).\qquad(2\mid N,\ N\geq 2).$$

For the second inequality, write $n=rs$, $r\leq s$, with both factors prime. If $n$ is excluded from $G_{b}$, then

(4.8) $$N^{b}\leq r\leq\sqrt{n}<\sqrt{N},\qquad g_{b}(n)\geq 1.\qquad(2\mid N,\ N\geq 2,\ \varepsilon N>1).$$

Prime squares are counted with their actual multiplicity.

Define

(4.9) $$\mathcal{E}=\{n\in\mathcal{A}_{N}:P^{-}(n)\geq z_{s},\ n\text{ is not squarefree or }(n,N)>1\}.\qquad(2\mid N,\ N\geq 2).$$

For $z_{s}=N^{\alpha}\geq 2$,

(4.10) $$\begin{aligned} \#\{n\in\mathcal{E}:n\text{ is not squarefree}\}&\leq N\sum_{k\geq\lceil z_{s}\rceil}k^{-2}\leq\frac{2N}{z_{s}},\\ \#\{n\leq N:P^{-}(n)\geq z_{s},\ (n,N)>1\}&\leq\sum_{\begin{subarray}{c}\ell\in\mathbb{P},\ \ell\mid N\\ \ell\geq z_{s}\end{subarray}}\frac{N}{\ell}\leq\frac{11N}{z_{s}},\\ |\mathcal{E}|&\leq 13N^{1-\alpha}.\end{aligned}\qquad(2\mid N,\ N\geq 2^{1/\alpha}).$$

The prime divisors in the middle sum are distinct. Their number is at most $\lfloor 1/\alpha\rfloor=11$.

Set $N_{\mathrm{pt}}=1+\left\lceil\max\{2^{1/\alpha},\varepsilon^{-3},\varepsilon^{-1/(1-\lambda-\beta)}\}\right\rceil$.

Lemma 4.1 (Pointwise source weight bound).

For all $N\geq N_{\mathrm{pt}}$, the weight bounds are

(4.11) $$\begin{gathered}w(n)=0\quad(P^{-}(n)<z_{s}),\qquad w(n)\leq 4\,1_{\Omega(n)\leq 2}\quad(n\in\mathcal{A}_{N}\setminus\mathcal{E}),\\ \max\{0,w(n)-4\,1_{\Omega(n)\leq 2}\}\leq 169\quad(n\in\mathcal{E}),\qquad E_{\mathrm{pt}}=0.\end{gathered}\qquad(2\mid N,\ N\geq N_{\mathrm{pt}}).$$

Proof.

Start from the roughness observation above. If $\Omega(n)\geq 3$, then $w(n)=\min\{W_{13}(n),0\}\leq 0$. If $(n,N)>1$, the coprimality conditions force every row occurrence to vanish. For $\Omega(n)\leq 2$ and $n\in\mathcal{E}$ with $(n,N)=1$, only $n=p^{2}$ occurs; only $S_{1},S_{2}$ contribute, giving $w(n)=4$. For $n\notin\mathcal{E}$ with $\Omega(n)\leq 2$, a prime has weight $4$. For $n=rs$, with primes $r<s$, the bands $[\alpha,\beta)$, $[\beta,\mu_{*})$, $[\mu_{*},\lambda)$, $[\lambda,1/3)$ and $[1/3,1/2)$ for $\log_{N}r$ give weights $1,2,2,3,4$, respectively, directly from the row table. This agrees with Paper II, (1.11) [27, (1.10)–(1.11)]. Here $N_{\mathrm{pt}}$ is chosen so that $\varepsilon N>N^{2/3}$ and $\varepsilon N>N^{\lambda+\beta}$. Then the larger prime exceeds $N^{1/3}$. The two-prime divisor rows vanish, since their products are below $N^{\lambda+\beta}$ for $S_{8}$–$S_{10}$ and below $N^{2/3}$ for $S_{11}$–$S_{13}$. These observations prove all three bounds, including the stronger exceptional bound with right-hand side zero. ∎

Put

(4.12) $$E_{\mathrm{pt}}=\frac{1}{4}\sum_{n\in\mathcal{E}}\max\{0,w(n)-4\,1_{\Omega(n)\leq 2}\}.\qquad(2\mid N,\ N\geq 2).$$

The pointwise lemma gives $E_{\mathrm{pt}}=0$ for $N\geq N_{\mathrm{pt}}$, within its stated scope. We still record cruder bounds. Summation gives

(4.13) $$\begin{gathered}0\leq E_{\mathrm{pt}}\leq\frac{169}{4}|\mathcal{E}|\leq\frac{2197}{4}N^{1-\alpha},\\ \mathcal{W}\leq 4P_{2}+4E_{\mathrm{pt}},\qquad 4G_{b}(N)\geq\mathcal{W}-4B_{b}-4E_{\mathrm{pt}}.\end{gathered}\qquad(2\mid N,\ N\geq N_{\mathrm{pt}}).$$

The transfer to $r\leq q^{3/4}$ is (1.4).

4.2. Signed arithmetic sums

Put

(4.14) $$\begin{aligned} R_{\mathrm{fix}}={}&3S_{1}-2S_{3}-S_{4}-S_{5}-S_{7}+S_{8}+S_{9}+S_{10}\\ &-2S_{11}-S_{12}-S_{13}.\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

Then (4.5) gives

(4.15) $$\mathcal{W}-4B_{b}=R_{\mathrm{fix}}+S_{2}-S_{6}-S_{\mathrm{sw}}-4B_{b}.\qquad(2\mid N,\ N\geq 2).$$

Define

(4.16) $$\begin{aligned} G_{\mathrm{rows}}&=2S_{11}+S_{12}+S_{13},\\ \mathcal{C}_{\mathrm{ret}}&=3S_{1}+S_{2}-S_{4}-S_{5}-S_{7}+S_{10}-G_{\mathrm{rows}},\\ \mathcal{Q}&=-2S_{3}+S_{8}+S_{9}.\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

Coefficient comparison gives

(4.17) $$\begin{gathered}\mathcal{C}_{\mathrm{ret}}+\mathcal{Q}=R_{\mathrm{fix}}+S_{2},\\ \mathcal{W}-4B_{b}=\mathcal{C}_{\mathrm{ret}}+\mathcal{Q}-S_{6}-S_{\mathrm{sw}}-4B_{b}.\end{gathered}\qquad(2\mid N,\ N\geq 2).$$

Substitution proves both equalities coefficient by coefficient. The coefficients $3S_{1}$ and $+S_{10}$ are fixed. The integer supports may overlap. These coefficient identities do not depend on later estimates. Take the signed normalized chains $Q_{j,k}$, their index set $\mathcal{I}$, and the stage index $k\in\{0,\ldots,10\}$ from Section 5. See its “original signed chains” paragraph and (5.4). Each $Q_{j,k}$ already includes its signed normalized coefficient, so no extra multiplicity factor enters. Finite telescoping gives

(4.18) $$\sum_{j\in\mathcal{I}}Q_{j,0}=\sum_{j\in\mathcal{I}}Q_{j,10}+\sum_{j\in\mathcal{I}}\sum_{k=1}^{10}(Q_{j,k-1}-Q_{j,k}).\qquad(2\mid N,\ N\geq 2).$$

This is the normalized-chain identity cited above. The arithmetic coefficient identities above keep their unnormalized definitions. Paper II uses joint endpoints for $S_{1},S_{2}\text{-}T_{8}/4,G_{\mathrm{rows}}$ [27, (6.4)–(6.4d), pp. 70–72]. Here $T_{8}$ is Paper II’s stage-eight endpoint in the $S_{2}$ chain. The notation $S_{2}\text{-}T_{8}/4$ is a joint-chain label; it defines no new arithmetic sum. Their joint endpoint is kept before the differences are split.

4.3. A finite upper decomposition

Each original representation occurrence uses the shifted-prime set of $\eqref{eq:intro-count}$:

(4.19) $$\mathcal{A}_{N}=\{N-p:p\in\mathbb{P},\ p<(1-\varepsilon)N\}.\qquad(2\mid N,\ N\geq 2).$$

The two arithmetic sources stay separate. Take $A_{b},E_{\mathrm{top}}$ from (7.2) and $S_{b}$ from the Section 7 definition (7.13); $A$ from (8.3); $\mathcal{P}_{N}$ from (8.1); and $b_{A},P_{N},S_{v},S_{A},E_{\mathrm{src}}$ from (8.6). The ordered prime coordinates $a_{i}$ and their prefix domains are those of (8.9) and the paragraph before it. Define

(4.20) $$\begin{gathered}h_{b}(n)=A_{b}(N-n),\qquad\mathscr{L}_{b}=\{p\in\mathbb{P}:p\nmid 2N\},\qquad E_{b}=E_{\mathrm{top}},\\ h_{\mathrm{sw}}(n)=A(N-n),\qquad\mathscr{L}_{\mathrm{sw}}=\mathcal{P}_{N},\qquad E_{\mathrm{sw}}^{\mathrm{src}}=E_{\mathrm{src}},\end{gathered}\qquad(2\mid N,\ N\geq 2).$$

Here $A_{b},E_{\mathrm{top}}$ are the Section 7 source objects in $\eqref{eq:k2-source}$. The objects $A,\mathcal{P}_{N},E_{\mathrm{src}}$ are the Section 8 source objects in $\eqref{eq:l3-source}$, $\eqref{eq:l3-constants}$, and $\eqref{eq:l3-sieved-source}$. Both sequences are nonnegative. No extra $1_{(e,2N)=1}$ factor enters. The script-font symbols $\mathscr{L}_{b}$, $\mathscr{L}_{\mathrm{sw}}$, $\mathscr{L}_{X}$, and $\mathscr{U}_{X}$ denote the variable sieve families and their sifted sums. The fixed prime support is written $\mathcal{P}_{N}$. For $X\in\{b,\mathrm{sw}\}$, write $h_{X},\mathscr{L}_{X},E_{X}$ for the corresponding line of $\eqref{eq:source-local-sequences}$, and put

(4.21) $$\begin{aligned} P_{X}(y)&=\prod_{\begin{subarray}{c}\ell<y\\ \ell\in\mathscr{L}_{X}\end{subarray}}\ell,\\ \mathscr{U}_{X}(d;y)&=\sum_{d\mid n}h_{X}(n)1_{(n/d,P_{X}(y))=1}.\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

Take $I_{L},I_{H},J,J_{\mathrm{tail}},J_{t},z_{s},\eta_{\mathrm{m}},v_{\mathrm{RA}}$ from the Root–A definition (6.1). Take $c_{R,K}$ from (6.4), and $R_{L},R_{\mathrm{tail}},B_{L},B_{\mathrm{tail}},B,c_{\mathrm{lo}},\Gamma$ from (6.6) and its preceding choice of fixed logarithmic exponents. Take $D_{\mathrm{m}}$ from (6.52). Write $Q_{\mathrm{RA}}$ for the level called $Q$ in (6.6). This alias is distinct from every signed chain $Q_{j,k}$. We do not redefine these objects here. Both source comparisons require the finite threshold $N_{6}$ of Section 2. It enters the operative $T_{\mathrm{RA}}$, the maximum of the sixth-sum comparison threshold and $N_{6}$. We do not redefine it here.

Use the common choice $\sigma=\sigma_{*}$ from (2.4). It lies in the Section 7 range of (7.1), and the switched comparison uses it. Take $\tau,\rho$ from Section 7 (7.1); Section 8 uses the same values. We make no new choice or definition of $\sigma,\tau,\rho$ here. Partition the logarithmic prime ranges into half-open intervals $[t_{i},\rho t_{i})$, with $t_{i}=u\rho^{k_{i}}$, $k_{i}\geq 0$. Clip the intervals at $\mu_{*}$. Each prime has one cell. For $X=b$, use the retained cells of the Section 7 $\eqref{eq:k2-retention}$ retained-cell definition and the cutoff in $\eqref{eq:k2-four-rows}$. Write them as $\mathcal{R}_{b},z_{b,c},\chi_{b,c}$. For $X=\mathrm{sw}$, use the Section 8 definitions $\eqref{eq:l3-buchstab-peel}$ and $\eqref{eq:l3-row3-level}$. Write them as $\mathcal{R}_{\mathrm{sw}},z_{\mathrm{sw},c},\chi_{\mathrm{sw},c}$. These definitions come from the two source sections. They form no common envelope in this section. We use no equality between the retained collections of the three sections. The closed-cell infima of Sections 7 and 8 and their endpoint rules apply only within their own source. For $X=b$, the packing function is $H$, as specified in (7.19). For $X=\mathrm{sw}$, it is $\widetilde{H}$, as specified in (8.10) and its preceding definition of $\Theta_{c}$. The two functions are defined in Paper II, equation (2.17d), and the paragraph after it. We have $\widetilde{H}=H$ only on the stated largest-coordinate range, and use no equality outside it. For either source write $z_{X,c}=N^{t_{3}}$, with $t_{3}$ the lower endpoint of the third coordinate of its cell $c$. Here $z_{\mathrm{sw},c}$ is the object called $z_{c}$ in Section 8; we use only the source-indexed name. The Section 7 cutoff is defined in (7.18), with the lower-endpoint convention stated before (7.31).

The source index here is always $X\in\{b,\mathrm{sw}\}$. For the $b$-source the underlying prime support is $\mathcal{P}_{N}$, so $P_{b}(y)=P_{N}(y)$. Both sieve products use primes below $y$ not dividing $2N$. For the $b$-source, the substitution $e=N-n$ identifies the generic sieve with $S_{b}$ of Section 7 on that source:

(4.22) $$\mathscr{U}_{b}(d;y)=\sum_{d\mid n}A_{b}(N-n)1_{(n/d,P_{b}(y))=1}=\sum_{d\mid N-e}A_{b}(e)1_{((N-e)/d,P_{b}(y))=1}=S_{b}(d;y).\qquad(2\mid N,\ N\geq 2).$$

For $X=\mathrm{sw}$, identify the generic cofactor sieve with the Section 8 sieve on precisely the following rows:

• $d=1$ with cutoff $y=N^{\mu_{*}}$, the source-comparison row $\mathscr{U}_{\mathrm{sw}}(1;N^{\mu_{*}})=S_{A}$ from Section 8 (8.6);

• $v=a_{1}\cdots a_{j}$, $0\leq j\leq 2$, with cutoff $y=N^{u}$;

• $v=a_{1}a_{2}a_{3}$, with dynamic cutoff $y=a_{3}$, or retained-cell cutoff $y=z_{\mathrm{sw},c}=N^{t_{3}}$;

• $vr=a_{1}a_{2}a_{3}r$ on a retained cell, with $z_{\mathrm{sw},c}\leq r<a_{3}$, at cutoff $y=r$ or $y=z_{\mathrm{sw},c}$.

Here the empty prefix is $v=1$, and all primes belong to $\mathcal{P}_{N}$ and have the ordering in (8.9) and (8.11). Let $d$ be the displayed prefix $v$ or product $vr$. Every prime divisor of $d$ is at least $y$, so $(d,P_{\mathrm{sw}}(y))=1$. The definitions give $P_{\mathrm{sw}}(y)=P_{N}(y)$. Write $S_{d}(y)$ for the $S_{v}(y)$ of Section 8 from (8.6) with $v=d$. For $d\mid n$, the Section 8 definition (8.6) gives

(4.23) $$\mathscr{U}_{\mathrm{sw}}(d;y)=\sum_{d\mid n}h_{\mathrm{sw}}(n)1_{(n/d,P_{N}(y))=1}=\sum_{d\mid n}b_{A}(n)1_{(n,P_{\mathrm{sw}}(y))=1}=S_{d}(y).\qquad(2\mid N,\ N\geq 2).$$

This identity holds on these rows only, not for arbitrary $d,y$. These $E$-terms are unnormalized source-restoration costs. The two source upper comparisons are stated separately: Here $B_{b}^{\prime}$ is the prime-occurrence part of $B_{b}$, and $E_{\mathrm{top}}$ is the removed-top cost.

Lemma 4.2 (Upper-source comparisons).

(4.24) $$\begin{gathered}B_{b}\leq E_{b}+\mathscr{U}_{b}(1;N^{\mu_{*}}),\qquad S_{\mathrm{sw}}\leq E_{\mathrm{sw}}^{\mathrm{src}}+\mathscr{U}_{\mathrm{sw}}(1;N^{\mu_{*}})\\ \text{for even }N\geq N_{6}.\end{gathered}$$

Proof.

The first inequality follows from Section 7 (7.16) and $E_{b}=E_{\mathrm{top}}$; the second is the Section 8 source comparison (8.7). We do not identify their hypotheses or their restoration costs. ∎

We invoke the source comparison once, with the named hypotheses displayed above.

For real cutoffs $1\leq y\leq w_{\mathrm{cut}}$, the least allowed prime gives

(4.25) $$\mathscr{U}_{X}(d;w_{\mathrm{cut}})=\mathscr{U}_{X}(d;y)-\sum_{\begin{subarray}{c}y\leq r<w_{\mathrm{cut}}\\ r\in\mathscr{L}_{X}\end{subarray}}\mathscr{U}_{X}(dr;r).\qquad(2\mid N,\ N\geq 2).$$

Write $n=dm$. If $m$ survives at $y$ but fails at $w_{\mathrm{cut}}$, choose its least allowed prime in $[y,w_{\mathrm{cut}})$. Exactly one summand records that prime. Repeated copies remain in the cofactor. Thus the identity also counts prime powers correctly.

Three applications give

(4.26) $$\begin{aligned} \mathscr{U}_{X}(1;N^{\mu_{*}})&=T_{X,0}-T_{X,1}+T_{X,2}-T_{X,3},\\ T_{X,j}&=\sum_{\begin{subarray}{c}N^{u}\leq a_{j}<\cdots<a_{1}<N^{\mu_{*}}\\ a_{i}\in\mathscr{L}_{X}\end{subarray}}\mathscr{U}_{X}(a_{1}\cdots a_{j};N^{u})\quad(0\leq j\leq 2),\\ T_{X,3}&=\sum_{\begin{subarray}{c}N^{u}\leq a_{3}<a_{2}<a_{1}<N^{\mu_{*}}\\ a_{i}\in\mathscr{L}_{X}\end{subarray}}\mathscr{U}_{X}(a_{1}a_{2}a_{3};a_{3}).\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

For $j=0$, the sum is $\mathscr{U}_{X}(1;N^{u})$. The primes are ordered; no factorial occurs. Every multiplicity remains in $h_{X}$.

For $d=a_{1}a_{2}a_{3}$ in a retained cell of source $X$, put

(4.27) $$\begin{aligned} -\mathscr{U}_{X}(d;a_{3})&=-\mathscr{U}_{X}(d;z_{X,c})+\sum_{\begin{subarray}{c}z_{X,c}\leq r<a_{3}\\ r\in\mathscr{L}_{X}\end{subarray}}\mathscr{U}_{X}(dr;r)\\ &\leq-\mathscr{U}_{X}(d;z_{X,c})+\sum_{\begin{subarray}{c}z_{X,c}\leq r<a_{3}\\ r\in\mathscr{L}_{X}\end{subarray}}\mathscr{U}_{X}(dr;z_{X,c}).\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

The equality is (4.25). Nonnegativity proves the inequality. Actual cell membership gives $z_{X,c}\leq a_{3}$. The closed infimum creates no additional occurrences.

Let $\sum_{\mathrm{ret},X}$ restrict the ordered triples to $\mathcal{R}_{X}$, equivalently to $\chi_{X,c}=1$. For either source define the peel terms

(4.28) $$\begin{aligned} R_{X,3}&=\sum_{\mathrm{ret},X}\mathscr{U}_{X}(a_{1}a_{2}a_{3};z_{X,c}),\\ P_{X,3}&=\sum_{\mathrm{ret},X}\sum_{\begin{subarray}{c}z_{X,c}\leq r<a_{3}\\ r\in\mathscr{L}_{X}\end{subarray}}\mathscr{U}_{X}(a_{1}a_{2}a_{3}r;z_{X,c}).\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

For each source $X\in\{b,\mathrm{sw}\}$, the non-retained depth-three terms are nonnegative and occur with a minus sign, so we drop them. The retained terms use the dynamic inequality above:

(4.29) $$-T_{X,3}\leq-R_{X,3}+P_{X,3}.\qquad(2\mid N,\ N\geq 2).$$

For even $N\geq N_{6}$, define the two upper expressions and their slacks, one per source. The next comparison proves their nonnegativity:

(4.30) $$\begin{aligned} U_{b}&:=T_{b,0}-T_{b,1}+T_{b,2}-R_{b,3}+P_{b,3},&Z_{b}&:=E_{b}+U_{b}-B_{b}\geq 0,\\ U_{\mathrm{sw}}&:=T_{\mathrm{sw},0}-T_{\mathrm{sw},1}+T_{\mathrm{sw},2}-R_{\mathrm{sw},3}+P_{\mathrm{sw},3},&Z_{\mathrm{sw}}&:=E_{\mathrm{sw}}^{\mathrm{src}}+U_{\mathrm{sw}}-S_{\mathrm{sw}}\geq 0.\end{aligned}\qquad(2\mid N,\ N\geq N_{6}).$$

Hence

(4.31) $$\mathscr{U}_{X}(1;N^{\mu_{*}})\leq\begin{cases}U_{b},&X=b,\\ U_{\mathrm{sw}},&X=\mathrm{sw}.\end{cases}\qquad(2\mid N,\ N\geq N_{6}).$$

These inequalities assume the two source comparisons in (4.24). They give no estimate for either restoration cost.

4.4. The conditional lower expression

Take $A_{\mathrm{mix}}$ from the Root–A definition (6.8), and $Z_{6}$ with its comparison from Section 6 (6.9). We redefine neither object. This sum uses the ordinary upper Rosser weights $\mu_{d}^{+}(D)$ with the odd-index cubic gates of $\eqref{eq:ra-rosser-weight}$. Their upper-sieve property is the Iwaniec upper-bound lemma [14, Lemma 1, (3.2), (3.6)] [13, Lemmas 2–3, p. 313]. This section introduces no new weight. The Section 6 comparison gives

(4.32) $$S_{6}\leq A_{\mathrm{mix}},\qquad Z_{6}\geq 0\qquad(2\mid N,\ N\geq N_{6}).$$

For a mark $p$, put $m=(N-e)/p$ and $P_{\mathrm{RA}}=\prod_{\ell<z_{s},\,\ell\nmid 2N}\ell$. For $D>1$, the cited Iwaniec upper-bound lemmas give, for the upper Rosser divisor sum, [14, Lemma 1, (3.2), (3.6)] [13, Lemmas 2–3, p. 313],

$$\sum_{d\mid(m,P_{\mathrm{RA}})}\mu_{d}^{+}(D)\geq 1_{(m,P_{\mathrm{RA}})=1}\geq 1_{P^{-}(m)\geq z_{s}}\qquad(2\mid N,\ N\geq 2,\ D>1).$$

Dropping $(m,Np)=1$ enlarges the original nonnegative occurrence count. For even $N\geq N_{6}$, each of $R_{L},R_{\mathrm{tail}},Q_{\mathrm{RA}}$ exceeds $N^{1/3}$. Hence take $D=R_{L}/p,R_{\mathrm{tail}}/p,Q_{\mathrm{RA}}/p$ on, respectively, $I_{L}\times J$, $I_{H}\times J_{\mathrm{tail}}$, $I_{H}\times J_{t}$. These pieces partition the original source: $e=N/4$ belongs to $I_{L}$, and the shared mark endpoint belongs to $J_{\mathrm{tail}}$. Since $p\geq N^{t_{0}}>z_{s}$, the mark is unique in $q=pd$. The condition $d\mid m$ is exactly $e\equiv N\pmod{pd}$. The original gates give $c_{R,K}(pd)=\mu_{d}^{+}(R/p)$. Summation gives precisely $S_{6}\leq A_{\mathrm{mix}}$ and $Z_{6}\geq 0$, with all mark multiplicities. A product-cell partition only splits these exact sums. It must keep the original $pd<R$, the cubic gates and every clipped endpoint. In particular $c_{\mathrm{lo}}$ and $\Gamma$ keep the gates at $Q_{\mathrm{RA}}$. Subdivide each of the three marked pieces in (6.7) into cells in its two displayed coordinates. Let $\delta=L^{-D_{\mathrm{m}}}$, and use half-open multiplicative intervals of relative width at most $1+\delta$ in each coordinate. Insert all row cuts, the points $N/4$, $1/3-\eta_{\mathrm{m}}$, $1/3$, and the two endpoints of $I_{H}$ as cell boundaries. Denote the resulting finite, disjoint cell families by $\Theta^{\sharp}_{L},\Theta^{\sharp}_{\mathrm{tail}},\Theta^{\sharp}_{t}$. Use the separate piece index $Y\in\{L,\mathrm{tail},t\}$ and the mapping

(4.33) $$(I_{Y},K_{Y},R_{Y})=\begin{cases}(I_{L},J,R_{L}),&Y=L,\\ (I_{H},J_{\mathrm{tail}},R_{\mathrm{tail}}),&Y=\mathrm{tail},\\ (I_{H},J_{t},Q_{\mathrm{RA}}),&Y=t.\end{cases}\qquad(2\mid N,\ N\geq N_{6}).$$

Thus each $C\in\Theta^{\sharp}_{Y}$ satisfies $C\subseteq(\mathbb{P}\cap I_{Y})\times(\mathbb{P}\cap N^{K_{Y}})$ and $(p,N)=1$, with $0<(N-e)/p$ on every included occurrence. Their unions are exactly the corresponding coprime marked source pieces $I_{L}\times J$, $I_{H}\times J_{\mathrm{tail}}$, and $I_{H}\times J_{t}$. A cell touching a boundary is kept as a boundary cell. For $Y\in\{L,\mathrm{tail},t\}$ and $C\in\Theta^{\sharp}_{Y}$, define

(4.34) $$\begin{aligned} S_{6,Y,C}&:=\sum_{(e,p)\in C}\mathbf{1}_{p\in\mathbb{P}}\mathbf{1}_{(p,N)=1}\mathbf{1}_{p\mid N-e}\mathbf{1}_{0<(N-e)/p}\mathbf{1}_{e\in\mathbb{P}}\mathbf{1}_{P^{-}((N-e)/p)\geq z_{s}}\mathbf{1}_{((N-e)/p,Np)=1},\\ A_{Y,C}&:=\sum_{(e,p)\in C}\mathbf{1}_{p\in\mathbb{P}}\mathbf{1}_{(p,N)=1}\mathbf{1}_{p\mid N-e}\mathbf{1}_{0<(N-e)/p}\mathbf{1}_{e\in\mathbb{P}}\sum_{\begin{subarray}{c}d\mid(N-e)/p\\ d\mid P_{\mathrm{RA}}\end{subarray}}c_{R_{Y},K_{Y}}(pd)\mathbf{1}_{e\equiv N\ (pd)},\end{aligned}\qquad(2\mid N,\ N\geq 2).$$

Here $R_{Y}$ is $R_{L}$, $R_{\mathrm{tail}}$, or $Q_{\mathrm{RA}}$. The corresponding coefficient gates stay in the inner sum. The half-open families make every marked occurrence belong to exactly one $C$, including $e=N/4$ and the shared mark endpoint. The one-term Rosser inequality at each occurrence gives $S_{6,Y,C}\leq A_{Y,C}$, cell by cell. Reindexing the finite sums gives

(4.35) $$S_{6}=\sum_{Y}\sum_{C\in\Theta^{\sharp}_{Y}}S_{6,Y,C},\qquad\sum_{Y}\sum_{C\in\Theta^{\sharp}_{Y}}A_{Y,C}=A_{\mathrm{mix}}.\qquad(2\mid N,\ N\geq N_{6}).$$

The finite comparison uses these cellwise Rosser inequalities. Thus the displayed comparison is the sum of finitely many explicit cell inequalities. No product cell, endpoint, or boundary strip is omitted. The later complete-cell estimates and boundary-strip bound are quantitative inputs only and do not alter this finite identity. The three slacks below are cited or defined by the source comparisons above, before this substitution. Substitution in (4.15) gives

(4.36) $$\begin{aligned} \mathcal{W}-4B_{b}-4E_{\mathrm{pt}}={}&R_{\mathrm{fix}}+S_{2}-A_{\mathrm{mix}}-U_{\mathrm{sw}}-4U_{b}\\ &-E_{\mathrm{sw}}^{\mathrm{src}}-4E_{b}-4E_{\mathrm{pt}}+Z_{6}+Z_{\mathrm{sw}}+4Z_{b}.\end{aligned}\qquad(2\mid N,\ N\geq N_{6}).$$

This equality follows from the source identity and the cited and local definitions. The Section 6 comparison (6.9) gives $Z_{6}\geq 0$. The comparisons in (4.30) give $Z_{\mathrm{sw}},Z_{b}\geq 0$. For even $N\geq\max\{N_{6},N_{\mathrm{pt}}\}$, (4.13) and nonnegativity of these slacks yield

(4.37) $$\begin{aligned} 4G_{b}(N)\geq{}&R_{\mathrm{fix}}+S_{2}-A_{\mathrm{mix}}\\ &-(T_{\mathrm{sw},0}-T_{\mathrm{sw},1}+T_{\mathrm{sw},2}-R_{\mathrm{sw},3}+P_{\mathrm{sw},3})\\ &-4(T_{b,0}-T_{b,1}+T_{b,2}-R_{b,3}+P_{b,3})\\ &-E_{\mathrm{sw}}^{\mathrm{src}}-4E_{b}-4E_{\mathrm{pt}}.\end{aligned}\qquad(2\mid N,\ N\geq\max\{N_{6},N_{\mathrm{pt}}\}).$$

The two expanded signs are

(4.38) $$\begin{aligned} -U_{\mathrm{sw}}&=-T_{\mathrm{sw},0}+T_{\mathrm{sw},1}-T_{\mathrm{sw},2}+R_{\mathrm{sw},3}-P_{\mathrm{sw},3},\\ -4U_{b}&=-4T_{b,0}+4T_{b,1}-4T_{b,2}+4R_{b,3}-4P_{b,3}.\end{aligned}\qquad(2\mid N,\ N\geq N_{6}).$$

Use one bound for the complete $S_{\mathrm{sw}}$. Paper II gives its coefficient-eight estimate [27, (1.9), (5.29)–(5.43)]; the Buchstab upper bound concerns the same occurrence. Sections 7 and 8 estimate these two five-term upper expressions (four principal rows and their peel term) separately. The displayed expansion fixes each peel term’s sign. The finite switched-source domination is already in the local $Z_{\mathrm{sw}}$ identity above. The remaining distribution and quantitative estimates for this occurrence are separate hypotheses of the later bill. They lie outside the finite decomposition of this section.

4.5. The lower expression with the original $S_{2}$

Keep the original $S_{2}$ throughout. Use $\overline{H}$, $M_{5,\mathrm{internal}}$ and $\Delta D_{L3}$ from (1.8). Here $\overline{H}$ is the switched integral bound. The distinct constant $153/256$ bounds the packing envelope. The complete normalized switched error is $E_{\mathrm{sw}}(N;\sigma)$. The source-restoration error $E_{\mathrm{sw}}^{\mathrm{src}}$ is unnormalized. The main route takes $g_{6}:=0$ and excludes the optional six-endpoint improvement. Fix this common parameter and an even $N\geq\max\{N_{6},N_{\mathrm{pt}}\}$ throughout the following conditional lemma. Take $W,E_{\mathrm{van}},\Lambda^{0}_{j,N},\Lambda_{N},\Delta_{j},\Delta_{G1}$ from (2.2), (1.12), and Section 5, and $g_{\mathrm{RA,lo}},E_{\mathrm{RA}}$ from (6.72). Use the complete switched and exterior errors from (8.72) and (7.67). The fixed constants are defined in (1.8) and (1.7).

Lemma 4.3 (Addition of the four comparisons).

Assume (4.11) and the following four inequalities. Write

(4.39) $$\begin{aligned} \frac{R_{\mathrm{fix}}+S_{2}}{W(N)}&\geq D_{\mathrm{II}}+\overline{J}_{6}+2\overline{H}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}-E^{\prime}(N;\sigma),\\ E^{\prime}(N;\sigma)&=E_{\mathrm{van}}(N;\sigma)+\sum_{j\in\mathcal{H}}\bigl{(}|\Lambda^{0}_{j,N}|+|\Lambda_{N}(\Delta_{j})|\bigr{)}+|\Lambda_{N}(\Delta_{G1})|.\end{aligned}\qquad(2\mid N,\ N\geq\max\{N_{6},N_{\mathrm{pt}}\}\ \text{(assumed)}).$$

Here $D_{\mathrm{II}}$ denotes the original Paper II floor. The main route does not use the six endpoint replacements. Here $\mathcal{H}=\{4,5,7,10\}$, with $5$ denoting the lower part of $S_{5}$ and $10$ its retained $S_{10}$ part. Section 5 gives the hypotheses for this estimate.

The other three comparisons have the forms

(4.40) $$\begin{aligned} \frac{S_{6}}{W(N)}&\leq\overline{J}_{6}-g_{\mathrm{RA,lo}}+E_{\mathrm{RA}}(N;\sigma),\\ \frac{S_{\mathrm{sw}}}{W(N)}&\leq 2\overline{H}-\Delta D_{L3}+E_{\mathrm{sw}}(N;\sigma),\\ \frac{4B_{b}}{W(N)}&\leq c_{K2,G0}I(b)+E_{K}(N;b,\sigma).\end{aligned}\qquad(2\mid N,\ N\geq\max\{N_{6},N_{\mathrm{pt}}\}\ \text{(assumed)}).$$

All error terms are nonnegative.

Then the following normalized lower bound holds:

(4.41) $$\begin{aligned} \frac{4G_{b}(N)}{W(N)}&\geq D_{\mathrm{II}}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}+g_{\mathrm{RA,lo}}+\Delta D_{L3}-c_{K2,G0}I(b)-\mathcal{E}_{B}(N;b,\sigma),\\ \mathcal{E}_{B}(N;b,\sigma)&=E^{\prime}(N;\sigma)+E_{\mathrm{RA}}(N;\sigma)+E_{\mathrm{sw}}(N;\sigma)+E_{K}(N;b,\sigma)+\frac{4E_{\mathrm{pt}}}{W(N)}.\end{aligned}\qquad(2\mid N,\ N\geq\max\{N_{6},N_{\mathrm{pt}}\}).$$

Proof.

Substitute the four assumed estimates into (4.15) and use (4.13). The terms $\overline{J}_{6}$ and $2\overline{H}$ cancel exactly. Each complete upper source is subtracted once. In particular, no additional switched majorant is added.

Using the same retained constant as in Section 9, put

(4.42) $$\begin{aligned} D_{B}&=D_{\mathrm{II}}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}+g_{\mathrm{RA,lo}}-\frac{1}{10000}+\Delta D_{L3},\\ \mathcal{E}_{B}(N;b,\sigma)\leq\frac{1}{10000}&\quad\Longrightarrow\quad\frac{4G_{b}(N)}{W(N)}\geq D_{B}-c_{K2,G0}I(b).\end{aligned}\qquad(2\mid N,\ N\geq\max\{N_{6},N_{\mathrm{pt}}\}).$$

∎

5. The retained side

Keep every original source of Paper II, including the complete sum $S_{2}$. Fix $0<\sigma\leq 10^{-10}$ satisfying Paper II, (1.27). Put

(5.1) $$L=\log N,\qquad\tau=\sigma^{9},\qquad\rho=1+\tau,\qquad h=h_{G}=\frac{9}{5000},\qquad\eta_{L}=\frac{1}{10000}.$$

The choice of $\sigma$ precedes the limit. Use $W$ from (2.2). Write $R_{\mathrm{fix}}$ for (4.14).

The comparison required in this section is

(5.2) $$\frac{R_{\mathrm{fix}}+S_{2}}{W} \geq D_{\mathrm{II}}+g_{6}+\overline{J}_{6}+2\overline{H}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}-E^{\prime}(N;\sigma),$$

$$E^{\prime}(N;\sigma) =E_{\mathrm{van}}(N;\sigma)+\sum_{j\in\mathcal{H}}\bigl{(}|\Lambda^{0}_{j,N}|+|\Lambda_{j,N}|\bigr{)}+|\Lambda_{G,N}|,\qquad\mathcal{H}=\{4,5,7,10\}.$$

In $\mathcal{H}$, indices $5$ and $10$ denote the lower $S_{5}$ and retained $S_{10}$ parts. Every displayed error is nonnegative. The notation agrees with (1.12):

(5.3) $$\Lambda_{j,N}=\Lambda_{N}(\Delta_{j}),\qquad\Lambda_{G,N}=\Lambda_{N}(\Delta_{G1}).$$

The required limit is $E^{\prime}(N;\sigma)\to 0$. Proposition 5.10 proves (5.2). Its inputs are listed in Remark 5.9. The six-endpoint lemma is optional: without it, the same proof gives (5.2) with $g_{6}$ replaced by $0$. We display the unchanged comparisons first. Then we specify the five incremental and four saved-array conversion terms.

5.1. The original signed chains

Let $\mathcal{I}=\mathcal{I}_{\mathrm{row}}\sqcup\{\mathrm{T5}\}$ index the signed chains of $R_{\mathrm{fix}}+S_{2}$. Here $\mathcal{I}_{\mathrm{row}}$ is the set of Paper II’s nineteen row chains. The chain $\mathrm{T5}$ is Paper II’s outer chain for the upper part of $S_{5}$ [27, (2.28), (5.10), (6.1) and §6.1, condition 1]. For $j\in\mathcal{I}$, let $Q_{j,k}$ include its normalized coefficient. Paper II uses ten stages [27, (6.4)–(6.4d)]:

(5.4) $$Q_{j,0}=Q_{j,10}+\sum_{k=1}^{10}(Q_{j,k-1}-Q_{j,k}).$$

The initial expression is the actual arithmetic source. The last expression is a directed rational endpoint. For $j=\mathrm{T5}$ the stages are those of Lemma 5.1: $Q_{\mathrm{T5},k}$ is defined there for $k\leq 3$, and $Q_{\mathrm{T5},k}=Q_{\mathrm{T5},3}$ for $3\leq k\leq 10$. For the three joint groups, use

(5.5) $$\mathcal{O} \in\{S_{1},S_{2}\text{-}T_{8}/4,G_{1}\},\qquad Q_{\mathcal{O},9}=\sum_{j\in\mathcal{O}}Q_{j,9},$$

$$\sum_{j\in\mathcal{O}}Q_{j,0} =E_{\mathcal{O}}+(Q_{\mathcal{O},9}-E_{\mathcal{O}})+\sum_{j\in\mathcal{O}}\sum_{k=1}^{9}(Q_{j,k-1}-Q_{j,k}),$$

$$Q_{\mathcal{O},9}-E_{\mathcal{O}} \geq 0.$$

The joint signs are kept before rounding. The $G_{1}$ weights are $(2,1,1)$.

Signed sources and parent assignment.

Here “parent” records the assignment of an arithmetic occurrence and its source-restoration share. It does not assert that each Buchstab child is numerically bounded by the unsplit parent. Let $T_{j,r}$ be the nonnegative plain-source terms of Paper II, (2.19) and (3.2). Let $\Delta_{j}^{\rm src}$ be its signed source correction. Then

$$3S_{1}=3(T_{1,0}-T_{1,1}+T_{1,2}-T_{1,3})+3\Delta_{1}^{\rm src},\qquad S_{2}=T_{2,0}-T_{2,1}+T_{2,2}-T_{2,3}+\Delta_{2}^{\rm src}.$$

For every third-depth S1 occurrence, $E_{3}\leq p_{3}$ gives $-3S_{\rm pl}(b,p_{3})\geq-3S_{\rm pl}(b,E_{3})$. For S2 we make this replacement only on Paper II’s selected whole-$K$ blocks $\mathcal{F}$, including $PF$ and $\mathcal{F}\cap R1$. There $-S_{\rm pl}(b,p_{3})\geq-S_{\rm pl}(b,E_{3})$. On the strict complement $\mathcal{B}$ keep the actual $p_{3}$ and Paper II’s strict-BV comparison. The last child of S2 keeps its negative sign on all three pieces $PF$, $\mathcal{F}\cap R1$, and $\mathcal{B}$. The strict BV piece stays in S2.

For the fixed-cutoff rows use their literal divisor domains. Their signed sources are $-2S_{3}$, $-S_{4}$, $-S_{5,\rm low}$, $-S_{7}$, $+S_{8}$, and $+S_{9}$, respectively. On each source cell, an upper sieve bound $S\leq U$ gives $cS\geq cU$ when $c<0$. A lower sieve bound $L\leq S$ gives $cS\geq cL$ when $c>0$. These comparisons keep the source corrections and the row’s allocated errors. Paper II, (2.28), gives the exact assignment identity

$$-S_{5}=-S_{5,\rm low}-S_{5,\rm top},\quad S_{5,\rm low}=\sum_{1/4\leq\log_{N}p<t_{*}}S(p,N^{\alpha}),\quad S_{5,\rm top}=\sum_{t_{*}\leq\log_{N}p<\tau_{S}}S(p,N^{\alpha}).$$

The endpoint $t_{*}$ belongs to the upper/BV part. The split precedes the structured sieve or Heath–Brown expansion and keeps parent labels. S5-low and T5 both have parent S5. S5-low gets no restoration address of its own. For S10 the source partition $S_{10}=S_{10,\rm Ret}+S_{10,\rm Drop}$ is exact, with $S_{10,\rm Ret},S_{10,\rm Drop}\geq 0$. Hence $+S_{10}\geq+S_{10,\rm Ret}$ before its lower sieve bound.

The joint source keeps the representation weights $(2,1,1)$. Paper II, (4.14), gives, with its stated exceptional-source and small-prime allowance $E_{G}\geq 0$,

$$G_{1}\leq 2S(\mathcal{B}_{11},N^{1/2})+S(\mathcal{B}_{12},N^{1/2})+S(\mathcal{B}_{13},N^{1/2})+E_{G}.$$

Make its cut-measure replacement with the original signed correction before applying (4.18). The four signed descendants of $-G_{1}$ then have coefficients $(-1,+1,-1,+1)$ on the single weighted source. Their last positive term is bounded from below by its retained cells. The change from actual $a_{3}$ to label base $E$ keeps the signed peel correction of Paper II, (4.45)–(4.57). The four rows share one joint source parent and one restoration allocation. Thus S1, S2, S4, S5, S7, S10 and G1 are the seven parents of the sixteen listed retained rows. S3, S8 and S9 keep their original single-comparison parents, and T5 remains the upper part of S5. These are sign-correct source reductions for the complete parent chains. They do not discharge the distinct S6, bad-source or switched-source comparisons in Section 4.

The nineteen original row chains of $\mathcal{I}_{\mathrm{row}}$ are the sixteen rows of Appendix B.7 and Paper II’s original $S_{3},S_{8},S_{9}$ comparisons. The strict BV piece of $S_{2}$ lies inside the joint $S_{2}$ group. The source $-S_{5}$ of $R_{\mathrm{fix}}$ is split at $N^{t_{*}}$, $t_{*}=7109/28000$, before any sieve step. Its lower part is the row $S_{5}$-low of $\mathcal{I}_{\mathrm{row}}$. Its upper part is the chain $\mathrm{T5}$. The chains of $S_{6}$ and $S_{\mathrm{sw}}$ are excluded. Thus the endpoint sum over $\mathcal{I}$ is

(5.6) $$\sum_{j\in\mathcal{I}}Q_{j,10}=B_{\mathrm{print}}-\delta_{G1}+\overline{J}_{6}+2\overline{H},\qquad\delta_{G1}=4.98\cdot 10^{-54}.$$

This follows from Paper II’s printed thirteen-term sum [27, §6.4, immediately before (6.6)]. Its joint endpoint is $Q_{G1,10}=L_{G1}-\delta_{G1}$; the printed sum uses $L_{G1}$. The term $-\overline{J}_{5}$ of (5.7) is $Q_{\mathrm{T5},10}$, by Lemma 5.1. Apart from $-\overline{J}_{6}-2\overline{H}$, every other term of (5.7) is an endpoint. It ends a chain of $\mathcal{I}_{\mathrm{row}}$ or a joint group. This is a finite coefficient identity. The printed sum is

(5.7) $$B_{\mathrm{print}}={} \underline{D}_{1}+\underline{d}_{2}-\frac{U_{3,A}}{2}-\frac{U_{4}+U_{5,\mathrm{low}}+U_{7}}{4}$$

$$+\frac{L_{8,A}+L_{9,A}+L_{10}}{4}+L_{G1}-\overline{J}_{5}-\overline{J}_{6}-2\overline{H}$$

$$> 1.83403742109641331345957914542194044932304169727869681073064.$$

The letters denote the rational endpoints printed in Paper II, Appendix A.2–A.6 and (A.7), (A.10), (A.13)–(A.14). The constant $\overline{J}_{6}$ keeps its rational definition:

(5.8) $$\overline{J}_{6}=1.3148581959157321533513247276782285310758770468886237619585.$$

The upper part of $S_{5}$ has the following outer chain. Put $t_{*}=7109/28000$, $\tau_{S}=57/224$, and let

(5.9) $$\overline{J}_{5}=\frac{1}{4}\bigl{(}c_{5}+8.84\cdot 10^{-57}\bigr{)},\qquad c_{5}=0.03658166423843258896399704402764682632822246022922592990,$$

which is Paper II’s directed upper endpoint for $J_{5}(0)=\int_{t_{*}}^{\tau_{S}}\frac{dt}{t(1/2-t)}$ [27, (5.16), (5.20), (A.14)–(A.14a)].

Lemma 5.1 (The outer chain of the upper part of $S_{5}$).

Define

$$Q_{\mathrm{T5},0} =-\frac{1}{W}\sum_{t_{*}\leq\log_{N}p_{1}<\tau_{S}}S(p_{1},N^{\alpha}), Q_{\mathrm{T5},1} =-\frac{T_{5}}{W},$$

$$Q_{\mathrm{T5},2} =-J_{5}(0), Q_{\mathrm{T5},3} =-\overline{J}_{5},$$

where the first sum is the second term of Paper II, (2.28), and $T_{5}$ is Paper II, (5.10). Then

$$Q_{\mathrm{T5},0}-Q_{\mathrm{T5},1}\geq-E_{\mathrm{T5},1}(N),\qquad Q_{\mathrm{T5},1}-Q_{\mathrm{T5},2}\geq-E_{\mathrm{T5},2}(N),\qquad Q_{\mathrm{T5},2}-Q_{\mathrm{T5},3}\geq 0,$$

with $E_{\mathrm{T5},1},E_{\mathrm{T5},2}\geq 0$ tending to $0$ at fixed $\sigma$. Both are entries of $E_{\mathrm{van}}$. In particular $Q_{\mathrm{T5},0}\geq-\overline{J}_{5}-E_{\mathrm{T5},1}(N)-E_{\mathrm{T5},2}(N)$.

Proof.

Paper II splits $-S_{5}$ at $N^{t_{*}}$ before the structured sieve or HB. The endpoint $t_{*}$ belongs to the upper part [27, (2.28) and the three lines after it]. The coefficient is $-1$.

Stage one. For a prefix $p_{1}\in\mathcal{P}_{N}$, dropping the condition $(m,Nb)=1$ gives $S(p_{1},N^{\alpha})\leq S_{p_{1}}(N^{\alpha})$ in the notation of Paper II, (5.9) [27, the sentence after (5.9)]. Prefixes $p_{1}\mid N$ are removed at their first failure. This is the error-table entry “Exceptional source factors”, of size $O_{\sigma}(L^{3}/N+N^{-\alpha/2+o(1)})$ [27, §6.3]. A prime equal to $N^{t_{*}}$ or $N^{\tau_{S}}$ divides $N$ [27, after (5.20)]. With the coefficient $-1$ this gives the first inequality.

Stage two. Paper II applies the upper sieve (5.4) to each $\mathcal{A}_{b}$, $b=p_{1}$. Its parameters are $X_{\mathcal{A}}=\operatorname{Li}(X)/\varphi(b)$, level $Q=N^{1/2}L^{-B}$ ($\kappa=0$), $D_{b}=Q/b$ and $z_{b}=N^{\alpha}$ [27, (5.12)]. For $t=\log_{N}p_{1}\in[t_{*},\tau_{S})$ one has $a_{0}(t)=1/2-t\geq 55/224$, $t_{*}-\alpha=5368241/32172000>0$, and

$$\frac{a_{0}}{\alpha}\in\Bigl{[}\frac{12639}{4480},\frac{7917759}{2800000}\Bigr{]}\subset(2,3)$$

[27, table after (5.12), row T5]. Hence eventually $s_{b}=(a_{0}-\delta_{N})/\alpha\in(2,3)$, where $F(s)=2e^{\gamma}/s$ [27, between (5.4) and (5.5)]. By (5.14) the remainders are $o(U_{C})$. The main terms are (5.15), and (5.16) gives $T_{5}\leq 4(1-\varepsilon)U_{C}J_{5}(0)+o(U_{C})$ with the vanishing logarithmic loss (5.21) [27, (5.14)–(5.16), (5.21) and the sentence “S5 keeps its vanishing logarithmic loss”]. Since $W=4(1-\varepsilon)U_{C}$, this is the second inequality. Its error is the error-table entry “S5 BV remainder” plus the prime-measure and Euler conversion of that row [27, §6.3].

Stage three. By Paper II, (A.14)–(A.14a), $4J_{5}(0)$ lies in the ball of centre $c_{5}$ and radius $8.84\cdot 10^{-57}$, so $J_{5}(0)\leq\overline{J}_{5}$. Paper II also gives the independent rational enclosure $P_{5}<4J_{5}(0)<P_{5}+R_{5}$ inside the same ball [27, after (A.14a)].

The chain uses $\kappa=0$, so the increment $J_{5}(\kappa)-J_{5}(0)$ does not occur. Paper II’s allowance $E_{5,\kappa}=4\cdot 10^{-8}$ is a summand of $g_{\mathrm{BV}}=0.00000575$ in (5.10) and stays there once. Paper II, Appendix A, records the hypothetical increment and its once-booked decomposition. They are in the paragraph opening For comparison, if S5-top used $\kappa=10^{-6}$ and in the BV-allowance line right after it [27, Appendix A, comparison paragraph and following decomposition]. Equation (A.14a) gives only the nominal value at $\kappa=0$. ∎

The chain $\mathrm{T5}$ does not meet the changes of this section. Its three stages use no sieve level $\ell_{c}$ and no saved profile. They use no Class II, packing or Class I term. Paper II’s prepared Class II, packing and Class I contributions of this row are zero [27, §2.4, paragraph “S5-top and S6 use BV separately”]. The lift below changes the kernels on $I_{4}$, $I_{5}=[1/4,t_{*})$, $I_{7}$ and $\mathcal{D}_{10}$ only. The interval $I_{5}$ is disjoint from $[t_{*},\tau_{S})$. The joint shift changes the $G_{1}$ cone only. Hence Lemma 5.1 holds unchanged after both changes.

Keep the whole fixed deduction of Paper II:

(5.10) $$F_{\mathrm{net}}={} 0.000006484294227956901+0.00000575+0.00002853328$$

$$+4.98\cdot 10^{-54}+2\cdot 10^{-24}+10^{-390}$$

$$<0.000040767574227956903.$$

These are Paper II’s six deductions [27, (6.6)]. The first remains paid although its source is excluded. The joint $G_{1}$ reconciliation also remains paid. Every adverse row difference has an absolute bound. Keep their whole vanishing envelope, $E_{\mathrm{van}}$. The row estimates required here are

(5.11) $$\sum_{j\in\mathcal{I}}\sum_{k=1}^{10}(Q_{j,k-1}-Q_{j,k})\geq-\bigl{(}F_{\mathrm{net}}-\delta_{G1}\bigr{)}-E_{\mathrm{van}}(N;\sigma),\qquad E_{\mathrm{van}}(N;\sigma)\longrightarrow 0.$$

The $\delta_{G1}$ part is already in the endpoint sum, so the remaining chain deduction is $F_{\mathrm{net}}-\delta_{G1}$. Their total is exactly $F_{\mathrm{net}}$. The fixed AA37 share is paid under the admissible condition $0<\sigma\leq 4g_{\mathrm{AA}}/(C_{E}W_{*})$ of Paper II, (1.27) [27]. See Paper II, §6.5. This condition is part of the fixed-parameter choice made before the limit. The analytic inputs are Paper II, (6.4b)–(6.5), and its direct and $G_{1}$ estimates [27, (4.66)–(4.69), (6.4b)–(6.5)]. Their logarithmic parameters are

(5.12) $$K_{0}=8042,\qquad A^{\prime}=8047,\qquad A_{\mathrm{work}}=8051,\qquad K_{0}+2-A^{\prime}=-3.$$

These are Paper II, (4.66), for the original coefficient families. This section leaves them unchanged.

Lemma 5.2 (Inherited chain errors).

The bound (5.11) holds, and $E_{\mathrm{van}}(N;\sigma)\to 0$ as $N\to\infty$ at fixed $\sigma$. Each row or separately listed stage difference used below has an allocated nonnegative share in Paper II’s error table. Denote it by $\mathcal{E}_{j}$. Paper II, §6, charges these shares once. An entry may be grouped across rows, so the lemma claims nothing for arbitrary subfamilies and does not claim pairwise disjoint entries. The only omission bounds used are those for the whole chains or stage differences named in the proof below.

Proof.

The signed-assembly theorem of Paper II proves $D_{\mathrm{src}}\geq B_{\mathrm{main}}-F_{\mathrm{pers}}-E_{\mathrm{van}}$ with $E_{\mathrm{van}}\to 0$ for the source (6.1) [27, §6.1, (6.1)–(6.2), (6.4)–(6.4d)]. Its proof bounds every adverse stage difference by an allocated share of Paper II’s error table, §6.3. Each entry is a nonnegative bound or a proved zero; some entries are grouped across rows. Paper II assigns each tuple to its first failed condition and charges each difference or part once below [27, §6.3]. The fixed entries are the six deductions of (5.10), that is $F_{\mathrm{pers}}=F_{\mathrm{net}}$ [27, (6.6)]. The others tend to zero at fixed $\sigma$, and their sum is $E_{\mathrm{van}}$. The rows of $\mathcal{I}$ are rows of (6.1). The $S_{6}$ level increment is still deducted when that chain is omitted. The $\delta_{G1}$ part is already in (5.6). This gives (5.11). For $\mathrm{T5}$ the differences and entries are those of Lemma 5.1. Separately listed entries cover the omissions used in this chapter. The whole $S_{6}$ and $S_{\mathrm{sw}}$ chains use the S6 BV remainder, S6 level increment and $E_{S_{\mathrm{sw}}}$ entries. The whole $G_{1}$ chain uses the §4.6 table and (4.85) of Paper II [27, §4.6, (4.85)]. The stage-nine strip for the retained part of $S_{10}$ uses the S10 selector entry. These bounds apply to the named whole chains or stage difference after allocating their shares. No arbitrary omitted subfamily is asserted. For the direct rows, the stage-nine entry “Profile comparison” is zero adverse. The stage-ten entry “Directed enclosure” is zero additional [27, §6.3]. For the retained part of $S_{10}$, the strip between the retained cells and $\mathcal{D}_{10}$ is the entry “S10 selector”, $7.160\tau$. The term $g_{\tau}=2\cdot 10^{-24}$ replaces it [27, §6.3, rows “S10 selector” and $g_{\tau}$].

The entries rest on the following inputs, all used by Paper II on these same sources. The linear-sieve main terms and remainders are Paper I, Theorem 2.6, (2.30)–(2.33b). This is the structured form of Iwaniec’s Theorem 4 [13, Theorem 4, p. 311], applied with $D=L_{c}$, $\epsilon_{s}=\sigma$ and $\delta_{s}=\tau$ [27, §6.2]. Before the exterior sign, applied once, each block remainder splits as

$$\frac{C_{c}}{M_{c}}\langle\gamma_{c},\Delta_{P}\rangle+NC_{\mathrm{sf},c}-\langle\Xi_{U,c},\Delta_{P}\rangle+\langle\Xi_{A,c},\Delta_{0}-\Delta_{P}\rangle,$$

with $\Xi_{A,c}=O_{c}\mathbf{1}_{q\ \mathrm{not\ squarefree}}$ [27, (6.4a)–(6.4b), (4.23)]. The term $NC_{\mathrm{sf}}$ vanishes on the direct rows, where $\Delta_{0}=\Delta_{P}$. On $G_{1}$ it is $O_{\sigma}(N^{-\alpha}L^{5})$ by Paper II, (4.64)–(4.65). The terms $\Xi_{U}$ and $\Xi_{A}$ are bounded jointly, once, by $232\Lambda_{\sigma}C_{\mathrm{ret},\sigma}N^{-\ell_{0}\sigma^{2}}(1+L)^{4}$, with $\ell_{0}=264897/2060000$ [27, (B.1)–(B.7)]. Here $C_{\mathrm{ret},\sigma}$ is the retained-side combinatorial constant; it depends on the fixed $\sigma$ and not on $N$. The persistent part $C_{E}\sigma W_{\mathrm{AA}}/4$ of Paper II, (B.8), lies in $g_{\mathrm{AA}}$; only its secondary part belongs to $E_{\mathrm{van}}$. The Class II estimate (6.5) of Paper II applies Paper I, Theorem 3.2, (3.13), with the precisions (5.12). The spectral input of that theorem is Paper I, Theorem 2.8; see Fact 3.22. In the large-norm case of Paper I, (3.48), its mean-square input is Lemma 6.5(ii), in place of [3, Theorem 0(a)]. Lemma 6.5(ii) applies under the Paper I hypotheses and selected parameter box stated there. It uses neither the optional six-endpoint lemma nor any broader BFI86 scope. Its premises are checked for these sources as follows. In Paper I, conditions (3.1), (3.2), (3.6), (3.8) with the length condition, and (3.11) display its hypotheses (H1)–(H5). Paper II verifies them for these same sources, coefficients, boxes and scales [27, §6.2, hypothesis table, row “C1–C5, LEN”]. Paper I proves condition (3.25) for every selected factor of this form. The proof uses the prime Siegel–Walfisz theorem (2.21) [26, (3.25) and its proof]. Conditions (3.26)–(3.27) are the definition of $\lambda_{r}$ and the Cauchy–Schwarz step of that proof. They hold under (H1)–(H5), with $|\lambda_{r}|\leq\tau(r)$ from the roughness in (H2) [26, (3.26)–(3.27)]. Stages one to eight are unchanged, so these are Paper II’s own checks. The prime-measure and Euler conversions are the prime number theorem, Mertens’ theorem and Paper II, (4.76)–(4.78). The profile lift and the shifted joint profile below act at stages nine and ten only. They change no source, cell, level, mask, box, norm or precision at stages one to eight. Hence $E_{\mathrm{van}}$ is Paper II’s own envelope for these rows. ∎

Paper II’s rational floor satisfies

(5.13) $$B_{\mathrm{print}}-F_{\mathrm{net}}>D_{\mathrm{II}}=\frac{45849916338009170593}{25000000000000000000};$$

see [27, (6.3a), (6.6)]. The selected S2 endpoint used in this floor is $d_{T8}$. Paper II’s tail reserve is

(5.14) $$\delta_{T8}=\frac{1}{20000},\qquad d_{T8}=\frac{P_{-}-(2/e^{\gamma})V_{\mathrm{full}}-(2/e^{\gamma})J_{+}+C_{-}}{4}-\delta_{T8}.$$

Thus $\delta_{T8}$ is deducted once inside $d_{T8}$, as in Paper II, (A.7)–(A.8). It is not a second deduction from $D_{\mathrm{II}}$. The separate fixed term $275009680\tau$ remains outside $d_{T8}$ in the S2 estimate [27, (3.22)–(3.23), (A.7)]. Combining the preceding estimates gives

(5.15) $$\frac{R_{\mathrm{fix}}+S_{2}}{W}\geq D_{\mathrm{II}}+\overline{J}_{6}+2\overline{H}-E_{\mathrm{van}}(N;\sigma).$$

This deduction uses the individual signed chains. Its analytic input is (5.11).

5.2. Six improved endpoints

Let

(5.16) $$\mathcal{J}=\{1,2,4,5,7,10\}.$$

For $j\in\mathcal{J}$, let $P_{j}$ be its Paper II profile integral. The indices $5$ and $10$ keep the preceding conventions. Let $O_{j}$ and $N_{j}$ be directed lower endpoints. The required numerical comparisons are

(5.17) $$P_{j}\geq N_{j}\geq O_{j},\qquad P_{j}-O_{j}=(N_{j}-O_{j})+(P_{j}-N_{j}).$$

The old and improved directed endpoints are the following exact terminating decimals. They are given in the normalized D contributions used by the signed sum. For S4, S5-low and S7 the sign $-1/4$ is included; for S10 the sign $+1/4$ is included.

Endpoint source	Directed contribution in D units
S1, [27, (A.4), (A.5a)]	$O_{1}=9.67653874894124412908238749318694090949698021569432709634$
	$N_{1}=9.676557035420731097796376157212899872020518126053458552042$
S2, [27, (A.8), (A.8a)]	$O_{2}=1.6195813569890835255913340433210572922007671408165$
	$N_{2}=1.62063469556829$
S4, [27, Appendix A.2–A.6]	$O_{4}=-0.8342438370724665$
	$N_{4}=-0.8341817006889064605174981431060625$
S5-low, [27, Appendix A.2–A.6]	$O_{5}=-0.05972210829776175$
	$N_{5}=-0.05971885506770766576732140865035175$
S7, [27, Appendix A.2–A.6]	$O_{7}=-0.574850630119163$
	$N_{7}=-0.574815106890481420929612425345365$
S10-Ret, [27, Appendix A.2–A.6]	$O_{10}=0.214711863944485$
	$N_{10}=0.21559427153309934685978873940373255$

The rational endpoint differences sum to

(5.18) $$g_{6} =\sum_{j\in\mathcal{J}}(N_{j}-O_{j})$$

$$=0.002054945489603492768011383006854970322770769542631455702.$$

Exact rational addition of the six entries above gives the displayed sum, and each entry satisfies $N_{j}\geq O_{j}$. The values $O_{j}$ are Paper II’s directed endpoints, and $P_{j}\geq O_{j}$ is proved there [27, (A.4), (A.5a), (A.8), (A.8a), Appendix A.2–A.6]. The values $N_{j}$ are lower bounds read from Paper II’s printed independent enclosures. The following optional lemma states them.

Lemma 5.3 (Six directed endpoints from Paper II; optional).

For each $j\in\mathcal{J}=\{1,2,4,5,7,10\}$, the Paper II profile integral $P_{j}$, with its signed normalized coefficient, satisfies $P_{j}\geq N_{j}$, with $N_{j}$ as tabulated above.

Proof.

For $j=1$, Paper II, (A.5a), gives the printed $3/4$ ball; its lower endpoint is $N_{1}$. For $j=2$, Paper II, (A.8a), gives the printed interval whose lower endpoint is $N_{2}$. Paper II states that this interval covers the continuous functional and the infinite tail. The $d_{T8}$ deduction in (6.6a) is released once in $N_{2}-O_{2}$. The term $F_{\mathrm{net}}$ in (6.6b) contains no $\delta_{T8}$ [27, (6.6a)–(6.6b)]. For $j=4,5,7$, the Appendix A.3 balls include the adverse $\tau=10^{-90}$ shifts. Since the profile function is nonincreasing, the shifted raw integrals are at least the $\tau$-free ones. Their upper endpoints with the negative coefficient therefore give $P_{j}\geq N_{j}$. For $j=10$, the Appendix A.3 $S_{10}$-Ret ball encloses the $\tau$-free actual-coordinate profile on $\mathcal{D}_{10}$. Its lower endpoint gives $P_{10}\geq N_{10}$. The separate $D_{*}\tau$ strip is paid in the $S_{10}$ selector estimate. These are the printed enclosures in Paper II, (A.5a), (A.8a), and Appendix A.3. Its numerical-certificate section records the adverse-shift and actual-coordinate conventions [27, (A.5a), (A.8a), Appendix A.3]. The remaining directed endpoints $O_{j}$ are unchanged. ∎

With $O_{j}$ in place of $N_{j}$, (5.17) is Paper II’s own endpoint comparison and $g_{6}$ is replaced by $0$. Only the endpoint changes in (5.4). All earlier signed differences remain unchanged. Thus (5.17) adds $g_{6}$ once. The $S_{5}$ upper part remains separate. The joint $G_{1}$ endpoint is absent from $\mathcal{J}$.

5.3. Four stronger profile comparisons

Use the sieve functions of Paper I [26, (2.23), immediately before Theorem 2.6]. Put

(5.19) $$\widehat{F}=\frac{F}{2e^{\gamma}},\qquad\widehat{f}=\frac{f}{2e^{\gamma}},\qquad K=\frac{57}{64}-u.$$

Define

(5.20) $$s(x) =\frac{K-5x/2}{\alpha},\qquad q(x,y)=\frac{K-5x/2-y}{\alpha},$$

$$B_{0}(x) =\min\{\beta,\max\{\alpha,K-11\alpha/5-5x/2\}\}.$$

The domains are

(5.21) $$\begin{gathered}I_{4}=[\mu_{*},1/4),\quad I_{5}=[1/4,7109/28000),\quad I_{7}=[\mu_{*},\lambda),\\ \mathcal{D}_{10}=\{(x,y):\mu_{*}\leq x<\lambda,\ \alpha\leq y<B_{0}(x)\}.\end{gathered}$$

These are Paper II’s domains [27, (2.27), (2.29), (2.31), (2.38)–(2.40)]. Set

(5.22) $$\Delta_{j} =\int_{I_{j}}\frac{\widehat{F}(s(x))-\widehat{F}(s(x)+h/\alpha)}{\alpha x}\,dx (j=4,5,7),$$

$$\Delta_{10} =\iint_{\mathcal{D}_{10}}\frac{\widehat{f}(q(x,y)+h/\alpha)-\widehat{f}(q(x,y))}{\alpha xy}\,dx\,dy.$$

The three upper rows carry a negative sign. The lower $S_{10}$ row carries a positive sign. Their original coefficients are included in these gains, so there is no further factor $1/4$.

Paper II’s actual levels give

(5.23) $$\ell_{c} \geq\alpha s(x)+h+\frac{1}{10000}-3\tau (j=4,5,7),$$

$$\ell_{c} \geq\alpha q(x,y)+h+\frac{1}{10000}-\left(\frac{5\lambda}{2}+\beta\right)\tau (j=10).$$

See [27, §2.4, proof of (2.26)–(2.31); §2.5, proof of (2.40)]. The exact split behind the displayed reserve is

(5.24) $$u-\eta_{L}=\frac{19}{10000}=\frac{9}{5000}+\frac{1}{10000}=h_{G}+\frac{1}{10000}.$$

After the $h_{G}$ lift, Paper II’s S4, S5-low (including its B branch) and S7 comparisons leave $1/10000-3\tau$. Its S10-Ret comparison leaves $1/10000-\tau(5\lambda/2+\beta)$. Both are positive for $\tau\leq 10^{-90}$. The G1-3 shift leaves $g_{s}=1/10000-(225/256)\tau>2\mu_{*}\tau$ and the $\mu_{*}$ crossing leaves $1/10000-(75/256)\tau>0$, by [27, (4.72)]. These are pointwise profile-numerator inequalities. They do not change the actual level $\ell_{c}$, the level $D_{c}$, or the masks, boxes, Class-II and AA37 inputs. The other level conditions of Paper II remain pointwise and unmodified. They include the G1 guard, S1–S3, S2/S3/S8/S9, and the whole-K margin. See Paper II, (2.21), (2.40), (4.72), and (4.19). Decreasing $F$ gives the first three comparisons. Increasing $f$ gives the fourth comparison. These operations leave the actual levels unchanged. The H6 lift and the M5 shift act only at stages 9–10, after Class II at stage 7. They change no Class-II, AA37, box, mask or precision input: $\ell_{c}=\Theta_{c}-\eta_{L}-\sum_{i}t_{i}$ contains no saved profile, and (5.12) remains unchanged.

Proposition B.1 gives finite integral enclosures. With the directed calculation of the computational appendix, they prove the following. Its exact rational interval is

$[0.005720047094327988714222991186379229149381887590042773120871489672965022499\mathbin{\pm}7.09\cdot 10^{-76}]$. This is a ball for the sum of the four rowwise lower bounds, not for $\sum_{j}\Delta_{j}$. We use the terminating credit $572004/10^{8}$. It lies below that exact lower endpoint and leaves the existing $10^{-20}$ allowance intact. This allowance is conservative; it is not an additional gain in the final constant.

(5.25) $$\frac{572004}{10^{8}}<\Delta_{4}+\Delta_{5}+\Delta_{7}+\Delta_{10}<\frac{606486}{10^{8}},\qquad\frac{572004}{10^{8}}-10^{-20}>\frac{5719}{10^{6}}.$$

The unchanged $S_{10}$ strip also satisfies

(5.26) $$\frac{10987/10000}{4\alpha^{2}}\left(\frac{5\lambda}{2}+\beta\right)\frac{\lambda-\mu_{*}}{\mu_{*}}=\frac{21137254516197687}{4120000000000000}<5.137<7.160.$$

The lifted $S_{10}$ argument satisfies $q+h/\alpha\leq 11519839/3200000+h/\alpha<4$. Thus $f\leq f(4)=2e^{\gamma}\log 3/4$ throughout the strip, as used in this density bound. These are finite bounds, including their stated domains. The retained cells also leave a reserve. Put

(5.27) $$\begin{gathered}R_{10}=\Bigl{[}\frac{1}{5},\frac{21}{100}\Bigr{]}\times\Bigl{[}\frac{9}{100},\frac{1}{10}\Bigr{]},\qquad\delta_{10}=\frac{1}{20000\alpha},\\ c_{10}=\frac{1}{\alpha}\cdot\frac{7\delta_{10}}{480}\cdot\frac{1}{210}=\frac{146689}{320000000000},\qquad c_{10}-\frac{21137254516197687}{4120000000000000}\,\tau>10^{-7}.\end{gathered}$$

Lemma 5.5 uses (5.26) and (5.27), so its comparison over the whole of $\mathcal{D}_{10}$ needs no deduction. Paper II’s $S_{10}$ selector entry $7.160\tau$ is replaced in Paper II by $g_{\tau}=2\cdot 10^{-24}$, a summand of (5.10). The fixed $g_{\tau}$ amount is deducted once. The retained reserve above eliminates the $S_{10}$ selector deduction in this comparison. The inherited $G_{1}$ peel and its label replacement are jointly covered by that same summand.

We next define the finite profile arrays. Let $X_{0}=[u,1]$. On this compact interval, use the zero extension of the profile outside its displayed domain, and put

(5.28) $$m_{N}=\sum_{\begin{subarray}{c}p\in\mathbb{P},\ p>2,\ p\nmid N\\ N^{u}\leq p\leq N\end{subarray}}\frac{\delta_{\log_{N}p}}{p-1},\qquad m(dt)=\frac{dt}{t}.$$

The symbol $\delta_{x}$ denotes unit mass at $x$. Define

(5.29) $$V_{N}(z)=\prod_{\begin{subarray}{c}p<z,\ p\in\mathbb{P}\\ p\nmid N\end{subarray}}\left(1-\frac{1}{p-1}\right),$$

and

(5.30) $$X_{D} =\sum_{v\in\mathbb{P}}\mathbf{1}_{N/L^{30}<v<(1-\varepsilon)N},$$

$$r_{D,N} =\frac{2e^{\gamma}\alpha X_{D}V_{N}(N^{\alpha})}{W}.$$

For $j=4,5,7$, set

(5.31) $$k_{j}^{0}(x) =-\frac{\mathbf{1}_{I_{j}}(x)}{\alpha}\widehat{F}(s(x)),$$

$$k_{j}^{h}(x) =-\frac{\mathbf{1}_{I_{j}}(x)}{\alpha}\widehat{F}(s(x)+h/\alpha).$$

For $j=10$, set

(5.32) $$k_{10}^{0}(x,y) =\frac{\mathbf{1}_{\mathcal{D}_{10}}}{\alpha}\widehat{f}(q(x,y)),$$

$$k_{10}^{h}(x,y) =\frac{\mathbf{1}_{\mathcal{D}_{10}}}{\alpha}\widehat{f}(q(x,y)+h/\alpha).$$

Put $e_{j}=1$ for $j=4,5,7$, and $e_{10}=2$. The original and lifted arrays are

(5.33) $$P_{j} =\int k_{j}^{0}\,d(m^{e_{j}}), S_{j,N}^{0} =r_{D,N}\int k_{j}^{0}\,d(m_{N}^{e_{j}}),$$

$$S_{j,N}^{h} =r_{D,N}\int k_{j}^{h}\,d(m_{N}^{e_{j}}), B_{j,N} =S_{j,N}^{h}-S_{j,N}^{0}.$$

The powers denote product measures. These formulas use $P_{j}$ from (5.17). Write $d_{j}=k_{j}^{h}-k_{j}^{0}$. For $j=4,5,7$, and then $j=10$,

(5.34) $$d_{j}(x) =\frac{\mathbf{1}_{I_{j}}(x)}{\alpha}[\widehat{F}(s(x))-\widehat{F}(s(x)+h/\alpha)],$$

$$d_{10}(x,y) =\frac{\mathbf{1}_{\mathcal{D}_{10}}}{\alpha}[\widehat{f}(q(x,y)+h/\alpha)-\widehat{f}(q(x,y))].$$

Thus $\Delta_{j}=\int d_{j}\,d(m^{e_{j}})$. Define the two conversion errors:

(5.35) $$\Lambda^{0}_{j,N}=S^{0}_{j,N}-P_{j},\qquad\Lambda_{j,N}=B_{j,N}-\Delta_{j}\qquad(j\in\mathcal{H}).$$

Lemma 5.4 (Finite profile shift).

For the arrays above,

(5.36) $$S_{j,N}^{h}=P_{j}+\Lambda^{0}_{j,N}+\Delta_{j}+\Lambda_{j,N}\geq P_{j}+\Delta_{j}-|\Lambda^{0}_{j,N}|-|\Lambda_{j,N}|.$$

Proof.

Substitute the two definitions in (5.35). ∎

Both absolute conversion terms appear as separate summands of $E^{\prime}$. We do not assume that Paper II’s original envelope contains the saved-array error. Its stage 8 converts the actual-level array before stage 9 domination [27, §6.1, stage table]. The arithmetic application uses the preceding level domination.

Restoration.

The four rows of $\mathcal{H}$ are Paper II’s rows $S_{4}$, the lower part of $S_{5}$, $S_{7}$ and the retained part of $S_{10}$. Their stage-two source-restoration shares are assigned once to the source parents $S_{4}$, $S_{5}$, $S_{7}$ and $S_{10}$. The literal $S_{5}$ split at $7109/28000$ is made before the structured sieve or Heath–Brown expansion, as above. Both parts retain the parent $S_{5}$ for this restoration accounting [27, (2.28), §6.1, stage table and the list of S10, S5 and S4/S7 endpoints]. There is no separate restoration for the lower part of $S_{5}$. The upper part of $S_{5}$ is the outer chain $\mathrm{T5}$ of Lemma 5.1, which the lift does not change. Each tuple of an actual source is assigned to one entry by Paper II’s first-failure rule [27, (6.4d) and the two sentences after it]. This section changes no source and no stage from one to eight, so the assignment is Paper II’s. Appendix B.7 lists the finite correspondence between these rows and Paper II’s error entries. It is an index of that assignment, and each entry occurs once.

Lemma 5.5 (Lifted direct comparison).

Let $j\in\mathcal{H}$, and let $Q_{j,8}$ be Paper II’s stage-eight quantity for this row. It is the continuum mass of the actual-level main term after the prime-measure and Euler conversion. There each actual cell carries its actual relative level $\ell_{c}$. Put $Q^{h}_{j,9}=P_{j}+\Delta_{j}=\int k_{j}^{h}\,d(m^{e_{j}})$. Then $Q_{j,8}\geq Q^{h}_{j,9}$.

Proof.

Stages one to eight are Paper II’s. Paper II passes from $Q_{j,8}$ to $Q_{j,9}=P_{j}$ by pointwise domination of the actual-level integrand by the specified profile. The cells and the continuum measure are the same [27, §6.1, stage table; §2.4, after (2.31); §2.5, after (2.40)]. We repeat that domination with the lifted profile. For each B-labelled parent contributing to $S_{4}$, $S_{7}$ or $S_{5}$-low, we first rederive the buffer used below. Write

$$m(t)=\min\left\{\frac{1+t}{2},\frac{57}{64}-\frac{3\rho t}{2}\right\},\qquad t_{\rm dd}=\frac{25}{32(1+3\rho)}.$$

Paper II gives $\Theta_{c}\geq m(t_{1})$. The branch comparison is the exact identity

$$\frac{1+t_{1}}{2}-\left(\frac{57}{64}-\frac{3\rho t_{1}}{2}\right)=\frac{1+3\rho}{2}(t_{1}-t_{\rm dd}).$$

Thus the decreasing branch is active exactly when $t_{1}>t_{\rm dd}$. Equality is assigned to A, where both bounds agree. This active branch test is applied to every B-labelled parent. The branch is never inferred from the largest completed coordinate alone. Paper II keeps the inherited label through clipping. In S4 and S7 the B parents are separate intersections of their respective row domains with $t_{1}>t_{\rm dd}$. No common parent interval is asserted. S5-low has $1/4\leq r<t_{*}=7109/28000$ and $t_{1}\leq r<\rho t_{1}$, so $t_{1}>1/(4\rho)>t_{\rm dd}$ because $\rho<8$. It is therefore entirely B. Let a parent in S4 or S7 be A-labelled. Then $t_{1}\leq t_{\rm dd}$, and Paper II’s first branch in (2.17f) [27, (2.17f)] gives $\Theta_{c}\geq(1+t_{1})/2$. Take an actual exponent $r\in[t_{1},\rho t_{1})$. The S4 and S7 row domains are clipped by $r\geq\mu_{*}$ [27, §2.1, fixed-cutoff rows], and the retained bin gives $r\geq t_{1}$. The residual is therefore

$$2(t_{1}-\mu_{*})+\frac{5}{2}(r-t_{1})=2(r-\mu_{*})+\frac{1}{2}(r-t_{1})\geq 0.$$

$$\ell_{c}-\left({57\over 64}-u-{5r\over 2}\right)\geq u-\eta_{L}-{25\over 64}+2t_{1}+{5\over 2}(r-t_{1})={19\over 10000}+2(t_{1}-\mu_{*})+{5\over 2}(r-t_{1})\geq{19\over 10000}.$$

Here $\mu_{*}=25/128$ is the retained cutoff parameter; the symbol $\mu(n)$ elsewhere denotes the Möbius function. Thus the A pieces also have the required $19/10000-3\tau$ buffer. This is a row-local calculation. It does not merge the distinct S4 and S7 parent domains. For any B parent, Paper II’s floor gives

(5.37) $$\Theta_{c}\geq{57\over 64}-{3\over 2}\rho t_{1},\qquad\ell_{c}-\left({57\over 64}-u-{5r\over 2}\right)\geq(u-\eta_{L})+{5\over 2}(r-t_{1})-{3\over 2}\tau t_{1}\geq{19\over 10000}-3\tau.$$

The first bound is Paper II’s linear branch of the full-anatomy floor [27, (2.17d)–(2.17f)]. Using $\ell_{c}=\Theta_{c}-\eta_{L}-t_{1}$, $r\geq t_{1}$, $u-\eta_{L}=19/10000$, and $t_{1}<t_{*}<1$ proves the second line by direct subtraction. The B pieces of S4 and S7 use the same row-local calculation. Their parent intervals stay distinct.

For $j=4,5,7$ the row coefficient is negative. At a point $x$ of an actual cell, the actual sieve argument is $\ell_{c}/\alpha$ and, by (5.23),

$$\frac{\ell_{c}}{\alpha}\geq s(x)+\frac{h}{\alpha}+\frac{10^{-4}-3\tau}{\alpha}>s(x)+\frac{h}{\alpha}.$$

These are Paper II’s margins $19/10000-3\tau$ for $S_{4}$, $S_{7}$ and the $B$ branch of the lower part of $S_{5}$. Here $h$ has moved into the profile [27, §2.4, display before “The S4/S7 A/B levels give the same buffer”]. Since $F$ is decreasing, $-\widehat{F}(\ell_{c}/\alpha)\geq-\widehat{F}(s(x)+h/\alpha)$. For $j=10$ the coefficient is positive, and on the retained cells

$$\frac{\ell_{c}}{\alpha}\geq q(x,y)+\frac{h}{\alpha}+\frac{10^{-4}-(5\lambda/2+\beta)\tau}{\alpha}>q(x,y)+\frac{h}{\alpha},$$

which is Paper II’s $\alpha(s_{\mathrm{live}}-s_{\mathrm{saved}})\geq(u-\eta_{L})-\tau((5/2)\lambda+\beta)$ with $h$ moved into the profile [27, §2.5, proof after (2.40)]. Since $f$ is nondecreasing, $\widehat{f}(\ell_{c}/\alpha)\geq\widehat{f}(q(x,y)+h/\alpha)$. Paper II uses the same monotonicity of $F$ and $f$ for these comparisons [27, §2.4]. For $j=4,5,7$ the cells cover $I_{j}$, and integrating the pointwise inequality against $m$ gives $Q_{j,8}\geq\int k_{j}^{h}\,dm=P_{j}+\Delta_{j}$.

For $j=10$ the retained cells need not cover $\mathcal{D}_{10}$. Let $R_{\tau}\subset\mathcal{D}_{10}$ be their union, put $a_{10}=5\lambda/2+\beta$, and write $T_{h}(E)=\int_{E}k_{10}^{h}\,d(m^{2})$. Then

$$Q_{10,8}=\frac{1}{\alpha}\sum_{c\ \mathrm{retained}}\int_{c}\widehat{f}(\ell_{c}/\alpha)\,d(m^{2}).$$

Strip. Let $(x,y)\in\mathcal{D}_{10}\setminus R_{\tau}$ lie in a cell with labels $t_{1}\leq x$, $t_{2}\leq y$. The cell is dropped, so $y\geq t_{2}>\operatorname{sel}(t_{1})\geq B_{0}(t_{1})-a_{10}\tau\geq B_{0}(x)-a_{10}\tau$. This uses the two deficit bounds of Paper II, §2.5, and the fact that $B_{0}$ is nonincreasing [27, §2.5, proof of (2.40)]. Also $y<B_{0}(x)$. On $\mathcal{D}_{10}$ one has $11/5\leq q$ and $q+h/\alpha<4$, so $\widehat{f}(q+h/\alpha)\leq\widehat{f}(4)=\log 3/4$. With $y\geq\alpha$ and $\int_{\mu_{*}}^{\lambda}dx/x\leq(\lambda-\mu_{*})/\mu_{*}$, this gives $T_{h}(\mathcal{D}_{10}\setminus R_{\tau})\leq D_{*}\tau$, where $D_{*}$ is the constant in (5.26).

Reserve. On a retained cell the margin in (5.23) satisfies $10^{-4}-a_{10}\tau>1/20000$, so $\ell_{c}/\alpha\geq q+h/\alpha+\delta_{10}$. Here $2<q+h/\alpha<q+h/\alpha+\delta_{10}<4$. For $2<s<4$, $\widehat{f}(s)=\log(s-1)/s$ and

$$\widehat{f}^{\prime}(s)=\frac{s/(s-1)-\log(s-1)}{s^{2}}>\frac{4/3-11/10}{16}=\frac{7}{480},$$

since $\log 3<11/10$. As $f$ is nondecreasing, $\widehat{f}(\ell_{c}/\alpha)-\widehat{f}(q+h/\alpha)\geq 7\delta_{10}/480$ at every point of $R_{\tau}$.

A fixed retained rectangle. Let $(x,y)\in R_{10}$, in a cell with labels $t_{1}\in(x/\rho,x]$, $t_{2}\in(y/\rho,y]$. Then $\mu_{*}<t_{1}<\lambda$ and $\alpha<t_{2}<\beta$. On $[\mu_{*},21/100]$ we have $B_{0}=\beta$, because $57/64-u-11\alpha/5-(5/2)(21/100)>\beta$. Hence $\operatorname{sel}(t_{1})\geq\beta-a_{10}\tau>1/10\geq t_{2}$, and the cell is retained: $R_{10}\subset R_{\tau}$. Also $m^{2}(R_{10})=\log\frac{21}{20}\log\frac{10}{9}\geq\frac{1}{21}\cdot\frac{1}{10}=\frac{1}{210}$.

Combining the three steps with the pointwise inequality above,

$$Q_{10,8}\geq T_{h}(R_{\tau})+c_{10}\geq T_{h}(\mathcal{D}_{10})-D_{*}\tau+c_{10}>T_{h}(\mathcal{D}_{10})=P_{10}+\Delta_{10},$$

by (5.27). No deduction is made. The reserve is the part of Paper II’s level margin left after the lift $h$. It exists on cells of $S_{10}$ only and is used only here. ∎

The finite arrays $S^{0}_{j,N},S^{h}_{j,N}$ are not used in this inequality. The errors $|\Lambda^{0}_{j,N}|$ and $|\Lambda_{j,N}|$ are nonnegative and are kept in $E^{\prime}$ so that $E^{\prime}$ agrees with (1.12). They tend to zero by Lemma 5.8.

5.4. The internal joint shift

The joint occurrence is $2S_{11}+S_{12}+S_{13}$. Its six-branch function is Paper II, (4.70) [27, (4.70)–(4.72)]:

(5.38) $$v_{1}(t) =\min\left\{\frac{1+t}{2},\frac{32-34t}{39},\frac{57}{64}-\frac{3t}{2}\right\},$$

$$v(t) =v_{1}(t)-u,\qquad\psi(x,y)=x\mathbf{1}_{x\geq y},$$

$$\Theta_{*}(a,b,c) =\max\{v(a),v(b),v(a+b),v(a+b+c),$$

$$\hskip 96.73936pt\psi(v(a+c),a+2b+c),\psi(v(b+c),2a+b+c)\}.$$

Subtract $u$ before testing each guard. Set

(5.39) $$\mathcal{C}=\{(t_{1},t_{2},t_{3}):u\leq t_{3}<t_{2}<t_{1}<\mu_{*}\},\qquad a_{3}(\mathbf{t})=\frac{\Theta_{*}(\mathbf{t})-t_{1}-t_{2}-t_{3}}{t_{3}}.$$

Shift the output of $\Theta_{*}$ by $h$. Keep its guards and actual source unchanged:

(5.40) $$a_{3}^{s}(\mathbf{t}) =a_{3}(\mathbf{t})+\frac{h}{t_{3}},$$

$$J =\iiint_{\mathcal{C}}\frac{1-f(a_{3}(\mathbf{t}))}{t_{1}t_{2}t_{3}^{2}}\,d\mathbf{t},$$

$$J_{s} =\iiint_{\mathcal{C}}\frac{1-f(a_{3}^{s}(\mathbf{t}))}{t_{1}t_{2}t_{3}^{2}}\,d\mathbf{t}.$$

Use $f(s)=0$ for $s<2$. Keep its true values at every larger argument. The new margins are

(5.41) $$g_{s}=\frac{1}{10000}-\frac{225}{256}\tau>2\mu_{*}\tau,\qquad g_{s,\mu_{*}}=\frac{1}{10000}-\frac{75}{256}\tau>0.$$

The second margin concerns a parent interval crossing $\mu_{*}$. Both follow by subtracting $h$ in Paper II, (4.72); see Lemma 5.6. They hold throughout $0<\tau\leq 10^{-90}$. If $f(a_{3}^{s})>0$, the retained-cell condition follows:

(5.42) $$\ell_{c}\geq\Theta_{*}+h-t_{1}-t_{2}-t_{3}+g_{s}>2t_{3}+2\mu_{*}\tau\geq 2\rho t_{3}.$$

Thus the canonical restriction still loses zero. At fixed numerator, use the increase of $sf(s)$. This is the same comparison as Paper II, (4.72).

Lemma 5.6 (Shifted retention).

The margins (5.41) hold for $0<\tau\leq 10^{-90}$. On the original joint source, every positive shifted lower contribution lies in an original retained whole cell of Paper II, (4.19). On such a cell the actual lower main term dominates the shifted canonical one.

Proof.

Use an original G1-3 cell with inherited labels $t_{1}\geq t_{2}\geq t_{3}>0$, and order its distinct actual prefix primes as $a_{1}>a_{2}>a_{3}$. Put $x_{i}=\log_{N}a_{i}$. Unequal parent bins enforce the actual order, while within an equal-label group we order the primes. Hence $t_{i}\leq x_{i}<\rho t_{i}$, including clipped bins. No strict inequality between the labels is required.

Let $\Theta_{c}$ be the original full-anatomy infimum and $\ell_{c}=\Theta_{c}-\eta_{L}-\sum_{i}t_{i}$. Put $B_{s}(x)=\Theta_{*}(x)+h-\sum_{i}x_{i}$. The function $v_{1}$ is $3/2$-Lipschitz. On the closed ordered parent a coordinate changes by at most $\mu_{*}\tau$. A sum of at most three coordinates changes by at most $3\mu_{*}\tau$. If a saved guarded branch is active at $x$, its buffered guard has nonnegative gap, including equality. Removing its buffer adds $u$. Throughout the parent its unbuffered gap is therefore at least $u-(9/2)\mu_{*}\tau-4\mu_{*}\tau>0$. Thus it stays available in the full anatomy. Take an unguarded or active guarded saved branch. The corresponding full branch is at least its value at $x$ plus $u-(9/2)\mu_{*}\tau$. Taking their maximum and then the parent infimum preserves this lower bound. Ties between maximizing branches cause no loss. Therefore

$$\ell_{c}\geq B_{s}(x)+g_{s}+\sum_{i}(x_{i}-t_{i}),\qquad g_{s}=u-\eta_{L}-h-(9/2)\mu_{*}\tau=10^{-4}-(225/256)\tau.$$

For a parent crossing $\mu_{*}$, the largest exponent is within $\mu_{*}\tau$ of $\mu_{*}$, and $v_{1}(\mu_{*})=153/256$. Every saved branch is at most $153/256-u$, so the crossing comparison gives instead $10^{-4}-(75/256)\tau\geq g_{s}$. Clipping only shrinks the infimum domain. These arguments also apply when two or all three labels coincide. For $0<\tau\leq 10^{-90}$ we have $g_{s}>2\mu_{*}\tau>0$.

If $f(B_{s}(x)/x_{3})>0$, then $B_{s}(x)>2x_{3}$ and hence

$$\ell_{c}>2x_{3}+2\mu_{*}\tau\geq 2t_{3}+2\tau t_{3}=2\rho t_{3}.$$

Thus the whole original cell is retained by Paper II, (4.19). On positive support $\ell_{c}>B_{s}(x)>0$ and $0<t_{3}\leq x_{3}$ imply

$$\frac{\ell_{c}}{t_{3}}\geq\frac{B_{s}(x)}{x_{3}},\qquad f(\ell_{c}/t_{3})\geq f(B_{s}(x)/x_{3}),\qquad\frac{f(\ell_{c}/t_{3})}{t_{3}}\geq\frac{f(B_{s}(x)/x_{3})}{x_{3}}.$$

These use only nonnegativity and monotonicity of $f$. Equal labels invalidate none of the weak inequalities. Also $E=N^{t_{3}}\leq a_{3}$, so $V_{N}(E)\geq V_{N}(a_{3})$. At zero shifted contribution the required comparison follows from nonnegativity without any assertion that $B_{s}(x)>0$.

Keep the original orbit normalization. If the equal-label groups are $G_{h}$, put $M_{c}=\prod_{h}|G_{h}|!\in\{1,2,6\}$. For a full labelled realization $\mathbf{a}$, write $\mathbf{a}^{\downarrow}$ for its decreasing rearrangement within the equal-label groups. The shifted comparison above is evaluated at $\mathbf{x}^{\downarrow}$ and is invariant under these group permutations. For every nonnegative, invariant main-mass weight $w_{c}$, therefore,

$$\frac{1}{M_{c}}\sum_{\rm full}K_{P}(\mathbf{a})w_{c}(\mathbf{a})V_{N}(E)f(\ell_{c}/t_{3})\geq\frac{1}{M_{c}}\sum_{\rm full}K_{P}(\mathbf{a})w_{c}(\mathbf{a})V_{N}(a_{3}^{\downarrow})f(B_{s}(\mathbf{x}^{\downarrow})/x_{3}^{\downarrow}).$$

Each distinct ordered triple has exactly $M_{c}$ labelled realizations. The prefix mask $K_{P}$ and its correction are kept exactly as in Paper II, (4.22)–(4.23). No repeated-prefix realization is reclassified. The tail mask and all remainder coefficients are likewise unchanged. This proves the finite lower-main-term comparison with the cell mass divided by $M_{c}$. The corresponding continuum comparison uses the preceding inequality with the $1/t_{3}$ and $1/x_{3}$ factors. Equal-label bins are kept. Their strictly ordered actual triples lie in the original cone. The orbit identity converts the full-block sum to that ordered sum exactly. Only actual coordinate diagonals have zero continuum measure. The proof imposes no $N\to\infty$ disappearance of equal-bin regions. ∎

Put $H_{\log}=\log(\mu_{*}/u)$. Paper II’s remaining terms satisfy

(5.43) $$0\leq R_{G}\leq R_{\mathrm{up}}={} 500(F(30)-1)+500H_{\log}(1-f(30))$$

$$+250H_{\log}^{2}(F(30)-1).$$

Its constant terms cancel exactly [27, (4.73)–(4.75)]:

(5.44) $$500\left(1-H_{\log}+\frac{H_{\log}^{2}}{2}\right)-\iiint_{\mathcal{C}}\frac{d\mathbf{t}}{t_{1}t_{2}t_{3}^{2}}=\frac{1}{\mu}_{*}.$$

The shifted upper comparison is therefore

(5.45) $$G_{1}^{s}=\frac{1}{\mu}_{*}+R_{G}+J_{s},\qquad\Phi_{G_{1}}^{s}=-K_{G}G_{1}^{s},\qquad K_{G}=\frac{2I_{11}+I_{12}+I_{13}}{2e^{\gamma}}.$$

The source-mass integrals are [27, (4.77)]

(5.46) $$I_{11}={} \int_{\lambda}^{1/3}\frac{\log((1-2t)/t)}{t(1-t)}\,dt$$

$$+\int_{\lambda}^{1/4}\frac{dt}{t}\int_{t}^{(1-t)/3}\frac{\log((1-t-2v)/v)}{v(1-t-v)}\,dv,$$

$$I_{12}={} \int_{\alpha}^{1/3}\frac{\log(2-3t)}{t(1-t)}\,dt,$$

$$I_{13}={} \int_{\beta}^{\lambda}\frac{\log((1-t-\lambda)/\lambda)}{t(1-t)}\,dt.$$

We have $I_{11},I_{12},I_{13}\geq 0$ and $I_{12}>0$. Thus $K_{G}>0$. The finite shifted enclosure gives

(5.47) $$0\leq G_{1}^{s}\leq\frac{6073714261}{10^{9}}\quad\text{(shifted mesh $n_{\mathrm{mesh}}=512$)}.$$

Paper II’s mass enclosure gives $0\leq K_{G}\leq K_{\mathrm{hi}}$ [27, (A.10d)]. The exact constants are

(5.48) $$K_{\mathrm{hi}}=\frac{3989480791070578466390443009830933753748442947478110112223}{10^{58}}.$$

Paper II, (4.84), gives

(5.49) $$L_{G1}=-\frac{243950428265194656254206921939947762868871361874409552095097}{100433627766186892221372630771322662657637687111424552206336}.$$

Define

(5.50) $$M_{5,\mathrm{internal}}=-K_{\mathrm{hi}}\frac{6073714261}{10^{9}}-L_{G1}.$$

Their exact substitution gives

(5.51) $$\begin{gathered}0.0058749499450263575220<M_{5,\mathrm{internal}},\\ M_{5,\mathrm{internal}}<0.0058749499450263575221.\end{gathered}$$

The finite enclosure and identity are proved. They concern the shifted profile integral. The required actual-source inequality is

(5.52) $$-\frac{2S_{11}+S_{12}+S_{13}}{W}\geq L_{G1}+M_{5,\mathrm{internal}}-E^{D}_{G1}-|\Lambda_{G,N}|.$$

Proof of (5.52).

Paper II proves

$$\frac{2S_{11}+S_{12}+S_{13}}{W}\leq K_{G}G_{1}+E^{D}_{G1},\qquad G_{1}=\frac{1}{\mu_{*}}+R_{G}+J,$$

from (4.73)–(4.77) [27, (4.75), (4.78)]. Run the same argument with $J$ replaced by $J_{s}$. The first three residual rows and the bound $R_{G}\leq R_{\mathrm{up}}$ do not change, and neither does the cancellation (5.44). In the fourth row, Lemma 5.6 replaces Paper II’s comparison after (4.72). The canonical restriction loses zero, and the actual lower main term dominates the shifted one. The source and its weights $(2,1,1)$ do not change, and neither do the peel (4.45)–(4.57) and the mass conversion (4.76)–(4.78). Hence

$$\frac{2S_{11}+S_{12}+S_{13}}{W}\leq K_{G}G_{1}^{s}+E^{D}_{G1}$$

with the same $E^{D}_{G1}$. By Proposition B.2, $0\leq G_{1}^{s}\leq 6073714261/10^{9}$, which is (5.47), with the residual bound supplied by (5.43). By Paper II, (A.10d), $0\leq K_{G}\leq K_{\mathrm{hi}}$. Both factors are nonnegative, so by (5.50)

$$-K_{G}G_{1}^{s}\geq-K_{\mathrm{hi}}\frac{6073714261}{10^{9}}=L_{G1}+M_{5,\mathrm{internal}}.$$

This gives (5.52) even without $|\Lambda_{G,N}|$. That term is nonnegative and is kept so that the error agrees with (1.12). ∎

The existing $E^{D}_{G1}$ is a disjoint component of the original $F_{\mathrm{net}}+E_{\mathrm{van}}$ allowance. Its fixed peel $145664566.4567\tau$ is covered by $g_{\tau}$, and its fixed $C_{E}\sigma$ share by $g_{\mathrm{AA}}$. Its vanishing part belongs to $E_{\mathrm{van}}$. When replacing the joint chain, carry these same components once. The reconciliation in (5.10) remains paid. These components are the $G_{1}$ comparator term $NC_{\mathrm{sf}}$ of Paper II, (4.64)–(4.65), and the $G_{1}$ share of the joint bound. That bound treats $\Xi_{U}$ and $\Xi_{A}$ in Paper II, (B.1)–(B.7). Both are attached to the original source measure and not to the profile. Hence the shift neither creates nor removes them.

5.5. Conversion of the finite arrays

We give the remaining definitions before stating the lemma. All representation variables in the following table are prime. Their product is $e$, with $\varepsilon N<e<N$. Let $R_{11,3}(e)$ and $R_{11,4}(e)$ count the first two forms. Let $R_{12}(e)$ and $R_{13}(e)$ count the last two forms. Put $R_{11}=R_{11,3}+R_{11,4}$. These are Paper II’s representation weights [27, (4.4), source table preceding (4.4)].

Table 5.1. The internal representation counts

Count	Product	Restrictions
$R_{11,3}$	$p_{1}p_{2}p_{3}$	$N^{\lambda}\leq p_{1}<N^{1/3}$, $p_{1}<p_{2}$, $p_{1}p_{2}^{2}<N$, $p_{3}\geq p_{2}$.
$R_{11,4}$	$p_{1}p_{2}p_{3}p_{4}$	The same conditions on $p_{1},p_{2}$; $p_{2}\leq p_{3}\leq p_{4}$.
$R_{12}$	$p_{1}p_{2}p_{3}$	$N^{\alpha}\leq p_{1}<N^{1/3}\leq p_{2}$, $p_{1}p_{2}^{2}<N$, $p_{3}\geq p_{2}$.
$R_{13}$	$p_{1}p_{2}p_{3}$	$N^{\beta}\leq p_{1}<N^{\lambda}\leq p_{2}$, $p_{1}p_{2}^{2}<N$, $p_{1}p_{2}p_{3}^{2}\geq N$.

For the last form the endpoint orders are distinct. Every permitted representation keeps its multiplicity. Define the upper cut and masses by

(5.53) $$N_{\mathrm{cut}} =N(1-L^{-333}),$$

$$X_{j}^{\mathrm{cut}} =\sum_{e}R_{j}(e)\mathbf{1}_{\varepsilon N<e\leq N_{\mathrm{cut}}} (j=11,12,13),$$

$$X_{G} =2X_{11}^{\mathrm{cut}}+X_{12}^{\mathrm{cut}}+X_{13}^{\mathrm{cut}},$$

$$r_{G,N}(t) =\frac{tX_{G}V_{N}(N^{t})}{WK_{G}}.$$

The cut precedes the sieve decomposition [27, (4.16)–(4.17)]. The source-mass input used below is

(5.54) $$X_{D} =\left(1-\varepsilon+o(1)\right)\frac{N}{L},$$

$$X_{G} =\left((1-\varepsilon)(2I_{11}+I_{12}+I_{13})+o(1)\right)\frac{N}{L}.$$

Lemma 5.7 (Source masses).

The formulas (5.54) hold.

Proof.

The sum $X_{D}$ counts the primes in $(N/L^{30},(1-\varepsilon)N)$, so the prime number theorem gives $X_{D}=(1-\varepsilon+o(1))N/L$. For the representation weights of Table 5.1, Paper II proves $X_{j}=((1-\varepsilon)I_{j}+o(1))N/L$ for the uncut masses, $j=11,12,13$ [27, (4.76)–(4.77)]. The cut strip $N_{\mathrm{cut}}<e<N$ carries joint mass at most $100(NL^{-333}+1)$ [27, (4.16)–(4.17)], which is $o(N/L)$. Summing with the weights $(2,1,1)$ gives the formula for $X_{G}$. ∎

On $X_{0}^{3}$, define the zero-extended kernel

(5.55) $$d_{G}(\mathbf{t})=\mathbf{1}_{\mathcal{C}}(\mathbf{t})K_{G}\frac{f(a_{3}^{s}(\mathbf{t}))-f(a_{3}(\mathbf{t}))}{t_{3}}.$$

Its arrays and integral are

(5.56) $$B_{G,N} =\iiint r_{G,N}(t_{3})d_{G}(\mathbf{t})\,d(m_{N}^{3})(\mathbf{t}),$$

$$\Delta_{G} =\iiint d_{G}\,d(m^{3})=K_{G}(J-J_{s}),$$

$$\Lambda_{G,N} =B_{G,N}-\Delta_{G}.$$

Equivalently,

(5.57) $$B_{G,N} =\frac{X_{G}}{W}\sum_{\begin{subarray}{c}p_{3}<p_{2}<p_{1},\ p_{i}\in\mathbb{P},\ p_{i}\nmid N\\ u\leq\log_{N}p_{3}<\log_{N}p_{2}<\log_{N}p_{1}<\mu_{*}\end{subarray}}\frac{V_{N}(p_{3})[f(a_{3}^{s}(\mathbf{t}))-f(a_{3}(\mathbf{t}))]}{(p_{1}-1)(p_{2}-1)(p_{3}-1)},$$

$$\hskip 108.12047ptt_{i}=\log_{N}p_{i}.$$

These definitions specify all five additional conversion errors.

Lemma 5.8 (Conversion of the five profile differences).

Fix $0<\sigma\leq 10^{-10}$. For $m_{N},m$ in (5.28), the finite arrays (5.33) and (5.56) satisfy

(5.58) $$\Lambda_{j,N}\longrightarrow 0\quad(j\in\mathcal{H}),\qquad\Lambda_{G,N}\longrightarrow 0,$$

and $\Lambda^{0}_{j,N}\to 0$ for $j\in\mathcal{H}$. Every limit ranges over even $N$ tending to infinity; the thresholds depend on $\sigma$.

Proof.

Step 1: convergence of the harmonic measures. Let $u\leq a<b\leq 1$. Mertens’ theorem and partial summation give

(5.59) $$m_{N}([a,b))=\log\frac{b}{a}+o(1).$$

Replacing $1/(p-1)$ by $1/p$ costs $O(N^{-u})$. There are at most $L/\log 2$ prime divisors of $N$. Each omitted divisor above $N^{u}$ has mass $O(N^{-u})$. Their total cost is $O(N^{-u}L)$. The largest endpoint atom is $O(N^{-u})$. The total masses therefore satisfy

(5.60) $$c_{N}:=m_{N}(X_{0})\longrightarrow c_{0}:=m(X_{0})=\log(1/u)>0.$$

Let $g$ be continuous on $X_{0}$. Choose a finite interval partition and a step function on it, uniformly within a prescribed error of $g$. The integrals of the step function converge by (5.59). The integral errors are at most the uniform error times $c_{N}+c_{0}$, which is bounded by (5.60). Hence $m_{N}\Rightarrow m$ [2, Theorem 2.2 and Example 2.3]. For $k=2,3$, rectangle masses are products of interval masses. The same argument with rectangular step functions gives

(5.61) $$m_{N}^{k}\Rightarrow m^{k}\qquad(k=1,2,3).$$

For a coordinate diagonal $D_{ij}=\{t_{i}=t_{j}\}$,

(5.62) $$m_{N}^{k}(D_{ij})\leq\left(\max_{p\geq N^{u}}\frac{1}{p-1}\right)c_{N}^{k-1}=O(N^{-u})\qquad(k\geq 2).$$

Thus strict prime order costs nothing in the limit, and no orbit factor enters.

Step 2: the source and Euler normalizations. For $p>2$,

(5.63) $$1-\frac{1}{p-1}=\left(1-\frac{1}{p}\right)\left(1-\frac{1}{(p-1)^{2}}\right).$$

Mertens’ product formula and the convergent second product yield

(5.64) $$V_{N}(N^{t})=\frac{2e^{-\gamma}C(N)}{tL}(1+o(1)),\qquad u/\rho\leq t\leq 1,$$

uniformly in $t$. The prime $2$ is omitted because $N$ is even. The missing divisors $p\mid N$ above $N^{t}$ contribute $\prod(1+1/(p-2))=1+O(N^{-u/\rho}L)$. The ordinary product error is uniform for $N^{t}\geq N^{u/\rho}$. By Lemma 5.7 and (2.2),

(5.65) $$r_{D,N}\longrightarrow 1,\qquad\sup_{u\leq t\leq\mu_{*}}|r_{G,N}(t)-1|\longrightarrow 0.$$

For example, $WK_{G}=2e^{-\gamma}(1-\varepsilon)(2I_{11}+I_{12}+I_{13})C(N)N/L^{2}$. This is exactly the main factor in $tX_{G}V_{N}(N^{t})$. Thus both the truncation factor and the representation multiplicities cancel once.

Step 3: restriction and bounded kernels. Write $\widehat{m}_{N}=m_{N}/c_{N}$ and $\widehat{m}=m/c_{0}$. These are probability measures, and their products converge weakly. Let $D$ have $m^{k}(\partial D)=0$, and let $g\mathbf{1}_{D}$ be bounded and $m^{k}$-almost everywhere continuous. Apply the mapping form of the portmanteau theorem [2, Theorems 2.1 and 2.7] to $g\mathbf{1}_{D}$ and the probability measures. Then multiply by $c_{N}^{k}\to c_{0}^{k}$. This gives

(5.66) $$\int_{D}g\,d(m_{N}^{k})\longrightarrow\int_{D}g\,d(m^{k}).$$

The intervals $I_{j}$ have finitely many boundary points. The boundary of $\mathcal{D}_{10}$ is piecewise affine, since $B_{0}$ is continuous and piecewise affine. Both have zero $m^{e_{j}}$-measure. On $I_{j}$ $(j=4,5,7)$ the argument $s(x)$ lies in a compact subinterval of $(1,\infty)$, where $F$ is continuous. On $\mathcal{D}_{10}$, $y<K-11\alpha/5-5x/2$ gives $q(x,y)>11/5$, and $q+h/\alpha<4$; there $f$ is continuous. Hence $k_{j}^{0},k_{j}^{h}$ and $d_{j}$ are bounded and continuous off a null set, so (5.66) applies to them. With (5.65), it gives

(5.67) $$S_{j,N}^{0}\longrightarrow P_{j},\qquad B_{j,N}\longrightarrow\Delta_{j}\qquad(j\in\mathcal{H}).$$

Step 4: the ordered three-dimensional kernel. The cone $\mathcal{C}$ has affine boundary faces, so its boundary is $m^{3}$-null. The function $\Theta_{*}$ is piecewise affine. Its possible jumps lie on finitely many guard hyperplanes, and its branch switches add finitely many affine faces. On every remaining open piece, $a_{3}$ and $a_{3}^{s}$ are continuous. The denominator satisfies $t_{3}\geq u>0$, so the arguments of $f$ remain in a fixed compact interval. On it $f$, extended by zero below $2$, is continuous. Since $0\leq f\leq 1$, $|d_{G}|\leq K_{G}/u$. Thus $d_{G}$ is bounded and $m^{3}$-almost everywhere continuous. Apply (5.66) to the zero extension on $X_{0}^{3}$:

(5.68) $$\iiint d_{G}\,d(m_{N}^{3})\longrightarrow\iiint d_{G}\,d(m^{3})=\Delta_{G}.$$

The normalization costs at most

(5.69) $$\left|B_{G,N}-\iiint d_{G}\,d(m_{N}^{3})\right|\leq\sup_{[u,\mu_{*}]}|r_{G,N}-1|\,\|d_{G}\|_{\infty}c_{N}^{3}\longrightarrow 0.$$

The cone’s coordinate-order inequalities are strict, so $B_{G,N}$ counts strictly ordered triples, as in (5.57). Hence $B_{G,N}\to\Delta_{G}$, and with Step 3 all limits in (5.58) and $\Lambda^{0}_{j,N}\to 0$ follow. ∎

Whole cells, retained cells and the label cutoff.

For the four direct rows, the arithmetic identification is Lemma 5.5. For the joint row it is the proof of (5.52). Both are made at stage eight on Paper II’s actual cells, with their inherited signed coefficients. Hence no further whole-cell kernel is needed. For the retained part of $S_{10}$, the actual whole cells cover $\mathcal{D}_{10}$ up to a strip. Lemma 5.5 absorbs this strip by the retained reserve (5.27). For $G_{1}$, the retained cells are those of Paper II, (4.19). By Lemma 5.6, the shifted profile has no positive support outside them. The arrays (5.33) and (5.56) enter only through the nonnegative errors in $E^{\prime}$, which vanish by Lemma 5.8.

The array (5.57) carries $V_{N}(p_{3})$ at the actual coordinate. The arithmetic lower main term carries $V_{N}(E)$, $E=N^{t_{3}}$, at the inherited label. Since $E\leq a_{3}$, $V_{N}(E)\geq V_{N}(a_{3})$. In the lower part of the joint upper bound this replacement is favourable. We nevertheless bound it absolutely and include it once in Paper II’s peel allowance (4.57). No fixed cost is added. By (5.64), uniformly for actual label/coordinate pairs in the same cell with $u/\rho\leq t_{3}\leq x_{3}\leq 1$ and $x_{3}<\rho t_{3}$,

$$\frac{V_{N}(N^{t_{3}})}{V_{N}(N^{x_{3}})}=\frac{x_{3}}{t_{3}}(1+o(1))\leq\rho(1+o(1)).$$

For a weight $0\leq w\leq 1$ on the strictly ordered cone, the replacement therefore costs at most

(5.70) $$\begin{gathered}C_{\mathrm{cut},N}\leq(\tau+o_{\sigma}(1))T_{N},\qquad T_{N}=\frac{X_{G}}{W}\sum_{\mathcal{C}}\frac{V_{N}(p_{3})}{(p_{1}-1)(p_{2}-1)(p_{3}-1)}\longrightarrow K_{G}I_{0},\\ I_{0}=\iiint_{\mathcal{C}}\frac{d\mathbf{t}}{t_{1}t_{2}t_{3}^{2}}=500\Bigl{(}1-H_{\log}+\frac{H_{\log}^{2}}{2}\Bigr{)}-\frac{1}{\mu_{*}},\end{gathered}$$

where the sum runs over the primes of (5.57). The limit follows from Steps 2–4 of the proof of Lemma 5.8 applied to the bounded kernel $\mathbf{1}_{\mathcal{C}}K_{G}/t_{3}$. We evaluate $I_{0}$ by (5.44). Since $1<H_{\log}<459/100$ and the quadratic increases for $H_{\log}>1$,

$$I_{0}<\frac{693381}{200},\qquad K_{G}I_{0}\leq K_{\mathrm{hi}}\frac{693381}{200}<1400.$$

At fixed $\tau$, eventually $o_{\sigma}(1)\leq\tau$ and $T_{N}\leq 1400$, so $C_{\mathrm{cut},N}\leq 2800\tau<4000\tau$. We include $4000\tau$ once. Paper II’s peel bound before its final rounding is

$$P_{\mathrm{peel}}\tau=\frac{3\cdot 502002\,(459/100)^{3}}{1-\varepsilon}\tau=\frac{8090814677193}{55550}\tau$$

[27, (4.47), (4.56)], and its printed allowance is $P_{\mathrm{cap}}\tau=(485500000000/3333)\tau$ [27, (4.57)]. The per-layer coefficient 502002 in Paper II, (4.53), includes the Selberg remainder. The printed allowance (4.57) uses that coefficient; we therefore use its slack with the Selberg remainder already included. Exact subtraction gives

(5.71) $$P_{\mathrm{cap}}-P_{\mathrm{peel}}=\frac{2555968421}{166650},\qquad P_{\mathrm{cap}}-P_{\mathrm{peel}}-4000=\frac{1889368421}{166650}>0.$$

Hence the peel and the label replacement together stay below $P_{\mathrm{cap}}\tau<145664566.4567\tau$, the entry already contained in $g_{\tau}$. The fixed deduction $F_{\mathrm{net}}$ is unchanged.

5.6. The resulting lower comparison

Remark 5.9 (Inputs).

The proof of Proposition 5.10 uses the cited statements of Paper I and Paper II. They are applied to Paper II’s unchanged sources. It uses Paper I, Theorem 2.8, through Fact 3.22, and the prime Siegel–Walfisz theorem, Paper I, (2.21). It uses Lemma 6.5(ii) at Paper I, (3.48), and the finite enclosures of Propositions B.1 and B.2. The optional endpoint improvement also uses Lemma 5.3. The table of Appendix B.7 is an index, not an input.

Proposition 5.10 (Retained comparison).

Fix $0<\sigma\leq 10^{-10}$ satisfying Paper II, (1.27) and assume Lemma 5.3. Then (5.2) holds, and $E^{\prime}(N;\sigma)\to 0$ as $N\to\infty$. Without that lemma, (5.2) holds with $g_{6}$ replaced by $0$; the same limit for $E^{\prime}(N;\sigma)$ holds.

Proof.

Write each chain of $\mathcal{I}$, including $\mathrm{T5}$, by (5.4) and each joint group by (5.5). Lemma 5.2 covers the differences at stages one to eight and the remaining Paper II differences at stages nine and ten. Their sum is at least $-(F_{\mathrm{net}}-\delta_{G1})-E_{\mathrm{van}}$. For $j\in\mathcal{H}$, replace the stage-nine object by $Q^{h}_{j,9}=P_{j}+\Delta_{j}$. Then Lemma 5.5 gives $Q_{j,8}-Q^{h}_{j,9}\geq 0$. For $j\in\mathcal{J}$, Lemma 5.3 gives the stage-ten difference $P_{j}-N_{j}\geq 0$, so the endpoint rises by $N_{j}-O_{j}$. Only the endpoint changes, so $g_{6}$ enters once. By Proposition B.1, $\sum_{j\in\mathcal{H}}\Delta_{j}>572004/10^{8}>5719/10^{6}$, which is (5.25). For the joint occurrence, (5.52) replaces Paper II’s $G_{1}$ bound. Its $E^{D}_{G1}$ is the $G_{1}$ part of $F_{\mathrm{net}}+E_{\mathrm{van}}$ and is counted once. These modifications concern disjoint profile terms, so their gains enter once. The endpoint sum (5.6) and the floor (5.13) give $D_{\mathrm{II}}+\overline{J}_{6}+2\overline{H}$. Hence

(5.72) $$\frac{R_{\mathrm{fix}}+S_{2}}{W}\geq{} D_{\mathrm{II}}+g_{6}+\overline{J}_{6}+2\overline{H}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}$$

$$-E_{\mathrm{van}}(N;\sigma)-\sum_{j\in\mathcal{H}}\bigl{(}|\Lambda^{0}_{j,N}|+|\Lambda_{j,N}|\bigr{)}-|\Lambda_{G,N}|.$$

The nine conversion errors are nonnegative, so (5.72) is weaker than the same bound without them. This is (5.2). By Lemmas 5.2 and 5.8, $E^{\prime}(N;\sigma)\to 0$; the threshold depends on the fixed $\sigma$. No fixed increment is added. The only new fixed cost is the label replacement, and by (5.71) it lies inside Paper II, (4.57). Without Lemma 5.3, take $N_{j}=O_{j}$; then the same argument gives (5.72) with $g_{6}$ replaced by $0$. ∎

6. The sixth sum at $N/4$

Put $L=\log N$ and fix $0<\sigma\leq 10^{-10}$ for the common error allocation. No estimate in the sixth-sum construction depends on $\sigma$. Fix

(6.1) $$\begin{gathered}\alpha=\frac{100}{1149},\qquad t_{0}=\frac{57}{224},\qquad v_{\mathrm{RA}}=\frac{5227}{10000},\qquad\eta_{\mathrm{m}}=\varepsilon=\frac{1}{10000},\\ J=[t_{0},1/3),\quad J_{t}=[t_{0},1/3-\eta_{\mathrm{m}}),\quad J_{\mathrm{tail}}=[1/3-\eta_{\mathrm{m}},1/3),\\ I_{L}=[2,N/4],\qquad I_{H}=(N/4,(1-\varepsilon)N),\qquad z_{s}=N^{\alpha}.\end{gathered}$$

Write $p\in N^{K}$ for $p\in\mathbb{P}$ with $\log p/L\in K$. Paper II, (1.7), gives the original sum [27, (1.7)]:

(6.2) $$S_{6}=\sum_{\begin{subarray}{c}p\in N^{J}\\ (p,N)=1\end{subarray}}\ \sum_{\begin{subarray}{c}m\geq 1,\ e=N-pm\in\mathbb{P}\\ 2\leq e<(1-\varepsilon)N\end{subarray}}\mathbf{1}_{P^{-}(m)\geq z_{s}}\mathbf{1}_{(m,Np)=1}.$$

Each prime mark remains a separate occurrence.

6.1. Three upper weights

For squarefree $d=\ell_{1}\cdots\ell_{k}$, order its primes decreasingly. The ordinary upper Rosser weight is

(6.3) $$\mu_{d}^{+}(D)=\mu(d)\prod_{\begin{subarray}{c}1\leq j\leq k\\ j\ \mathrm{odd}\end{subarray}}\mathbf{1}_{\ell_{1}\cdots\ell_{j-1}\ell_{j}^{3}<D},\qquad\mu_{1}^{+}(D)=1.$$

The weight is zero outside its stated squarefree support. Its divisor sum is a nonnegative upper sieve weight [14, Lemma 1, (3.2), (3.6)] [13, Lemmas 2–3, p. 313].

For $K\subseteq J$ and $p\in N^{K}$, define

(6.4) $$c_{R,K}(pd)=\mu(d)\mathbf{1}_{pd<R}\prod_{\begin{subarray}{c}1\leq j\leq k\\ j\ \mathrm{odd}\end{subarray}}\mathbf{1}_{p\ell_{1}\cdots\ell_{j-1}\ell_{j}^{3}<R}.$$

Require $\ell_{j}<z_{s}$ and $(pd,2N)=1$. Set the coefficient to zero otherwise. Since $t_{0}>\alpha$, the mark is unique. Thus $|c_{R,K}(q)|\leq 1$. On supported moduli with $p<R$, the gates give $d<R/p$. For $d=1$ the gate product is empty, and $p<R$ suffices. Every marked prime has $p<N^{1/3}$. Each sieve level below is $>N^{1/3}$ for large $N$. Hence

(6.5) $$c_{R,K}(pd)=\mu_{d}^{+}(R/p),\qquad\varphi(pd)=(p-1)\varphi(d).$$

The equality is claimed on supported moduli only. Off that support the coefficient is zero.

Choose fixed logarithmic exponents. Put

(6.6) $$R_{L} =N^{1/2}L^{-B_{L}}, R_{\mathrm{tail}} =N^{1/2}L^{-B_{\mathrm{tail}}},$$

$$Q =N^{v_{\mathrm{RA}}}L^{-B}, Q_{0} =N^{1/2}L^{-B_{0}},$$

$$b_{\mathrm{all}} =c_{R_{L},J}, b_{\mathrm{tail}} =c_{R_{\mathrm{tail}},J_{\mathrm{tail}}},$$

$$c_{\mathrm{trim}} =c_{Q,J_{t}}, c_{\mathrm{lo}}(q) =c_{\mathrm{trim}}(q)\mathbf{1}_{q\leq Q_{0}},$$

$$\Gamma(q) =c_{\mathrm{trim}}(q)\mathbf{1}_{q>Q_{0}}.$$

The last two arrays keep the gates at $Q$.

First omit $(m,Np)=1$ in (6.2). Sieve the larger sequence by primes $\ell\nmid 2N$, $\ell<z_{s}$. Use the three weights on

(6.7) $$I_{L}\times J,\qquad I_{H}\times J_{\mathrm{tail}},\qquad I_{H}\times J_{t}.$$

Their divisor sum is the finite weighted count

(6.8) $$A_{\mathrm{mix}}={} \sum_{q}b_{\mathrm{all}}(q)\sum_{\begin{subarray}{c}e\in I_{L}\cap\mathbb{P}\\ e\equiv N\ (q)\end{subarray}}1$$

$$+\sum_{q}\{b_{\mathrm{tail}}(q)+c_{\mathrm{trim}}(q)\}\sum_{\begin{subarray}{c}e\in I_{H}\cap\mathbb{P}\\ e\equiv N\ (q)\end{subarray}}1.$$

The upper-weight comparison

(6.9) $$S_{6}\leq A_{\mathrm{mix}},\qquad Z_{6}:=A_{\mathrm{mix}}-S_{6}\geq 0,$$

is proved in Lemma 6.11(a) below. Positivity is used before expanding the divisor sums.

6.2. The exact discrepancy identity

For a bounded interval $I\subset[2,\infty)$ and a finitely supported modulus array $a(q)$, put $\operatorname{Li}(I)=\int_{I}dt/\log t$ and define

(6.10) $$\Pi_{I}(q) =\sum_{e\in I\cap\mathbb{P}}\mathbf{1}_{e\equiv N\pmod{q}},$$

$$B_{I}(q) =\frac{1}{\varphi(q)}\sum_{e\in I\cap\mathbb{P}}\mathbf{1}_{(e,q)=1},$$

$$D_{I}(a) =\sum_{q}a(q)\left\{\Pi_{I}(q)-\frac{\operatorname{Li}(I)}{\varphi(q)}\right\},$$

$$P_{I}(a) =\sum_{q}a(q)\{\Pi_{I}(q)-B_{I}(q)\},\qquad C_{I}(a)=D_{I}(a)-P_{I}(a).$$

Also put

(6.11) $$M_{\mathrm{mix}}={} \operatorname{Li}(I_{L})\sum_{q}\frac{b_{\mathrm{all}}(q)}{\varphi(q)}$$

$$+\operatorname{Li}(I_{H})\sum_{q}\frac{b_{\mathrm{tail}}(q)+c_{\mathrm{trim}}(q)}{\varphi(q)}.$$

Finite expansion gives

(6.12) $$A_{\mathrm{mix}}={} M_{\mathrm{mix}}+D_{I_{L}}(b_{\mathrm{all}})+D_{I_{H}}(b_{\mathrm{tail}})$$

$$+C_{I_{H}}(c_{\mathrm{trim}})+P_{I_{H}}(c_{\mathrm{lo}})+P_{I_{H}}(\Gamma).$$

Since $c_{\mathrm{trim}}=c_{\mathrm{lo}}+\Gamma$,

(6.13) $$C_{I_{H}}(c_{\mathrm{trim}})+P_{I_{H}}(c_{\mathrm{lo}})=D_{I_{H}}(c_{\mathrm{lo}})+C_{I_{H}}(\Gamma).$$

Therefore

(6.14) $$A_{\mathrm{mix}}={} M_{\mathrm{mix}}+D_{I_{L}}(b_{\mathrm{all}})+D_{I_{H}}(b_{\mathrm{tail}})+D_{I_{H}}(c_{\mathrm{lo}})$$

$$+C_{I_{H}}(\Gamma)+P_{I_{H}}(\Gamma).$$

No low natural-coprime term is estimated separately.

For large $N$, each prime in $I_{H}$ exceeds $Q$. Hence

(6.15) $$C_{I_{H}}(\Gamma)=\{\pi(I_{H})-\operatorname{Li}(I_{H})\}\sum_{q}\frac{\Gamma(q)}{\varphi(q)}.$$

The elementary identity

(6.16) $$\frac{q}{\varphi(q)}=\sum_{d\mid q}\frac{\mu(d)^{2}}{\varphi(d)},\qquad\sum_{q\leq Q}\frac{1}{\varphi(q)}\ll\log(2Q)$$

follows by multiplicativity and a convergent Euler product. Thus Paper I, (2.21), bounds (6.15) by $O_{A}(NL^{-A})$ for each fixed $A$ [26, (2.21)].

For the three $D$-terms we use the maximal-endpoint prime form of Bombieri–Vinogradov. Here it is conditional on the prime Siegel–Walfisz input [26, (2.21)]. Define the prime progression error by

(6.17) $$\Delta_{\pi}(y;q,a)=\pi(y;q,a)-\frac{\operatorname{Li}(y)}{\varphi(q)}.$$

Fact 3.2 gives, for each fixed $S>0$,

(6.18) $$\sum_{q\leq X^{1/2}(\log X)^{-(S+60)}}\max_{(a,q)=1}\sup_{0\leq y\leq X}|\Delta_{\pi}(y;q,a)|\ll_{S}X(\log X)^{-S}.$$

Here $\operatorname{Li}(y)=0$ for $0\leq y<2$. The level sits at $X$, not at the variable endpoint $y$, so intervals with endpoints chosen separately for each modulus cost at most twice this bound. This is the maximal prime statement. We do not substitute the fixed-endpoint von Mangoldt statement of Paper I, (2.20). The source and the fixed $60$ logarithmic buffer are those of Fact 3.2.

Set $X=N$, $S=A+10$, and choose the half-level buffers so that

(6.19) $$B_{L},\ B_{\mathrm{tail}},\ B_{0}\geq A+72.$$

Lemma 6.1 (The three low discrepancies).

Under (6.19),

(6.20) $$|D_{I_{L}}(b_{\mathrm{all}})|+|D_{I_{H}}(b_{\mathrm{tail}})|+|D_{I_{H}}(c_{\mathrm{lo}})|\ll_{A}NL^{-A-10}.$$

Proof.

The arrays $b_{\mathrm{all}}$, $b_{\mathrm{tail}}$ and $c_{\mathrm{lo}}$ are supported on $q<R_{L}$, $q<R_{\mathrm{tail}}$ and $q\leq Q_{0}$, respectively. Each is one-bounded. Since $B_{L},B_{\mathrm{tail}},B_{0}\geq A+72>S+60$, every such $q$ has $q\leq N^{1/2}L^{-(S+60)}$. This is the level of (6.18) at $X=N$. Every supported $q$ is coprime to $N$, so $N\bmod q$ is a reduced class. Since $\pi(y;q,N)=\operatorname{Li}(y)=0$ for $y<2$,

$$\Pi_{I_{L}}(q)-\frac{\operatorname{Li}(I_{L})}{\varphi(q)}=\Delta_{\pi}(N/4;q,N).$$

Write $y^{-}$ for the left limit at $y$. Then

$$\Pi_{I_{H}}(q)-\frac{\operatorname{Li}(I_{H})}{\varphi(q)}=\Delta_{\pi}\bigl{(}((1-\varepsilon)N)^{-};q,N\bigr{)}-\Delta_{\pi}(N/4;q,N).$$

A left limit is a limit of values at points of $[0,N]$. Hence each value on the right is at most $\sup_{0\leq y\leq N}|\Delta_{\pi}(y;q,N)|$. Therefore the left side of (6.20) is at most five times the left side of (6.18), with $X=N$ and $S=A+10$. This is the maximal prime statement of Fact 3.2. ∎

6.3. The high coefficient

Put

(6.21) $$\begin{gathered}U=V=N^{1/3},\qquad z_{0}=N^{1/(\log\log N)^{3}},\\ \mathcal{R}(n)=\mathbf{1}_{P^{-}(n)>z_{0}},\qquad\mathcal{R}(1)=1,\qquad\mathcal{A}(a)=\mu(a)\mathcal{R}(a),\\ \mathcal{C}(t)=\mathcal{R}(t)\sum_{\begin{subarray}{c}ab=t\\ a\leq U,\ b\leq V\end{subarray}}\mu(a)\Lambda(b),\qquad\mathcal{B}(s)=\mathcal{R}(s)\sum_{\begin{subarray}{c}b\mid s\\ b>V\end{subarray}}\Lambda(b),\\ \mathcal{K}_{q}(e)=\mathbf{1}_{e\equiv N\pmod{q}}-\frac{\mathbf{1}_{(e,q)=1}}{\varphi(q)}.\end{gathered}$$

The von Mangoldt identity gives three sums:

(6.22) $$T_{1} =\sum_{q}\Gamma(q)\sum_{\begin{subarray}{c}a\leq U,\ c\geq 1\\ ac\in I_{H}\end{subarray}}\mathcal{A}(a)\mathcal{R}(c)\frac{\log c}{\log(ac)}\mathcal{K}_{q}(ac),$$

$$T_{2} =-\sum_{q}\Gamma(q)\sum_{\begin{subarray}{c}t\leq UV,\ c\geq 1\\ tc\in I_{H}\end{subarray}}\frac{\mathcal{C}(t)\mathcal{R}(c)}{\log(tc)}\mathcal{K}_{q}(tc),$$

$$T_{\mathrm{II}} =\sum_{q}\Gamma(q)\sum_{\begin{subarray}{c}a>U,\ s>V\\ as\in I_{H}\end{subarray}}\frac{\mathcal{A}(a)\mathcal{B}(s)}{\log(as)}\mathcal{K}_{q}(as).$$

These come from

(6.23) $$\Lambda=\mu_{\leq U}*\log-\mu_{\leq U}*\Lambda_{\leq V}*1+\Lambda_{\leq V}+\mu_{>U}*\Lambda_{>V}*1.$$

Roughness is multiplicative, including shared prime factors. The short $\Lambda$-term vanishes on $I_{H}$.

The detector retains proper prime powers whose bases exceed $z_{0}$. For $e\in I_{H}$, hence $e>z_{0}$,

(6.24) $$\frac{\mathcal{R}(e)\Lambda(e)}{\log e}=\mathbf{1}_{e\in\mathbb{P}}+\sum_{\begin{subarray}{c}P\in\mathbb{P},\ j\geq 2\\ e=P^{j}\end{subarray}}\frac{\mathbf{1}_{P>z_{0}}}{j}.$$

Consequently

(6.25) $$P_{I_{H}}(\Gamma) =T_{1}+T_{2}+T_{\mathrm{II}}-E_{\mathrm{pp}},$$

$$E_{\mathrm{pp}} =\sum_{q}\Gamma(q)\sum_{\begin{subarray}{c}P\in\mathbb{P},\ j\geq 2\\ P^{j}\in I_{H}\end{subarray}}\frac{\mathbf{1}_{P>z_{0}}}{j}\mathcal{K}_{q}(P^{j}).$$

Let $e\in I_{H}$. A supported $q$ with $e\equiv N\pmod{q}$ divides $N-e$, and $|\Gamma(q)|\leq 1$ by unique marking. The comparator part is at most $\sum_{q\leq Q}1/\varphi(q)$. Hence (6.16) gives

(6.26) $$\sum_{q}|\Gamma(q)||\mathcal{K}_{q}(e)|\leq d(N-e)+O(L),\qquad E_{\mathrm{pp}}\ll_{\nu}N^{1/2+\nu}L^{2}.$$

Here $N-e\geq\varepsilon N$, and $0<\nu<1/2$ is fixed. The second bound follows from the first, $d(N-e)\ll_{\nu}N^{\nu}$ and $\sum_{j\geq 2}\#\{P\in\mathbb{P}:P^{j}\leq N\}\ll N^{1/2}$. The high-coefficient estimate uses the $P>z_{0}$ mask of $E_{\mathrm{pp}}$ in (6.25). That mask excludes bases $P\leq z_{0}$, so the displayed decomposition needs no separate add-back.

Proposition 6.12(3) proves the high-coefficient estimate

(6.27) $$P_{I_{H}}(\Gamma)\ll_{A,B,B_{0}}NL^{-A}\qquad(A>0\ \hbox{fixed}).$$

Its modulus coefficient is exactly (6.6). The estimate permits no joint arithmetic mask. Table 6.1 gives its decomposition. Each power boundary belongs to the interval on its right.

Table 6.1. The high-coefficient decomposition

Sum and range	Tested array	Required estimate
$T_{1}$, all $a\leq U$	$\mathcal{R}(c)$	Rough-factor estimate
$T_{2}$, $c<N^{.46}$	$\mathcal{R}(c)$	Two-factor convolution
$T_{2}$, $N^{.46}\leq c<N^{.50}$	$\mathcal{C}(t)/(2L)$	Central estimate
$T_{2}$, $N^{.50}\leq c<N^{.54}$	$\mathcal{R}(c)$	Central estimate
$T_{2}$, $c\geq N^{.54}$	$\mathcal{R}(c)$	Rough-factor estimate
$T_{\mathrm{II}}$, $a<N^{.46}$	$\mathcal{A}(a)$	Two-factor convolution
$T_{\mathrm{II}}$, $N^{.46}\leq a<N^{.50}$	$\mathcal{B}(s)/(2L)$	Central estimate
$T_{\mathrm{II}}$, $N^{.50}\leq a<N^{.54}$	$\mathcal{A}(a)$	Central estimate
$T_{\mathrm{II}}$, $a\geq N^{.54}$	$\mathcal{B}(s)/(2L)$	Two-factor convolution

6.4. The three distribution estimates

Lemma 6.2 (Rough-factor distribution under parent-scale and strict-roughness conditions).

Let $L=\log N$, $z_{0}=N^{1/(\log\log N)^{3}}$, and $Q=N^{5227/10000}L^{-B}$, with fixed $B$. Assume the two explicit conditions used by the direct rows:

1. The original parent satisfies $cN^{.54}\leq H\leq N$, for a fixed $c>0$. In the $T_{1}$ row $H=N/a\leq N$; in the right $T_{2}$ row $H=N/t\leq N$.

2. Use the strict roughness convention $\mathcal{R}(n)=\mathbf{1}_{P^{-}(n)>z_{0}}$, as in (6.21). It is not replaced by $P^{-}(n)\geq z_{0}$.

For every fixed $T_{\mathrm{save}}>0$, uniformly in real intervals $I\subset(0,2H]$ (so $|I|\leq 2H$; open, closed or one-point), supported $q\leq Q$ with $(q,N)=1$, and reduced residues $(b,q)=1$,

(6.28) $$\sum_{n\in I}\mathcal{R}(n)\left\{\mathbf{1}_{n\equiv b\pmod{q}}-\frac{\mathbf{1}_{(n,q)=1}}{\varphi(q)}\right\}\ll_{T_{\mathrm{save}}}\frac{H}{qL^{T_{\mathrm{save}}}}.$$

The interval may be arbitrarily short. The saving is measured at its original parent. The estimate is an integer-sieve statement and uses no prime distribution input.

Proof.

Put $\ell=\log\log N$ and $\mathcal{P}_{q}=\{p\leq z_{0}:p\nmid q\}$. The Euler product at $s=1+1/\log z_{0}$ gives $h_{q}=\sum_{p\in\mathcal{P}_{q}}p^{-1}\leq 3\ell$ for sufficiently large $N$. Choose the fixed constant $K$ after $T_{\mathrm{save}}$, with

(6.29) $$K>3e,\qquad K\log\!\left(\frac{K}{3e}\right)>T_{\mathrm{save}}+4.$$

Let $k=2\lceil K\ell/2\rceil$. The two adjacent Bonferroni truncations bracket the strict exclusion of all primes in $\mathcal{P}_{q}$. Their normalized gap satisfies

(6.30) $$k=2\lceil K\log\log N/2\rceil,\qquad E_{k}\leq\frac{h_{q}^{k}}{k!}\leq\left(\frac{3e}{K}\right)^{K\ell}\leq L^{-T_{\mathrm{save}}-4},\qquad(k+1)z_{0}^{k}=N^{O(K/\ell^{2})}.$$

Every divisor in the truncation is at most $z_{0}^{k}$. By CRT, each divisor contributes its expected interval mass with error $O(1)$. This is uniform for open, closed, or singleton intervals. The same expansion applies to the comparator after using

$$\mathbf{1}_{(n,q)=1}=\sum_{\begin{subarray}{c}d\mid\operatorname{rad}(q)\\ d\mid n\end{subarray}}\mu(d).$$

Thus the main terms cancel. The accumulated lattice error is $N^{O(K/\ell^{2})}q/\varphi(q)$, and $q/\varphi(q)\leq\omega(q)+1\ll L$. Under (W1), $H/q\gg N^{.0173}L^{B}$. The Bonferroni remainder is at most $(|I|/q)E_{k}L^{2}\leq 2HL^{-T_{\mathrm{save}}-2}/q$ because $|I|\leq 2H$. Hence both are absorbed by $H/(qL^{T_{\mathrm{save}}})$ once the fixed truncation precision is raised. This proves (6.28).

The strict convention in (W2) has no threshold atom. An application might instead use $P^{-}(n)\geq z_{0}$. Then the only possible difference is the atom $z_{0}=P$ when $z_{0}$ is prime. That weak convention requires its separate CRT boundary correction. It is not used in (6.28). ∎

The positive gap $.54-v_{\mathrm{RA}}=173/10000$ absorbs the lattice error in (6.30). For $T_{1}$, the free scale is at least $N^{2/3}/4$. For the right $T_{2}$-part, it exceeds a fixed multiple of $N^{.54}$. Partial summation then gives

(6.31) $$\sum_{m\leq M_{\mathrm{out}}}\sum_{q\leq Q}\frac{N}{mqL^{T_{\mathrm{save}}}}\ll NL^{2-T_{\mathrm{save}}}.$$

Both kernel terms remain in this estimate. Here $M_{\mathrm{out}}$ is the outer endpoint of the current row. It is $U$ in the $T_{1}$ row and the corresponding low-row endpoint in the $T_{2}$ decomposition.

The other tested arrays need uniformity under coprimality with an auxiliary integer. Write $\tau(h)=\sum_{d\mid h}1$ and $\omega(h)=\#\{p\in\mathbb{P}:p\mid h\}$. For any sequence $f(n)$ on its original parent $(H,2H]$, define

(6.32) $$\Delta_{f}(I;k,b;h)={} \sum_{\begin{subarray}{c}n\in I,\ n\equiv b\pmod{k}\\ (n,h)=1\end{subarray}}f(n)$$

$$-\frac{1}{\varphi(k)}\sum_{\begin{subarray}{c}n\in I\\ (n,kh)=1\end{subarray}}f(n).$$

Lemma 6.3 (Prime-variable transfer).

Let $N^{1/3}/8\leq H\leq 2N$, $L=\log N$, and $z_{0}=N^{1/(\log\log N)^{3}}$. Let $y_{0}\geq z_{0}$ and

$$f(n)=\sum_{\begin{subarray}{c}mp=n\\ p\in J_{m}\end{subarray}}\alpha(m)w(p),$$

where $p$ is prime and $|\alpha(m)|\leq 1$. For each $m$ the allowed primes lie in a real interval $J_{m}\subset[y_{0},\infty)$. Assume $\|w\|_{\infty}+\operatorname{Var}(w)\leq W$ on $[y_{0},2H]$ and $|f(n)|\leq 1$ on $(H,2H]$. For every fixed $P>0$, uniformly in intervals $I\subset(H,2H]$, $k,h\geq 1$, and $(b,k)=1$, prime Siegel–Walfisz, Paper I (2.21), implies

(6.33) $$|\Delta_{f}(I;k,b;h)|\ll_{P}WH\left(L^{-P}+\frac{\omega(h)L}{y_{0}}\right)\ll_{P}WH\tau(h)L^{-P}.$$

Proof.

Put $K=2P+4$. If $k>L^{K}$, the two terms in the discrepancy are bounded by a constant times $\|f\|_{\infty}(H/k+1+H/\varphi(k))$. For each represented $n$, a prime divisor $p\geq y_{0}$ determines at most one pair $(m,p)$, since $m=n/p$. Hence

$$|f(n)|\leq W\frac{\log(2H)}{\log y_{0}}\ll W(\log L)^{3},$$

using $y_{0}\geq z_{0}$, $H\leq 2N$, and $\log z_{0}=L/(\log L)^{3}$. Since $\varphi(k)\gg\sqrt{k}$ and $H\geq N^{1/3}/8$, the resulting discrepancy is $\ll_{P}WHL^{-P}$. The two spare powers in $K/2=P+2$ absorb $(\log L)^{3}$. The additive $1$ is absorbed because $H\geq N^{1/3}/8\gg L^{P}$.

Now let $k\leq L^{K}$. Terms with $(m,kh)>1$ vanish from both sides. Otherwise the prime progression is $p\equiv bm^{-1}\pmod{k}$ on $J_{m}\cap I/m$. Primes dividing $h$ are excluded on the progression side and in the comparator. For each $m$ the two exclusions differ by at most $2W\omega(h)$. There are at most $2H/y_{0}$ possible values of $m$, so these terms contribute $\ll WH\omega(h)/y_{0}$.

For the remaining primes define

$$E(t)=\pi(t;k,c)-\frac{1}{\varphi(k)}\sum_{\begin{subarray}{c}a\bmod k\\ (a,k)=1\end{subarray}}\pi(t;k,a).$$

The reduced-class sum equals $\pi(t)-\#\{p\leq t:p\mid k\}$. Thus this is the centered prime discrepancy without the primes dividing $k$. Apply (2.21) to the class $c$ and average its uniform error over all reduced classes. This gives $E(t)\ll_{A^{\prime}}t(\log t)^{-A^{\prime}}$, with no factor of $k$. Since $t\geq y_{0}\geq z_{0}$, $\log t\geq L/(\log L)^{3}$. Hence, for all sufficiently large $N$, $k\leq(\log t)^{2K}$. Therefore (2.21), with modulus exponent $2K$ and fixed saving $A^{\prime}$, applies throughout each interval of integration. Stieltjes partial summation bounds each $m$-contribution by $O(W(H/m)(\log y_{0})^{-A^{\prime}})$. Sum over $m\leq 2H/y_{0}$ and take $A^{\prime}=2(P+K+2)$. This gives $O_{P}(WHL^{-P})$. The estimate is measured at the original parent $H$, including for short intervals. Finally, $L/y_{0}\ll_{P}L^{-P}$ because $y_{0}\geq z_{0}$, and $\omega(h)\leq\tau(h)$. This proves (6.33). ∎

Lemma 6.4 (Four-family auxiliary-coprime Siegel–Walfisz).

By prime Siegel–Walfisz, Paper I (2.21), for each of $\mathcal{R},\mathcal{A},\mathcal{C}/(2L),\mathcal{B}/(2L)$, the bound

(6.34) $$|\Delta_{f}(I;k,b;h)|\ll_{P}H\tau(h)L^{-P},\qquad(b,k)=1,$$

holds uniformly for every $I\subset(H,2H]$ and $k,h\geq 1$. The parents are the central ranges in Table 6.1 and the convolution parents $N^{1/3}/8\leq H\leq 2N^{.46+O(1/L)}$. The proof uses the lower bound only to absorb the large-modulus lattice term and to exclude $n=1$. It uses $H\leq 2N^{.54}$ for the remaining weight bounds. The divisor exponent is the single fixed value $C_{\mathrm{SW}}=1$, independent of $P$. The implied constant may depend on $P$ and the fixed family parameters, but not on $N,H,I,k,h,b$. The error is measured at $H$, even when $I$ is short.

Proof.

For sufficiently large $N$, every parent has $H\geq N^{1/3}/8>1$, so $n=1$ is excluded. The rough and signed-rough arrays have exact prime-variable decompositions. For $\mathcal{R}(n)$, take $\alpha(m)=\mathcal{R}(m)$, $J_{m}=[P^{+}(m),\infty)\cap(z_{0},\infty)$, and $w(p)=1$. Use $P^{+}(1)=1$. The largest prime factor gives the unique term. For $\mathcal{A}(n)=\mu(n)\mathcal{R}(n)$, take $\alpha(m)=-\mu(m)\mathcal{R}(m)$, $J_{m}=(\max(P^{+}(m),z_{0}),\infty)$, and $w(p)=1$. The strict inequality excludes repeated largest primes, where $\mu(n)=0$.

For $\mathcal{B}(s)/(2L)$, the $j=1$ terms in $b=p^{j}\mid s$, $b>V$, use $\alpha(m)=\mathcal{R}(m)$, $J_{m}=(V,\infty)$, and $w(p)=\log p/(2L)$. For $\mathcal{C}(t)/(2L)$, the $j=1$ terms use $\alpha(a)=\mu(a)\mathcal{R}(a)\mathbf{1}_{a\leq U}$, $J_{a}=(z_{0},V]$, and the same weight. In each case $\|w\|_{\infty}+\operatorname{Var}(w)\leq 1$. The four tested arrays are one-bounded on their parent ranges. For the last two arrays this follows from $\sum_{d\mid n}\Lambda(d)=\log n$ and $\log(2H)<2L$. On the central ranges $H\leq 2N^{.54}$. There the weight bound follows from $y_{0}=V=N^{1/3}$ for $\mathcal{B}$ and $y_{0}=z_{0}$ for $\mathcal{C}$, since $\log(2H)\leq.54L+O(1)$.

For $\mathcal{B}$ and $\mathcal{C}$, write the remaining $j\geq 2$ terms as $f_{\geq 2}$. They need not be nonnegative. For instance, the $\mu(a)$ coefficient in $\mathcal{C}$ may change sign. Take absolute values before summing:

$$\sum_{n\leq 2H}|f_{\geq 2}(n)|\ll\frac{H}{L}\sum_{p>z_{0}}\sum_{j\geq 2}\frac{\log p}{p^{j}}\ll\frac{H\log z_{0}}{Lz_{0}}\ll\frac{H}{z_{0}}.$$

For any sequence $g$, $|\Delta_{g}|\leq 2\sum_{n\leq 2H}|g(n)|$. Hence this remainder is $O_{P}(HL^{-P})$ for every fixed $P$. Applying Lemma 6.3 to each prime term proves (6.34) with the same exponent $C_{\mathrm{SW}}=1$. ∎

The mean-square estimate below has five uses. Section 7 uses it for the zero mode of the exterior sum. Section 5 uses it for the Class II estimate of Paper II inside the retained side. Section 8 uses it for the row-three dispersion estimate of the switched sum. These three apply Paper I, Theorem 3.2. In the large-norm case of Paper I, (3.48), part (ii) below replaces [3, Theorem 0(a)]. The other two uses are the central rectangular estimate of Lemma 6.9 and the small-modulus norm-relative step of Lemma 6.8. Part (i) derives the conclusion of that published theorem (Paper I, Theorem 2.4). Its inputs are the norm-relative condition (6.35) and an elementary large sieve, so the argument is self-contained at this step. We do not correct [3]. Paper I’s use of it is unaffected. The norm-relative premise is not deduced from the absolute parent-scale estimate (6.34) alone. Let $\xi$ be complex and supported in $[1,Z]$, $Z\geq 3$. Fix $B_{\mathrm{div}}\geq 0$. Consider the following condition. For every $P>0$, uniformly in integers $h,k\geq 1$ and reduced classes $b\bmod k$,

(6.35) $$\left|\sum_{\begin{subarray}{c}n\equiv b\ (k)\\ (n,h)=1\end{subarray}}\xi_{n}-\frac{1}{\varphi(k)}\sum_{(n,kh)=1}\xi_{n}\right|\ll_{P}\|\xi\|_{2}Z^{1/2}\tau(h)^{B_{\mathrm{div}}}(\log(2Z))^{-P},$$

The constants are common to the family of pairs $(Z,\xi)$ under consideration. Write $\Delta_{\xi}(k,b)$ for the centered discrepancy at $h=1$. The norm conclusion is

(6.36) $$\sum_{k\leq Z(\log Z)^{-B_{1}(A)}}\sum_{b\bmod k}^{*}|\Delta_{\xi}(k,b)|^{2}\ll_{A}\|\xi\|_{2}^{2}Z(\log Z)^{-A}.$$

In the selected-factor form of Paper I, (3.48), $\xi_{n}=\alpha_{n}1_{(n,r)=1}$, $Z=\lceil 2T_{\mathrm{int}}\rceil$, and

(6.37) $$\sum_{q\leq 3Q}\sum_{b\bmod q}^{*}|\Delta_{\xi}(q,b)|^{2}\ll N_{\mathcal{I}}T_{\mathrm{int}}L^{-A_{3}},\qquad 3Q\leq Z(\log Z)^{-B_{F}},\quad B_{F}=A_{3}+15,$$

where $N_{\mathcal{I}}$ is the interval mass in Paper I, (3.8).

Lemma 6.5 (Transferred mean-square estimate).

1. If (6.35) holds, then (6.36) holds for every fixed $A>0$. Here $B_{1}(A)$ and the implied constant depend only on $A$, $B_{\mathrm{div}}$ and the common constants.

2. Assume the hypotheses (3.1), (3.2), (3.6), (3.8) with its length condition, (3.11) and (3.25)–(3.27) of Paper I. Then the estimate (6.37) holds uniformly over the parameter boxes and selected factors of Paper I, Theorem 3.2. Hence that theorem holds with this input in place of [3, Theorem 0(a)] at (3.48).

Proof.

(i) Expand $\Delta_{\xi}(k,b)$ in Dirichlet characters modulo $k$ and pass to primitive characters. For conductors up to a fixed power of $\log Z$, apply (6.35). Take the auxiliary divisor $h$ to be the imprimitive part of the modulus. For larger conductors use the multiplicative large sieve inequality. Its elementary proof uses finite Fourier identities and the spacing of reduced fractions. Summing over the moduli gives (i). (ii) If $\|\xi\|_{2}^{2}\leq N_{\mathcal{I}}L^{-A_{3}-1}$, the elementary bound in Paper I, (3.48), applies. Otherwise (3.25), with auxiliary divisor $dr$, and (3.26) give (6.35) for $\xi_{n}=\alpha_{n}1_{(n,r)=1}$ with divisor exponent one. Also (3.6) gives the modulus range in (6.37). Part (i) then applies. The complete argument is in Appendix  A.2, with its row index in Appendix  B.8. We assert no implication from (6.34) alone to the norm bound. ∎

For the three convolution rows, let

(6.38) $$D_{1}=HN^{-\sigma_{\mathrm{p}}},\qquad D_{2}=QN^{w_{\mathrm{RA}}+\sigma_{\mathrm{p}}}/H,\qquad\sigma_{\mathrm{p}}=10^{-6},\quad w_{\mathrm{RA}}=9/1000,$$

where $N^{1/3}/8\leq H\leq 2N^{.46+O(1/L)}$. A supported modulus has the unique form $q=pd$ of (6.4). Here $p$ is the marked prime and $d$ the squarefree cofactor.

(6.39) $$r<D_{1},\qquad s<D_{2},\qquad rs=q<Q.$$

Lemma 6.6 (Deterministic allocation).

For each supported $q=pd$, the following rule defines a unique coprime split $q=rs$ satisfying (6.39). Put $p$ in $r$. Order the cofactor primes exceeding $N^{w_{\mathrm{RA}}}$ as $b_{1}>\cdots>b_{m}$. Here $m\leq(v_{\mathrm{RA}}-t_{0})/w_{\mathrm{RA}}=37553/1260<30$. Process them in this order. Put $b_{i}$ in $r$ exactly when the current $rb_{i}<D_{1}$, and put it in $s$ otherwise. Let $r_{\rm big}$ and $s_{\rm big}$ be the resulting factors. Order the remaining cofactor primes as $l^{\prime}_{1}>\cdots>l^{\prime}_{k}$. Set $r_{0}=r_{\rm big}\prod_{j=1}^{k}l^{\prime}_{j}$. If $r_{0}<D_{1}$, take $r=r_{0}$, $s=s_{\rm big}$. Otherwise let $j_{*}$ be the least index for which $r_{0}/(l^{\prime}_{1}\cdots l^{\prime}_{j_{*}})<D_{1}$, and take

$$r=r_{0}/(l^{\prime}_{1}\cdots l^{\prime}_{j_{*}}),\qquad s=s_{\rm big}l^{\prime}_{1}\cdots l^{\prime}_{j_{*}}.$$

Call the resulting pair $\operatorname{Sel}_{H}(q)=(r,s)$.

Proof.

The marked prime fits, since $p<N^{1/3-\eta_{\rm m}}<D_{1}$, even at the fixed lower endpoint $H\geq N^{1/3}/8$. The accepted big-prime partial products stay below $D_{1}$. Thus $r_{\rm big}<D_{1}$, and the least index $j_{*}$ exists when needed. Write all cofactor primes in decreasing order as $l_{1}>l_{2}>\cdots$. The odd cube gates in (6.4) imply

$$pl_{1}\cdots l_{i-1}l_{i}^{2}<Q\qquad\text{for every }i:$$

For odd $i$, drop one factor from the cube gate. For even $i$, use the preceding odd gate and $l_{i}\leq l_{i-1}$.

Suppose $r_{0}<D_{1}$, and let $b_{i}$ be the last big prime rejected into $s$. The rejection test gives $r^{(i)}b_{i}\geq D_{1}$. The two-gate inequality then gives

$$s=s_{\rm big}=\frac{pb_{1}\cdots b_{i}}{r^{(i)}}\leq\frac{pb_{1}\cdots b_{i-1}b_{i}^{2}}{D_{1}}<\frac{Q}{D_{1}}<D_{2}.$$

If no big prime is rejected, $s=1$. In the prefix case, minimality gives $rl^{\prime}_{j_{*}}\geq D_{1}$, hence $r\geq D_{1}N^{-w_{\mathrm{RA}}}$ and

$$s=q/r<QN^{w_{\mathrm{RA}}}/D_{1}=D_{2}.$$

The factors partition the squarefree prime factors of $q$. Thus $(r,s)=1$, $rs=q<Q$, and $r<D_{1}$. ∎

Define the exact joint coefficient on each dyadic Cartesian modulus shell by

(6.40) $$\begin{gathered}F_{H}(r,s)=\begin{cases}\Gamma(rs),&rs\text{ is supported and }\operatorname{Sel}_{H}(rs)=(r,s),\\ 0,&\text{otherwise},\end{cases}\\ \sum_{q}\Gamma(q)\Phi(q)=\sum_{r,s}F_{H}(r,s)\Phi(rs),\\ F_{H}(r,s)=\sum_{\nu}c_{\nu}f_{\nu}(r)g_{\nu}(s),\qquad|f_{\nu}|,|g_{\nu}|\leq 1,\qquad\sum_{\nu}|c_{\nu}|\ll L^{C_{\mathrm{tensor}}},\quad C_{\mathrm{tensor}}=2^{30}+91.\end{gathered}$$

Here $\operatorname{Sel}_{H}$ is the rule in Lemma 6.6. The expansion is exact on the full Cartesian parent, including zero values off the selected pairs.

Lemma 6.7 (Global tensor for the selector).

The coefficient $F_{H}$ has a rank-one expansion satisfying (6.40), with the sharper bound

(6.41) $$\sum_{\nu}|c_{\nu}|\leq 4(2^{30}-1)(3+L)^{78}\ll L^{79}.$$

Proof.

Fix a pair of dyadic shells. A branch fixes the merge order $\pi\in\{R,S\}^{m}$ of the at most 29 big cofactor primes. It also fixes the final large-prime/prefix case. In the prefix case it fixes the parity of the number of factors assigned to $s$. There are at most $4\sum_{m=0}^{29}2^{m}=4(2^{30}-1)$ branches. Within a branch, unary conditions impose squarefreeness, coprimality with $2N$ and the unique marked prime $p$ in $r$. They also impose the bounds $P^{+}(r/p)<z_{s}$ and $P^{+}(s)<z_{s}$, assigned big-prime counts, $r<D_{1}$, $s<D_{2}$, and $r_{\rm big}<D_{1}$. The remaining unary data are the case indicator, the parity of the small-prime count, and the sign $\Gamma(rs)=\mu(rs/p)=-\mu(r)\mu(s)$.

Every cross-condition is a comparison $\phi(r)<\psi(s)$ or its weak counterpart. Let $K_{0}$ be the number of distinct values of $\phi$ on the $r$-shell. For ranks $1\leq a\leq K_{0}$, $0\leq b\leq K_{0}$, put $M=2K_{0}+1$ and

$$c_{t}=M^{-1}\sum_{u=0}^{K_{0}}e(-tu/M).$$

Finite Fourier orthogonality gives the exact separator

(6.42) $$\mathbf{1}_{a\leq b}=\sum_{t\bmod M}c_{t}e(tb/M)e(-ta/M),\qquad\sum_{t\bmod M}|c_{t}|\leq 3+\log M.$$

Rank $\phi(r)$ among its shell values, and count the values below $\psi(s)$. Each shell has at most $Q$ values, so $\log M\ll L$. The cross-comparisons are exactly:

1. the adjacent strict comparisons in the fixed merge order of the big primes (at most 28);

2. for each $S$-letter $b_{i}$, the greedy rejection test $p(\prod_{b_{j}>b_{i},\ b_{j}\in r}b_{j})b_{i}\geq D_{1}$ (at most 29);

3. the odd cube gates $pl_{1}\cdots l_{i-1}l_{i}^{3}<Q$ at odd indices among the big primes (at most 15);

4. the remaining odd cube gates on the small-prime part, grouped by the fixed parity into the two maxima of the $r$- and $s$-side partial products (two comparisons);

5. in the prefix case, the prefix-order test $\min P>\max R_{\rm small}$ and the least-prefix test $r\min P\geq D_{1}$ (two comparisons);

6. the support range $Q_{0}<rs<Q$ (two comparisons).

There are at most 78 comparisons in total. Once a merge branch is fixed, each displayed test compares an $r$-only quantity with an $s$-only quantity. The small-prime maxima combine all gates of the same parity and side. The branch conditions are exactly the greedy tests, the cube gates and the least-prefix condition. Hence their union is precisely $F_{H}$. Multiplying the separators in (6.42) gives (6.41). The unary factors remain one-bounded. The bound in (6.41) is per dyadic shell pair. The resulting $O(L^{2})$ shell pairs are absorbed by the chosen logarithmic precision. ∎

For the convolution rows, split $s=dt$ with $P^{+}(d)\leq z<P^{-}(t)$, where $z=N^{1/(\log\log N)^{3}}$ and $y=N^{1/\log\log N}$. The aggregate with $d>y$ is bounded by Rankin’s method. The mean-square modulus in the proof of Lemma 6.8 is $k=rd$, not the full $q$. Its scale is

(6.43) $$Q^{\prime}=D_{1}y=HN^{-\sigma_{\mathrm{p}}+1/\log\log N},\qquad k\leq 5Q^{\prime},\qquad Z=\lceil 2H\rceil.$$

The inner factor $t$ is $z$-rough and is removed in $L^{2}$, with cost $H^{2}L^{C}/z$. No bound for $\tau(r)$ and no roughness condition on $r$ is used. The required range is

(6.44) $$5Q^{\prime}\leq Z(\log Z)^{-B_{F}},\qquad B_{F}=P_{\mathrm{MS}}+15,$$

where $P_{\mathrm{MS}}$ is the fixed source-level precision in the supplementary Lemma 6.5(i). This holds since $Q^{\prime}/H=N^{-\sigma_{\mathrm{p}}+o(1)}$. The full-modulus range $3Q\leq Z(\log Z)^{-B_{F}}$ is not asserted for these parents.

Lemma 6.8 (Two-factor convolution estimate).

Let $N^{1/3}/8\leq H\leq 2N^{.46+O(1/L)}$ and $HM\asymp N$. Let $a_{n}$, $n\sim H$, be one-bounded, supported on $P^{-}(n)>z$, and satisfy (6.34) at the original parent $H$, where $z=N^{1/(\log\log N)^{3}}$. Let $b_{m}$, $m\sim M$, be one-bounded. Let $U_{1},V_{1}$ be dyadic modulus scales, allowing the singleton scale $1/2$ for the integer $1$, such that

$$U_{1}\leq D_{1},\qquad V_{1}\leq D_{2},\qquad U_{1}V_{1}\leq N^{v_{\mathrm{RA}}}.$$

Let $f(r)$, $r\sim U_{1}$, and $g(s)$, $s\sim V_{1}$, be arbitrary one-bounded arrays. Keep the unary restrictions $r<D_{1}$, $s<D_{2}$, $(r,N)=(s,N)=1$. Neither a pointwise condition $rs<Q$ nor a joint condition $(r,s)=1$ is imposed on the pair. Then, for every fixed $A>0$, uniformly in these occupied-parent scales, $f$, $g$, unary interval cuts of $a_{n}$ and $b_{m}$, and intervals $I\subset[N/8,2N]$,

(6.45) $$\sum_{r,s}f(r)g(s)\sum_{\begin{subarray}{c}n,m\\ nm\in I\end{subarray}}a_{n}b_{m}\mathcal{K}_{rs}(nm)\ll_{A}NL^{-A}.$$

Proof.

We use the dispersion method of Paper I, Theorem 3.2. It runs on the whole Cartesian shell $r\sim U_{1}$, $s\sim V_{1}$, with residue class $N\bmod rs$. Its occupied-parent scale conditions are $U_{1}\leq HN^{-\sigma_{\mathrm{p}}}$, $V_{1}\leq N^{v_{\mathrm{RA}}+w_{\mathrm{RA}}}/(HN^{-\sigma_{\mathrm{p}}})$, and $U_{1}V_{1}\leq N^{v_{\mathrm{RA}}}$. The second follows from the displayed bound by $D_{2}$ after dropping $L^{-B}$. The tested sequence is strictly $z$-rough. The unary supports carry $(rs,N)=1$ and the individual caps. No pointwise product cap is needed for the rank-one arrays. After Cauchy’s inequality in $m$, the split $s=dt$ above reduces the modulus of the mean square to $k=rd$ in the range (6.44). The mean square is split at a fixed power of $L$ below its target. Below it the trivial bound suffices. Above it (6.34) gives (6.35) with a loss of that power, so Lemma 6.5(i) applies. The prime Siegel–Walfisz input is Paper I, (2.21). No joint condition $(r,s)=1$ is used, because $r,s$ are not the moduli of Paper I, Theorem 2.8. Write $s=dt$ as above and $k=rd$. A prime $p\mid(r,s)$ with $p\leq z$ divides $d$ and only enlarges $k$. The Chinese remainder step needs only that $k,t_{1},t_{2}$ are pairwise coprime. The smooth–rough split gives $(d,t_{1}t_{2})=1$. A prime $p\mid(r,s)$ with $p>z$ divides $k$ and a rough $t_{i}$. Such tuples are the bad type $(k,t_{1}t_{2})>1$. If $p\mid t_{2}$, the second progression $kt_{2}\mid mn_{2}-N$ places $n_{2}$ in one class modulo $kp$. If $p\mid t_{1}$ only, exchange the two copies first; the absolute square majorant is symmetric, and the same statement then applies. We therefore remove these tuples before the Chinese remainder step. The cost in the square is $\ll MHL^{C}[H/(zK)+1]$. After the factor $KM$ of the first Cauchy step this is $\ll NL^{C}(z^{-1/2}+(K/H)^{1/2})$. That is $O_{A}(NL^{-A})$, since $z=N^{1/(\log\log N)^{3}}$ and $K/H\leq N^{-\sigma_{\mathrm{p}}/2}$. The zero-mode identity holds for all tuples, including those with gcd failures. After the second Cauchy step, Paper I, Theorem 2.8 bounds the nonzero frequencies. It is applied with $r_{\mathrm{int}}=tt^{\prime}$, the product of two $z$-rough inner variables, and $s_{\mathrm{int}}=p_{0}$, the least prime not dividing $N$. Since $p_{0}\ll\log N<z$, $(r_{\mathrm{int}},s_{\mathrm{int}})=1$, exactly as in Paper I, (3.82)–(3.83). This is the form of Paper I, Theorem 2.8 (Sections 5.3 and 7.4) used here. Its statement allows arbitrary one-bounded $f,g$ with only $(rs,N)=1$. The product restriction $nm\in I$ is kept as in Paper I, Theorem 3.2. The finite product partition used there also handles the endpoints. Neither (6.37) at the full modulus $3Q$ nor a norm-relative consequence of (6.34) for small norms is used. The complete argument is in Appendix  A.3, with its row index in Appendix  B.8. ∎

The central rows use $Y=N^{y}$, $y=341/5000$. Choose the largest divisor $s\mid d$ with $s\leq Y$. Since $p<N^{1/3-\eta_{\mathrm{m}}}$ and $q=pd>Q_{0}$, $d>Q_{0}/p\gg N^{1/6+\eta_{\mathrm{m}}}L^{-B_{0}}>Y$. Hence a next divisor exists. The cube gates control adjacent divisors and give

(6.46) $$\frac{Yq}{Q}<s\leq Y,\qquad r=q/s<Q/Y.$$

This allocation keeps a one-bounded joint modulus coefficient. The required central estimate has

(6.47) $$\begin{gathered}H_{\mathrm{c}}M_{\mathrm{c}}\asymp N,\qquad N^{.50}\ll H_{\mathrm{c}}\ll N^{.54},\\ Q_{1}\ll N^{.4545},\qquad Q_{2}\ll N^{.0682}.\end{gathered}$$

Lemma 6.9 (Central rectangular distribution).

Let $x$ be even, $H_{\mathrm{c}}M_{\mathrm{c}}\asymp x$, and let the bounds in (6.47) hold with $N$ replaced by $x$, with fixed comparison constants. Let $a_{n},b_{m}$ be one-bounded on the complete dyadic rectangle $n\sim H_{\mathrm{c}},m\sim M_{\mathrm{c}}$. Let the tested $a_{n}$ satisfy (6.34) at its original parent. The divisor exponent is fixed, and the constants are uniform at every requested precision. Write

$$\mathcal{D}(q,a)=\sum_{n\sim H_{\mathrm{c}},m\sim M_{\mathrm{c}}}a_{n}b_{m}\left(1_{nm\equiv a\ (q)}-\frac{1_{(nm,q)=1}}{\varphi(q)}\right).$$

Then, for every fixed $A>0$,

(6.48) $$\sum_{r\sim Q_{1}}\max_{(b,r)=1}\sum_{\begin{subarray}{c}s\sim Q_{2}\\ \mu(rs)^{2}=1,\ (rs,x)=1\end{subarray}}\max_{\begin{subarray}{c}a\bmod rs,\ (a,rs)=1\\ a\equiv b\ (r)\end{subarray}}|\mathcal{D}(rs,a)|\ll_{A}x(\log x)^{-A}.$$

The constant is uniform over these families, rectangles, squarefree modulus supports and unary interval restrictions of $a_{n}$ and $b_{m}$. All errors are measured at the original parents. One outer residue $b$ is chosen before $s$ varies. In the application it is $N\bmod r$.

Proof.

This is the rectangular dispersion theorem proved in Appendix  A.4 and indexed in Appendix B.8. The restriction $(rs,x)=1$ only removes nonnegative terms. Its proof opens the inner modulus $s$ by dispersion with the outer residue fixed. It uses the mean square over all moduli of Lemma 6.5(i), which gives Paper I, (2.17), under its hypothesis (2.16). The supplementary proof applies the mean square twice. The second application is repeated over two correlations. At each application a constant sequence on the same parent is added. Its sign is chosen by the parallelogram identity to make the tested sequence norm-large. The resulting sequence and the constant sequence satisfy (6.35) uniformly. Their squared norms are at most the original parent times a fixed power of $L$. Fix the requested logarithmic modulus buffer $B_{1}$ first. Then choose the source-level parameter so that the moduli lie in the range of (6.36). Apply part (i) to both sequences, subtract by linearity, and absorb that fixed power by increasing the requested precision. This uses neither part (ii) nor a mean-square conclusion for an arbitrary small-norm sequence from absolute Siegel–Walfisz alone. The other inputs are Paper I, (2.21) (prime Siegel–Walfisz), (2.35) (Weil–Estermann) and Lemma 3.4 (completed hyperbola count). It does not use Paper I, Theorem 2.8. For the four tested arrays of Table 6.1, Lemma 6.4 supplies (6.34) with divisor exponent one. ∎

Lemma 6.9 concerns complete dyadic rectangles and one-bounded sequences. Lemma 6.10 below reduces the four central rows, with their product cuts, to it. The raw exponent checks are

(6.49) $$7v_{\mathrm{RA}}+5y =3.9999<4,$$

$$3v_{\mathrm{RA}}-y-1 =.4999<.50,$$

$$1-v_{\mathrm{RA}}+y =.5455>.54.$$

The three margins verify the numerical range only. They do not establish the rectangular-support or family hypotheses for a selected product region.

Product cuts require an additional finite reduction. For $e\in I_{H}$, put

(6.50) $$w(e) =\frac{2L}{\log e},$$

$$\mathcal{A}(a)\mathcal{R}(c)\frac{\log c}{\log(ac)} =\left(\mathcal{A}(a)\mathcal{R}(c)\frac{\log c}{2L}\right)w(ac),$$

$$-\frac{\mathcal{C}(t)\mathcal{R}(c)}{\log(tc)} =\left(-\frac{\mathcal{C}(t)}{2L}\mathcal{R}(c)\right)w(tc),$$

$$\frac{\mathcal{A}(a)\mathcal{B}(s)}{\log(as)} =\left(\mathcal{A}(a)\frac{\mathcal{B}(s)}{2L}\right)w(as).$$

Here $\log c/(2L)$ is unary, and $\|w\|_{\infty}\leq 2L/(L-\log 4)=O(1)$, and

$$\operatorname{Var}_{I_{H}}(w)=\frac{2L}{L-\log 4}-\frac{2L}{L+\log(1-\varepsilon)}=O(1).$$

Thus uniform product-prefix estimates restore the exact weights by Stieltjes summation, including the minus sign of $T_{2}$. Use multiplicative cells of relative width $L^{-D_{\mathrm{m}}}$. The original-source boundary strips satisfy

(6.51) $$E_{\mathrm{strip}}\ll\bigl{(}NL^{-D_{\mathrm{m}}/2}+N^{1/2}\bigr{)}L^{15/2}.$$

Each unnormalized source multiplicity is at most $3d(e)$. Cauchy and $d(n)^{4}\leq d_{16}(n)$ give (6.51). Sum the boundary strips before the modulus expansion. For a final saving $A$, fix

(6.52) $$D_{\mathrm{m}}=2A+40,\qquad T_{\mathrm{save}}>A+C_{\mathrm{tensor}}+2D_{\mathrm{m}}+40.$$

Then fix every Siegel–Walfisz and half-level precision. Choose the common threshold last.

Lemma 6.10 (Product-region reduction for the central rows).

Fix $A>0$ and choose $D_{\mathrm{m}}$ and $T_{\mathrm{save}}$ as in (6.52). Take one of the four central rows of Table 6.1. Keep its unary cuts, its normalized arrays from (6.50) and the coefficient $\Gamma$. For $N/4<u\leq(1-\varepsilon)N$ let $\mathcal{S}(u)$ be its normalized discrepancy sum over $N/4<nm\leq u$, without the factor $w$. Then

(6.53) $$\sup_{N/4<u\leq(1-\varepsilon)N}|\mathcal{S}(u)|\ll_{A}NL^{-A},$$

and the same bound holds for the sum over $nm\in I_{H}$ with the factor $w(nm)$.

Proof.

Selector. Let $q=pd$ be supported, with the primes of $d$ in decreasing order $\ell_{1}>\cdots>\ell_{k}$. As in the proof of Lemma 6.6, the cube gates give $p\ell_{1}\cdots\ell_{i-1}\ell_{i}^{2}<Q$ for every $i$, hence

$$\frac{\ell_{i}}{\ell_{i+1}\cdots\ell_{k}}<\frac{Q}{q}\qquad(1\leq i\leq k).$$

Consecutive divisors of a squarefree $d$ have ratio at most $\max_{i}\ell_{i}/(\ell_{i+1}\cdots\ell_{k})$. Write $d=\ell_{1}M$. The divisors of $d$ form the two chains $\mathcal{D}(M)$ and $\ell_{1}\mathcal{D}(M)$. If $\ell_{1}>M$, the only new gap is $\ell_{1}/M$. If $\ell_{1}\leq M$, the ranges $[1,M]$ and $[\ell_{1},\ell_{1}M]$ overlap. A gap of the union below $M$ lies inside a gap of $\mathcal{D}(M)$. One above $M$ lies inside a gap of $\ell_{1}\mathcal{D}(M)$. Induction on $k$ gives the claim. Since $d>Y$, the largest divisor $s\leq Y$ of $d$ has a successor $s^{\prime}>Y$ with $s^{\prime}/s<Q/q$. This proves (6.46). The pair $(r,s)$ is determined by $q$, and $q=rs$. Put $F(r,s)=\Gamma(rs)$ on selected pairs and $F(r,s)=0$ otherwise. This coefficient is one-bounded and supported on squarefree $rs$ with $(rs,2N)=1$. Its dyadic shells satisfy $Q_{1}\ll N^{v_{\mathrm{RA}}-y}$ and $Q_{2}\ll N^{y}$, where $v_{\mathrm{RA}}-y=909/2000$ and $y=341/5000$. There are $O(L^{2})$ shell pairs.

Parents. In the first central $T_{2}$ row the tested variable and array are $n=t$ and $\mathcal{C}(t)/(2L)$. In the second they are $n=c$ and $\mathcal{R}(c)$. In the central $T_{\mathrm{II}}$ rows they are $n=s$ and $\mathcal{B}(s)/(2L)$ in the first. In the second they are $n=a$ and $\mathcal{A}(a)$. The other variable carries the remaining one-bounded array, with the sign of $T_{2}$. On $I_{H}$, every relevant pair of dyadic parents has $H_{\mathrm{c}}M_{\mathrm{c}}\asymp N$ and $N^{.50}\ll H_{\mathrm{c}}\ll N^{.54}$ with fixed constants. There are $O(L^{2})$ such pairs. Lemma 6.4 gives (6.34) for the tested arrays at these parents.

Restriction. Let $a_{n}$ be a tested array on its parent and $J$ an interval. On the same parent, $a_{n}\mathbf{1}_{J}(n)$ satisfies $\Delta_{a\mathbf{1}_{J}}(I;k,b;h)=\Delta_{a}(I\cap J;k,b;h)$ for every interval $I$. It therefore satisfies (6.34) with the same constants, measured at the original parent. The other array may be restricted to any interval.

Mesh. Put $\delta=L^{-D_{\mathrm{m}}}$. Divide each parent into $O(\delta^{-1})$ intervals with endpoint ratio at most $1+\delta$. All unary cut points of the row are endpoints. For fixed $u$, call a cell complete if it lies inside $\{N/4<nm\leq u\}$. Any other cell meeting this region has all its products within a factor $(1+\delta)^{2}$ of $N/4$ or of $u$. The part of the region it covers lies in the set $\mathcal{E}_{u}$ of integers $e\in(N/4,(1-\varepsilon)N]$ with $|e-N/4|\leq 3\delta N$ or $|e-u|\leq 3\delta N$. Here $\#\mathcal{E}_{u}\ll\delta N+1$. For a complete cell apply Lemma 6.9 with $x=N$, saving $T_{\mathrm{save}}$, the restricted arrays, $b=N\bmod r$ and $a=N\bmod rs$. Both classes are reduced since $(rs,N)=1$, and $a\equiv b\pmod{r}$. Thus

$$\Bigl{|}\sum_{r,s}F(r,s)\mathcal{D}_{\mathrm{cell}}(rs,N)\Bigr{|}\leq\sum_{r\sim Q_{1}}\max_{(b,r)=1}\sum_{\begin{subarray}{c}s\sim Q_{2}\\ \mu(rs)^{2}=1,\ (rs,N)=1\end{subarray}}\max_{\begin{subarray}{c}a\bmod rs\\ a\equiv b\ (r)\\ (a,rs)=1\end{subarray}}|\mathcal{D}_{\mathrm{cell}}(rs,a)|\ll_{T_{\mathrm{save}}}NL^{-T_{\mathrm{save}}}.$$

There are $O(L^{2D_{\mathrm{m}}+4})$ choices of cell and shell pair, so the complete cells contribute $O(NL^{-T_{\mathrm{save}}+2D_{\mathrm{m}}+4})$.

Strips. Sum the incomplete cells together before expanding the modulus coefficient. For fixed $e$, the factorizations $e=nm$ in the row carry total normalized weight at most $d(e)$. Also (6.26) bounds the modulus sum by $d(N-e)+O(L)$. Their contribution is therefore

$$\ll\sum_{e\in\mathcal{E}_{u}}d(e)\{d(N-e)+L\}\ll(\#\mathcal{E}_{u})^{1/2}N^{1/2}L^{15/2}\ll\bigl{(}N\delta^{1/2}+N^{1/2}\bigr{)}L^{15/2},$$

by Cauchy’s inequality twice, $d(n)^{4}\leq d_{16}(n)$ and $\sum_{n\leq N}d_{16}(n)\ll NL^{15}$. This is (6.51).

Conclusion. With $D_{\mathrm{m}}=2A+40$ and $T_{\mathrm{save}}>A+C_{\mathrm{tensor}}+2D_{\mathrm{m}}+40$, the two contributions are $O(NL^{-A-C_{\mathrm{tensor}}-36})$ and $O(NL^{-A-25/2}+N^{1/2}L^{15/2})$, uniformly in $u$. This proves (6.53). Since $\|w\|_{\infty}+\operatorname{Var}_{I_{H}}(w)=O(1)$, Stieltjes summation against $\mathcal{S}$ gives the weighted bound, with the sign of $T_{2}$ kept. ∎

6.5. The main coefficient

For primes $\ell\nmid 2N$, set $g_{N}(\ell)=1/(\ell-1)$, and put $g_{N}(\ell)=0$ for $\ell\mid 2N$. Extend $g_{N}$ multiplicatively to squarefree integers, with $g_{N}(\ell^{j})=0$ for $j\geq 2$. For primes $e\equiv N\pmod{p}$ in the fixed reduced class modulo $p$, and primes $\ell\nmid 2Np$, the condition $\ell\mid(N-e)/p$ has density $1/\varphi(\ell)=g_{N}(\ell)$. Put

(6.54) $$V_{N}(z_{s})=\prod_{\begin{subarray}{c}2<\ell<z_{s}\\ \ell\nmid N\end{subarray}}\left(1-\frac{1}{\ell-1}\right)=\frac{2e^{-\gamma}C(N)}{\alpha L}(1+o(1)),$$

with $C(N)$ as in Paper II, (1.2). For a marked prime $p\in N^{K}$ and a level $R$, put $D=R/p$ and $s_{p}=\log D/(\alpha L)$. Let $P=P_{N}(z_{s})$ be the product of the primes $\ell<z_{s}$ with $\ell\nmid 2N$.

Lemma 6.11 (Upper-sieve application).

1. The comparison (6.9) holds.

2. There is an absolute constant $K_{\mathrm{dim}}$ such that, for every even $N$,

   (6.55) $$\prod_{w\leq\ell<z}(1-g_{N}(\ell))^{-1}\leq\frac{\log z}{\log w}\left(1+\frac{K_{\mathrm{dim}}}{\log w}\right)\quad(z>w\geq 2).$$

3. For the three pairs $(R,K)\in\{(R_{L},J),(R_{\mathrm{tail}},J_{\mathrm{tail}}),(Q,J_{t})\}$, uniformly in the mark and in even $N$,

   (6.56) $$\sum_{d\mid P}\frac{\mu_{d}^{+}(R/p)}{\varphi(d)}\leq V_{N}(z_{s})\{F(s_{p})+O(L^{-1/3})\},$$

   and, with $R=N^{r}L^{-B_{r}}$,

   (6.57) $$\sum_{q}\frac{c_{R,K}(q)}{\varphi(q)}\leq\frac{2e^{-\gamma}C(N)}{\alpha L}\left\{\int_{K}F\left(\frac{r-t}{\alpha}\right)\frac{dt}{t}+o(1)\right\}.$$

Proof.

We use Iwaniec’s upper bound for the linear sieve [13]. Let $\mathcal{A}$ be a finite sequence and $D>1$ a level. Let $P(z)$ be the product of the sifting primes below $z$. The upper weight $\mu_{d}^{+}$ of [13, p. 313] is $(-1)^{k}$ when $d=\ell_{1}\cdots\ell_{k}\mid P(z)$, $\ell_{1}>\cdots>\ell_{k}$, and $\ell_{1}\cdots\ell_{2l}\ell_{2l+1}^{3}<D$ for every $0\leq l\leq(k-1)/2$. Otherwise it is $0$. With $j=2l+1$ these are the gates of (6.3), so (6.3) is this weight on $d\mid P$. By the definition of $W^{+}$ [13, p. 310] and Lemma 2 there [13, p. 313],

$$\sum_{d\mid P(z)}\mu_{d}^{+}|\mathcal{A}_{d}|=W^{+}\geq S(\mathcal{A},P,z)$$

for every $D>1$, since $W^{+}$ is $S(\mathcal{A},P,z)$ plus nonnegative sifted counts. If the density satisfies (6.55) with a constant $K$ and $z\leq D$, Lemma 3, (12), of [13, p. 313] gives, with $s=\log D/\log z$,

$$\sum_{d\mid P(z)}\mu_{d}^{+}g(d)\leq V(z)\{F(s)+O(e^{\sqrt{K}-s}(\log D)^{-1/3})\}.$$

(a) Let $e$ be a prime in one of the pieces (6.7) and $p$ a mark with $pm=N-e$. If $P^{-}(m)\geq z_{s}$, then $(m,P)=1$. The inequality $W^{+}\geq S$ for the one-term sequence $\{m\}$ at level $D=R/p$ gives

(6.58) $$\sum_{\begin{subarray}{c}d\mid P\\ d\mid m\end{subarray}}\mu_{d}^{+}(D)\geq\mathbf{1}_{(m,P)=1}\geq\mathbf{1}_{P^{-}(m)\geq z_{s}}.$$

This holds for every $D>1$, in particular when $s_{p}=383/200<2$. Omitting $(m,Np)=1$ in (6.2) only enlarges the count. The condition $d\mid m$ is $e\equiv N\pmod{pd}$, and $c_{R,K}(pd)=\mu_{d}^{+}(R/p)$ by (6.5). The pieces (6.7) partition $[2,(1-\varepsilon)N)\times J$. The point $e=N/4$ lies in $I_{L}$ and $t=1/3-\eta_{\mathrm{m}}$ lies in $J_{\mathrm{tail}}$. Summing (6.58) over $e$ and $p$, each mark with its multiplicity, gives $S_{6}\leq A_{\mathrm{mix}}$ by (6.8).

(b) For odd $\ell$,

$$\Bigl{(}1-\frac{1}{\ell-1}\Bigr{)}^{-1}=\Bigl{(}1-\frac{1}{\ell}\Bigr{)}^{-1}\Bigl{(}1-\frac{1}{(\ell-1)^{2}}\Bigr{)}^{-1}.$$

The factors with $\ell\mid 2N$ equal one. Omitting them only lowers the product. Mertens’ theorem, with its uniform error term, bounds the product of the first factors over $w\leq\ell<z$ by $(\log z/\log w)(1+K_{1}/\log w)$. The product of the second factors over $\ell\geq\max(w,3)$ is at most $1+K_{2}/\log w$. The constants $K_{1},K_{2}$ are absolute. This gives (6.55) with one $K_{\mathrm{dim}}$, independent of $N$ and of $\omega(N)$.

(c) The limiting ranges of $s_{p}$ for the three pairs are

$$\Bigl{(}\frac{383}{200},\frac{12639}{4480}\Bigr{]},\qquad\Bigl{(}\frac{383}{200},\frac{1916149}{1000000}\Bigr{]},\qquad\Bigl{(}\frac{544243}{250000},\frac{43148397}{14000000}\Bigr{]},$$

all inside $[383/200,3.083)\subset(1,\infty)$. The factors $L^{-B_{r}}$ move $s_{p}$ by $O(\log L/L)$. Hence $z_{s}\leq D$ and $\log D\asymp L$ for $N\geq N_{0}(B_{r})$. Then (12) gives (6.56), with an error $O(L^{-1/3})$ uniform in the mark, the pair and $N$. For (6.54), use the factorization in (b) for odd $\ell<z_{s}$ and restore the factors with $\ell\mid N$. The primes $\ell\mid N$ with $\ell\geq z_{s}$ number at most $1/\alpha$ and change the product by a factor $1+O(1/z_{s})$. By (6.5),

$$\sum_{q}\frac{c_{R,K}(q)}{\varphi(q)}=\sum_{\begin{subarray}{c}p\in N^{K}\\ p\nmid N\end{subarray}}\frac{1}{p-1}\sum_{d\mid P}\frac{\mu_{d}^{+}(R/p)}{\varphi(d)}.$$

The outer weights are positive, so (6.56) may be summed termwise. Since $F$ is Lipschitz on the compact range above, the prime number theorem and partial summation give (6.57). Marks dividing $N$ number at most $1/t_{0}$, and endpoint atoms and the logarithmic buffers contribute $o(1)$. ∎

The delayed branch $s_{p}>3$ of $F$ occurs only for $c_{\mathrm{trim}}$, when $t<t_{\mathrm{RA}}$, where $t_{\mathrm{RA}}=v_{\mathrm{RA}}-3\alpha$.

Define

(6.59) $$J_{K}(r)=\frac{1}{2e^{\gamma}\alpha}\int_{K}F\left(\frac{r-t}{\alpha}\right)\frac{dt}{t}.$$

The rational endpoint $\overline{J}_{6}$ is as in Paper II [27, (A.13)]:

(6.60) $$\begin{gathered}\overline{J}_{6}=\frac{13148581959157321533513247276782285310758770468886237619585}{10^{58}},\\ J_{J}(1/2)=2\log(110/57)\leq\overline{J}_{6}.\end{gathered}$$

The inequality is certified by outward interval evaluation. No central approximation is rounded. Since $W=4(1-\varepsilon)C(N)N/L^{2}$, the interval masses give

(6.61) $$\frac{M_{\mathrm{mix}}}{W} \leq w_{L}J_{J}(1/2)+w_{H}\{J_{J_{\mathrm{tail}}}(1/2)+J_{J_{t}}(v_{\mathrm{RA}})\}+o(1),$$

$$w_{L} =\frac{2500}{9999},\qquad w_{H}=\frac{7499}{9999}.$$

Here $w_{L}+w_{H}=1$. No mark multiplicity or singular-series factor is suppressed.

On $J_{\mathrm{tail}}$, the explicit upper-sieve formula gives

(6.62) $$0 \leq w_{H}\{J_{J_{\mathrm{tail}}}(1/2)-J_{J_{\mathrm{tail}}}(v_{\mathrm{RA}})\}$$

$$=w_{H}\int_{1/3-\eta_{\mathrm{m}}}^{1/3}\frac{v_{\mathrm{RA}}-1/2}{t(1/2-t)(v_{\mathrm{RA}}-t)}\,dt$$

$$\leq\frac{w_{H}\eta_{\mathrm{m}}(v_{\mathrm{RA}}-1/2)}{(1/3)(1/2-1/3)(v_{\mathrm{RA}}-1/3)}$$

$$=:d_{\mathrm{m}}=\frac{5106819}{31557955000}.$$

The last integrand is increasing on this interval. Put

(6.63) $$\mathfrak{g}_{\mathrm{RA}}=w_{H}\{J_{J}(1/2)-J_{J}(v_{\mathrm{RA}})\}-d_{\mathrm{m}}.$$

Combining (6.61) with (6.60) gives

(6.64) $$\frac{M_{\mathrm{mix}}}{W}\leq\overline{J}_{6}-\mathfrak{g}_{\mathrm{RA}}+o(1).$$

The remaining integral is elementary. The linear-sieve functions $F$ and $f$ are as in [14, (1.8)–(1.9), pp. 172–173]. Set $t_{\mathrm{RA}}=v_{\mathrm{RA}}-3\alpha$. On the ranges here,

(6.65) $$\frac{F(s)}{2e^{\gamma}}=\begin{cases}1/s,&1\leq s\leq 3,\\[2.0pt] \displaystyle\frac{1}{s}\left\{1+\int_{2}^{s-1}\frac{\log(\xi-1)}{\xi}\,d\xi\right\},&3<s<5.\end{cases}$$

This is the linear-sieve normalization [14, p. 202][26, (2.23)]. Write the delayed contribution as

(6.66) $$\mathcal{I}_{\mathrm{RA}}=\int_{t_{0}}^{t_{\mathrm{RA}}}\frac{dt}{t(v_{\mathrm{RA}}-t)}\int_{2}^{(v_{\mathrm{RA}}-t)/\alpha-1}\frac{\log(\xi-1)}{\xi}\,d\xi.$$

Substitution gives

(6.67) $$\mathfrak{g}_{\mathrm{RA}}+d_{\mathrm{m}}=w_{H}\Bigg{\{} \int_{t_{0}}^{1/3}\left(\frac{1}{t(1/2-t)}-\frac{1}{t(v_{\mathrm{RA}}-t)}\right)dt$$

$$-\mathcal{I}_{\mathrm{RA}}\Bigg{\}}.$$

Swap the order in (6.66) to obtain

(6.68) $$\mathcal{I}_{\mathrm{RA}} =\frac{1}{v_{\mathrm{RA}}}\int_{3}^{s_{\max}}h(\zeta)\,d\zeta,\qquad s_{\max}=\frac{v_{\mathrm{RA}}-t_{0}}{\alpha},$$

$$h(\zeta) =\frac{\log(\zeta-2)}{\zeta-1}\log\frac{(v_{\mathrm{RA}}-\alpha\zeta)(v_{\mathrm{RA}}-t_{0})}{\alpha\zeta t_{0}}.$$

For $n=20000$, set $d=(s_{\max}-3)/n$. The composite midpoint sum has error at most

$$\left|\int_{3}^{s_{\max}}h(\zeta)\,d\zeta-d\sum_{j=0}^{n-1}h(3+(j+1/2)d)\right|\leq\frac{(s_{\max}-3)d^{2}}{24}\sup|h^{\prime\prime}|.$$

Directed interval differentiation on $400$ covering boxes gives $\sup|h^{\prime\prime}|\leq 0.720711111$. An independent directed-interval replay encloses $\mathfrak{g}_{\mathrm{RA}}$ in $[0.09895983533531812,0.09895983533543704]$. Outward rounding therefore gives

(6.69) $$0.0989598353353\leq\mathfrak{g}_{\mathrm{RA}}\leq 0.0989598353355.$$

Thus we may use the certified rational floor

(6.70) $$g_{\mathrm{RA,lo}}=\frac{989598353}{10^{10}}=0.0989598353,\qquad\mathfrak{g}_{\mathrm{RA}}\geq g_{\mathrm{RA,lo}}.$$

The exact integral $\mathfrak{g}_{\mathrm{RA}}$ and its certified floor $g_{\mathrm{RA,lo}}$ are distinct.

Proposition 6.12 (Sixth-sum source applications).

On the exact arrays above, the following hold.

1. The comparison (6.9), and the density bound (6.57) for the three pairs $(R,K)$.

2. The three low discrepancy bounds (6.20).

3. For every fixed $A>0$, each of the nine rows of Table 6.1, with its exact weight, is $O_{A}(NL^{-A})$. Consequently the high-coefficient estimate (6.27) holds.

Proof.

Items 1 and 2 are Lemmas 6.11 and 6.1. For item 3 fix $A$. Choose in this order: $D_{\mathrm{m}}=2A+40$; $T_{\mathrm{save}}>A+C_{\mathrm{tensor}}+2D_{\mathrm{m}}+40$; the saving $A+C_{\mathrm{tensor}}$ in Lemma 6.8 and the saving $T_{\mathrm{save}}$ in Lemma 6.9. Then fix the Siegel–Walfisz precisions that these require, the buffers (6.19) and $B$, and the constant $K_{\mathrm{dim}}$. Choose the common threshold last. The rows partition $T_{1}+T_{2}+T_{\mathrm{II}}$: $T_{1}$ is all of $a\leq U$. The sums $T_{2}$ in $c$ and $T_{\mathrm{II}}$ in $a>U$ are split at $N^{.46}$, $N^{.50}$ and $N^{.54}$. Each boundary goes to the interval on its right.

Direct rows. These are $T_{1}$ and the part of $T_{2}$ with $c\geq N^{.54}$. Fix the outer variable $a$, respectively $t$. Terms with $(a,q)>1$, respectively $(t,q)>1$, vanish in $\mathcal{K}_{q}$. Otherwise $\mathcal{K}_{q}(ac)$ is the kernel of (6.28) for $c$ in the reduced class $N\bar{a}\bmod q$, and similarly for $t$. The parent is $H=N/a$, respectively $H=N/t$. The free variable runs over $H/4<c<(1-\varepsilon)H$, intersected with $c\geq N^{.54}$ in the $T_{2}$-row and with any prefix cut. This interval lies in $(0,2H]$. Condition (W1) holds. For $a\leq U$ we have $N^{2/3}\leq H\leq N$. If the right $T_{2}$-row is nonempty, then $N^{.54}<N/t=H\leq N$, since $N^{.54}\leq c<(1-\varepsilon)N/t$. Thus Lemma 6.2 applies with this $H$. (Equivalently, split at $H/2$ into $(H/4,H/2]$ and $(H/2,(1-\varepsilon)H)$. Each lies inside a dyadic parent $(H^{\prime},2H^{\prime}]$ with $H^{\prime}\in\{H/4,H/2\}\geq N^{.54}/4$. This only doubles the constant.) The weights $\log c/\log(ac)$ and $\mathcal{C}(t)/\log(tc)$ are bounded and monotone in $c$. Partial summation and (6.31) with $T_{\mathrm{save}}>A+2$ bound both rows by $O(NL^{-A})$.

Convolution rows. These are $T_{2}$ with $c<N^{.46}$ (tested $\mathcal{R}(c)$, $c>N^{1/3}/4$), $T_{\mathrm{II}}$ with $a<N^{.46}$ (tested $\mathcal{A}(a)$, $a>N^{1/3}$), and $T_{\mathrm{II}}$ with $a\geq N^{.54}$ (tested $\mathcal{B}(s)/(2L)$, $N^{1/3}<s<2N^{.46+O(1/L)}$). Each tested array contains the strict roughness factor $\mathcal{R}(n)=1_{P^{-}(n)>z}$. This is explicit for $\mathcal{R}(c)$. It follows from $\mathcal{A}(a)=\mu(a)\mathcal{R}(a)$ and from $\mathcal{B}(s)=\mathcal{R}(s)\sum_{b\mid s,\ b>V}\Lambda(b)$. Lemma 6.4 gives (6.34) on these original parents. Write $\sum_{q}\Gamma(q)\Phi(q)=\sum_{r,s}F_{H}(r,s)\Phi(rs)$ as in (6.40). Lemma 6.6 places the joint support in (6.39). Discard a dyadic shell pair $r\sim U_{1}$, $s\sim V_{1}$ if $F_{H}$ vanishes identically there. Otherwise a supported pair in it satisfies $U_{1}V_{1}<rs<Q$, $U_{1}<D_{1}$, and $V_{1}<D_{2}$. Hence this occupied shell satisfies all scale conditions of Lemma 6.8. Lemma 6.7 expands $F_{H}$ on the whole Cartesian shell into rank-one terms with $\sum_{\nu}|c_{\nu}|\ll L^{79}$. The unary caps and coprimality are kept. Each rank-one term is estimated on that whole shell. Its factors need not preserve the joint condition $rs<Q$ or the condition $(r,s)=1$, since Lemma 6.8 imposes neither. Apply Lemma 6.8 with saving $A+C_{\mathrm{tensor}}$ to each term on each occupied shell pair and each product prefix $I=(N/4,u]$. Since $C_{\mathrm{tensor}}\geq 83$, the total is $O(NL^{-A})$, uniformly in $u$. The factor $w$ is then restored by Stieltjes summation, as in (6.50).

Central rows. The four central rows are bounded by Lemma 6.10.

Proper powers. By (6.25) and (6.26), $P_{I_{H}}(\Gamma)=T_{1}+T_{2}+T_{\mathrm{II}}+O(N^{1/2+\nu}L^{2})$. This proves (6.27). ∎

Proposition 6.13 (Sixth-sum estimate).

Fix $\sigma$ in the range stated above before taking $N\to\infty$. Then, along even integers,

(6.71) $$\frac{S_{6}}{W}\leq\overline{J}_{6}-\mathfrak{g}_{\mathrm{RA}}+o(1).$$

In particular, the certified floor in (6.70) gives

(6.72) $$\frac{S_{6}}{W}\leq\overline{J}_{6}-g_{\mathrm{RA,lo}}+E_{\mathrm{RA}}(N;\sigma),\qquad E_{\mathrm{RA}}(N;\sigma)\geq 0,\qquad E_{\mathrm{RA}}(N;\sigma)\longrightarrow 0.$$

Proof.

By Proposition 6.12(1), $S_{6}\leq A_{\mathrm{mix}}$. Use (6.14), obtained from (6.12) and (6.13). Fix $A>2$, so that $NL^{-A}=o(W)$. The three low discrepancies are $O(NL^{-A-10})$ by Proposition 6.12(2). The term (6.15) is $O_{A}(NL^{-A})$. Also $P_{I_{H}}(\Gamma)=O_{A}(NL^{-A})$ by Proposition 6.12(3). Hence $S_{6}/W\leq M_{\mathrm{mix}}/W+o(1)$, and (6.64) gives (6.71). Enlarge the vanishing error to make it nonnegative. The certified floor (6.70) gives (6.72). ∎

The precisions are fixed in the order (6.52). All constants precede the common threshold. Their dependence through prime Siegel–Walfisz may be ineffective. The final section uses only (6.72). It uses the certified rational floor in (9.2). Its gain is independent of the common coefficient used in the balanced-semiprime and switched comparisons.

7. The balanced semiprime term

We bound the complete occurrence sum $B_{b}(N)$. The two prime orientations keep their multiplicities. In this section, fix

(7.1) $$\begin{gathered}\frac{1}{3}<b<\frac{1}{2},\qquad\varepsilon=\eta_{L}=10^{-4},\qquad u=\frac{1}{500},\qquad\mu_{*}=\frac{25}{128},\qquad L=\log N,\\ 0<\sigma\leq 10^{-10},\qquad\tau=\sigma^{9},\qquad\rho=1+\tau,\\ N^{\prime}=N(1-L^{-333}),\\ \mathcal{N}_{K}=(1-\varepsilon)U_{C}=\frac{W(N)}{4},\\ I(b)=\int_{b}^{1/2}\frac{dt}{t(1-t)}\\ =\log\frac{1-b}{b}.\end{gathered}$$

The final argument uses (1.2). Every limit below fixes $b$ and $\sigma$ first. For row three, write $\ell_{0}=53/412-\eta_{L}/2=264897/2060000>0$. This is the fixed floor of Paper I, §3.1.

7.1. The convolution source

Define

(7.2) $$A_{b}(e) =\#\{(m,p^{\prime})\in\mathbb{P}^{2}:N^{b}\leq m\leq\sqrt{N},\quad\varepsilon N<mp^{\prime}=e\leq N^{\prime}\},$$

$$X_{b} =\sum_{e}A_{b}(e),\qquad B_{b}^{\prime}=\sum_{e}A_{b}(e)1_{N-e\in\mathbb{P}},$$

$$E_{\mathrm{top}} =B_{b}-B_{b}^{\prime},\qquad 0\leq A_{b}(e)\leq 2,\qquad 0\leq E_{\mathrm{top}}\leq 2(NL^{-333}+1).$$

The last bound counts integers in the removed interval. A prime square contributes one occurrence. No new ordering or coprimality restriction is imposed.

For a finite sequence $F$, put

(7.3) $$\Delta_{F}(q,a)=\sum_{e}F(e)\left(1_{e\equiv a\ (q)}-\frac{1_{(e,q)=1}}{\varphi(q)}\right),\qquad(a,q)=1.$$

Both terms use the full support. For this source, PNT and partial summation give

(7.4) $$X_{b}=\frac{(1-\varepsilon)N}{L}I(b)+o_{b}(N/L).$$

This is the calculation of Paper II, [27, (5.44)–(5.51)]. The moving top removes at most $2(NL^{-333}+1)$ mass.

The required convolution estimate is

(7.5) $$\sum_{q\leq Q}\max_{(a,q)=1}|\Delta_{A_{b}}(q,a)|\ll_{A,b}NL^{-A},\qquad Q\leq N^{1/2}L^{-B_{\mathrm{BV}}(A,b)},$$

for every fixed $A>0$. Its proof keeps the complete occurrence source and the moving top. Fix a top $T\in\{N,N^{\prime}\}$. Before the discrepancy estimate, compare the exact cell sources (7.10) with their upper and lower convolutions. Each signed child keeps the natural coprimality comparator $1_{(e,q)=1}/\varphi(q)$. It is not replaced after summing the blocks. The step from $T=N$ to $T=N^{\prime}$ costs the endpoint bands (7.11). It also costs the removed top mass in (7.2). The actual moving-top application is proved below.

We use the following convolution input. Let $a_{m},b_{n}$ have supports $m\sim M,n\sim Y$. Put $x=MY\geq 3$, and fix $\eta,B_{\mathrm{div}}>0$. Assume

(7.6) $$\begin{gathered}x^{\eta}\leq M,Y\leq x^{1-\eta},\\ \left|\sum_{\begin{subarray}{c}n\equiv l\ (k)\\ (n,h)=1\end{subarray}}b_{n}-\frac{1}{\varphi(k)}\sum_{(n,hk)=1}b_{n}\right|\ll_{A_{2}}\|b\|_{2}Y^{1/2}d(h)^{B_{\mathrm{div}}}(\log(2Y))^{-A_{2}}\end{gathered}$$

uniformly for every fixed $A_{2}>0$, $h,k\geq 1$, and $(l,k)=1$. Here $d(n)$ is the divisor function. Then Fact 3.3 gives

(7.7) $$\sum_{q\leq Q}\max_{(a,q)=1}|\Delta_{a*b}(q,a)|\ll_{A}\|a\|_{2}\|b\|_{2}x^{1/2}(\log x)^{-A},\qquad Q\leq x^{1/2}(\log x)^{-B_{\mathrm{BV}}}.$$

The constants depend on all family constants in (7.6). This specializes Fact 3.3. It is Paper II, Lemma 5.1 [27, (5.23)–(5.28)]. It applies to complete rectangular supports.

We make the interval reduction explicit. Fix a target $A>2$, and put $H_{g}=A+4$. Choose

(7.8) $$\begin{gathered}R_{g}=\left\lceil\frac{(1/2-b)L}{\log(1+L^{-H_{g}})}\right\rceil,\qquad\xi=N^{(1/2-b)/R_{g}},\qquad y_{j}=N^{b}\xi^{j},\\ J_{j}=[y_{j-1},y_{j}),\qquad Y_{j}=y_{j}.\end{gathered}$$

Close the final upper endpoint. For the fixed top $T\in\{N,N^{\prime}\}$, put

(7.9) $$b_{j}(m) =1_{m\in\mathbb{P}\cap J_{j}},$$

$$a_{j,T}^{+}(p^{\prime}) =1_{p^{\prime}\in\mathbb{P}}1_{\varepsilon N/Y_{j}\leq p^{\prime}\leq T\xi/Y_{j}},$$

$$a_{j,T}^{-}(p^{\prime}) =1_{p^{\prime}\in\mathbb{P}}1_{\varepsilon N\xi/Y_{j}<p^{\prime}<T/Y_{j}}.$$

For $F,G$ on the positive integers write $(F*G)(e)=\sum_{mn=e}F(m)G(n)$, and put $A_{j,T}^{\pm}=a_{j,T}^{\pm}*b_{j}$. The exact cell source is

(7.10) $$A_{j,T}(e)=1_{\varepsilon N<e\leq T}\sum_{\begin{subarray}{c}m\in\mathbb{P}\cap J_{j},\ p^{\prime}\in\mathbb{P}\\ mp^{\prime}=e\end{subarray}}1.$$

The definitions give the pointwise envelopes $A_{j,T}^{-}\leq A_{j,T}\leq A_{j,T}^{+}$. Their difference is supported in multiplicative boundary bands at $\varepsilon N$ and $T$. Put $\mathcal{E}_{Q}(F)=\sum_{q\leq Q}\max_{(a,q)=1}|\Delta_{F}(q,a)|$. Let $B_{T}(e)=\sum_{j}(A_{j,T}^{+}(e)-A_{j,T}^{-}(e))$. Then $B_{T}\geq 0$, with support in $[\varepsilon N/\xi,\varepsilon N\xi]\cup[T/\xi,T\xi]$. Each integer has at most two ordered factorizations into primes. Since $1<\xi\leq 1+L^{-H_{g}}$ and $T\in\{N,N^{\prime}\}$ satisfies $T\leq N$,

$$\sum_{e}B_{T}(e)\leq 2\{\varepsilon N(\xi-\xi^{-1})+T(\xi-\xi^{-1})+4\}\ll NL^{-H_{g}}+1.$$

Since $\xi-\xi^{-1}\leq 2(\xi-1)\leq 2L^{-H_{g}}$, counting integers in the two complete boundary bands gives the displayed mass bound. Paper II, (5.47), gives the aggregate boundary discrepancy

(7.11) $$\mathcal{E}_{Q}\!\left(\sum_{j}(A_{j,T}^{+}-A_{j,T}^{-})\right)\ll NL^{-H_{g}+1}+Q.$$

The $Q$ term includes the bounded-multiplicity changes at exact endpoints. Split the second factor into dyadic blocks $(M,2M]$:

$$a_{j,T,M}^{\pm}(p^{\prime})=a_{j,T}^{\pm}(p^{\prime})1_{M<p^{\prime}\leq 2M}.$$

For each block meeting the support, $\varepsilon N/(2Y_{j})<M\leq T\xi/Y_{j}$, so $M\asymp_{\varepsilon}N/Y_{j}$, with only $O_{\varepsilon}(1)$ blocks. Here $n\sim Z$ means $Z<n\leq 2Z$, as in Section 3. For the short factor, the theorem-scale parent is $Y=Y_{j}/2$; since $\xi<2$ eventually, $J_{j}\subset(Y_{j}/2,Y_{j}]$ lies in that parent. The norms are

(7.12) $$\|a_{j,T,M}^{+}\|_{2}\ll\sqrt{M},\qquad\|a_{j,T,M}^{-}\|_{2}\leq\|a_{j,T,M}^{+}\|_{2},\qquad\|b_{j}\|_{2}\ll\sqrt{Y_{j}}L^{-(H_{g}+1)/2}.$$

The prime number theorem on these intervals also gives a matching lower bound. It is $\|b_{j}\|_{2}\gg\sqrt{Y_{j}}L^{-(H_{g}+1)/2}$ for all sufficiently large $N$. This lower bound normalizes the conversion of the absolute prime Siegel–Walfisz error to (7.6). The prime-interval form of (7.6) holds uniformly in $j,T,M$. For a fixed target $A>2$, write $K_{\mathrm{SW}}(A)$ for the chosen logarithmic level. For auxiliary moduli $k\leq L^{K_{\mathrm{SW}}(A)}$, this is prime Siegel–Walfisz at arbitrarily high fixed logarithmic precision. The divisor loss from excluding primes dividing the auxiliary integers is absorbed into $d(h)^{B_{\mathrm{div}}}$. For $k>L^{K_{\mathrm{SW}}(A)}$, use

$$|\Delta|\leq|J|/k+1+\pi(J)/\varphi(k)$$

and $\sup_{k>L^{K_{\mathrm{SW}}(A)}}\varphi(k)^{-1}\ll L^{-K_{\mathrm{SW}}(A)}\log\log(3L^{K_{\mathrm{SW}}(A)})$, taking $K_{\mathrm{SW}}(A)$ larger than the requested saving. For every fixed target $A>2$, choose $H_{g}=A+4$. The grid has $O_{b}(L^{H_{g}+1})$ intervals. The prime number theorem gives the norm bounds (7.12). With $A_{\mathrm{block}}=A+H_{g}/2+2$, Fact 3.3 and Paper II, (5.23)–(5.28), bound the rectangular blocks by $O_{A,b}(NL^{-A-3/2})$. The boundary term in (7.11) is $O(NL^{-A-3}+Q)$. Increasing the single fixed logarithmic level buffer absorbs the $x\asymp_{b}N$ constants. Thus the complete moving-top source, including the actual top $T=N^{\prime}$, satisfies (7.5). No fixed-width truncation estimate is substituted.

7.2. Four directed sieve rows

Let $\mathcal{P}_{N}=\{p\in\mathbb{P}:p\nmid 2N\}$. Put

(7.13) $$S_{b}(d;z)=\sum_{d\mid N-e}A_{b}(e)\,1_{((N-e)/d,\,P_{N}(z))=1},\qquad P_{N}(z)=\prod_{p<z,\ p\in\mathcal{P}_{N}}p.$$

There is no prefix–cofactor coprimality condition. Both source primes exceed $N^{\mu_{*}}$ eventually. Thus, for every supported modulus $d$,

(7.14) $$\sum_{d\mid N-e}A_{b}(e)=\frac{X_{b}}{\varphi(d)}+\Delta_{A_{b}}(d,N),\qquad(d,2N)=1,\quad P^{+}(d)<N^{\mu_{*}}.$$

Here $P^{+}(d)$ is the largest prime factor, with $P^{+}(1)=1$. For the squarefree sieve divisors, use the multiplicative density

(7.15) $$g(p)=\frac{1}{p-1}\quad(p\in\mathcal{P}_{N}),\qquad g(p^{j})=0\quad(j\geq 2),\qquad g(p)=0\quad(p\notin\mathcal{P}_{N}).$$

This is the squarefree-divisor density of the linear-sieve application. That input requires its zero prime-power tail. The full arithmetic density is $g(p^{j})=1/\varphi(p^{j})$ for $p\in\mathcal{P}_{N}$ and $j\geq 1$. No estimate uses its prime-power tail. The two densities agree at active primes. These are the only local factors used by the main-mass calculation. In Paper I, Theorem 2.6 notation, this density is

$$\omega(p)=\frac{p}{p-1}\quad(p\in\mathcal{P}_{N}),\qquad\omega(p)=0\quad(p\notin\mathcal{P}_{N}),\qquad\omega(p^{a})=0\quad(a\geq 2).$$

Thus $\omega(p)/p=g(p)$, and condition (2) has zero left side. For $p\geq 3$,

$$\left(1-\frac{1}{p-1}\right)^{-1}=\left(1-\frac{1}{p}\right)^{-1}\left(1+\frac{1}{p(p-2)}\right).$$

The product of the second factors over $p\geq w$ is $1+O(1/w)=1+O(1/\log w)$. Hence Mertens’ product estimate gives condition (1) with exponent one and one fixed $K_{\mathrm{dim}}\geq 2$. We may take $L_{\omega}=2$. These are the dimension and prime-power inputs of Iwaniec’s linear sieve [13].

If $N-e$ is prime, it is at least $NL^{-333}$. It therefore exceeds $N^{\mu_{*}}$ eventually. Hence

(7.16) $$B_{b}^{\prime}\leq S_{b}(1;N^{\mu_{*}}).$$

Use the ordered primes $N^{u}\leq a_{j}<\cdots<a_{1}<N^{\mu_{*}}$ in $\mathcal{P}_{N}$. The exact decomposition of (4.26)–(4.30) gives

(7.17) $$B_{b}\leq E_{\mathrm{top}}+T_{0}-T_{1}+T_{2}-R_{3}+P_{3}.$$

Here

(7.18) $$T_{0} =S_{b}(1;N^{u}),$$

$$T_{j} =\sum_{N^{u}\leq a_{j}<\cdots<a_{1}<N^{\mu_{*}}}S_{b}(a_{1}\cdots a_{j};N^{u})\quad(j=1,2),$$

$$R_{3} =\sum_{\mathrm{ret}}S_{b}(a_{1}a_{2}a_{3};N^{t_{3}}),$$

$$P_{3} =\sum_{\mathrm{ret}}\sum_{\begin{subarray}{c}N^{t_{3}}\leq r<a_{3}\\ r\in\mathcal{P}_{N}\end{subarray}}S_{b}(a_{1}a_{2}a_{3}r;N^{t_{3}}).$$

Every prime in these sums belongs to $\mathcal{P}_{N}$. The last sum keeps the half-open cells it inherits. The symbol $\sum_{\mathrm{ret}}$ restricts by

(7.19) $$\Theta_{\mathfrak{c}}=\inf_{\overline{\mathfrak{c}}}H,\qquad\ell_{\mathfrak{c}}=\Theta_{\mathfrak{c}}-\eta_{L}-t_{1}-t_{2}-t_{3},\qquad\ell_{\mathfrak{c}}\geq 2\rho t_{3}.$$

Here $H$ is the full eight-branch packing function, (3.29). The infimum uses the complete closed ordered cell. It does not use a quadrature node.

The four directed applications are

(7.20) $$\begin{array}[]{c|c|c|c}\text{term}&\text{row sign}&\text{sieve direction}&\text{level exponent}\\ \hline\cr T_{0}&+&\text{upper }F&1/2-\eta_{L}\\ T_{1}&-&\text{lower }f&1/2-\eta_{L}-\rho t_{1}\\ T_{2}&+&\text{upper }F&1/2-\eta_{L}-\rho(t_{1}+t_{2})\\ R_{3}&-&\text{lower }f&\ell_{\mathfrak{c}}\end{array}$$

The first three cutoffs are $N^{u}$. The fourth cutoff is $N^{t_{3}}$. The canonical functions are Iwaniec’s linear-sieve functions [13]. Extend $f$ by zero for $s\leq 2$. No shifted beta-sieve profile is substituted.

For each prefix product $h$, let $\mathcal{A}_{h}$ be the cofactor sequence indexed by $n=(N-e)/h$, with multiplicity $A_{b}(e)$ when $h\mid N-e$. Put $X_{h}=X_{b}/\varphi(h)$. Write $z_{h}$ for the row cutoff, $D_{h}$ for its level, and $s_{h}=\log D_{h}/\log z_{h}$. Every prefix prime is at least $z_{h}$, and each squarefree sieve modulus $t\mid P_{N}(z_{h})$ has prime factors below $z_{h}$. Thus $(h,t)=1$, and the local identity (7.14) gives the exact remainder identity

(7.21) $$r(\mathcal{A}_{h},t)=\Delta_{A_{b}}(ht,N).$$

Since $\omega(t)/t=1/\varphi(t)$ and $(h,t)=1$, the local main term is $X_{h}\omega(t)/t=X_{b}/\varphi(ht)$, as required by Theorem 2.6. For rows $j=0,1,2$, put $S=t_{1}+\cdots+t_{j}$, with $S=0$ for row zero, and $\theta=1/2-\eta_{L}$. The clipped exponent cells give

$$h\leq N^{\rho S},\qquad D_{h}=N^{\theta-\rho S},\qquad S\leq 2\mu_{*}.$$

Hence $\log D_{h}/\log N\geq\theta-2\rho\mu_{*}>0.1092>2u$. The upper rows satisfy $N^{u}\leq D_{h}$. The lower row satisfies $N^{u}\leq D_{h}^{1/2}$. For sufficiently large $N$, with $\sigma$ fixed, $D_{h}\geq 2$. Apply Paper I, Theorem 2.6, (2.30), with density $g(p)=1/(p-1)$ on $\mathcal{P}_{N}$, $g(p^{a})=0$ for $a\geq 2$, and sieve parameter $\epsilon_{s}=\sigma$. With $\overline{E}=2C_{\mathrm{AA37}}\sigma$, the resulting directed inequalities are

(7.22) $$S(\mathcal{A}_{h},z_{h}) \leq X_{h}V_{N}(z_{h})\{F(s_{h})+E^{+}_{s_{h}}\}+R_{1}^{+}(h)+R^{+}(h),$$

$$S(\mathcal{A}_{h},z_{h}) \geq X_{h}V_{N}(z_{h})\{f(s_{h})-E^{-}_{s_{h}}\}-R_{1}^{-}(h)-R^{-}(h),$$

$$|E^{\pm}_{s_{h}}| \leq\overline{E}.$$

Only the upper bound $|E^{\pm}_{s_{h}}|\leq\overline{E}$ is used. No sign of $E^{\pm}_{s_{h}}$ is assumed. Since $X_{h}V_{N}(z_{h})\geq 0$, replacing $E^{\pm}_{s_{h}}$ by $\overline{E}$ weakens each inequality. The actual dimension and prime-power constants, the level lower bound, and the fixed-$\sigma$ threshold are in (7.40).

The superscript $\mathrm{AA}$ below marks errors from this linear-sieve theorem. It does not denote a separate family of profiles. For row $j$, sum over its actual prefix set $\mathcal{H}_{j}$. Let $\mathcal{F}_{j}$ be $F$ on the upper rows and $f$ on the lower rows. The definitions used in the signed equalities are

(7.23) $$m_{j} =\sum_{h\in\mathcal{H}_{j}}X_{h}V_{N}(z_{h})\mathcal{F}_{j}(s_{h}),$$

$$\mathfrak{M}_{j} =\sum_{h\in\mathcal{H}_{j}}X_{h}V_{N}(z_{h}),$$

$$\mathcal{A}_{j}^{\mathrm{AA}} =\overline{E}\,\mathfrak{M}_{j},$$

$$r_{j} =\sum_{h\in\mathcal{H}_{j}}\{R_{1}^{\nu_{j}}(h)+R^{\nu_{j}}(h)\},\qquad\nu_{j}=+\text{ on upper rows},\quad\nu_{j}=-\text{ on lower rows},$$

$$z_{j} =m_{j}+\mathcal{A}_{j}^{\mathrm{AA}}+r_{j}-T_{j}\quad\text{on upper rows},$$

$$z_{j} =T_{j}-m_{j}+\mathcal{A}_{j}^{\mathrm{AA}}+r_{j}\quad\text{on lower row }j=1,$$

$$z_{3} =R_{3}-m_{3}+\mathcal{A}_{3}^{\mathrm{AA}}+r_{3}.$$

Summing (7.22) over prefixes proves $z_{j}\geq 0$. Also $\mathfrak{M}_{j}\geq 0$, so $\mathcal{A}_{j}^{\mathrm{AA}}\geq 0$. These nonnegativity statements are proved at source level for the exact signed remainders in (7.21). No sign of these remainders is assumed. Write the directed equalities as

(7.24) $$T_{0} =m_{0}+\mathcal{A}_{0}^{\mathrm{AA}}+r_{0}-z_{0}, T_{1} =m_{1}-\mathcal{A}_{1}^{\mathrm{AA}}-r_{1}+z_{1},$$

$$T_{2} =m_{2}+\mathcal{A}_{2}^{\mathrm{AA}}+r_{2}-z_{2}, R_{3} =m_{3}-\mathcal{A}_{3}^{\mathrm{AA}}-r_{3}+z_{3},\qquad\mathcal{A}_{j}^{\mathrm{AA}},z_{j}\geq 0.$$

Lemma 7.1 (Signed row identity).

With $M_{K}=m_{0}-m_{1}+m_{2}-m_{3}$,

$$\Delta_{\mathrm{arith}}:=T_{0}-T_{1}+T_{2}-R_{3}-M_{K}=\sum_{j=0}^{3}\mathcal{A}_{j}^{\mathrm{AA}}+\sum_{j=0}^{3}r_{j}-\sum_{j=0}^{3}z_{j},$$

and every $\mathcal{A}_{j}^{\mathrm{AA}}$ occurs with coefficient $+1$.

Proof.

Substitute (7.24). The lower rows carry the row sign $-$, which turns $-\mathcal{A}_{j}^{\mathrm{AA}}-r_{j}+z_{j}$ into $+\mathcal{A}_{j}^{\mathrm{AA}}+r_{j}-z_{j}$. ∎

For rows $j=0,1,2$, the total moduli in these remainders fit the level of (7.5). A tail tuple satisfies $\prod_{i}D_{i}<D_{h}$. In the parity ending in a defining cubic condition, remove two copies of the last label. In the other parity, use the preceding cubic condition and the ordering of the last two labels. The empty and one-label cases follow from the tuple definitions. Each $p_{i}$ lies in the interval $D_{i}\leq p_{i}<\min(D_{i}^{\rho},z_{h})$, and $d<D_{h}^{\sigma}$. Thus, with the same $S,\theta$ as above,

$$q=hd\prod_{i}p_{i}\leq hD_{h}^{\sigma+\rho}\leq N^{\rho S+(\sigma+\rho)(\theta-\rho S)}=N^{\theta+(\sigma+\tau)(\theta-\rho S)}\leq N^{\theta+(\sigma+\tau)\theta}<N^{1/2-\eta_{L}/2}.$$

This power buffer is below $N^{1/2}L^{-B}$ for every fixed $B$, once $N$ is large enough.

The factors of such a modulus that are at least $z_{h}$ are exactly the ordered prefix primes. They determine $h$ and its clipped cell. Factoring the remaining part at $D_{h}^{\sigma^{2}}$ determines the squarefree smooth divisor and the tail-prime multiset. The number of tail labels is at most $J_{\sigma}=\lceil\sigma^{-2}\rceil$. Equal-label permutations contribute at most $J_{\sigma}!$. Each AA coefficient has modulus at most one. The number of label vectors is less than $\exp(8\sigma^{-3})$. Hence the total coefficient at each modulus is at most $C_{\mathrm{K2},\sigma}=1+\exp(8\sigma^{-3})J_{\sigma}!$, independent of $N$. Apply (7.5) with target $A=4$. Divide by $\mathcal{N}_{K}=(1-\varepsilon)C(N)N/L^{2}$; here $C(N)\geq C_{2}>0$ by Paper II, (1.2). This proves

(7.25) $$\frac{|r_{0}|+|r_{1}|+|r_{2}|}{\mathcal{N}_{K}}\ll_{b,\sigma}L^{-2}.$$

The low-row bound uses Fact 3.3, the prime Siegel–Walfisz theorem, the prime number theorem and the Paper II convolution inputs. It gives no estimate for the principal masses or row three.

The $m_{j}$ are the directed main terms. The $r_{j}$ are signed arithmetic remainders. Thus, with $M_{K}=m_{0}-m_{1}+m_{2}-m_{3}$,

(7.26) $$B_{b}\leq{} E_{\mathrm{top}}+P_{3}+M_{K}+\sum_{j=0}^{3}\mathcal{A}_{j}^{\mathrm{AA}}+\sum_{j=0}^{3}r_{j}-\sum_{j=0}^{3}z_{j}$$

$$\leq{} E_{\mathrm{top}}+P_{3}+M_{K}+\sum_{j=0}^{3}\mathcal{A}_{j}^{\mathrm{AA}}+\sum_{j=0}^{3}|r_{j}|.$$

This step is finite algebra. The directed source comparisons are proved separately.

The canonical row masses are explicit. For a prefix cell $\mathfrak{c}$, put

$$H_{j}(\mathfrak{c})=\sum_{\begin{subarray}{c}a_{1}>\cdots>a_{j}\\ a_{i}\text{ in the cell}\end{subarray}}\frac{1}{\varphi(a_{1}\cdots a_{j})}.$$

Since the prefix primes exceed every sieve prime, the product over sieve primes at each row is $V_{N}(z_{h})$. With the ordered cell sets $D_{j}$ from (7.31),

(7.27) $$m_{0} =X_{b}V_{N}(N^{u})F(s_{0}),$$

$$m_{1} =X_{b}V_{N}(N^{u})\sum_{\mathfrak{c}\in D_{1}}H_{1}(\mathfrak{c})f(s_{1}(\mathfrak{c})),$$

$$m_{2} =X_{b}V_{N}(N^{u})\sum_{\mathfrak{c}\in D_{2}}H_{2}(\mathfrak{c})F(s_{2}(\mathfrak{c})),$$

$$m_{3} =X_{b}\sum_{\mathfrak{c}\in D_{3}}\chi_{\mathfrak{c}}V_{N}(N^{t_{3}})H_{3}(\mathfrak{c})f(\ell_{\mathfrak{c}}/t_{3}).$$

The masses $\mathfrak{M}_{j}$ are these sums with $F$ and $f$ replaced by $1$. In particular, the row-three retention indicator $\chi_{\mathfrak{c}}$ appears in both sums.

7.3. The principal functional

We display the numerical functional used for $M_{K}$. Set

(7.28) $$v_{1}(s) =\min\left\{\frac{1+s}{2},\frac{32-34s}{39},\frac{57}{64}-\frac{3}{2}s\right\},$$

$$v(s) =v_{1}(s)-u,\qquad\psi(x,y)=x1_{x\geq y},$$

$$\Theta_{*}(x,y,z) =\max\{v(x),v(y),v(x+y),v(x+y+z),$$

$$\hskip 79.66771pt\psi(v(x+z),x+2y+z),\psi(v(y+z),2x+y+z)\}.$$

The subtraction of $u$ precedes both guards. This is Paper II, (4.70)–(4.72) [27, (4.70)–(4.72)]. Put

(7.29) $$H_{\log} =\log(\mu_{*}/u),$$

$$R_{\mathrm{up}} =500(F(30)-1)+500H_{\log}(1-f(30))+250H_{\log}^{2}(F(30)-1),$$

$$J_{*} =\int_{u}^{\mu_{*}}\frac{dx}{x}\int_{u}^{x}\frac{dy}{y}\int_{u}^{y}\frac{1-f((\Theta_{*}(x,y,z)-x-y-z)/z)}{z^{2}}\,dz,$$

$$G_{*} =\frac{1}{\mu_{*}}+R_{\mathrm{up}}+J_{*}.$$

We keep the true delay-function tail. Paper II, (4.73)–(4.75), gives the constant cancellation

(7.30) $$\frac{1-H_{\log}+H_{\log}^{2}/2}{u}-\int_{u}^{\mu_{*}}\frac{dx}{x}\int_{u}^{x}\frac{dy}{y}\int_{u}^{y}\frac{dz}{z^{2}}=\frac{1}{\mu_{*}}.$$

The finite-parent comparison also needs its direction. Fix a finite clipped partition $\Pi$ of $[u,\mu_{*})$. Its cells have endpoint ratios at most $\rho$. Each clipped cell is a product of half-open exponent intervals $[t_{i},t_{i}^{\prime})$ with $t_{i}^{\prime}\leq\rho t_{i}$. Here $t_{i}$ is the lower endpoint, so $N^{t_{i}}\leq a_{i}<N^{\rho t_{i}}$, i.e. $t_{i}\leq x_{i}<\rho t_{i}$. The closure $\overline{\mathfrak{c}}$ is used only for infima and lower bounds. The cutoffs $N^{t_{3}}$ in $R_{3},P_{3}$ and the factors $\rho t_{1},\rho(t_{1}+t_{2})$ in $s_{1},s_{2}$ use this convention. For a clipped cell $\mathfrak{c}$, let $w_{\mathfrak{c}}=\int_{\mathfrak{c}}\prod_{i}dx_{i}/x_{i}$. Define

(7.31) $$s_{0} =(1/2-\eta_{L})/u,$$

$$s_{1}(\mathfrak{c}) =(1/2-\eta_{L}-\rho t_{1})/u,$$

$$s_{2}(\mathfrak{c}) =(1/2-\eta_{L}-\rho(t_{1}+t_{2}))/u,$$

$$E_{\Pi} =\frac{F(s_{0})-1}{u}+\sum_{\mathfrak{c}\in D_{1}}\frac{w_{\mathfrak{c}}}{u}(1-f(s_{1}(\mathfrak{c})))$$

$$\qquad+\sum_{\mathfrak{c}\in D_{2}}\frac{w_{\mathfrak{c}}}{u}(F(s_{2}(\mathfrak{c}))-1),$$

$$P_{\Pi} =\frac{1-H_{\log}+H_{\log}^{2}/2}{u}+E_{\Pi}-\sum_{\mathfrak{c}\in D_{3}}w_{\mathfrak{c}}1_{\ell_{\mathfrak{c}}\geq 2\rho t_{3}}\frac{f(\ell_{\mathfrak{c}}/t_{3})}{t_{3}}.$$

Here $D_{j}=\{u\leq x_{j}<\cdots<x_{1}<\mu_{*}\}$. Each cell is intersected with this ordered domain. Its weight includes the actual ordered multiplicity. The required comparisons are

(7.32) $$0\leq E_{\Pi}\leq R_{\mathrm{up}},\qquad P_{\Pi}\leq\frac{1}{\mu_{*}}+E_{\Pi}+J_{*}\leq G_{*}.$$

The last line is an upper comparison. It asserts no equality between finite sums and integrals.

Compare the six canonical branches of $\Theta_{*}$ with their matching branches among the eight parent branches of the packing function $H$ in (3.29). Adding $u$ to an unguarded canonical branch gives the matching parent branch. For an active canonical guarded branch, the parent guard is also active, and adding $u$ again gives the matching parent branch. An inactive canonical guarded branch is zero and cannot maximize the canonical ceiling, since its first unguarded branch is positive on the ordered domain. The two extra parent branches can only increase $H$. Hence $H(x,y,z)\geq\Theta_{*}(x,y,z)+u$ pointwise on that domain. On a closed anchored cell, each coordinate moves by at most $\mu_{*}\tau$. Fix a point and a maximizing canonical branch there, and use its matching parent branch. If it is guarded, its guard has reserve $u-7\mu_{*}\tau>0$ throughout the cell. This selected branch changes by at most $(9/2)\mu_{*}\tau$ across the cell. Therefore

(7.33) $$\Theta_{\mathfrak{c}} \geq\Theta_{*}(x,y,z)+u-(9/2)\mu_{*}\tau,$$

$$\ell_{\mathfrak{c}} \geq\Theta_{*}(x,y,z)-x-y-z+g,$$

$$g =u-\eta_{L}-(9/2)\mu_{*}\tau>2\mu_{*}\tau.$$

Monotonicity of $f(s)$, together with this reserve, gives the directed kernel comparison

(7.34) $$\chi_{\mathfrak{c}}\frac{f(\ell_{\mathfrak{c}}/t_{3})}{t_{3}}\geq\frac{f((\Theta_{*}(x,y,z)-x-y-z)/z)}{z}.$$

Summing over cells gives the upper comparison in (7.32). It does not identify a finite sum with the integral.

Put $[x]_{+}=\max\{x,0\}$. We need the arithmetic-to-integral estimate

(7.35) $$\frac{[M_{K}-\mathcal{N}_{K}\,2e^{-\gamma}I(b)G_{*}]_{+}}{\mathcal{N}_{K}}=o_{b,\sigma}(1).$$

The quantity $M_{K}$ contains only the canonical values of $F$ and $f$. The linear-sieve errors $E^{\pm}_{s_{h}}$ and the remainders $r_{j}$ lie outside it. They are paid in (7.38) and (7.47). Fix $(b,\sigma)$. Three inputs enter. The first is the two-orientation arithmetic-mass estimate for the source (7.2). The second is its four signed canonical AA main masses (7.27). The third is the finite functional $P_{\Pi}$ in (7.31). For all sufficiently large even $N$ they give

$$M_{K}/\mathcal{N}_{K}\leq 2e^{-\gamma}I(b)G_{*}+o_{b,\sigma}(1).$$

Here the error tends to zero with $b,\sigma$ and the cell count fixed, as $N$ tends to infinity. This estimate is for the finite functional in (7.31). It gives (7.35) directly, so $N_{\mathrm{mass}}(b,\sigma)$ can be the threshold of this fixed-$(b,\sigma)$ statement. This bound controls only the principal-mass term. It does not control the AA remainders, the optional numerical certificate, the complete exterior comparison or the endpoint terms. No sharper arithmetic-mass estimate is used here.

For the exterior comparison we use the common $G_{0}$ coefficient

(7.36) $$\begin{gathered}G_{0}=\frac{6088440363}{10^{9}},\qquad G_{*}\leq G_{0},\\ 2e^{-\gamma}G_{0}\leq c_{K2,G0}:=\frac{6836825163875517756286846321650754372931094689}{10^{45}}.\end{gathered}$$

For $G_{*}\leq G_{0}$, use the tight endpoint $J_{*}\leq 0.967120$ of Paper II, (A.10a). It holds for the exact profile (7.29) [27, (4.73), (4.75), (A.10a)]. Its weaker displayed endpoint is not used. The Taylor-model delay certificate is used for this coefficient comparison and is not an input to the shifted replay itself. It gives the explicit bounds

$$0\leq F(30)-1<3\cdot 10^{-49},\qquad 0\leq 1-f(30)<3\cdot 10^{-49}.$$

Since $0<H_{\log}=\log(3125/32)<5$, (7.29) then gives

$$R_{\mathrm{up}}<(500+500\cdot 5+250\cdot 25)3\cdot 10^{-49}<10^{-44},\qquad G_{*}<\frac{128}{25}+\frac{967120}{10^{6}}+10^{-44}<\frac{6088440363}{10^{9}}=G_{0}.$$

The finite delay calculation and this rational budget do not replace the source-to-profile comparison or its fixed-parameter hypotheses. For the scalar inequality, define

(7.37) $$\begin{gathered}\gamma_{-}:=\frac{57721566490153286060651209008240243104215933593992}{10^{50}},\\ \gamma_{+}:=\frac{57721566490153286060651209008240243104215933593993}{10^{50}},\\ \gamma_{-}<\gamma<\gamma_{+},\\ c_{K2,G0}-2G_{0}\sum_{j=0}^{60}\frac{(-\gamma_{-})^{j}}{j!}>\frac{75307}{10^{50}}>0.\end{gathered}$$

The rational enclosure follows from Euler–Maclaurin for $H_{n}$ at $n=10^{5}$, through $B_{16}$. The next-term remainder is bounded by $|B_{18}|/(18n^{18})$. Both $\log n$ and $H_{n}$ are enclosed outwards by the certified exact-input record. For $0<x<1$, the even Taylor partial sum of $e^{-x}$ is an upper bound. Thus the exact rational comparison in (7.37) proves $2e^{-\gamma}G_{0}<c_{K2,G0}$.

The switched comparison uses the same coefficient. At $b=2999993/7000000$, the common final-margin inequality (9.24) has slack $0.000228954311267289\ldots>1/5000$ on the stated route $g_{6}:=0$. Both comparisons use the same $G_{0}$ bound. Scalar positivity does not establish the required source estimates.

7.4. The complete error bound

The analytic hypotheses are stated below. The sums are over the original applications. The four principal linear-sieve errors satisfy

(7.38) $$\begin{gathered}\frac{\sum_{j=0}^{3}\mathcal{A}_{j}^{\mathrm{AA}}}{\mathcal{N}_{K}}\leq A_{K}(b)C_{\mathrm{AA}}\sigma,\\ A_{K}(b)=2e^{-\gamma}I(b)W_{\mathrm{AA}},\qquad W_{\mathrm{AA}}=500\left(1+H_{\log}+\frac{H_{\log}^{2}}{2}+\frac{H_{\log}^{3}}{6}\right).\end{gathered}$$

Let $C_{\mathrm{AA37}}$ be the absolute constant in Paper I, Theorem 2.6, (2.31), and define

(7.39) $$C_{\mathrm{AA}}=4C_{\mathrm{AA37}}.$$

The coefficient four is the factor two in (7.43) times the factor two in the common bound $|E_{s}^{\pm}|\leq 2C_{\mathrm{AA37}}\sigma$. The sum over the four rows is already included in $W_{\mathrm{AA}}$. It is independent of $\sigma,N$. It is unrelated to the global tensor constant in (6.40). All four actual levels satisfy $D\geq N^{1/250}$. We use (2.31) with the actual fixed dimension constant $K_{\mathrm{dim}}$. For the squarefree density in (7.15), the left side of (2.27) is zero. Hence the actual prime-power constant is $L_{\omega}=2$. This value is used in the exponential below.

(7.40) $$|E_{s}^{\pm}|\leq C_{\mathrm{AA37}}\{\sigma+\sigma^{-8}e^{K_{\mathrm{dim}}+2}(\log D)^{-1/3}\}\leq 2C_{\mathrm{AA37}}\sigma$$

once $L\geq 250\exp(3(K_{\mathrm{dim}}+2))\sigma^{-27}$. Name the corresponding even threshold

$$N_{\mathrm{AA,dec}}(\sigma):=2\left\lceil\frac{1}{2}\exp\!\left(250\exp(3(K_{\mathrm{dim}}+2))\sigma^{-27}\right)\right\rceil.$$

It is chosen at fixed $\sigma$. Every $N\geq N_{\mathrm{AA,dec}}(\sigma)$ satisfies the displayed decay bound. The four ordered harmonic row masses, before the common Euler and source factor, are bounded by

(7.41) $$\begin{gathered}\frac{1}{u},\qquad\frac{H_{\log}}{u},\qquad\frac{H_{\log}^{2}}{2u},\\ \text{row-3 mass}\leq\rho\int_{u}^{\mu_{*}}\frac{dx}{x}\int_{u}^{x}\frac{dy}{y}\int_{u}^{y}\frac{dz}{z^{2}}\leq\frac{\rho H_{\log}^{2}}{2u}<\frac{H_{\log}^{3}}{6u}.\end{gathered}$$

For the last inequality use $3\rho<H_{\log}=\log(3125/32)$. The factor $\rho$ allows the anchored-bin endpoint enlargement. The first three identities are the volumes of ordered logarithmic simplexes. For the fourth, bound $\int_{u}^{y}dz/z^{2}\leq 1/u$ and integrate over the ordered $(x,y)$ simplex. The sum of the four displayed upper envelopes is $W_{\mathrm{AA}}$, the four-row envelope used here. It is distinct from Paper II’s coarser global sieve-error bound. Let $H=H_{\log}=\log(3125/32)$. PNT and Mertens’ theorems, uniformly over the finitely many fixed-$\sigma$ cells, give

$$V_{N}(N^{a})=\frac{2e^{-\gamma}C(N)}{aL}(1+o(1)),\qquad H_{j}(\mathfrak{c})=w_{\mathfrak{c}}+o(1).$$

For the retained row, $1/t_{3}\leq\rho/x_{3}$ on each cell. Dropping the retention restriction only increases its mass. Consequently

(7.42) $$\frac{\sum_{j=0}^{3}\mathfrak{M}_{j}}{\mathcal{N}_{K}} \leq 2e^{-\gamma}I(b)\left\{\frac{1}{u}+\frac{H}{u}+\frac{H^{2}}{2u}+\rho J_{4}\right\}+o(1),$$

$$J_{4} =\frac{H^{2}/2-H+1}{u}-\frac{1}{\mu_{*}}\leq\frac{H^{2}}{2u}.$$

The bracket is smaller than $W_{\mathrm{AA}}=u^{-1}(1+H+H^{2}/2+H^{3}/6)$, with relative gap greater than $0.2842$ for $0<\sigma\leq 10^{-10}$. The coarser bound $\rho J_{4}\leq\rho H^{2}/(2u)$ still leaves relative gap greater than $0.1723$. Consequently, for each fixed $b,\sigma$, there is a threshold $N_{\mathrm{AA,dens}}(b,\sigma)$ such that, for every even $N\geq N_{\mathrm{AA,dens}}(b,\sigma)$, $\sum_{j}\mathfrak{M}_{j}\leq\mathcal{N}_{K}\,2e^{-\gamma}I(b)W_{\mathrm{AA}}$. Define

$$N_{\mathrm{AA}}(b,\sigma):=\max\{N_{\mathrm{AA,dens}}(b,\sigma),N_{\mathrm{AA,dec}}(\sigma)\}.$$

Thus both the density estimate and (7.40) hold above $N_{\mathrm{AA}}(b,\sigma)$, as required for (7.38). The maxima defining $T_{K}$ and $T_{r}$ below include this combined threshold. The displayed source-density bound below is weaker by a factor of two. Let $\mathfrak{M}_{j}$ be the nonnegative principal mass multiplying the relative sieve error in row $j$. The resulting bound is

(7.43) $$\sum_{j=0}^{3}\mathfrak{M}_{j}\leq 2\mathcal{N}_{K}\,2e^{-\gamma}I(b)W_{\mathrm{AA}}.$$

Multiplying it by $2C_{\mathrm{AA37}}\sigma$ gives (7.38) with (7.39). The Euler factors and the actual-source mass identification are included above. The secondary decay is paid once.

The positive peel is bounded by the two-form argument of Paper II, [27, (4.46)–(4.56)], applied to the original source. Put $\mathcal{B}_{m}=\{e:\ m\mid N-e\}$ with multiplicity $A_{b}(e)$. For $m=a_{1}a_{2}a_{3}r$ in $P_{3}$, put $E=N^{t_{3}}$ and $w=N^{u}\leq E$. Write $A_{b}=\sum_{h=1}^{2}1_{\{A_{b}\geq h\}}$. Every $e$ in the support is a product of two primes at least $N^{b}$. Hence $e$ is odd and $w$-rough. Write $N-e=mk$. Then $k$ is odd. A prime $p<w$ dividing $k$ and $N$ would divide $e$, which is impossible. Hence both forms $k$ and $N-mk$ are free of all primes below $w$. Therefore $S_{b}(m;E)$ is at most the two-form Selberg majorant of [27, (4.48)–(4.53)] on $1\leq k\leq N/m$, once for each layer. All prime factors of $m$ are at least $E\geq w$ and lie in $\mathcal{P}_{N}$. Hence $(m,N)=1$, and $\nu_{N}(p)$ is as in [27, (4.48)]. Retention gives, as in [27, (4.55)],

(7.44) $$\log_{N}m\leq\rho(t_{1}+t_{2}+2t_{3})\leq\rho(\Theta_{\mathfrak{c}}-\eta_{L})<\tfrac{153}{256}+2\tau,\qquad\log_{N}(w^{2}m)<1-0.0116,\qquad\log_{N}w=u\geq\tfrac{2}{1001}.$$

Consequently [27, (4.47), (4.53), (4.54)] give, eventually and uniformly in $m$,

(7.45) $$S_{b}(m;E)\leq 2\cdot\frac{4.002}{u^{2}}\cdot\frac{U_{C}}{m}\leq\frac{2008008\,U_{C}}{m},\qquad\max_{h}M_{h}\leq 2\tau,\qquad\sum_{h}M_{h}\leq\frac{459}{100},$$

since one layer is at most $(N/m)/\mathcal{G}_{N}(w)+O(w^{2}\log^{5}w)$ with $\mathcal{G}_{N}(w)=(1+o(1))u^{2}L^{2}/(4C(N))$, and $4.002/u^{2}=1000500<1004004=2\cdot 502002$. The bin sums are [27, (4.56)]: $\log(\mu_{*}/u)=\log(3125/32)<4.59$, and $a_{3},r$ lie in one parent bin. Therefore

(7.46) $$\frac{P_{3}}{U_{C}}\leq 2008008\cdot 2\left(\frac{459}{100}\right)^{3}\tau=\frac{24272444031579}{62500}\tau<388400000\,\sigma^{9},$$

and $P_{3}/\mathcal{N}_{K}\leq P_{K}\sigma^{9}$ with $P_{K}=388400000/(1-\varepsilon)=3884000000000/9999$, for $N\geq N_{\rm peel}(\sigma)$. The coefficient $2008008$ includes the two layers. No orientation factor is added.

The remaining estimates are

(7.47) $$\frac{|r_{0}|+|r_{1}|+|r_{2}|}{\mathcal{N}_{K}} \ll_{b,\sigma}L^{-2},$$

$$\frac{|r_{3}|}{\mathcal{N}_{K}} \ll_{b,\sigma}N^{-\ell_{0}\sigma^{2}}L^{3}+N^{-b/4}L^{7}+L^{-260}+L^{-120}+L^{-2}+N^{-c_{\sigma}}L^{4024+C}=O_{b,\sigma}(L^{-2}).$$

The first line is (7.25). The second is Proposition 7.3 below. Its displayed terms come, in order, from the squareful mask, proper powers, logarithm freezing and the boundary. The last two are the combined logarithmic Type-II and Type-I-2 bound and the Type-II power-saving term.

7.5. The row-three remainder

We bound the complete third-row remainder of the actual source (7.2) by Paper I, Theorem 3.2. Every other contribution is bounded separately. Fix $b,\sigma$. Write $\Xi^{\sharp}_{3}$ for the retained cells $\mathfrak{c}\in D_{3}$ with $\chi_{\mathfrak{c}}=1$. Its size is $O_{\sigma}(1)$. For $\mathfrak{c}\in\Xi^{\sharp}_{3}$ put $D_{\mathfrak{c}}=N^{\ell_{\mathfrak{c}}}$, $z_{\mathfrak{c}}=N^{t_{3}}$ and $u_{\mathfrak{c}}=D_{\mathfrak{c}}^{\sigma^{2}}$. Let $I_{i}$ be the actual half-open exponent interval of coordinate $i$ of the clipped cell. Then $\mathfrak{c}\subset I_{1}\times I_{2}\times I_{3}$. Put $\mathcal{P}_{i}=\{a\in\mathcal{P}_{N}:\log_{N}a\in I_{i}\}$. For every $a_{i}\in\mathcal{P}_{i}$, $N^{t_{i}}\leq a_{i}<N^{\rho t_{i}}$ and $a_{i}<N^{\mu_{*}}$, also for a clipped interval. Write $t_{i}^{\prime}\leq\min(\rho t_{i},\mu_{*})$ for the upper endpoint of $I_{i}$. On $\overline{\mathfrak{c}}$ the actual coordinates satisfy $x_{1}\geq x_{2}\geq x_{3}\geq u$. Two of the intervals are equal or disjoint.

Exact occurrence map.

By (7.21) and Paper I, (2.30), (2.32)–(2.33), applied to $\mathcal{A}_{h}$ at level $D_{\mathfrak{c}}$, cutoff $z_{\mathfrak{c}}\leq D_{\mathfrak{c}}^{1/2}$ and $\epsilon_{s}=\sigma$,

(7.48) $$r_{3}=\sum_{\mathfrak{c}\in\Xi^{\sharp}_{3}}\ \sum_{k\in\mathcal{K}_{\mathfrak{c}}}\sum_{q}O_{\mathfrak{c},k}(q)\,\Delta_{A_{b}}(q,N),\qquad\mathcal{K}_{\mathfrak{c}}=\{\mathrm{R1}\}\cup\mathcal{D}^{-}(D_{\mathfrak{c}}),$$

$$O_{\mathfrak{c},k}(q)=\sum_{\begin{subarray}{c}a_{1}>a_{2}>a_{3}\\ a_{i}\in\mathcal{P}_{i}\end{subarray}}\sum_{\begin{subarray}{c}d<D_{\mathfrak{c}}^{\sigma}\\ d\mid P_{N}(u_{\mathfrak{c}})\end{subarray}}\lambda^{(k)}_{d}\sum_{\begin{subarray}{c}(p_{j})_{j\leq\rho_{k}}\\ p_{j}\in\mathcal{Q}_{k,j}\\ p_{i}\neq p_{j}\ ((i,j)\in I(\mathbf{D}))\end{subarray}}1_{q=a_{1}a_{2}a_{3}d\,p_{1}\cdots p_{\rho_{k}}}.$$

Here $\lambda^{(\mathrm{R1})}_{d}=\phi_{d}^{-}$ with no tail ($\rho_{\mathrm{R1}}=0$). For $k=\mathbf{D}$, $\lambda^{(k)}_{d}=\lambda_{d}^{-}(\mathbf{D})$ and $\mathcal{Q}_{k,j}=\mathcal{P}_{N}\cap[D_{j},\min(D_{j}^{1+\tau},z_{\mathfrak{c}}))$. All coefficients have modulus at most one and depend only on $(\mathfrak{c},k)$. The identity is exact at every supported modulus. Let $M_{\mathfrak{c}}\in\{1,2,6\}$ be the number of orderings of the equal intervals among $I_{1},I_{2},I_{3}$. Let $\Gamma_{\mathfrak{c},k}$ be $M_{\mathfrak{c}}^{-1}$ times the same sum, with $a_{i}\in\mathcal{P}_{i}$ independent and no tail distinctness. Then

(7.49) $$O_{\mathfrak{c},k}=\Gamma_{\mathfrak{c},k}-\Xi_{T,\mathfrak{c},k}-\Xi_{P,\mathfrak{c},k},$$

where $\Xi_{T}$ collects unrestricted terms with $p_{i}=p_{j}$ for some forbidden pair $(i,j)\in I(\mathbf{D})$. $\Xi_{P}$ collects the remaining unrestricted terms with $a_{i}=a_{j}$ for an equal-interval pair $I_{i}=I_{j}$. A prefix prime is at least $z_{\mathfrak{c}}$, and every tail prime is below it. The primes of $d$ are below $u_{\mathfrak{c}}\leq D_{j}$. Unequal grid intervals are disjoint. Hence every modulus in the support of $\Xi_{T}$ or $\Xi_{P}$ is divisible by $p^{2}$ for a prime $p\geq u_{\mathfrak{c}}\geq u_{0}$. Here $u_{0}=N^{\ell_{0}\sigma^{2}}$.

Support, cutoffs and constants.

Every supported modulus is coprime to $2N$. The cells are clipped to $[u,\mu_{*})$, and $v_{\rm pk}(x)=(1+x)/2$ on $[0,\mu_{*}]$. Thus $\Theta_{\mathfrak{c}}\geq(1+t_{1})/2$. The actual coordinate bound $x_{1}\geq x_{3}\geq u$ on $\overline{\mathfrak{c}}$ also gives $\Theta_{\mathfrak{c}}\geq(1+u)/2$. Hence

(7.50) $$\ell_{\mathfrak{c}}\geq\max\{A_{3}-t_{3},2\rho t_{3}\}\geq\tfrac{2}{3}A_{3}=\tfrac{33109}{240000},\qquad A_{3}=\tfrac{1}{2}-\eta_{L}-\tfrac{3}{2}\mu_{*},$$

and $\ell_{\mathfrak{c}}-\ell_{0}\geq 231463/24720000$. By the cubic tail conditions, $\sum_{j}s_{j}<\ell_{\mathfrak{c}}$ for $s_{j}=\log_{N}D_{j}$. Thus every supported modulus satisfies

(7.51) $$q<N^{\rho(t_{1}+t_{2}+t_{3})+\sigma\ell_{\mathfrak{c}}+\rho\ell_{\mathfrak{c}}}\leq N^{e_{Q}},\qquad e_{Q}=(\rho+\sigma)(H_{\max}-\eta_{L})<\tfrac{3}{5},\qquad H_{\max}=\tfrac{153}{256}.$$

Since the cell ceiling gives $0<\ell_{\mathfrak{c}}\leq H_{\max}=153/256<1$ and $0<\sigma\leq 10^{-10}$, $\ell_{\mathfrak{c}}\sigma^{2}<\sigma^{2}<u=1/500$. Thus $u_{\mathfrak{c}}=N^{\ell_{\mathfrak{c}}\sigma^{2}}<N^{u}$. The cutoff comparison uses only this exponent inequality.

Every labelled prime is at least $u_{\mathfrak{c}}$, so the number of labels is below $3/(5\ell_{0}\sigma^{2})$. With $M_{\sigma}=\lceil 3/(5\ell_{0}\sigma^{2})\rceil$ and the vector count $\exp(8\sigma^{-3})$, one has $|\Gamma_{\mathfrak{c},k}|,|\Xi_{T}|,|\Xi_{P}|\leq M_{\sigma}!$, and $\sum_{\mathfrak{c},k}\bigl{(}|\Gamma_{\mathfrak{c},k}|+|\Xi_{T}|+|\Xi_{P}|\bigr{)}\leq C^{\prime}_{\mathrm{K2},\sigma}$.

The squareful terms.

Since $|\Delta_{A_{b}}(q,N)|\leq 2(N/q+1)+2N/\varphi(q)$, Paper I, (B.5), with $Y=u_{0}$ and $Q_{\rm supp}=N^{e_{Q}}$ gives

(7.52) $$\sum_{\mathfrak{c},k}\sum_{q}|\Xi_{T,\mathfrak{c},k}(q)+\Xi_{P,\mathfrak{c},k}(q)|\,|\Delta_{A_{b}}(q,N)|\ll_{\sigma}N^{1-\ell_{0}\sigma^{2}}L.$$

Prime detection and the two Heath–Brown expansions.

Put $z_{1}=N^{1/(\log L)^{3}}$, so $z_{1}\geq z_{0}(X)$ for every occurring box scale $\varepsilon N/2\leq X\leq N$. Both source primes exceed $z_{1}$. Apply Paper I, (B.7), to $1_{\mathbb{P}}(m)$ and to $1_{\mathbb{P}}(p^{\prime})$. Apply Paper I, (B.9), with roughness $z_{1}$, to $\Lambda_{z_{1}}(m)$ with cap $U_{1}=2N^{1/6}$ and to $\Lambda_{z_{1}}(p^{\prime})$ with cap $U_{2}=2N^{(1-b)/3}$. The identities apply since $m\leq N^{1/2}<U_{1}^{3}$ and $p^{\prime}\leq N^{1-b}<U_{2}^{3}$. For patterns $j,j^{\prime}\in\{1,2,3\}$ the tested index is a labelled tuple $\mathbf{v}=(\mathbf{v}^{(1)},\mathbf{v}^{(2)})\in\mathcal{R}_{z_{1}}^{2j+2j^{\prime}}$, at most twelve coordinates, with

$$m=\textstyle\prod\mathbf{v}^{(1)},\quad p^{\prime}=\prod\mathbf{v}^{(2)},\quad e=mp^{\prime},\quad N^{b}\leq m\leq N^{1/2},\quad\varepsilon N<e\leq N^{\prime},$$

the Möbius coordinates of $\mathbf{v}^{(i)}$ at most $U_{i}$, and coefficient $c_{j}(\mathbf{v}^{(1)})c_{j^{\prime}}(\mathbf{v}^{(2)})/(\log m\log p^{\prime})$. Here $c_{j}$ is the coefficient after (B.10). For a fixed product $e$, put

(7.53) $$\mathcal{S}_{e}(q;N):=\Delta_{\delta_{e}}(q,N)=1_{e\equiv N\ (q)}-\frac{1_{(e,q)=1}}{\varphi(q)}.$$

Thus the Heath–Brown tuple map uses the point-mass discrepancy $\mathcal{S}_{e}(q;N)$. The exact weighted point-mass expansion is

$$\Delta_{A_{b}}(q,N)=\sum_{e}A_{b}(e)\mathcal{S}_{e}(q;N).$$

The weight $A_{b}(e)$ keeps both eligible ordered orientations, as in (7.2). A prime square is counted once. The proper-power remainder $\mathcal{E}_{\rm pp}$ consists of the pairs in which $m$ or $p^{\prime}$ is a proper prime power. For every fixed $(\mathfrak{c},k)$,

(7.54) $$\sum_{q}\Gamma_{\mathfrak{c},k}(q)\Delta_{A_{b}}(q,N)=\mathcal{E}_{\rm pp}+\sum_{j,j^{\prime}}\sum_{\mathbf{v}}\frac{c_{j}(\mathbf{v}^{(1)})c_{j^{\prime}}(\mathbf{v}^{(2)})}{\log m\log p^{\prime}}\sum_{q}\Gamma_{\mathfrak{c},k}(q)\mathcal{S}_{e}(q;N),$$

exactly. Each tuple keeps the natural discrepancy of its own product. The per-cell coefficient bound is the factorial bound above; the sum over cells uses the separate aggregate bound defined there. Since $e<N$, $|c_{j}c_{j^{\prime}}|/(\log m\log p^{\prime})\leq 9$ and $\sum_{q}|\Gamma(q)||\mathcal{S}_{e}(q;N)|\leq C_{\mathrm{K2},\sigma}F(e)$ with $F(e)=\tau(N-e)+2(1+L)$. The tuple multiplicity is at most $\tau_{12}(e)$. The divisor bounds $\tau_{12}^{2}\leq\tau_{144}$, $\tau^{4}\leq\tau_{16}$ and Cauchy–Schwarz give

(7.55) $$\sum_{e<N}\tau_{12}(e)F(e)\ll NL^{73},\qquad\sum_{e<N}\tau_{12}(e)F(e)^{2}\ll NL^{79}.$$

There are $\ll N^{1-b/2}L$ proper-power pairs, and $\sum_{e<N}\tau(e)F(e)^{2}\ll NL^{9}$. Therefore

(7.56) $$|\mathcal{E}_{\rm pp}|\ll_{\sigma}(N^{1-b/2}L)^{1/2}(NL^{9})^{1/2}=N^{1-b/4}L^{5}.$$

Boxes, logarithm freezing and faces (H6).

Put $\delta=L^{-334}$. Use the equal-ratio mesh of Paper I, (B.12), with $\delta_{X}$ replaced by $\delta$, on every ambient coordinate axis. Keep $\{1\}$ as an atom. A box $\mathcal{B}$ is a Cartesian product of cells. Its lower endpoints are $V_{i}$, and $X=\prod_{i}V_{i}$. Let $\lambda_{1},\lambda_{2}$ be the sums of $\log V_{i}$ over the two sides. Then $\varepsilon N/2\leq X\leq N$ on every box meeting the source, so $K_{\rm scale}=1$. Replacing $1/(\log m\log p^{\prime})$ by $1/(\lambda_{1}\lambda_{2})$ changes the weight by $O(L^{-335})$, since $\lambda_{1}\geq bL/2$. By (7.55),

(7.57) $$|E_{\rm LOG}|\ll_{\sigma}NL^{72-334}.$$

The faces are $m=N^{b}$, $m=N^{1/2}$, $e=\varepsilon N$, $e=N^{\prime}$, and the cap faces $v_{i}=U_{1}$, $v_{i}=U_{2}$ on Möbius coordinates. All are singletons. On each, an active coordinate is at least $N^{1/20}$. Paper I, Lemma (B.14), with $s\leq 12$, $K_{\gamma}=1$ and $\eta_{\rm dist}\leq 12\delta$, gives at most $\ll NL^{11-334}$ boundary tuples. Cauchy–Schwarz with (7.55) gives

(7.58) $$|E_{\rm bdry}|\ll_{\sigma}NL^{45-334/2}.$$

Thus (H6) holds with $c_{\log}=72$, $c_{R}=45$. On a nonboundary box every source cut and cap is constant. The coefficient is the outer scalar $c_{\mathcal{B}}=(-1)^{j+j^{\prime}}\binom{3}{j}\binom{3}{j^{\prime}}(\log X)^{2}/(\lambda_{1}\lambda_{2})$, times $w(\mathbf{v})=\prod_{i}w_{i}(v_{i})$ with $w_{i}\in\{\mu_{z_{1}},1_{\mathcal{R}_{z_{1}}},(\log v/\log X)1_{\mathcal{R}_{z_{1}}}\}$. Each side carries exactly one logarithmic coordinate, and $(\log X)^{2}/(\lambda_{1}\lambda_{2})\leq 1/(b(1-b))+o(1)\leq 4.5+o(1)$, hence the scalar norm bound $\lVert c_{\mathcal{B}}\rVert_{\infty}\leq 45$. Each coordinate meets at most $2L^{335}$ cells, so

(7.59) $$\sum_{\mathfrak{c},k}\sum_{j,j^{\prime}}\sum_{\mathcal{B}}|c_{\mathcal{B}}|\ll_{\sigma}L^{4020}\leq L^{4022},\qquad K_{0}=4022\leq 24C_{\rm box}+26=8042.$$

Classification.

Let $e_{i}=\log_{N}V_{i}$, $\mathcal{W}=[60\sigma,1/3+14\sigma]$, and $\kappa=(\log 2+12\log(1+\delta))/L\leq\sigma/202$. A box is Type II if some $e_{i}\in\mathcal{W}$. Select the largest such $e_{i}$, breaking ties by the fixed coordinate order. Otherwise every Möbius coordinate is small, since $1/6+\kappa$ and $(1-b)/3+\kappa$ are below $1/3+14\sigma$. On the $m$-side, $\sum e_{i}\geq b-\kappa>360\sigma$. On the $p^{\prime}$-side, $\sum e_{i}\geq 1/2-\kappa-\log(1/\varepsilon)/L>360\sigma$. These are the only side bounds needed to rule out an all-small side. Together with the global Heath–Brown sum bound $\sum e_{i}<1+\kappa$, the threshold $1/3+14\sigma$ permits at most two large variables. Hence each side has exactly one large smooth coordinate. Every other box has exactly two large smooth coordinates $N_{1}\geq N_{2}\geq X^{1/3}$. It has at most ten small coordinates, of product $M$, with

(7.60) $$\log_{X}M\leq\frac{(600+1/202)\sigma}{1-\sigma/101}<6.02\cdot 10^{-8},\qquad N_{1}N_{2}M\leq 2^{30}X,\qquad q<X^{0.5977}.$$

There are no Type-I-1 boxes.

Type-II boxes: Paper I, (H1)–(H5) and LEN.

Fix $(\mathfrak{c},k)$ and a Type-II box. Let its selected lower endpoint be $T$, put $\xi=\log_{N}T$, and let the other lower endpoints have product $M=X/T$.

H1.:

$C_{\mathcal{B}}=\Gamma_{\mathfrak{c},k}$ has the form (3.4). Here $J$ is the three prefix labels and the tail labels. The prime sets are $\mathcal{P}_{i}$ and $\mathcal{Q}_{k,j}$, $u_{L}=u_{\mathfrak{c}}\geq u_{0}$, and $d\mid P_{N}(u_{\mathfrak{c}})$ is squarefree. Repeated primes are kept. We have $q\leq X^{0.6}$ by (7.51) once $(3/5-e_{Q})L\geq(3/5)\log(2/\varepsilon)$. Also $(q,2N)=1$. The selector $\chi_{\mathfrak{c}}$ is constant on the block. The removed corrections are $\Xi_{T},\Xi_{P}$ and $\mathcal{E}_{\rm pp}$. There is no comparator correction, since the comparator is natural on every tuple.

H2.:

Take the upper exponents $t_{1}^{\prime}\geq t_{2}^{\prime}\geq t_{3}^{\prime}$ and $\min(\rho s_{j},t_{3})$. For $d_{*}=\rho(\Theta_{\mathfrak{c}}-\eta_{L})$, the merged list satisfies $\mathrm{WF}(d_{*})$. The reserves are $\rho(1/2-\eta_{L}-3\mu_{*}/2)=33109\rho/160000$, $\rho(1/2-\eta_{L}-5\mu_{*}/2)=1859\rho/160000$, $\rho(\ell_{\mathfrak{c}}-t_{3})\geq\rho(2\rho-1)t_{3}$. For a tail rank, the reserve is $\rho$ times the strict inequality $\sum_{i<j}s_{i}+2s_{j}<\ell_{\mathfrak{c}}$ from the $\mathcal{D}^{-}$ conditions. Put $h=d_{*}+60\sigma$, $\delta_{\rm pk}=30\sigma$, $g=\xi-30\sigma$. Then $h-1/2\geq 9\cdot 10^{-4}$, $\Theta_{\mathfrak{c}}-h>0.0000999939$, $h\leq H_{\max}$,

$$30\sigma\leq g\leq 1/3-16\sigma,\qquad\delta_{\rm pk}=30\sigma,\qquad g\leq 1/3-\delta_{\rm pk}/2.$$

The upper requirement has reserve at least $\sigma$, since $(1/3-15\sigma)-(1/3-16\sigma)=\sigma$. Also $h\leq\Theta_{\mathfrak{c}}\leq\widetilde{H}(t_{1}^{\prime},t_{2}^{\prime},t_{3}^{\prime})$ because $(t_{1}^{\prime},t_{2}^{\prime},t_{3}^{\prime})\in\overline{\mathfrak{c}}$ and $t_{1}^{\prime}\leq\mu_{*}$. Paper I, Theorem 4.1, gives $J=J_{1}\sqcup J_{2}\sqcup J_{3}$. Put $d$ in slot one, with $\gamma_{s}$ as in Paper I, (3.5). The tuple map gives $\Gamma_{\mathfrak{c},k}=M_{\mathfrak{c}}^{-1}C_{\rm slot}(\gamma_{1}*\gamma_{2}*\gamma_{3})$, $C_{\rm slot}\leq M_{\sigma}!$, $|\gamma_{s}|\leq 1$. Slots two and three are $u_{0}$-rough, and empty slots carry the atom $1_{v=1}$. The scalar $M_{\mathfrak{c}}^{-1}$ is an outer coefficient.

H3.:

Let $Q,R,S$ be the scales of $\gamma_{1},\gamma_{3},\gamma_{2}$. The loads (4.3), the slot-one load $\sigma\ell_{\mathfrak{c}}\leq\sigma H_{\max}$ of $d$, and $A_{0}=X^{1+\sigma/20}$ give reductions $(7527/256,\,60,\,114421/1280,\,26247/256)\sigma$. They are, in turn, in $Q/T$, $T^{2}R^{2}S$, $A_{0}^{7/32}TQR^{2}S^{5}$, $(TR/S)^{7/16}TQR^{2}S^{5}$. The required reductions are $29,8,15,15$, with losses at most $2\omega$, $\omega=\log_{N}(N/X)\leq\sigma/202$. We also have $\log_{N}(QRS)\leq e_{Q}$, so $QRS<M_{I}X^{-89\sigma}$ with reserve $>0.0691$. Next, $N_{I}\geq QX^{\sigma/3}$ with reserve $29.069\sigma$, and $M_{I}\geq X^{1/2}$. Thus (3.6)–(3.7) hold. The selected coefficient is $\alpha_{n}=w_{\rm sel}(n)1_{I_{T}}(n)\in\{\mu(n),1,\log n/\log X\}1_{P^{-}(n)\geq z_{1}}1_{I_{T}}(n)$, with $\zeta=z_{1}\geq z_{0}(X)$ and $I_{T}=(T,(1+\Delta)T]$. The complementary coefficient $\beta_{m^{\sharp}}$, the product of the other at most eleven weights, satisfies $|\beta_{m^{\sharp}}|\leq\tau_{11}(m^{\sharp})\leq\tau_{23}(m^{\sharp})$. Here $\tau_{23}=d_{23}$ is the 23-fold divisor function, with $d_{23}(n)=\sum_{n=n_{1}\cdots n_{23}}1$. As in Paper I, (B.22)–(B.23), with at most eleven complementary cells, $N_{I}\geq T/(6L^{334})$, $M_{I}=M((1+\Delta)^{r}-1)\asymp ML^{-334}$, $m_{0}=M(1+(1+\Delta)^{r})/2\leq 2X/T$, so $C_{\rm ctr}=2$. On the doubled support, $mn\leq X(1+\Delta)(1+\tfrac{3}{2}((1+\Delta)^{11}-1))\leq N^{\prime}(1+20\delta)$, because $X\leq e\leq N^{\prime}$ for every tested product of the box. Hence $N-mn\geq NL^{-334}(L-20)\geq 1$ and $|mn-N|\leq N<X^{3}$, which is (3.10). The integer convention of $I_{T}$ is that of Paper I, Lemma B.22, since the mesh and the roughness are the same. LEN follows from the condition

(7.61) $$\sigma L\geq 2^{130}(335)(\log L+1).$$

It gives $T^{2^{-126}}\geq 12L^{334}$ and $N_{I}\geq(2T)^{1-2^{-126}}$.

H4.:

$a=N$, $0<N\leq XL\leq A_{0}$, and every supported modulus is coprime to $2N$.

H5.:

$N^{60\sigma}\leq T\leq N^{1/3+14\sigma}$, and (3.99) gives $T\leq X^{1/3+15\sigma}$ and $T<X^{1/2}$.

The selected-factor Siegel–Walfisz bound is Paper I, (3.25). It is proved for these three coefficient classes with any $\zeta\geq z_{0}$. In the large-norm case of Paper I, (3.48), the zero mode uses Lemma 6.5(ii) in place of [3, Theorem 0(a)]. Its premises (3.1), (3.2), (3.6), (3.8) with LEN, (3.11) and (3.25)–(3.27) are those verified above. Theorem 2.8 of Paper I enters only inside the proof of Theorem 3.2, with the dictionary

(7.62) $$r_{T}=z,\quad s_{T}=p_{0},\quad n_{T}=p_{0}m,\quad a_{T}=\varepsilon_{y}N,\quad N_{T}=8p_{0}H_{*}SF,\quad A_{T}=A_{0},$$

both signs, all nine smooth weights and the full bracket. Paper I, Theorem 3.2, (3.13), gives, uniformly in the box, the block and the cell,

(7.63) $$|S_{\mathcal{B}}^{\rm PSDT}|\ll_{\sigma,A^{\prime}}XL^{-A^{\prime}}+X^{1-c_{\sigma}}L^{C},\qquad S_{\mathcal{B}}^{\rm PSDT}=\sum_{q}\Gamma_{\mathfrak{c},k}(q)\sum_{\mathbf{v}\in\mathcal{B}}w(\mathbf{v})\mathcal{S}_{e}(q;N).$$

With $A=4$, $C_{\rm box}=334$, $A^{\prime}=A+24C_{\rm box}+31=8051$ and (7.59), Paper I, (3.92), gives

(7.64) $$E_{\rm II}=\sum|c_{\mathcal{B}}|\,|S_{\mathcal{B}}^{\rm PSDT}|\ll_{b,\sigma}NL^{4022-8051}+N^{1-c_{\sigma}}L^{4022+C}.$$

Here $S_{\mathcal{B}}^{\rm PSDT}$ already contains $\Gamma_{\mathfrak{c},k}$, hence its factor $M_{\mathfrak{c}}^{-1}$ from (7.49). The coefficient norm (7.59) sums only $|c_{\mathcal{B}}|$, so the orbit factor is paid exactly once.

Lemma 7.2 (Relaxed Type-I-2 estimate).

The source hypotheses are the Type-I-2 conditions (SH1)–(SH5) stated in Paper I:

For a direct shell $[X,2X)$, assume the following source conditions (SH1)–(SH5).

• The coefficient $\gamma(q)$ is independent of tested variables, $(q,2N)=1$, $|\gamma(q)|\leq C_{\sigma}$, $q\leq N^{e_{Q}}$, $e_{Q}<153/256$, and $(a,q)=1$ on its support.

• The tested geometric source and cap restrictions are constant on each nonboundary box; the roughness sieve is separate as in (SH3). Smooth factor weights have $\sup|w|+V(w)\ll 1$.

• All HB factors are rough at the threshold $z^{*}=z_{0}(X)$ or $z_{0}(N)$.

• The discrepancy is exactly (3.3).

• The assignment priority is proper prime powers, boundary boxes, Type II, Type I-1, then Type I-2.

Assume (SH1)–(SH5) for a nonboundary K2 box. Let the two large smooth variables have nonnegative weights $w_{1},w_{2}$ and $N_{1},N_{2}\geq X^{1/3}$. Let the remaining small variables, at most ten, have product $M$. Assume

$$M\leq X^{6.02\cdot 10^{-8}},\qquad N_{1}N_{2}M\leq 2^{30}X,\qquad q<X^{0.5977},$$

and $|\alpha_{m}|\ll\tau_{10}(m)\leq\tau(m)^{9}$. Then, for every fixed $A>0$, uniformly in the supported unit residues,

$$\sum_{q\asymp Q}|\gamma(q)|\left|\sum_{m}\alpha_{m}\sum_{n_{i}\in I_{i},\ P^{-}(n_{i})\geq z^{*}}w_{1}(n_{1})w_{2}(n_{2})\mathcal{S}_{mn_{1}n_{2}}(q;a)\right|\ll_{\sigma,A}XL^{-A}.$$

The same statement holds for a fixed finite sum of signed weights. Split each bounded real weight into its two nonnegative parts.

Proof.

Restrict first to $(m,q)=1$, and write $\Delta=C-M_{q}$ for the positive congruence and comparator functionals. In the K2 application the inner product is $e=mn_{1}n_{2}$, the residue is $N$, and $(q,N)=1$. If $(m,q)>1$, then $(e,q)>1$, so both $1_{e\equiv N\ (q)}$ and the comparator $1_{(e,q)=1}/\varphi(q)$ vanish. The omitted range contributes zero. Apply the Brun–Bonferroni sandwich of Paper I, (3.102)–(3.104), at $D_{f}=X^{\epsilon_{I}/2}$, $\epsilon_{I}=7.5\sigma$. For a requested fixed logarithmic precision $A_{\rm sv}$, put

$$s_{f}=\frac{\log D_{f}}{\log z^{*}},\quad E=\sum_{p<z^{*}}\frac{1}{p},\quad J=2\lfloor s_{f}/2\rfloor,$$

and increase the common threshold until $J\geq\max\{2,e^{2}E,A_{\rm sv}\log L\}$. Since $s_{f}\asymp\sigma(\log\log N)^{3}$ and $E=O(\log\log N)$, this is eventual at fixed $\sigma$, and $E^{J}/J!\leq L^{-A_{\rm sv}}$.

For the two large factors put $R_{i}=1_{P^{-}(n_{i})\geq z^{*}}$ and $G_{i}=\Psi_{i}^{+}-\Psi_{i}^{-}\geq 0$. Positivity gives

$$\Psi_{1}^{+}\Psi_{2}^{+}-G_{1}\Psi_{2}^{+}-\Psi_{1}^{+}G_{2}\leq R_{1}R_{2}\leq\Psi_{1}^{+}\Psi_{2}^{+},$$

so the three signed gaps and the two nonnegative comparator gaps bound $|\Delta(R_{1}R_{2})|$, as in Paper I, (3.113). Expand the three gaps over $d_{i}\leq D_{f}$. The completed hyperbola estimate (3.109) gives $O_{\epsilon_{X}}(D_{f}^{2}q^{1/2}X^{\epsilon_{X}})$. Here we fix $\epsilon_{X}=10^{-4}$. Squareful moduli are included by the full-modulus Weil–Estermann bound.

The two one-dimensional sieve gaps satisfy

$$\sum_{n\in I_{i}}G_{i}(n)w_{i}(n)\ll_{A_{\rm sv}}N_{i}L^{-A_{\rm sv}}+D_{f},\qquad\sum_{n\in I_{i}}\Psi_{i}^{+}(n)w_{i}(n)\ll N_{i}+D_{f}.$$

Thus the total is bounded by

(I-2’) $$MD_{f}^{2}Q^{3/2}X^{\epsilon_{X}}L^{C}+XL^{C-A_{\rm sv}}+MD_{f}(N_{1}+N_{2})L^{C}+MD_{f}^{2}L^{C}.$$

The ten-variable check uses the following upper bound. The expression is increasing throughout $0<\sigma\leq 10^{-10}$, so its maximum is evaluated at $\sigma=10^{-10}$:

$$\frac{600\sigma+\kappa}{1-\omega}\leq\frac{(600+1/202)\sigma}{1-\sigma/101}\leq\frac{(600+1/202)10^{-10}}{1-10^{-10}/101}=\frac{121201}{2019999999998}\approx 6.0000495049564356926\cdot 10^{-8}<6.02\cdot 10^{-8},$$

where $\omega,\kappa\leq\sigma/202$ absorb all fixed dyadic factors, so $1-\omega\geq 1-\sigma/202\geq 1-\sigma/101$, consistently with (7.60). Eleven small variables would give $\frac{133321}{2019999999998}\approx 6.600049504957030\cdot 10^{-8}$, which exceeds the allowed bound. With $q<X^{0.5977}$, the first exponent in (I-2’) is at most

$$6.02\cdot 10^{-8}+7.5\cdot 10^{-10}+1.5(0.5977)+\epsilon_{X}=0.89665006095<0.897,$$

leaving slack $0.00034993905$. Also, using $N_{1}N_{2}M\leq 2^{30}X$, $N_{1}\geq N_{2}\geq X^{1/3}$, and $D_{f}=X^{3.75\sigma}$,

$$MD_{f}(N_{1}+N_{2})\leq 2^{31}D_{f}X/N_{2}\ll X^{2/3+3.75\cdot 10^{-10}}<X^{0.668},\qquad MD_{f}^{2}\leq X^{6.02\cdot 10^{-8}+7.5\cdot 10^{-10}}<X^{0.668}.$$

The coefficient bound uses $\tau_{10}(m)\leq\tau(m)^{9}$, so $B_{0}=9$ is fixed. Finally, a family of at most $O_{\sigma}(L^{K})$ boxes is absorbed by requesting $A_{\rm sv}\geq K+A+5$, after all fixed logarithmic losses. This proves the lemma. ∎

Type-I-2 boxes: the complementary estimate.

Lemma 7.2 is a relaxed form of Paper I, (3.111), for this source. Below we check each displayed hypothesis against the actual K2 source.

For the present K2 Type-I-2 box, the nontrivial SH2 check is the display

(7.65) $$\begin{gathered}1\leq v\leq X,\qquad w(v)\in\left\{1,\frac{\log v}{\log X}\right\},\qquad P^{-}(n_{i})\geq z^{*}\ \text{is a separate tested condition},\\ \sup|1|+V(1)=1,\qquad\sup\left|\frac{\log v}{\log X}\right|+V\!\left(\frac{\log v}{\log X}\right)\leq 1+\frac{2\log(1+L^{-334})}{\log X}=O(1).\end{gathered}$$

Thus the two tested large factors have the required nonnegative weights and bounded supremum plus variation. The roughness condition is not hidden inside the weight.

For the Type-I-2 boxes above, (SH1) has $\gamma=\Gamma_{\mathfrak{c},k}$. The support and bound are (7.51) and (7.49), $e_{Q}<153/256$, and the residue is the unit $a=N$. The partition construction proves the geometric constancy in (SH2) on each nonboundary box. In this two-large-variable case the two tested large factors are the selected smooth factors. The roughness restriction is kept as the separate condition $P^{-}(n_{i})\geq z^{*}$ in the inner sum. Their smooth weights are then individually $1$ or $\log v/\log X$, not $\mu_{z_{1}}$. Hence they are nonnegative with $\sup|w|+V(w)\ll 1$ as required by (SH2). The $\mu_{z_{1}}$ factor occurs only among the complementary small factors. Condition (SH3) is $z_{1}\geq z_{0}(X)$. Condition (SH4) is the natural full-product discrepancy fixed above. Condition (SH5) is the priority in the preceding boundary and Type-II classification.

On a large-coordinate cell $(V,(1+\delta)V]$, $\delta=L^{-334}$, one has $1\leq V\leq X$. The smooth weight $w(v)=1$ has supremum plus variation equal to 1. The alternative $w(v)=\log v/\log X$ is nonnegative and increasing, with

$$\sup|w|+V(w)\leq 1+\frac{2\log(1+\delta)}{\log X}=O(1).$$

The roughness restriction remains a separate tested condition under (SH3) and inside the sieve, and it is not claimed to be constant on a box.

Thus the relaxed estimate of this chapter applies with $z^{*}=z_{1}$, the two large smooth weights nonnegative, and

$$|\alpha_{m^{\sharp}}|\leq\tau_{10}(m^{\sharp})\leq\tau(m^{\sharp})^{9}.$$

Each of the at most ten small labels contributes one factor. Normalized logarithmic weights $\log v/\log X$ are at most one on the support. Thus the coefficient hypothesis uses the fixed divisor exponent $B_{0}=9$, independent of $N$. The first term of (I-2’) uses the following values. The slack is $0.00034993905$ at exponent $0.89665006095$. The two final terms have the bound $2/3+3.75\cdot 10^{-10}$. The absorption is $\omega,\kappa\leq\sigma/202$, and the divisor bound is $\tau_{10}$ with $B_{0}=9$. By (7.60), Lemma 7.2 applies to every K2 Type-I-2 box. Requesting $A_{I}=K+A+5$ for the fixed $O_{\sigma}(L^{K})$ box family (here $K=4022$) gives

(7.66) $$E_{\rm I2}\ll_{b,\sigma}NL^{-A}=NL^{-4}.$$

Proposition 7.3 (Row-three remainder).

Fix $b\in(1/3,1/2)$ and $0<\sigma\leq 10^{-10}$. Under Paper I, Theorems 2.8 and 3.2, with the relaxed Type-I-2 Lemma 7.2 and Lemma 6.5(ii) at (3.48),

$$\frac{|r_{3}|}{\mathcal{N}_{K}}\ll_{b,\sigma}N^{-\ell_{0}\sigma^{2}}L^{3}+N^{-b/4}L^{7}+L^{-260}+L^{-120}+L^{-2}+N^{-c_{\sigma}}L^{4024+C}.$$

Proof.

By (7.48) and (7.49), $r_{3}$ is the $\Gamma$-sum minus the squareful sum (7.52). By (7.54), the $\Gamma$-sum is $\mathcal{E}_{\rm pp}$ plus the tuple sum. Every tuple lies in exactly one box. The boundary tuples give (7.58). On the other boxes logarithm freezing costs (7.57). Each box is of Type II or Type I-2. Sum (7.52), (7.56), (7.57), (7.58), (7.64) and (7.66), with $O_{\sigma}(1)$ pairs $(\mathfrak{c},k)$. Dividing by $\mathcal{N}_{K}=(1-\varepsilon)C(N)N/L^{2}\geq(1-\varepsilon)C_{2}N/L^{2}\gg_{b}N/L^{2}$ gives the claim. ∎

Proposition 7.4 (Exterior source estimates).

Fix $b,\sigma$ before $N$. For the source (7.2):

1. (7.32) and (7.35) hold for the two-orientation source and four signed AA main terms in (7.27) and (7.31);

2. (7.43) and (7.38) hold for $N\geq N_{\mathrm{AA}}(b,\sigma)$;

3. (7.46) holds for $N\geq N_{\rm peel}(\sigma)$;

4. (7.47) holds.

Proof.

Item 1 is (7.33)–(7.34) and the derivation after (7.35). Item 2 is (7.40)–(7.42). Item 3 is (7.44)–(7.46). Item 4 is (7.25) and Proposition 7.3. ∎

Proposition 7.5 (Exterior comparison).

For fixed $b,\sigma$ there is a nonnegative $r_{K}(N;b,\sigma)\to 0$ such that, for $N\geq T_{r}(b,\sigma)$ with $T_{r}$ as defined below,

(7.67) $$\frac{4B_{b}(N)}{W(N)} \leq c_{K2,G0}I(b)+A_{K}(b)C_{\mathrm{AA}}\sigma+P_{K}\sigma^{9}+r_{K}(N;b,\sigma),$$

$$P_{K} =\frac{388400000}{1-\varepsilon}=\frac{3884000000000}{9999}.$$

Here $T_{K}=\max\{N_{\mathrm{mass}}(b,\sigma),N_{\mathrm{AA}}(b,\sigma),N_{\rm peel}(\sigma),T_{\mathrm{len}}\}$. The threshold $N_{\mathrm{mass}}$ is from the fixed-parameter estimate at (7.35). $N_{\mathrm{AA}}$ is from item 2 of Proposition 7.4. Both exist with $b,\sigma$ fixed. Finally, $T_{\mathrm{len}}$ is the named length and application threshold of Section 2. In addition,

(7.68) $$r_{K}=\frac{E_{\rm top}+[M_{K}-\mathcal{N}_{K}2e^{-\gamma}I(b)G_{*}]_{+}+\sum_{j=0}^{3}|r_{j}|}{\mathcal{N}_{K}}=o_{b,\sigma}(1).$$

In particular, let $\widehat{T}_{r}$ be the least even threshold above which $r_{K}\leq\delta_{\rm err}/10$, and set $T_{r}=\max\{N_{\mathrm{mass}}(b,\sigma),N_{\mathrm{AA}}(b,\sigma),T_{K},\widehat{T}_{r}\}$. These thresholds exist at fixed $b,\sigma$.

Remark 7.6 (Single payment of each error).

In [26, Theorem 2.6, (2.30)] the three summands $XVE_{s}$, $R_{1}^{\nu}$ and $R^{\nu}$ are distinct. The first is paid only by (7.38), the other two only by (7.47). The decaying part of (7.40) and the $o(1)$ in (7.42) are absorbed by thresholds at fixed $\sigma$. The peel term $P_{3}$ contains no linear-sieve application. Thus

$$r_{K}=\frac{E_{\mathrm{top}}+\sum_{j}|r_{j}|+[M_{K}-\mathcal{N}_{K}2e^{-\gamma}I(b)G_{*}]_{+}}{\mathcal{N}_{K}},$$

and no error is counted twice.

Proof.

Use (7.26) and divide by $\mathcal{N}_{K}=W/4$. The principal term is bounded by $\mathcal{N}_{K}2e^{-\gamma}I(b)G_{*}$ plus its positive part. Also $2e^{-\gamma}G_{*}\leq c_{K2,G0}$ by (7.36)–(7.37). Proposition 7.4(2),(3) pay the sieve and peel terms above $T_{r}$. By definition, $T_{r}$ includes $N_{\mathrm{mass}}(b,\sigma)$, $N_{\mathrm{AA}}(b,\sigma)$ and $T_{K}$. The remaining terms form $r_{K}$. First, $E_{\rm top}/\mathcal{N}_{K}\ll L^{-331}$. The positive part contributes $o_{b,\sigma}(1)$ by (7.35). The row remainders are $O_{b,\sigma}(L^{-2})$ by (7.47). Hence $r_{K}=o_{b,\sigma}(1)$. The convergence is ineffective through Siegel–Walfisz, so $T_{r}$ exists but is not computed. ∎

8. The switched sum

We estimate the single term $-S_{\mathrm{sw}}/W$. The normalization is (2.2). Fix $\sigma$ as in (2.4). Put

(8.1) $$\begin{gathered}\eta_{L}=10^{-4},\quad\varepsilon=10^{-4},\qquad H_{\max}=\frac{153}{256},\qquad\ell_{0}=\frac{264897}{2060000},\qquad\ell_{1}=\frac{109}{1000},\\ N_{\mathrm{cut}}=N(1-L^{-333}),\qquad\mathcal{P}_{N}=\{p\in\mathbb{P}:p\nmid 2N\}.\end{gathered}$$

All limits in this section fix $\sigma$ first.

8.1. A common nonnegative source

We use the prime-coordinate cells of Paper II [27, (5.29a)–(5.31)]. Their rules are

(8.2) $$\begin{gathered}I_{1}=[\alpha,\beta),\quad I_{2}=[\beta,\lambda),\quad I_{3}=[\lambda,1/3),\quad I_{4}=[1/3,1],\\ C_{\mathrm{L3}}=\frac{57}{64}-u-\frac{11\alpha}{5}=\frac{6408241}{9192000},\qquad v_{\beta}=\frac{2(C_{\mathrm{L3}}-\beta)}{5},\\ k_{1}+k_{2}+k_{3}+k_{4}=k,\quad 5\leq k\leq 11,\quad k_{1}\geq m_{1}\geq\cdots\geq m_{k_{2}}\geq 0,\\ v_{j}\leq v_{\beta}\ \Longrightarrow\ m_{j}=k_{1},\\ v_{j}>v_{\beta}\ \Longrightarrow\ x_{m_{j}}<C_{\mathrm{L3}}-\frac{5}{2}v_{j}\leq x_{m_{j}+1},\\ w_{C}=3+1_{k_{1}=0}-2k_{1}-2k_{2}-k_{3}+\binom{k_{1}}{2}+\sum_{j=1}^{k_{2}}m_{j}.\end{gathered}$$

The $x_{i}$ and $v_{j}$ increase in $I_{1}$ and $I_{2}$. The outer sentinels are infinite. Let $\mathcal{C}_{+}$ contain the cells with $w_{C}>0$. For such a cell, put

(8.3) $$a_{C}(e) =1_{\begin{subarray}{c}e=p_{1}\cdots p_{k},\ N^{\alpha}\leq p_{1}<\cdots<p_{k}\\ (\log_{N}p_{1},\ldots,\log_{N}p_{k})\in C\end{subarray}},$$

$$A(e) =\sum_{C\in\mathcal{C}_{+}}w_{C}a_{C}(e)1_{\varepsilon N<e<N_{\mathrm{cut}}}, 0\leq A(e) \leq 36,$$

$$X_{A} =\sum_{e}A(e), Y_{A}(q) =\sum_{(e,q)=1}A(e),$$

$$C_{A}(q;N) =\sum_{e}A(e)1_{q\mid N-e}.$$

Since $\sum_{j}m_{j}\leq k_{1}k_{2}$, we have

$$w_{C}\leq 3+1_{k_{1}=0}-2k_{1}-2k_{2}-k_{3}+\frac{k_{1}(k_{1}-1)}{2}+k_{1}k_{2}\leq 36.$$

For $k_{1}\geq 2$, maximize at $k_{3}=0$, $k_{2}=11-k_{1}$; the resulting concave quadratic has maximum $36$ at $k_{1}=10,11$. The cases $k_{1}=0,1$ give at most $4,1$. The strict cells are disjoint, so this proves the bound for $A$. Every prime is permitted in $A$. In particular, $A$ has no factor $1_{(e,2N)=1}$. Keep the selector $1_{\Omega(e)\geq 3}$ in the switched source. The required pointwise comparison is

(8.4) $$1_{\Omega(e)\geq 3}(W_{13}(e))_{+}\leq A(e)\quad\left(\begin{gathered}\mu_{\mathrm{Mob}}(e)^{2}=1,\quad P^{-}(e)\geq N^{\alpha},\\ \varepsilon N<e<N_{\mathrm{cut}}\end{gathered}\right).$$

Here $\mu_{\mathrm{Mob}}$ denotes the Möbius function. Paper II gives zero positive part when $\Omega(e)=3,4$ [27, (1.7), (4.9), (5.29a)–(5.31)].

Proof of (8.4).

Let $e$ satisfy the side conditions and write $e=p_{1}\cdots p_{k}$, $p_{1}<\cdots<p_{k}$, $x_{i}=\log_{N}p_{i}$. If $(e,N)>1$, every occurrence in (4.3) vanishes, because it requires $(d,N)=1$ and $(n/d,Nd)=1$ [27, (1.7)]. Then $W_{13}(e)=0$. Assume $(e,N)=1$. Since $N^{k\alpha}\leq e<N$ and $12\alpha=1200/1149>1$, we have $k\leq 11$. If $k\leq 2$, the selector vanishes. If $k\in\{3,4\}$, Paper II, (4.9), gives $W_{13}(e)\leq 0$. Its hypotheses are exactly squarefree $e$, $(e,N)=1$, $\varepsilon N<e<N$ and $P^{-}(e)\geq N^{\alpha}$ [27, (4.9)]. Let $5\leq k\leq 11$. The table after Paper II, (5.29a), shows that the three negative pair counts vanish, so

$$W_{13}(e)=3+1_{k_{1}=0}-2k_{1}-2k_{2}-k_{3}+\binom{k_{1}}{2}+\sum_{j=1}^{k_{2}}m_{B}(v_{j}),$$

where $m_{B}(v)$ counts first-band coordinates below $B_{\mathrm{O1}}(v)$ [27, (5.29a), (5.30)]. The bins determine $k_{1},\ldots,k_{4}$. If $v_{j}\leq v_{\beta}$, then $B_{\mathrm{O1}}(v_{j})=\beta$ exceeds every first-band coordinate, so $m_{B}(v_{j})=k_{1}$. If $v_{j}>v_{\beta}$, then $B_{\mathrm{O1}}(v_{j})=C_{\mathrm{L3}}-\frac{5}{2}v_{j}$, and $m_{B}(v_{j})$ is the unique $m_{j}$ with $x_{m_{j}}<C_{\mathrm{L3}}-\frac{5}{2}v_{j}\leq x_{m_{j}+1}$ (infinite sentinels). It is nonincreasing in $j$. Thus $x$ lies in exactly one cell $C$ of (8.2), and $W_{13}(e)=w_{C}$ [27, (5.31)]. If $w_{C}\leq 0$, the left side of (8.4) is zero. Otherwise $C\in\mathcal{C}_{+}$ and $a_{C}(e)=1$, so the left side equals $w_{C}=A(e)$. This proves (8.4). The same argument, without the upper cut, gives

(8.5) $$1_{\Omega(e)\geq 3}(W_{13}(e))_{+}\leq A^{0}(e):=\sum_{C\in\mathcal{C}_{+}}w_{C}a_{C}(e)\leq 36$$

for squarefree $e$ with $P^{-}(e)\geq N^{\alpha}$ and $\varepsilon N<e<N$.

Put

(8.6) $$\begin{gathered}b_{A}(n)=A(N-n),\qquad P_{N}(z)=\prod_{\begin{subarray}{c}p<z\\ p\in\mathcal{P}_{N}\end{subarray}}p,\\ S_{v}(z)=\sum_{\begin{subarray}{c}n\geq 1\\ v\mid n\end{subarray}}b_{A}(n)1_{(n,P_{N}(z))=1},\qquad S_{A}=S_{1}(N^{\mu_{*}}),\\ E_{\mathrm{src}}=(S_{\mathrm{sw}}-S_{A})_{+},\qquad Z_{\mathrm{src}}=(S_{A}-S_{\mathrm{sw}})_{+}.\end{gathered}$$

The definitions give the finite identity

(8.7) $$S_{\mathrm{sw}}=S_{A}+E_{\mathrm{src}}-Z_{\mathrm{src}},\qquad E_{\mathrm{src}},Z_{\mathrm{src}}\geq 0.$$

Under (8.4), the required bound is

(8.8) $$E_{\mathrm{src}}\leq 910N^{1-\alpha}+36(NL^{-333}+1)+36N^{\mu_{*}}.$$

There are at most $2N^{1-\alpha}$ rough squareful integers. Each costs at most $455$, by the pointwise weight bound of Paper II, proof of Lemma 1.1, immediately after (1.11) [27, Lemma 1.1, proof after (1.11)]. The last two terms cover the top strip and small primes.

Proof of (8.8).

Write $e=N-p$ for $p\in\mathbb{P}$, $p<(1-\varepsilon)N$. Every prime $p\geq N^{\mu_{*}}$ has $(p,P_{N}(N^{\mu_{*}}))=1$, and $A\geq 0$. Hence $S_{A}\geq\sum_{p\geq N^{\mu_{*}}}A(N-p)$, and

$$S_{\mathrm{sw}}-S_{A}\leq\sum_{p\geq N^{\mu_{*}}}\bigl{(}s_{\mathrm{sw}}(e)-A(e)\bigr{)}_{+}+\sum_{p<N^{\mu_{*}}}s_{\mathrm{sw}}(e).$$

Here $s_{\mathrm{sw}}(e)=0$ unless $(e,N)=1$ and $P^{-}(e)\geq N^{\alpha}$, and always $0\leq s_{\mathrm{sw}}(e)\leq|W_{13}(e)|\leq 455$ [27, Lemma 1.1, proof after (1.11)]. Split the summands into three classes; overlaps only enlarge the bound. (i) Non-squarefree $e<N$ with $P^{-}(e)\geq N^{\alpha}$ number at most $\sum_{\ell\geq N^{\alpha}}N/\ell^{2}\leq 2N^{1-\alpha}$; they cost at most $455\cdot 2N^{1-\alpha}=910N^{1-\alpha}$. (ii) Squarefree $e$ with $N_{\mathrm{cut}}\leq e<N$ number at most $NL^{-333}+1$; by (8.5) they cost at most $36(NL^{-333}+1)$. (iii) The primes $p<N^{\mu_{*}}$ with squarefree $e$ cost at most $36N^{\mu_{*}}$, again by (8.5). For the remaining squarefree $e$ with $\varepsilon N<e<N_{\mathrm{cut}}$ and $p\geq N^{\mu_{*}}$, (8.4) gives a zero summand. This proves (8.8). Since $W\gg N/L^{2}$, $E_{\mathrm{src}}/W\ll L^{-331}$.

8.2. Four rows and the lower-sieve cutoff

Use the clipped intervals $[N^{t},N^{\rho t})\cap[N^{u},N^{\mu_{*}})$. Let $\mathcal{A}_{j}$ comprise the ordered prefixes

$$N^{u}\leq a_{j}<\cdots<a_{1}<N^{\mu_{*}},\qquad a_{i}\in\mathcal{P}_{N}.$$

Write $v=a_{1}\cdots a_{j}$. The empty prefix has $v=1$. Define

(8.9) $$\begin{gathered}T_{0}=S_{1}(N^{u}),\quad T_{1}=\sum_{\mathcal{A}_{1}}S_{a_{1}}(N^{u}),\quad T_{2}=\sum_{\mathcal{A}_{2}}S_{a_{1}a_{2}}(N^{u}),\\ T_{3}^{\mathrm{dyn}}=\sum_{\mathcal{A}_{3}}S_{a_{1}a_{2}a_{3}}(a_{3}),\qquad S_{A}=T_{0}-T_{1}+T_{2}-T_{3}^{\mathrm{dyn}}.\end{gathered}$$

This is Buchstab’s finite identity [27, (4.18)]. Its coefficient is $1-k+\binom{k}{2}-\binom{k-1}{2}=0$ for $k\geq 1$. At $k=0$ it equals $1$.

At depth three, let $\Theta_{c}$ be the infimum of $\widetilde{H}$ on the closed clipped parent. The underlying prime cells stay half-open. The closure is used only for this infimum. Use the whole anatomy function before choosing a branch [27, (2.17d)–(2.17e), (4.19)]. Set

(8.10) $$\ell_{c}=\Theta_{c}-\eta_{L}-t_{1}-t_{2}-t_{3},\quad D_{c}=N^{\ell_{c}},\quad z_{c}=N^{t_{3}},\quad\chi_{c}=1_{\ell_{c}\geq 2\rho t_{3}}.$$

Retained prefixes have $\chi_{c}=1$. For these prefixes define

(8.11) $$B_{3} =\sum_{\mathcal{A}_{3}:\chi_{c}=1}S_{v}(z_{c}), D_{-} =\sum_{\mathcal{A}_{3}:\chi_{c}=0}S_{v}(a_{3}),$$

$$D_{+} =\sum_{\mathcal{A}_{3}:\chi_{c}=1}\{S_{v}(z_{c})-S_{v}(a_{3})\}, P_{3} =\sum_{\mathcal{A}_{3}:\chi_{c}=1}\sum_{\begin{subarray}{c}r\in\mathcal{P}_{N}\\ z_{c}\leq r<a_{3}\end{subarray}}S_{vr}(z_{c}),$$

$$Z_{\mathrm{peel}} =D_{-}+P_{3}-D_{+}\geq 0.$$

The union bound gives $D_{+}\leq P_{3}$. Therefore

(8.12) $$S_{A}=T_{0}-T_{1}+T_{2}-B_{3}+P_{3}-Z_{\mathrm{peel}}.$$

The favorable term enters exactly once [27, (4.45)].

8.3. Capping the first three levels

At depths $j=0,1,2$, replace the relative level by

(8.13) $$T_{c}=\sum_{i=1}^{j}t_{i},\quad\ell_{c}=\frac{1}{2}-\eta_{L}-\rho T_{c},\quad D_{c}=N^{\ell_{c}},\quad z_{j,c}=N^{u}.$$

Here $T_{c}=0$ at depth zero. Depth three keeps (8.10), with $z_{3,c}=z_{c}$. The elementary bounds are

(8.14) $$\begin{gathered}\ell_{c}\geq\frac{1}{2}-\eta_{L}-j\rho\mu_{*}\geq\ell_{1},\qquad s_{c}=\ell_{c}/u\geq 54.6375-10^{-87}>30,\\ \text{continuum reference values for depths }0,1,2\text{ (the first three rows)}\text{ at }(\eta_{L},\rho)=(0,1):250,\ \frac{4875}{32},\ \frac{875}{16},\\ q<N^{\rho T_{c}}D_{c}^{\rho+\sigma}\leq N^{1/2-\eta_{L}+(\sigma+\tau)/2}<N^{1/2-9\cdot 10^{-5}}.\end{gathered}$$

The modulus includes the smooth divisor and every tail prime. The exponent $\rho+\sigma$ is kept. The cap changes no row-three level.

The lower floor changes the repeated-prime argument. Put

(8.15) $$u_{1}=N^{\ell_{1}\sigma^{2}},\qquad M_{\sigma}=\left\lceil\frac{3}{5\ell_{0}\sigma^{2}}\right\rceil,\qquad C_{\mathrm{L3},\sigma}=M_{\sigma}!.$$

(P2.8b) Rough-coordinate and repeated-prime bounds.

Assume the support and rough-coordinate hypotheses stated below. These are the capped version of Paper II’s Lemmas B.1–B.4. The coefficients and masks are defined in (8.30). Every rough coordinate is at least $u_{c}=D_{c}^{\sigma^{2}}\geq u_{1}$. On an active block, count these coordinates by $m_{c}$. The capped modulus bound gives

(8.16) $$u_{1}^{m_{c}}\leq q<N^{1/2},\qquad m_{c}<\frac{500}{109\sigma^{2}}\leq\frac{3}{5\ell_{0}\sigma^{2}},\qquad\frac{m_{c}!}{M_{c}}\leq C_{\mathrm{L3},\sigma}.$$

The prefix orbit size is $M_{c}\in\{1,2,6\}$. The small-prime part of $q$ fixes the smooth divisor. The rough multiset fills at most $m_{c}!$ labelled positions. Thus

(8.17) $$\frac{|C_{\mathfrak{b}}(q)|}{M_{c}}\leq C_{\mathrm{L3},\sigma}.$$

Define $\mathcal{H}_{1}=\{q<N^{1/2}:p^{2}\mid q\text{ for some }p\geq u_{1}\}$. The forbidden and allowed-repeat masks satisfy

(8.18) $$|\Xi_{U,\mathfrak{b}}(q)|+|\Xi_{A,\mathfrak{b}}(q)|\leq C_{\mathrm{L3},\sigma}1_{\mathcal{H}_{1}}(q).$$

Writing $q=vp^{2}$ and overcounting gives

(8.19) $$\sum_{q\in\mathcal{H}_{1}}\frac{1}{q}\leq\frac{2(1+L)}{u_{1}},\qquad\#\mathcal{H}_{1}\leq\frac{2N^{1/2}}{u_{1}}.$$

The tested measure has $0\leq A\leq 36$ and support below $N$. For $N\geq N_{\mathrm{dom}}$, $q/\varphi(q)\leq 1+L$ throughout $q<N^{1/2}$. Consequently

(8.20) $$|\Delta_{P}(q)|,\ |d_{A}(q)|\leq 36\left(\frac{N}{q}+1+\frac{N}{\varphi(q)}\right)\leq 36\left(\frac{N(2+L)}{q}+1\right).$$

Per block, these bounds give

(8.21) $$\sum_{q}\bigl{(}|\Xi_{U}(q)\Delta_{P}(q)|+|\Xi_{A}(q)d_{A}(q)|\bigr{)}\leq\frac{216C_{\mathrm{L3},\sigma}N(1+L)^{2}}{u_{1}}.$$

There are finitely many blocks at fixed $\sigma$. Division by $W\gg N/L^{2}$ proves the conditional estimate

(8.22) $$|E_{\mathrm{sq}}|/W=O_{\sigma}(N^{-\ell_{1}\sigma^{2}}L^{4}).$$

These are Paper II, Lemmas B.1–B.4 and (B.3)–(B.7), with $u_{0}$ replaced by $u_{1}$ on rows $0,1,2$ [27, Appendix B.1, Lemmas B.1–B.4, (B.3)–(B.7)]. Row three keeps $u_{0}=N^{\ell_{0}\sigma^{2}}$, since its level floor in (8.43) exceeds $\ell_{0}$.

Verification for the capped rows.

Fix a block in row $j\leq 2$. Prefix primes lie in $[N^{t_{i}},N^{\rho t_{i}})\cap[N^{u},N^{\mu_{*}})$, so their product is below $N^{\rho T_{c}}$. The tail tuple satisfies $\prod_{i}D_{i}<D_{c}$ by its last cubic constraint, and the tail primes are below $D_{i}^{\rho}$; also $d<D_{c}^{\sigma}$. Thus

$$\log_{N}q<\rho T_{c}+(\rho+\sigma)\ell_{c}=(\rho+\sigma)(\tfrac{1}{2}-\eta_{L})-\rho(\rho+\sigma-1)T_{c}\leq(\rho+\sigma)(\tfrac{1}{2}-\eta_{L})<\tfrac{1}{2}-9\cdot 10^{-5}.$$

Every prime of $q$ lies in $\mathcal{P}_{N}$ and $d\mid P_{N}(u_{c})$, so $(q,2N)=1$. Tail primes are at least $D_{i}\geq D_{c}^{\sigma^{2}}=u_{c}$. Prefix primes are at least $N^{u}>u_{c}$, because $\ell_{c}\sigma^{2}<u$. Also $u_{c}=N^{\ell_{c}\sigma^{2}}\geq N^{\ell_{1}\sigma^{2}}=u_{1}$. For an active block choose a nonzero realization. Then $u_{1}^{m_{c}}\leq q<N^{1/2}$, so $m_{c}<1/(2\ell_{1}\sigma^{2})=500/(109\sigma^{2})$, and $1/(2\ell_{1})\leq 3/(5\ell_{0})$. This is (8.16). The primes of $q$ below $u_{c}$ determine $d$. The rough multiset fills the $m_{c}$ labelled positions in at most $m_{c}!$ ways, and $|\lambda_{d}|\leq 1$. This gives (8.17). It also gives (8.18). A forbidden tuple repeats a prime $\geq u_{c}$. An allowed non-squarefree tuple has $p^{2}\mid q$ with $p\geq u_{c}$, since $d$ is squarefree and coprime to the rough part. The grid count of Paper II, Lemma B.1(4), gives finitely many blocks at fixed $\sigma$. These are Paper II, Lemmas B.1–B.2, with $(u_{0},N^{3/5})$ replaced by $(u_{1},N^{1/2})$ [27, Appendix B.1, Lemmas B.1–B.2]. Lemma B.4 with $u_{1}$ and $Q=N^{1/2}$ gives (8.19). Lemma B.3, with $A\leq 36<A_{\max}$, support in $[1,N)$ and $q/\varphi(q)\leq 1+\log_{2}q\leq 1+L$, gives (8.20). Then $\sum_{q\in\mathcal{H}_{1}}(N(2+L)/q+1)\leq 6N(1+L)^{2}/u_{1}$ gives (8.21). Row three is Paper II, Lemmas B.1–B.4 as printed: its modulus bound $q<N^{e_{Q}}<N^{3/5}$ and floor $\ell_{c}>\ell_{0}$ are (8.43). Hence (8.22) holds for all four rows.

8.4. The signed sieve identity

Put $\nu_{j}=(-1)^{j}$. For all four rows, define

(8.23) $$s_{c}=\frac{\ell_{c}}{\log_{N}z_{j,c}}=\begin{cases}\ell_{c}/u,&j=0,1,2,\\ \ell_{c}/t_{3},&j=3.\end{cases}$$

Let $K_{j}=F$ for even $j$, and $K_{j}=f$ for odd $j$. For a prefix $v$, set

(8.24) $$\begin{gathered}\mathcal{X}_{v}=X_{A}/\varphi(v),\qquad\omega_{v}(p)=\frac{p}{p-1}1_{p\in\mathcal{P}_{N}},\qquad\omega_{v}(p^{h})=0\quad(h\geq 2),\\ V_{N}(z)=\prod_{\begin{subarray}{c}p<z\\ p\in\mathcal{P}_{N}\end{subarray}}\left(1-\frac{1}{p-1}\right),\qquad\Delta_{0}(q)=C_{A}(q;N)-\frac{X_{A}}{\varphi(q)}.\end{gathered}$$

All cutoff primes lie below the prefix primes. Thus the density product on the sieve support is $V_{N}$. The squarefree remainder is $\Delta_{0}(vd)$. Apply Fact 3.17 with $\epsilon_{s}=\sigma$, $\delta_{s}=\tau$, and $L_{\omega}=2$. Choose a common nonnegative majorant

(8.25) $$E_{s,c}:=C_{E}\{\sigma+\sigma^{-8}e^{K_{\mathrm{M}}+2}(\ell_{c}L)^{-1/3}\}.$$

The signed sieve error satisfies $|E_{s}|\leq E_{s,c}$.

Verification of the sieve data.

Fix a row, a cell and a prefix $v$, and sieve the finite multiset $\{n:v\mid n\}$ with multiplicity $b_{A}(n)\geq 0$ by $\mathcal{P}_{N}$ below $z=z_{j,c}$. Every prefix prime is at least $z$: on rows $0,1,2$ it is at least $N^{u}=z$, and on row three $a_{3}\geq N^{t_{3}}=z_{c}$. Hence $(d,v)=1$ for $d\mid P_{N}(z)$. In particular $\omega_{v}(d)=\prod_{p\mid d}p/(p-1)$ for these $d$; the primes $p\mid v$ are all $\geq z$ and are never sifted. For squarefree such $d$,

$$|\mathcal{A}_{d}|=\sum_{e}A(e)1_{vd\mid N-e}=C_{A}(vd;N),\qquad\frac{\omega_{v}(d)}{d}\mathcal{X}_{v}=\frac{X_{A}}{\varphi(v)\varphi(d)}=\frac{X_{A}}{\varphi(vd)},$$

so $r(\mathcal{A},d)=\Delta_{0}(vd)$. Each remainder modulus in (8.26) is squarefree. Here $d$ uses primes below $D_{c}^{\sigma^{2}}$. Tail primes are at least $D_{c}^{\sigma^{2}}$ and distinct (equal labels by the mask, unequal grid intervals disjoint). Prefix primes are at least $z$. Next, $0<\omega_{v}(p)=p/(p-1)<p$ for every $p\in\mathcal{P}_{N}$ (all such $p$ are odd), $\omega_{v}=0$ off $\mathcal{P}_{N}$, and $\omega_{v}(p^{h})=0$ for $h\geq 2$. Hence the second condition in (3.19) holds with $L_{\omega}=2$. For the first, $(1-\omega_{v}(p)/p)^{-1}=(1-1/(p-1))^{-1}$, and Paper II, (5.5), gives

$$\prod_{\begin{subarray}{c}w\leq p<z\\ p\in\mathcal{P}_{N}\end{subarray}}\left(1-\frac{1}{p-1}\right)^{-1}\leq\frac{\log z}{\log w}\left(1+\frac{K}{\log w}\right)\qquad(2\leq w<z)$$

with one absolute $K$, uniformly in $N$ [27, (5.5)]. Take $K_{\mathrm{M}}$ in (8.25) to be the maximum of this $K$ and the constant of (3.25). It remains absolute and is fixed before $\sigma$. The product $V(z)$ of Fact 3.17 is then $V_{N}(z)$. The levels satisfy $z\leq D_{c}$ on rows $0,2$ ($s_{c}>30$) and $z\leq D_{c}^{1/2}$ on rows $1,3$ ($s_{c}>30$, resp. $\chi_{c}=1$, i.e. $\ell_{c}\geq 2\rho t_{3}$). Fact 3.17 with $\epsilon_{s}=\sigma$, $\delta_{s}=\tau$, $D=D_{c}$ therefore applies to every actual sequence, and (8.28) follows with $E_{s}\leq E_{s,c}$, $\log D_{c}=\ell_{c}L$.

For each cell fix one coefficient construction. Use it for every prefix realization in that cell. Write its remainders as

(8.26) $$R_{1,v}^{\nu} =\sum_{\begin{subarray}{c}d<D_{c}^{\sigma}\\ d\mid P_{N}(D_{c}^{\sigma^{2}})\end{subarray}}\phi_{d}^{\nu}\Delta_{0}(vd),$$

$$R_{v}^{\nu} =\sum_{\mathbf{D}\in\mathcal{D}^{\nu}(D_{c})}\sum_{\begin{subarray}{c}d<D_{c}^{\sigma}\\ d\mid P_{N}(D_{c}^{\sigma^{2}})\end{subarray}}\lambda_{d}^{\nu}(\mathbf{D})\sum_{\begin{subarray}{c}D_{i}\leq p_{i}<\min(D_{i}^{\rho},z_{j,c})\\ p_{i}\in\mathcal{P}_{N},\ \text{equal-label entries distinct}\end{subarray}}\Delta_{0}(vd\textstyle\prod_{i}p_{i}).$$

The coefficients satisfy $|\phi_{d}^{\nu}|,|\lambda_{d}^{\nu}(\mathbf{D})|\leq 1$ [26, Theorem 2.6, (2.29)–(2.33b)]. Define the finite quantities

(8.27) $$M_{j} =\sum_{c}\sum_{v\in c}\frac{X_{A}}{\varphi(v)}V_{N}(z_{j,c})K_{j}(s_{c}),$$

$$E_{j} =\sum_{c}\sum_{v\in c}\frac{X_{A}}{\varphi(v)}V_{N}(z_{j,c})E_{s,c},\qquad R_{j}=\sum_{c}\sum_{v\in c}(R_{1,v}^{\nu_{j}}+R_{v}^{\nu_{j}}),$$

$$Q_{j} =M_{j}+\nu_{j}E_{j}+\nu_{j}R_{j},\qquad(S_{0},S_{1},S_{2},S_{3})=(T_{0},T_{1},T_{2},B_{3}),$$

$$Z_{\mathrm{AA}} =\sum_{j=0}^{3}\nu_{j}(Q_{j}-S_{j}),\qquad M_{\mathrm{prin}}=\sum_{j=0}^{3}\nu_{j}M_{j},$$

$$M_{\mathrm{err}} =\sum_{j=0}^{3}E_{j},\qquad R_{\mathrm{adv}}=\sum_{j=0}^{3}R_{j}.$$

The sums keep strict prefix order. Under the stated sieve hypotheses,

(8.28) $$Q_{0}\geq S_{0},\quad Q_{1}\leq S_{1},\quad Q_{2}\geq S_{2},\quad Q_{3}\leq S_{3},\quad Z_{\mathrm{AA}}\geq 0.$$

Multiplying the signs gives

(8.29) $$\sum_{j=0}^{3}\nu_{j}S_{j}=M_{\mathrm{prin}}+M_{\mathrm{err}}+R_{\mathrm{adv}}-Z_{\mathrm{AA}}.$$

Every remainder in $R_{\mathrm{adv}}$ has sign $+1$.

8.5. The natural coprime comparator

A block $\mathfrak{b}=(j,c,f,\mathbf{D})$ fixes a row, a cell, and either remainder type $f$. The tail vector is empty for $R_{1}$. Prefix realizations stay inside the coefficient. Let $K_{P},K_{T}$ impose prefix and tail distinctness. Let $C_{\mathfrak{b}}$ be the unrestricted labelled coefficient. Define

(8.30) $$C_{\mathfrak{b}}(q) =\sum_{\mathrm{full\ labelled\ tuples}}\widetilde{\lambda}_{d}1_{q=d\prod p_{i}},$$

$$O_{\mathfrak{b}}(q) =\frac{1}{M_{c}}\sum_{\mathrm{full\ labelled\ tuples}}\widetilde{\lambda}_{d}K_{P}K_{T}1_{q=d\prod p_{i}},$$

$$\Xi_{U,\mathfrak{b}}(q) =\frac{1}{M_{c}}\sum_{\mathrm{full\ labelled\ tuples}}\widetilde{\lambda}_{d}(1-K_{P}K_{T})1_{q=d\prod p_{i}},$$

$$O_{\mathfrak{b}} =C_{\mathfrak{b}}/M_{c}-\Xi_{U,\mathfrak{b}},\qquad\Xi_{A,\mathfrak{b}}=O_{\mathfrak{b}}1_{q\ \mathrm{not\ squarefree}}.$$

The two comparator discrepancies are

(8.31) $$\Delta_{P}(q)=C_{A}(q;N)-\frac{Y_{A}(q)}{\varphi(q)},\qquad d_{A}(q)=\Delta_{0}(q)-\Delta_{P}(q)=-\frac{X_{A}-Y_{A}(q)}{\varphi(q)}.$$

For $\langle f,g\rangle=\sum_{q}f(q)g(q)$, put

(8.32) $$E_{\mathrm{disp},f} =\sum_{\mathfrak{b}:\mathrm{type}=f}\frac{\langle C_{\mathfrak{b}},\Delta_{P}\rangle}{M_{c}}\quad(f=1,2),$$

$$E_{\mathrm{sq}} =\sum_{\mathfrak{b}}\{-\langle\Xi_{U,\mathfrak{b}},\Delta_{P}\rangle+\langle\Xi_{A,\mathfrak{b}},d_{A}\rangle\},$$

$$E_{\mathrm{nc}} =\sum_{\mathfrak{b}}\sum_{q\ \mathrm{squarefree}}O_{\mathfrak{b}}(q)d_{A}(q),$$

$$R_{\mathrm{adv}} =E_{\mathrm{disp},1}+E_{\mathrm{disp},2}+E_{\mathrm{sq}}+E_{\mathrm{nc}}.$$

This is a finite coefficient identity [27, (4.22)–(4.26), Appendix B.1–B.4]. All moduli satisfy $(q,2N)=1$. That restriction does not puncture $A$.

8.6. The full-support bilinear application

Split each prime coordinate into relative-width $L^{-334}$ intervals. An interior box has constant order and descriptor predicates. For $B=\prod_{i}[A_{i},B_{i})$, put

(8.33) $$\begin{gathered}T=\lceil A_{1}\rceil-1,\quad U_{J}=\lceil B_{1}\rceil-1,\quad I_{T}=(T,U_{J}],\quad X=\prod_{i}A_{i},\\ a_{n}=1_{\mathbb{P}}(n)1_{I_{T}}(n),\qquad b_{m}=\sum_{\begin{subarray}{c}p_{2}<\cdots<p_{k}\\ p_{i}\in[A_{i},B_{i})\end{subarray}}1_{m=p_{2}\cdots p_{k}},\\ b_{m}\in\{0,1\},\qquad N^{\alpha}/2\leq T<N^{1/5},\qquad\varepsilon N\leq X\leq N.\end{gathered}$$

The box contributes exactly $w_{C}a_{n}b_{m}$. For $T>2^{50}$ and $U_{J}\leq 2T$,

(8.34) $$a_{n}=1_{P^{-}(n)\geq T^{51/100}}1_{I_{T}}(n).$$

A composite $n\leq 2T$ has a smaller prime divisor. This is Paper I, (3.8), option 1 [26, (3.8)–(3.9)]. Its logarithmic replacement error is zero.

Define the full-support discrepancy

(8.35) $$\Delta_{B}(q,N) =\sum_{nm\equiv N\ (q)}a_{n}b_{m}-\frac{1}{\varphi(q)}\sum_{(nm,q)=1}a_{n}b_{m},$$

$$\mathcal{S}_{B}(\mathfrak{b}) =\sum_{q}C_{\mathfrak{b}}(q)\Delta_{B}(q,N).$$

There is no condition $nm\leq X$ in these sums. The exact interior decomposition is

(8.36) $$D_{\mathrm{int}}^{(j)} =\sum_{\mathfrak{b}:\mathrm{row}=j}\frac{1}{M_{c}}\sum_{C}\sum_{B\subset C\ \mathrm{interior}}w_{C}\mathcal{S}_{B}(\mathfrak{b}),$$

$$E_{\mathrm{disp},1}+E_{\mathrm{disp},2} =\sum_{j=0}^{3}D_{\mathrm{int}}^{(j)}+D_{\mathrm{bdry}}.$$

For rows $0,1,2$, apply Fact 3.3. Its Siegel–Walfisz sequence must be the selected prime sequence. The dictionary with Fact 3.3 is $b_{n}^{\mathrm{F2}}:=a_{n}$, $a_{m}^{\mathrm{F2}}:=b_{m}$, and $Y:=T$. Thus the prime sequence carries the Siegel–Walfisz hypothesis. We need Paper I, (3.25), in the form

(8.37) $$\sum_{\begin{subarray}{c}n\in I\\ n\equiv l\ (k)\\ (n,d)=1\end{subarray}}1_{\mathbb{P}}(n)-\frac{1}{\varphi(k)}\sum_{\begin{subarray}{c}n\in I\\ (n,dk)=1\end{subarray}}1_{\mathbb{P}}(n)\ll_{A_{\mathrm{SW}}}T\tau(d)L^{-A_{\mathrm{SW}}},\qquad(l,k)=1,\quad I\subset I_{T}.$$

The normalized hypothesis also needs

(8.38) $$\|a\|_{2}T^{1/2}\gg TL^{-168}.$$

Use Fact 3.1, specifically (3.2). Its endpoint estimate gives, for every fixed $A>335$,

(8.39) $$\pi(U_{J})-\pi(T)=\operatorname{Li}(U_{J})-\operatorname{Li}(T)+O_{A}(TL^{-A}).$$

Full mesh intervals satisfy $U_{J}-T\gg TL^{-334}$. Thus $\#(I_{T}\cap\mathbb{P})\gg TL^{-335}$ uniformly. This implies (8.38). The $k=1$ discrepancy in (8.37) is identically zero. The counting input is the endpoint estimate just displayed.

Proof of (8.37) and (8.38).

Let $N^{\alpha}/2\leq T<N^{1/5}$, $U_{J}\leq 2T$, $(l,k)=1$, and let $I=(y_{1},y_{2}]\subset I_{T}$ be an arbitrary subinterval, so $T\leq y_{1}\leq y_{2}\leq U_{J}$. (Paper I, (3.25), is quantified over all such $I$ [26, (3.25)].) The condition $(n,d)=1$ removes at most $\omega(d)$ primes from each sum. If $k\leq L^{A+2}$, apply (3.2) at $y=y_{2}$ and $y=y_{1}$, with $C=A+3$. Since $T\leq y\leq 2T$, $\log y\asymp L$ and the error is $O_{A}(TL^{-A})$ at each endpoint. Hence $\pi(y_{2};k,l)-\pi(y_{1};k,l)=(\operatorname{Li}(y_{2})-\operatorname{Li}(y_{1}))/\varphi(k)+O_{A}(TL^{-A})$. The same holds for the unrestricted count divided by $\varphi(k)$, up to the $\omega(k)$ primes dividing $k$. If $k>L^{A+2}$, since $y_{2}-y_{1}\leq T$ both terms are at most $T/k+1+T/\varphi(k)\ll TL^{-A}$, using $\varphi(k)\gg k/\log\log k$. This proves (8.37) with $\tau(d)$, ineffectively, uniformly in $I\subset I_{T}$. For the norm, $U_{J}-T\geq A_{1}L^{-334}-1\geq TL^{-334}/2$ on a full mesh interval, and (8.39) with $A=336$ gives $\#(I_{T}\cap\mathbb{P})\geq TL^{-335}/4$ eventually. Hence $\|a\|_{2}T^{1/2}\geq TL^{-335/2}/2\geq TL^{-168}$. Both bounds are uniform over all interior boxes, since only $T$ enters.

Assume (8.37) and (8.38) uniformly. Taking $A_{\mathrm{SW}}=A_{0}+168$ verifies (3.4) with divisor exponent one. Both factor lengths exceed $x_{B}^{1/12}$ eventually. Split each of the $k-1$ complementary prime intervals at any dyadic boundary. This gives at most $2^{k-1}$ pieces. Localize each resulting product on $m\sim M_{B}$. Since $(1+L^{-334})^{k-1}<2$ eventually, there are at most two product dyadic intervals per piece. This bounded splitting changes no logarithmic exponent. Set $x_{B}=TM_{B}\asymp N$. There are at most $T$ selected integers in $(T,2T]$ and at most $2M_{B}$ integers in a product dyadic interval. Unique factorization gives $\|a\|_{2}\|b\|_{2}\leq\sqrt{2x_{B}}$. The required application is

(8.40) $$\begin{gathered}\frac{|C_{\mathfrak{b}}(q)|}{M_{c}}\leq C_{\mathrm{L3},\sigma},\qquad Q<N^{1/2-9\cdot 10^{-5}}\leq x_{B}^{1/2}(\log x_{B})^{-B_{\mathrm{BV}}},\\ \|a\|_{2}\|b\|_{2}\leq\sqrt{2}\,x_{B}^{1/2},\qquad|\mathcal{S}_{B}(\mathfrak{b})|/M_{c}\ll_{\sigma}NL^{-3692}.\end{gathered}$$

Use $A=3692$, $B=1$, $\epsilon=1/12$ and $B_{\mathrm{BV}}=B_{\mathrm{BV}}(3692,1,1/12)$ in (3.5). The source has $O_{\sigma}(L^{3689})$ weighted boxes. There are finitely many blocks at fixed $\sigma$. Since $W\gg N/L^{2}$, summation gives

(8.41) $$\frac{|D_{\mathrm{int}}^{(0)}|+|D_{\mathrm{int}}^{(1)}|+|D_{\mathrm{int}}^{(2)}|}{W}\ll_{\sigma}L^{-1}.$$

Verification of (8.40).

The coefficient bound is (8.17), proved above, and the modulus range is (8.14). Every supported $q$ is prime to $2N$, so $a=N$ is a reduced class and the maximum in (3.5) covers it. The selected sequence $a_{n}$ is supported in $(T,2T]$, and $b_{m}$, after at most two dyadic pieces, in $(M_{B},2M_{B}]$. For $x_{B}=TM_{B}$, $x_{B}\in[\varepsilon N/8,2N]$ and $\log T/\log x_{B}\in[\alpha-o(1),1/5+o(1)]\subset[1/12,11/12]$, because $\alpha-1/12=17/4596>0$. The hypothesis (3.4) holds with $B=1$ by (8.37), (8.38) and $A_{\mathrm{SW}}=A_{0}+168$. Interior boxes have constant order predicates, so $nm=e$ has one labelled representation and $b_{m}\in\{0,1\}$. Thus $\|a\|_{2}\|b\|_{2}\leq(2x_{B})^{1/2}$. On each interior box, $\mathcal{S}_{B}(\mathfrak{b})=\sum_{q}C_{\mathfrak{b}}(q)\Delta_{B}(q,N)$ with $\Delta_{B}=\sum_{\mathrm{pieces}}\Delta_{\mathrm{fs}}$, and Fact 3.3 gives $|\mathcal{S}_{B}(\mathfrak{b})|/M_{c}\ll_{\sigma}NL^{-3692}$. Fact 3.3 is the full-support bilinear estimate derived from Bombieri–Friedlander–Iwaniec, Theorem 0(b). We verified its hypotheses above for this source.

8.7. The source-level row-three estimate

Row three keeps the full anatomy level. Its factorization keeps every labelled occurrence. The smooth divisor stays in the first factor. For $d_{*}=\rho(\Theta_{c}-\eta_{L})$ and $e_{i}=\rho t_{i}$, the prefix inequalities are

(8.42) $$d_{*}-2e_{1} \geq\frac{33109}{160000}\rho,$$

$$d_{*}-e_{1}-2e_{2} \geq\frac{1859}{160000}\rho,$$

$$d_{*}-e_{1}-e_{2}-2e_{3} \geq\frac{33109}{480000}\rho\quad(\chi_{c}=1).$$

The anatomy branch gives $\Theta_{c}\geq(1+t_{1})/2$. The condition $\chi_{c}=1$ gives the last inequality. In particular, with $A_{3}=33109/160000$,

(8.43) $$\begin{gathered}\ell_{c}\geq\max\{A_{3}-t_{3},2\rho t_{3}\}\geq\frac{33109}{240000},\qquad\ell_{c}-\ell_{0}\geq\frac{231463}{24720000}=0.0093633899676\ldots,\\ \Theta_{c}\leq H_{\max}=\frac{153}{256},\qquad e_{Q}=(\rho+\sigma)(153/256-\eta_{L})<\frac{3}{5},\\ \frac{3}{5}-e_{Q}\geq\frac{391}{160000}-\frac{95609}{160000}(\sigma+\tau)>0.0024437499.\end{gathered}$$

Here $e_{Q}$ bounds the full modulus exponent $\rho(t_{1}+t_{2}+t_{3})+(\rho+\sigma)\ell_{c}$. The inherited anatomy and rough-factor estimates are Paper II, (4.20) and Appendix B, Lemma B.1(1)–(2) [27, (4.20), Appendix B, Lemma B.1(1)–(2)]. For the tail exponents $s_{i}$, the cubic conditions give

(8.44) $$\sum_{i<l}s_{i}+2s_{l}<\ell_{c},\qquad h=d_{*}+60\sigma,\qquad\Theta_{c}-h\geq\eta_{L}-\tau H_{\max}-60\sigma>0.$$

Theorem 4.1 of Paper I then supplies the load surpluses

(8.45) $$\left(\frac{7527}{256},\ 60,\ \frac{114421}{1280},\ \frac{26247}{256}\right)\sigma.$$

The required length data are

(8.46) $$\begin{gathered}1\leq|mn-N|\leq N<X^{3},\quad N_{I}\geq\frac{T}{6L^{334}}\geq(2T)^{1-2^{-126}},\\ X^{60\sigma}\leq T\leq X^{1/3+15\sigma},\quad QX^{\sigma/3}\leq N_{I},\quad QRS<M_{I}X^{-89\sigma},\quad M_{I}\geq X^{1/2}.\end{gathered}$$

Here $N_{I}$ is the selected integer interval length. The complement has length scale $M_{I}$. The letters $Q,R,S$ are the three factor scales. Each condition holds only above its fixed-$\sigma$ application threshold. The floor $2\lceil e^{2^{150}}/2\rceil$ is imposed separately. See Paper I, (3.6)–(3.9) and (4.9) [26, (3.6)–(3.9), Theorem 4.1, (4.9)]. We verify them for this source below.

The remaining analytic estimate is

(8.47) $$|\mathcal{S}_{B}(\mathfrak{b})|\ll_{\sigma,A_{\mathrm{loc}}}XL^{-A_{\mathrm{loc}}}+X^{1-\ell_{0}\sigma^{2}/4}L^{C_{\mathrm{loc}}}\quad(\mathfrak{b}\text{ in row }3).$$

It is uniform over the permitted boxes and blocks. Here $A_{\mathrm{loc}}>0$ is arbitrary; $C_{\mathrm{loc}}$ is fixed by the local theorem. Paper I, Theorem 3.2, (3.13), supplies the local dispersion theorem. Its Theorem 2.8 enters only in the inner spectral stage. There we use the form (P1.7), Fact 3.22, with the full bracket (3.38). The inner substitution is

(8.48) $$r_{T}=z,\quad s_{T}=p_{0},\quad n_{T}=p_{0}m,\quad a_{T}=\epsilon_{y}N\quad(\epsilon_{y}=\pm 1),\quad A_{T}=X^{1+\sigma/20}.$$

The auxiliary $z$ and $p_{0}$ are those of Paper I’s dispersion proof. Both signs and the mixed cusp pair are kept. Let $\chi$ be smooth, zero below zero and one above one. For the fixed bump $\psi_{0}$, define

(8.49) $$\psi_{0,1}(v) =\psi_{0}(v)\{1-\chi((v-0.8)/0.1)\},$$

$$\psi_{0,2}(v) =\psi_{0}(v)\chi((v-0.8)/0.1)\{1-\chi((v-1.4)/0.2)\},$$

$$\psi_{0,3}(v) =\psi_{0}(v)\chi((v-1.4)/0.2).$$

They sum to $\psi_{0}$. Their strict dyadic windows are $(0.45,0.90]$, $(0.80,1.60]$, and $(1.40,2.80]$. The two smooth modulus coordinates use all nine products $\psi_{0,i}(b/D)\psi_{0,j}(c/C)$, $1\leq i,j\leq 3$. Multiply each by the fixed cutoffs on the supported $n_{T},z,p_{0}$. These cutoffs equal one on the actual coefficient support. The derivative hypothesis is (3.36) with $\epsilon_{0}=\sigma/2000$ and $\epsilon=\sigma/100$.

Write $\theta_{\mathrm{sp}}=7/32$ and $V_{\mathrm{sp}}=T^{4}R^{4}S^{6}$. The coefficient in (P1.7) is zero outside the actual indices $n_{T}=p_{0}m$, $r_{T}=z$, $s_{T}=p_{0}$. Use its supported norm

(8.50) $$\begin{gathered}\sum_{n^{\prime\prime}\mid N^{\infty}}(n^{\prime\prime})^{\theta_{\mathrm{sp}}}\|\widetilde{b}(n^{\prime\prime})\|_{2}^{2},\\ \|\widetilde{b}(n^{\prime\prime})\|_{2}^{2}=\sum_{\begin{subarray}{c}n^{\prime},r,s\\ (n^{\prime},N)=1\end{subarray}}|b_{n^{\prime}n^{\prime\prime},r,s}|^{2}.\end{gathered}$$

Here $b_{n,r,s}$ is the coefficient of (P1.7). It is distinct from the complementary sequence $b_{m}$. The norm includes no coefficients outside that support. After substitution, the four bracket contributions divided by $V_{\mathrm{sp}}$ are bounded by

(8.51) $$K_{A} =A_{T}^{\theta_{\mathrm{sp}}}\frac{TQR^{2}S^{5}}{X^{2}}, K_{2} =\frac{A_{T}}{X}K_{A},$$

$$K_{R} =\left(\frac{TR}{S}\right)^{2\theta_{\mathrm{sp}}}\frac{TQR^{2}S^{5}}{X^{2}}, K_{4} =\frac{TQR^{2}S^{5}}{X^{2}}\frac{T^{2}R^{2}}{X}.$$

Only uniform $X^{o(1)}$ factors and powers of $p_{0}$ are suppressed. These are precisely Paper I, (3.81), (3.85), and (3.87) [26, (3.69), (3.81)–(3.87)]. Each term carries the actual-use hypotheses.

8.8. The row-three local dispersion estimate

The source-wide estimate also has boundary and comparator contributions. The local argument gives only its interior row-three block.

We treat one row-three block with $\chi_{c}=1$ and one full interior prime-coordinate mesh box. We do not assert the source-wide boundary, comparator, or row-zero-to-two estimates. Fix $\sigma>0$, $\tau=\sigma^{9}$, $\rho=1+\tau$, and a retained closed clipped parent $c$, used only for the displayed infimum and endpoint checks. The actual prime cells are half-open. Keep its inherited labels $t_{1}\geq t_{2}\geq t_{3}$, $\Theta_{c}=\inf_{c}\widetilde{H}$, $\ell_{c}=\Theta_{c}-\eta_{L}-t_{1}-t_{2}-t_{3}$, $D_{c}=N^{\ell_{c}}$, $z_{c}=N^{t_{3}}$ and $u_{c}=D_{c}^{\sigma^{2}}$. On the whole block $\chi_{c}=1$. The inequalities (8.42)–(8.43) apply with the inherited labels, including clipped endpoints. Here an interior source box is a full relative-width $x=(\log N)^{-334}$ mesh box whose closure lies on one side of every source, order and descriptor face. A clipped terminal mesh fragment belongs to the boundary contribution. LEN is not applied to it.

The prefix constants can be checked without an unclipped replacement. Put $A_{3}=33109/160000$. Since $t_{1},t_{2}\leq 25/128$ and $\Theta_{c}\geq(1+t_{1})/2$, the first two gaps are at least $\rho(1/2-3(25/128)/2-\eta_{L})=\rho A_{3}$ and $\rho(1/2-5(25/128)/2-\eta_{L})=\rho\,1859/160000$. For the third, $\ell_{c}-t_{3}\geq A_{3}-2t_{3}$ and retention gives $\ell_{c}-t_{3}\geq(2\rho-1)t_{3}$; their maximum is at least $A_{3}/3$. Likewise $\ell_{c}\geq\max\{A_{3}-t_{3},2\rho t_{3}\}\geq 2A_{3}/3>\ell_{0}$. Thus all three prefix WF inequalities have strict room. The whole anatomy infimum is taken before this branch bound is used.

Exact coefficient and labelled factorization (H1–H2).

Write $\mathfrak{b}=(3,c,f,\mathbf{D})$. Give the three prefix occurrences their independent prime sets $[N^{t_{i}},N^{\rho t_{i}})\cap[N^{u},N^{\mu_{*}})\cap\mathcal{P}_{N}$. For $f=2$, append each tail occurrence, independently, with prime set $[D_{j},\min(D_{j}^{\rho},z_{c}))\cap\mathcal{P}_{N}$. For $f=1$ the tail list is empty. In both cases put $J$ equal to this complete labelled list, and set $\lambda_{d}=\lambda_{d}^{-}(\mathbf{D})$ or $\phi_{d}^{-}$, respectively. Thus, exactly,

$$C_{\mathfrak{b}}(q)=\sum_{\begin{subarray}{c}d<D_{c}^{\sigma}\\ d\mid P_{N}(u_{c})\end{subarray}}\lambda_{d}\sum_{(p_{j})_{j\in J}}1_{q=d\prod_{j\in J}p_{j}},\qquad|\lambda_{d}|\leq 1.$$

Neither distinctness mask is inserted here. The orbit divisor $M_{c}$, the source weight $w_{C}$, and the two distinctness corrections remain outside this coefficient, as in (8.30)–(8.32). In particular its supports include repeated-prime moduli, and $(q,2N)=1$. The prime sets are the actual clipped sets, not their unclipped parents. All labelled primes are at least $u_{c}\geq u_{0}$ with $u_{0}=N^{\ell_{0}\sigma^{2}}$, since $\ell_{c}>\ell_{0}$ and $\ell_{c}\sigma^{2}<u$.

For packing use the actual upper exponents of the clipped prime sets. The first three are at most $\rho t_{i}$; all tail upper exponents are at most $t_{3}$. The prefix masks in (8.42) and the tail masks $\sum_{i<j}s_{i}+2s_{j}<\ell_{c}$ show that their merged decreasing list satisfies $\mathrm{WF}(d_{*})$, where $d_{*}=\rho(\Theta_{c}-\eta_{L})$. For a tail rank, this follows after adding at most $\rho(t_{1}+t_{2}+t_{3})$ to its tail inequality. The top three upper exponents are the prefix endpoints, belonging to the closed clipped parent. Hence their anatomy value is at least $\Theta_{c}$. Put $h=d_{*}+60\sigma$, $\delta=30\sigma$ and $g=\log_{N}T-\delta$. Every point $\mathbf{x}$ of the closed clipped parent satisfies $u\leq x_{1}\leq\mu_{*}=25/128$. On this interval the $v_{\rm pk}(x_{1})$ branch of Paper I gives $\widetilde{H}(\mathbf{x})\geq v_{\rm pk}(x_{1})=(1+x_{1})/2$; hence $\Theta_{c}\geq(1+u)/2$. Therefore $h\geq\Theta_{c}-\eta_{L}+60\sigma\geq\frac{1+u}{2}-\eta_{L}>\frac{1}{2}$; also $h<\Theta_{c}\leq 153/256$, since $\Theta_{c}-h\geq\eta_{L}-\tau H_{\max}-60\sigma>0$. Theorem 4.1 of Paper I applies to this same labelled list, with $\delta\leq g\leq 1/3-\delta/2$. Let its partition be $J=J_{1}\sqcup J_{2}\sqcup J_{3}$, with the smooth divisor attached to slot one. Define, using precisely the preceding prime sets,

$$\gamma_{1}(v)=\frac{1}{|J_{1}|!}\sum_{d,(p_{j})_{J_{1}}}\lambda_{d}1_{v=d\prod_{J_{1}}p_{j}},\qquad\gamma_{s}(v)=\frac{1}{|J_{s}|!}\sum_{(p_{j})_{J_{s}}}1_{v=\prod_{J_{s}}p_{j}}\quad(s=2,3).$$

Empty slots two and three mean the atom $1_{v=1}$. Slot one always keeps the $d$ sum, even if $J_{1}$ is empty. The finite tuple map gives

$$C_{\mathfrak{b}}=C_{\rm slot}(\gamma_{1}*\gamma_{2}*\gamma_{3}),\qquad C_{\rm slot}=\prod_{s}|J_{s}|!,\qquad|\gamma_{s}|\leq 1.$$

The primes below $u_{c}$ determine $d$ uniquely. At most $|J_{s}|!$ assignments have a fixed slot product. The count allows repeated primes. The rough coordinate count is less than $3/(5\ell_{0}\sigma^{2})$, so $C_{\rm slot}=O_{\sigma}(1)$ and $|C_{\mathfrak{b}}(q)|/M_{c}\leq M_{\sigma}!$, with $M_{\sigma}=\lceil 3/(5\ell_{0}\sigma^{2})\rceil$. For the tested boxes below, take $C_{\rm ctr}=2$. Uniformly for sufficiently large $N$,

$$\frac{N_{I}}{TL^{-334}}\in[1/2,2],\qquad\frac{M_{I}}{(X/T)L^{-334}}\in[2,20].$$

The same tested box (H3–H5).

Use exactly (8.33) and (8.35): $X=\prod_{i}A_{i}$, $T=\lceil A_{1}\rceil-1$, $U_{J}=\lceil B_{1}\rceil-1$, and $\alpha_{n}=1_{\mathbb{P}}(n)1_{T<n\leq U_{J}}$. No product truncation is added to $\Delta_{B}$. Take $C_{\rm box}=334$; $\log X=\log N+O(1)$ throughout this family. For sufficiently large $N$, $T>2^{50}$ and $U_{J}\leq 2T$, so

$$\alpha_{n}=1_{P^{-}(n)\geq T^{51/100}}1_{T<n\leq U_{J}}.$$

Thus (3.8), option $1$, holds with $\zeta=T^{51/100}\geq z_{0}(X)$. The complementary sequence is the actual $b_{m}\in\{0,1\}$ of (8.33); in particular $|b_{m}|\leq\tau_{23}(m)$. There is no Heath–Brown expansion, logarithm freezing, or proper-power replacement error for this selected direct-prime representation. The selected-factor Siegel–Walfisz bound is Paper I (3.25), uniform in this stronger cutoff and every endpoint and auxiliary coprimality. Its relative-norm alternative is kept as in (3.48). We assume no lower bound on any coprimality-truncated selected norm. In the large-norm case of (3.48), use Lemma 6.5(ii). Its premises (3.1), (3.2), (3.6), (3.8) with LEN, (3.11), (3.25), (3.26) and (3.27) are the data verified in this subsection.

Let $M=\prod_{i=2}^{k}A_{i}$, $M_{+}=\prod_{i=2}^{k}B_{i}$, $M_{I}=M_{+}-M$, $m_{0}=(M+M_{+})/2$, and $\psi_{I}(m)=\phi_{0}((m-m_{0})/M_{I})$, with the fixed bump of Paper I (3.9). Then $\psi_{I}=1$ on the full complementary product interval. Since $5\leq k\leq 11$ and the mesh boxes are full, $M_{I}\asymp ML^{-334}$, $m_{0}\leq 2X/T$, and $N_{I}=U_{J}-T\geq T/(6L^{334})$, where $L=\log N$. For fixed $\sigma$ enlarge the threshold so that $\sigma L\geq 2^{130}(\log L+1)$; then

$$N_{I}\geq(2T)^{1-2^{-126}},\quad X^{60\sigma}\leq T\leq X^{1/3+15\sigma},\quad M_{I}\geq X^{1/2}.$$

These follow from $N^{\alpha}/2\leq T<N^{1/5}$ and $\varepsilon N\leq X\leq N$. The integer convention is $(T,U_{J}]$; no continuous interval length replaces $N_{I}$.

The doubled support lies in $[M-M_{I}/2,M_{+}+M_{I}/2]$. Since $k-1\leq 10$, $(1+x)^{k-1}-1\leq 11x$ eventually. On this entire support and for every selected integer, $mn\leq N_{\rm cut}+25xN$. Consequently

$$1\leq N-mn,\qquad N-mn\geq(L-25)NL^{-334}>0,\qquad|mn-N|\leq N<X^{3}.$$

This verifies nonresonance on the full doubled rectangle, including both Cauchy–Schwarz copies. Checking only where $b_{m}\neq 0$ would not suffice.

After localization, $Q,R,S$ denote the scales of $\gamma_{1},\gamma_{3},\gamma_{2}$, respectively. Theorem 4.1 gives, before fixed dyadic rounding and base conversion, reductions

$$(7527/256,\ 60,\ 114421/1280,\ 26247/256)\sigma$$

in $Q/T$, $T^{2}R^{2}S$, $A_{0}^{7/32}TQR^{2}S^{5}$ and $(TR/S)^{7/16}TQR^{2}S^{5}$, respectively, relative to their bounds $1,N,N^{2},N^{2}$. Here $A_{0}=X^{1+\sigma/20}$. The first slot includes the additional load $\sigma\ell_{c}\leq\sigma H_{\max}$. Each listed reduction strictly exceeds the required losses $29,8,15,15$. Hence fixed dyadic constants and $\log_{N}(N/X)=O(1/L)$ are absorbed for fixed $\sigma$. In particular the exact requirements (3.6)–(3.7) hold:

$$Q<TX^{-29\sigma},\quad N_{I}\geq QX^{\sigma/3},\quad QRS<M_{I}X^{-89\sigma},\quad M_{I}\geq X^{1/2},$$

$$T^{2}R^{2}S<X^{1-8\sigma},\quad A_{0}^{7/32}TQR^{2}S^{5}<X^{2-15\sigma},\quad(TR/S)^{7/16}TQR^{2}S^{5}<X^{2-15\sigma}.$$

For $QRS<M_{I}X^{-89\sigma}$ use the positive fixed gap $4/5-153/256$ and $M_{I}\gg N^{4/5}L^{-334}$. The complete modulus, including $d$ and all labelled primes, obeys

$$q<N^{e_{Q}},\qquad e_{Q}=(\rho+\sigma)(153/256-\eta_{L})<3/5.$$

Because $X\geq\varepsilon N$, impose $(3/5-e_{Q})\log N\geq(3/5)\log(1/\varepsilon)$ to obtain $q\leq X^{3/5}$. Take the moving frequency $a=N$ and $K_{\rm scale}=1$: eventually $0<N\leq X\log N\leq A_{0}$ and every supported modulus is prime to $2a$. This proves H1–H5 with the support relaxation of Paper I Theorem 3.2.

Application and restoration.

Apply Paper I Theorem 3.2, (3.13), to this full unrestricted coefficient and this full-support natural-coprime discrepancy. For every fixed $A_{\rm loc}>0$ it gives exactly

$$|\mathcal{S}_{B}(\mathfrak{b})|\ll_{\sigma,A_{\rm loc}}X(\log X)^{-A_{\rm loc}}+X^{1-\ell_{0}\sigma^{2}/4}(\log X)^{C_{\rm loc}}.$$

Constants are uniform in these boxes and in blocks with $\chi_{c}=1$ after fixing $\sigma$ and source constants. The threshold is ineffective. Theorem 3.1’s smaller abstract support class is not used. The proof includes all local restorations (3.32)–(3.50): with $\epsilon=\ell_{0}\sigma^{2}/8$ and $A_{\rm work}=A_{\rm loc}+4$,

$$XL^{-A_{\rm work}}+XL^{C}z_{0}^{-1/2}+XL^{C}(Q/N_{I})^{1/2}+XX^{\epsilon/2}L^{C}\{u_{0}^{-1/2}+(Q/N_{I})^{1/2}\}.$$

Keep both terms in braces. They come from restoring the five cross-copy gcd failures of (3.30), the close-pair set, and the zero-mode main-term cancellation. The spectral estimate alone does not give them. The $z_{0}$ term joins the arbitrary logarithmic saving, and the remaining terms have savings exceeding $\ell_{0}\sigma^{2}/4$.

Theorem 2.8 enters only at Paper I (3.69)–(3.90). In that inner stage, with $p_{0}=\min\{p:p\nmid 2N\}$, set $r_{T}=z$, $s_{T}=p_{0}$, $n_{T}=p_{0}m$, $a_{T}=\varepsilon_{y}N$, $N_{T}=8p_{0}H_{*}SF$ and $A_{T}=A_{0}$. Both signs are kept. The coefficient is exactly $\beta_{z,m}^{\chi,\varepsilon_{y}}$ of (3.68) on these indices and zero elsewhere. In particular

$$\sum_{n^{\prime\prime}\mid N^{\infty}}(n^{\prime\prime})^{7/32}\|\widetilde{b}(n^{\prime\prime})\|_{2}^{2}\ll X^{o(1)}(H_{*}S)^{2}F,\quad\|\widetilde{b}(n^{\prime\prime})\|_{2}^{2}=\sum_{\begin{subarray}{c}n^{\prime},r,s\\ (n^{\prime},N)=1\end{subarray}}|b_{n^{\prime}n^{\prime\prime},r,s}|^{2},$$

with no coefficients added outside that support. The unweighted norm has the same bound. These are the endpoint norm bounds (3.70)–(3.81). When slot two is a unit, $S=1/2$ and every nonzero-frequency block is empty. Otherwise (3.82)–(3.85) identify the exact phase, permit $p_{0}\mid c$, and use all nine dyadic weights from (8.49), with $\epsilon_{0}=\sigma/2000$ and $\epsilon=\sigma/100$. The full bracket of (3.38), including its moving-frequency and mixed-cusp terms, yields all four quantities in (8.51). No bracket term is discarded. The full bounds (3.87)–(3.90) then control the inner spectral term. After that, the restoration terms above and $C_{\rm slot}$ are put back.

Finally, this estimate concerns $C_{\mathfrak{b}}\Delta_{B}$. The outer identity is still

$$R_{\rm adv}=E_{\rm disp,1}+E_{\rm disp,2}+E_{\rm sq}+E_{\rm nc}.$$

Thus $M_{c}^{-1}$ and $w_{C}$ enter once. The forbidden and allowed-repeat masks remain in $E_{\rm sq}$. The natural-to-$X_{A}$ comparator remains in $E_{\rm nc}$. The source cutoff, peel, boundary and terminal-mesh restorations, capped rows zero to two, and complete-family coefficient counts require their own estimates. This local estimate alone gives neither the complete L3 bound nor the RSW bound.

The outer application of Theorem 3.2 above supplies (8.47). The scoped spectral implications in Section 3 do not by themselves give it.

8.9. The complete box family

The required complete-family bounds are

(8.52) $$\#\mathcal{E}_{\mathrm{bdry}}\ll NL^{-334},\qquad D_{\mathrm{bdry}}\ll_{\sigma}NL^{-331/2},\qquad\sum_{B}|w_{B}|\ll_{\sigma}L^{3689}.$$

There are at most $122$ face families per $k$. Their integer normals have maximum coefficient five. Their coefficient sums are at most eleven.

Proof of the complete-family bounds.

Use the unclipped geometric mesh $[A^{(j)},A^{(j+1)})$, $A^{(j)}=(N^{\alpha}/2)(1+L^{-334})^{j}$, $j\geq 0$, in every coordinate. Boxes with a coordinate $\geq N^{1-4\alpha}$ carry no source mass, since $k\geq 5$ and every prime is at least $N^{\alpha}$. So there are no terminal fragments, and every interior box is a full mesh box. Box count. Each coordinate has at most $((1-\alpha)L+1)(L^{334}+1)+1\leq L^{335}$ mesh intervals for $L\geq 24$. With $k\leq 11$, finitely many cells and $w_{C}\leq 36$, $\sum_{B}|w_{B}|\leq 36|\mathcal{C}_{+}|\sum_{k=5}^{11}L^{335k}\ll L^{3685}\leq L^{3689}$. Boundary count. Let $h^{\prime}=\log(1+L^{-334})/L\leq L^{-335}$. A source integer $e$ in a boundary box has its vector within $h^{\prime}$ in each coordinate of a face. A radial face forces $e$ into $[y(1+L^{-334})^{-11},y(1+L^{-334})^{11}]$ with $y\in\{\varepsilon N,N_{\mathrm{cut}}\}$. There are at most $2(23NL^{-334}+1)$ such $e$. Every other face is $c_{i}x_{i}+c_{j}x_{j}=\theta$ with integers $|c_{i}|,|c_{j}|\leq 5$, $c_{i}\neq 0$ [27, (5.29a)–(5.31) and the facet discussion after (5.40)]. Write $e=p_{i}m^{\prime}$. For fixed $m^{\prime}$, $p_{i}$ lies in an interval of relative width at most $21L^{-334}$ below $N/m^{\prime}$, which contains at most $21L^{-334}N/m^{\prime}+1$ integers. Here $m^{\prime}$ is a product of at most ten primes in $[N^{\alpha},N)$, and $m^{\prime}<N^{1-\alpha}$. By Mertens, $\sum_{m^{\prime}}1/m^{\prime}\leq\exp(\sum_{N^{\alpha}\leq p<N}1/p)\leq e/\alpha<32$ eventually. Hence one face family contributes at most $700NL^{-334}+N^{1-\alpha}$. The number of face families is finite ($\leq 122$ for each $k$). Thus $\#\mathcal{E}_{\mathrm{bdry}}\leq C_{\mathrm{bd}}NL^{-334}$ with an absolute $C_{\mathrm{bd}}$. No numerical value is required below. Boundary discrepancy. Let $A_{\mathrm{bd}}$ be $A$ restricted to boundary boxes. Then $D_{\mathrm{bdry}}=\sum_{\mathfrak{b}}M_{c}^{-1}\langle C_{\mathfrak{b}},\Delta_{P,\mathrm{bd}}\rangle$. By $|C_{\mathfrak{b}}|/M_{c}\leq C_{\mathrm{L3},\sigma}$, $\sum_{q}1_{q\mid N-e}\leq\tau(N-e)$, Cauchy–Schwarz, $\tau^{2}\leq\tau_{4}$, $\sum_{n\leq N}\tau_{4}(n)\leq N(1+L)^{3}$ and $\sum_{q\leq N}1/\varphi(q)\leq 2(1+L)$,

$$|D_{\mathrm{bdry}}|\ll_{\sigma}36(\#\mathcal{E}_{\mathrm{bdry}})^{1/2}(N(1+L)^{3})^{1/2}+36\,\#\mathcal{E}_{\mathrm{bdry}}(1+L)\ll_{\sigma}NL^{-331/2},$$

with finitely many blocks at fixed $\sigma$. This proves (8.52), with the first bound in the form $\#\mathcal{E}_{\mathrm{bdry}}\ll NL^{-334}$. Under (8.47), take $A_{\mathrm{loc}}\geq 3692$. The $L^{3689}$ box count and $W\gg N/L^{2}$ then leave $L^{-1}$. Equations (8.41) and (8.52) then give

(8.53) $$(R_{\mathrm{adv}})_{+}/W=o_{\sigma}(1),$$

provided (8.22) and the coprime-comparator bound hold.

8.10. The principal term and its price

(P2.8d) The source-independent profile comparison.

We use Paper II, (4.70)–(4.75), in the following form. Put

(8.54) $$v_{1}(s) =\min\left\{\frac{1+s}{2},\frac{32-34s}{39},\frac{57}{64}-\frac{3}{2}s\right\},$$

$$v_{*}(s) =v_{1}(s)-u,\qquad\psi_{*}(x,y)=\begin{cases}x,&x\geq y,\\ -\infty,&x<y,\end{cases}$$

$$\Theta_{*}(x,y,z) =\max\{v_{*}(x),v_{*}(y),v_{*}(x+y),v_{*}(x+y+z),$$

$$\hskip 39.83385pt\psi_{*}(v_{*}(x+z),x+2y+z),\psi_{*}(v_{*}(y+z),2x+y+z)\}.$$

The subtraction of $u$ precedes both guards. This is the same guard convention as equation B.14. It agrees with Paper II’s zero guard because an unguarded branch is positive, as explained immediately after that display. Define

(8.55) $$\begin{gathered}H_{\log}=\log(\mu_{*}/u),\qquad a_{3}(\mathbf{t})=\frac{\Theta_{*}(\mathbf{t})-t_{1}-t_{2}-t_{3}}{t_{3}},\\ J_{*}=\int_{u}^{\mu_{*}}\frac{dt_{1}}{t_{1}}\int_{u}^{t_{1}}\frac{dt_{2}}{t_{2}}\int_{u}^{t_{2}}\frac{1-f(a_{3}(\mathbf{t}))}{t_{3}^{2}}\,dt_{3},\\ R_{\mathrm{up}}=500(F(30)-1)+500H_{\log}(1-f(30))+250H_{\log}^{2}(F(30)-1),\\ G_{*}=\frac{1}{\mu_{*}}+R_{\mathrm{up}}+J_{*}.\end{gathered}$$

Extend $f$ by zero below two. These are Paper II, (4.70)–(4.75) [27, (4.70)–(4.75)]. Their source-independent comparison uses

(8.56) $$g=(u-\eta_{L})-\frac{9}{2}\mu_{*}\tau>0.001899>2\mu_{*}\tau.$$

Hence a positive canonical lower term lies in a cell with $\chi_{c}=1$. The capped rows still satisfy $s_{c}>30$. Their residuals are bounded by the same $R_{\mathrm{up}}$. Thus the cap changes neither $G_{*}$ nor its price.

Let $\mathcal{P}_{\Pi}$ be the signed prefix profile:

(8.57) $$\mathcal{P}_{\Pi}={} \frac{1}{u}F(s_{0})-\frac{1}{u}\sum_{c:j=1}\operatorname{vol}_{c}f(s_{c})+\frac{1}{u}\sum_{c:j=2}\operatorname{vol}_{c}F(s_{c})$$

$$-\sum_{c:j=3,\chi_{c}=1}\frac{\operatorname{vol}_{c}}{t_{3}}f(\ell_{c}/t_{3}),\qquad\operatorname{vol}_{c}=\int_{c}\prod_{i}\frac{dt_{i}}{t_{i}}.$$

The required mass and comparison statements are

(8.58) $$\begin{gathered}X_{A}\leq\frac{N}{L}\{(1-\varepsilon)I_{\mathrm{HL}}+O(L^{-1})\},\\ M_{\mathrm{prin}}=X_{A}\frac{2e^{-\gamma}C(N)}{L}(\mathcal{P}_{\Pi}+o_{\sigma}(1)),\qquad\mathcal{P}_{\Pi}\leq G_{*}+o_{\sigma}(1).\end{gathered}$$

The first bound is Paper II, (5.34)–(5.42). The last comparison uses (4.72)–(4.75) [27, (4.72)–(4.75), (5.34)–(5.42)].

Proof of (8.58).

Mass. The cells $\mathcal{C}_{+}$ are Paper II’s cells with $w_{C}\geq 1$, and the radial range $\varepsilon N<e<N$ is Paper II’s. The outer-box majorant $f^{+}$ of Paper II, (5.34), counts labelled factorizations into all primes in $[N^{\alpha},N)$, and $c(\mathbf{j})\geq w_{C}$ for the box $\mathbf{j}$ containing the ordered factor vector of $e$, which lies in $C_{N}$. Hence $A\leq f^{+}$ pointwise directly (no comparison with the intermediate $g$ is used). Paper II, (5.34)–(5.42), then gives $X_{A}\leq X^{+}=\frac{N}{L}\{(1-\varepsilon)I_{\mathrm{HL}}+O(L^{-1})\}$ [27, (5.32), (5.34), (5.42)]. In particular $X_{A}\leq N/L$ eventually, since $I_{\mathrm{HL}}\leq\overline{H}<1$. Principal term. As in Paper II, Lemma B.9, $V_{N}(z)=(1+o(1))2e^{-\gamma}C(N)/\log z$ uniformly for $z\geq N^{h}$, $h=1/1000$ [27, (B.9) and its proof]. The proof of (B.9) gives an exact factorization. Mertens’ theorem gives the asymptotic. The primes $p\mid N$ with $p\geq z$ change the product by $1+O(\omega(N)/z)$. Every actual cutoff is at least $N^{u/\rho^{3}}>N^{h}$ [27, Lemma B.8]. At fixed $\sigma$ the cells are finitely many, with fixed exponent boxes. Mertens gives $\sum_{v\in c}1/\varphi(v)=\operatorname{vol}_{c}+o(1)$, the primes dividing $2N$ costing $O(\omega(N)N^{-u})$. Also $K_{j}(s_{c})$ is bounded. Summing $\nu_{j}M_{j}$ gives $M_{\mathrm{prin}}=X_{A}\frac{2e^{-\gamma}C(N)}{L}(\mathcal{P}_{\Pi}+o_{\sigma}(1))$, with $\log_{N}z_{3,c}=t_{3}$ supplying the factor $1/t_{3}$. Profile comparison. Write $F=1+(F-1)$, $f=1-(1-f)$. On rows $0,1,2$, $s_{c}>30$, and $F$ decreases and $f$ increases. With $\sum_{j=1}\operatorname{vol}_{c}=H_{\log}$ and $\sum_{j=2}\operatorname{vol}_{c}=H_{\log}^{2}/2$, the residuals are at most $R_{\mathrm{up}}$ [27, (4.73)]. For row three, if $f(a_{3}(\mathbf{x}))>0$ at an actual point $\mathbf{x}$ of a cell, then (8.56) gives $\chi_{c}=1$ and $\ell_{c}\geq\Theta_{*}(\mathbf{x})-x_{1}-x_{2}-x_{3}$ [27, (4.72) and the two paragraphs after it]. For $x_{3}\in[t_{3},\rho t_{3}]$, monotonicity of $sf(s)$ at fixed numerator $\ell_{c}$ gives $f(\ell_{c}/t_{3})/t_{3}\geq f(\ell_{c}/x_{3})/x_{3}\geq f(a_{3}(\mathbf{x}))/x_{3}$. Integrating over each retained cell, $\sum_{j=3,\chi_{c}=1}\operatorname{vol}_{c}f(\ell_{c}/t_{3})/t_{3}\geq\int\!\!\int\!\!\int f(a_{3})\,dt_{1}dt_{2}dt_{3}/(t_{1}t_{2}t_{3}^{2})$. The constant parts give $1/\mu_{*}$ by [27, (4.74)]. Hence $\mathcal{P}_{\Pi}\leq 1/\mu_{*}+R_{\mathrm{up}}+J_{*}+o_{\sigma}(1)=G_{*}+o_{\sigma}(1)$, which is [27, (4.75)] for this source.

Use the same coefficient bound as the exterior comparison:

(8.59) $$\begin{gathered}G_{*}\leq G_{0}=\frac{6088440363}{10^{9}},\qquad 2e^{-\gamma}G_{0}\leq c_{K2,G0}=\frac{6836825163875517756286846321650754372931094689}{10^{45}},\\ I_{\mathrm{HL}}\leq\overline{H}=\frac{389253}{5000000}.\end{gathered}$$

The functional bound $G_{*}\leq G_{0}$ is Paper II, [27, (4.73), (4.75), (4.79), (A.10a)], for the identical canonical functional (8.55). Its scalar rounding is the same as (7.36). The mass bound is Paper II, (A.11) [27, (A.11)]. The optional unshifted certificate in the unshifted-comparison subsection of Appendix B is outside this proof.

The constant $c_{\mathrm{O1}}$ is optional, non-certified appendix material outside the L3 proof. In the optional scalar comparisons below, $c$ denotes the rational from (1.6). No functional comparison with $c_{\mathrm{O1}}$ is needed here. Its rational definition is

(8.60) $$c_{\mathrm{O1}}=\frac{\sum_{i=0}^{7}n_{i}B^{i}}{\sum_{i=0}^{7}d_{i}B^{i}},\qquad B=10^{25},$$

with the integers in Table 8.1.

Table 8.1. Exact certificate constant

$i$	$n_{i}$	$d_{i}$
0	4790995369589327356095267	5033118760359274030825472
1	6963256824825772577191322	8266496527677353149933429
2	8996423827217290029713216	9138360880901719297497815
3	4916164251930472277375479	4671424972330288696699300
4	329572634628519073712385	939590108467331323910141
5	9317769528669888650546075	5634060699913223990647374
6	8280830054071768648850166	2330146450712536415611653
7	389762593301325772	57030507720277440

The exact rational comparison of $c$ and $c_{\mathrm{O1}}$ is in Section 9, (9.5). It uses no exponential comparison. It is independent of the coarse enclosure for $U_{0}$. The mass-integral bound is Paper II, (A.11) [27, (A.11)]. Put

(8.61) $$\begin{gathered}m_{\mathrm{sw}}=\frac{e^{-\gamma}}{2}I_{\mathrm{HL}}G_{*},\qquad\Delta_{\mathrm{cert}}=\frac{c_{K2,G0}}{4}\overline{H}-m_{\mathrm{sw}}\geq 0,\\ C_{\mathrm{sw}}=\frac{650000000}{1111}C_{E},\qquad P_{\mathrm{sw}}=\frac{48544888063158}{27775},\\ a_{\mathrm{sw}}=C_{\mathrm{sw}}\sigma,\qquad p_{\mathrm{sw}}=P_{\mathrm{sw}}\sigma^{9}.\end{gathered}$$

The remaining required estimates are

(8.62) $$\begin{gathered}(M_{\mathrm{prin}}/W-m_{\mathrm{sw}})_{+}=o_{\sigma}(1),\\ M_{\mathrm{err}}/W\leq C_{\mathrm{sw}}\sigma+o_{\sigma}(1),\qquad P_{3}/W\leq P_{\mathrm{sw}}\sigma^{9}+o_{\sigma}(1),\\ |E_{\mathrm{nc}}|/W=O_{\sigma}(N^{-\alpha}L^{4}).\end{gathered}$$

The exact numerical multipliers are

(8.63) $$\frac{4000\cdot 585}{4(1-10^{-4})}=\frac{650000000}{1111},\qquad\frac{36\cdot 502002(459/100)^{3}}{1-10^{-4}}=\frac{48544888063158}{27775}.$$

(P2.8c) Density product and prefix sums.

Put $h=1/1000$ and $H_{N}=\sum_{N^{h}\leq p\leq N}(p-1)^{-1}$. The two bounds used in the first multiplier are

(8.64) $$\begin{gathered}V_{N}(z)\leq\frac{4C(N)}{\log z}\leq\frac{4000C(N)}{L}\quad(z\geq N^{h}),\\ \sum_{\text{actual prefixes }v}\frac{1}{\varphi(v)}\leq\sum_{j=0}^{3}H_{N}^{j}\leq 1+8+64+512=585.\end{gathered}$$

These are Paper II, Appendix B, Lemmas B.9 and B.11, (B.9) and (B.11) [27, Appendix B, Lemmas B.9, B.11, (B.9), (B.11)]. Each half-open prefix cell is counted once. Using $X_{A}\leq N/L$ from (P2.8d) now gives

(8.65) $$M_{\mathrm{err}}\leq\frac{X_{A}4000C(N)}{L}\,585\{C_{E}\sigma+o_{\sigma}(1)\}.$$

Division by $W=4(1-\varepsilon)C(N)N/L^{2}$ gives $C_{\mathrm{sw}}$.

(P2.8e) The upper mass and the peel.

The source mass is (8.58), Paper II, (5.34)–(5.42). The native lower-sieve cutoff uses the finite peel (8.11) and the estimate

(8.66) $$\frac{P_{3}}{W}\leq\frac{36\cdot 502002(459/100)^{3}}{1-\varepsilon}\tau+o_{\sigma}(1).$$

For the uniform two-form calculation, put $m=vr$ and $w=\min(z_{c},N^{\alpha})$. The required ranges are

(8.67) $$\begin{gathered}\log_{N}m\leq 153/256+2\tau,\qquad\log_{N}w\geq 2/1001,\\ w^{2}m\leq N^{253/256+(153/64)\tau}<N^{1-0.0116}.\end{gathered}$$

For each sieve prime $p<w$, every prime factor of $m=vr$ is at least $z_{c}\geq w$, so $p\nmid m$. The local two-form root count is then $\nu_{N}(p)=1$ for $p\mid N$, and $\nu_{N}(p)=2$ otherwise. For squarefree $d$, define $g_{N}(d)$ multiplicatively by $g_{N}(p)=\nu_{N}(p)/(p-\nu_{N}(p))$. The denominator and one-layer bound are

(8.68) $$\begin{gathered}\mathcal{G}_{N}(w)=\sum_{\begin{subarray}{c}d\leq w\\ d\mid P(w)\end{subarray}}g_{N}(d)=\frac{\log^{2}w}{4C(N)}(1+o(1)),\\ S^{(1)}_{m}(w)\leq\frac{N/m}{\mathcal{G}_{N}(w)}+O(w^{2}\log^{5}w)\leq 502002\,\frac{2C(N)N}{mL^{2}}.\end{gathered}$$

Here $P(w)$ is the product of all primes below $w$. The last step uses $2.001/(2/1001)^{2}=501251.00025<502002$. The exponent saving in (8.67) absorbs its remainder. There are at most $36$ source layers. For the parent-bin prime sums $M_{h}$, the needed estimates are

(8.69) $$M_{h}=\sum_{p\ \mathrm{in\ bin}\ h}\frac{1}{p}\leq 2\tau,\qquad\sum_{h}M_{h}\leq\frac{459}{100},\qquad\sum_{a_{1},a_{2},a_{3},r}\frac{1}{a_{1}a_{2}a_{3}r}\leq 2\left(\frac{459}{100}\right)^{3}\tau.$$

The pair $a_{3},r$ belongs to one parent bin. Multiplication of (8.68) by $36$ and this sum, then division by $W$, gives (8.66). These are the exact steps of Paper II, (4.45)–(4.56), with the layer ceiling changed from three to $36$ [27, (4.45)–(4.56), (5.34)–(5.42)]. The uniform two-form estimates are proved here under the stated ranges.

Proof of (8.62).

Principal. Since $\mathcal{P}_{\Pi}\leq G_{*}+o_{\sigma}(1)$ and $G_{*}>0$, the mass bound gives $M_{\mathrm{prin}}/W\leq\frac{(1-\varepsilon)I_{\mathrm{HL}}2e^{-\gamma}(G_{*}+o(1))}{4(1-\varepsilon)}=m_{\mathrm{sw}}+o_{\sigma}(1)$. Secondary sieve. $E_{s,c}\leq C_{E}\sigma+o_{\sigma}(1)$ uniformly, since $\ell_{c}\geq\ell_{1}$. Lemma B.9, Lemma B.11 and $X_{A}\leq N/L$ give (8.65). Every prefix lies in one row, so the four depths together contribute at most $585$ [27, (B.9), (B.11)]. Peel. For $m=vr$ in $P_{3}$, all primes of $m$ are at least $z_{c}$, $(m,N)=1$ and $z_{c}\in[N^{u/\rho^{3}},N^{\rho\mu_{*}}]$. Since $\chi_{c}=1$, $\log_{N}m\leq\rho(t_{1}+t_{2}+2t_{3})\leq\rho(153/256-\eta_{L})<153/256+2\tau$ [27, (4.55)]. Write $A=\sum_{h=1}^{36}1_{\{A\geq h\}}$. Each layer consists of $N^{\alpha}$-rough $e$. With $N-e=mk$, both $k$ and $N-mk$ are $w$-rough, $w=\min(z_{c},N^{\alpha})$. A prime $p<w$ dividing $k$ and $N$ would divide $e$, and $k$ is odd. The two-form Selberg bound [27, (4.48)–(4.54)] gives (8.68) for each layer, with remainder absorbed by (8.67). The prime sums (8.69) are [27, (4.56)], since $r$ and $a_{3}$ lie in one parent bin. Multiplying by $36$ gives (8.66). Comparator. For squarefree $q$, $|d_{A}(q)|=\varphi(q)^{-1}\sum_{(e,q)>1}A(e)$. A common prime $\ell$ of $e$ and $q$ is at least $N^{\alpha}$, so with $q=\ell r$,

$$\sum_{q\ \mathrm{sf}}|O_{\mathfrak{b}}(q)d_{A}(q)|\leq C_{\mathrm{L3},\sigma}\sum_{\ell\geq N^{\alpha}}\frac{36N}{\ell(\ell-1)}\sum_{r\leq N}\frac{1}{\varphi(r)}\ll_{\sigma}N^{1-\alpha}L.$$

Summing finitely many blocks and dividing by $W$ gives $|E_{\mathrm{nc}}|/W=O_{\sigma}(N^{-\alpha}L^{3})$, which implies the stated $O_{\sigma}(N^{-\alpha}L^{4})$. This is the coprime-comparator bound used in (8.53).

8.11. The finite signed identity and its consequence

(P2.8a) Finite signed identities.

For the finite source (8.3) and stated sieve data, combine (8.7), (8.12) and (8.29). Their conditional signed identity is

(8.70) $$S_{\mathrm{sw}}=E_{\mathrm{src}}+P_{3}+M_{\mathrm{prin}}+M_{\mathrm{err}}+R_{\mathrm{adv}}-(Z_{\mathrm{src}}+Z_{\mathrm{AA}}+Z_{\mathrm{peel}}).$$

Define

(8.71) $$\delta_{P} =P_{3}/W-p_{\mathrm{sw}},\qquad\delta_{A}=M_{\mathrm{err}}/W-a_{\mathrm{sw}},\qquad\delta_{M}=M_{\mathrm{prin}}/W-m_{\mathrm{sw}},$$

$$R_{\mathrm{sw}} =E_{\mathrm{src}}/W+\delta_{P}+\delta_{A}+\delta_{M}+R_{\mathrm{adv}}/W,$$

$$V_{\mathrm{sw}} =(R_{\mathrm{sw}})_{+}\leq E_{\mathrm{src}}/W+(\delta_{P})_{+}+(\delta_{A})_{+}+(\delta_{M})_{+}+(R_{\mathrm{adv}})_{+}/W.$$

Under the stated sieve hypotheses, all three $Z$ terms are nonnegative. Consequently

(8.72) $$-\frac{S_{\mathrm{sw}}}{W} =-m_{\mathrm{sw}}-a_{\mathrm{sw}}-p_{\mathrm{sw}}-R_{\mathrm{sw}}+\frac{Z_{\mathrm{src}}+Z_{\mathrm{AA}}+Z_{\mathrm{peel}}}{W}$$

$$\geq-\frac{c_{K2,G0}}{4}\overline{H}-C_{\mathrm{sw}}\sigma-P_{\mathrm{sw}}\sigma^{9}-V_{\mathrm{sw}}.$$

The favorable $Z_{\mathrm{peel}}$ is used once. Every adverse remainder remains in (8.71).

Assume all estimates stated above for the same $A$. Then (8.8), (8.53) and (8.62) imply

(8.73) $$V_{\mathrm{sw}}=o_{\sigma}(1).$$

Proof of (8.73).

All estimates above are proved for the same $A$, the same cells and the same blocks. By (8.8), $E_{\mathrm{src}}/W\ll L^{-331}$. By (8.62), $(\delta_{M})_{+},(\delta_{A})_{+},(\delta_{P})_{+}$ are $o_{\sigma}(1)$. For $R_{\mathrm{adv}}$, (8.32) and (8.36) give

$$\frac{|R_{\mathrm{adv}}|}{W}\leq\frac{|D^{(0)}_{\mathrm{int}}|+|D^{(1)}_{\mathrm{int}}|+|D^{(2)}_{\mathrm{int}}|+|D^{(3)}_{\mathrm{int}}|+|D_{\mathrm{bdry}}|+|E_{\mathrm{sq}}|+|E_{\mathrm{nc}}|}{W}.$$

The terms are $O_{\sigma}(L^{-1})$ by (8.41); $O_{\sigma}(L^{-1})$ by (8.47) with $A_{\mathrm{loc}}=3692$, $|w_{C}|/M_{c}\leq 36$ and $\sum_{B}|w_{B}|\leq L^{3689}$; $O_{\sigma}(L^{-327/2})$ by (8.52); and $O_{\sigma}(N^{-\ell_{1}\sigma^{2}}L^{4})$, $O_{\sigma}(N^{-\alpha}L^{4})$. Hence $V_{\mathrm{sw}}=o_{\sigma}(1)$. After fixing $\sigma_{*}$, choose the eventual threshold so that

(8.74) $$V_{\mathrm{sw}}\leq\frac{1}{200000}.$$

That threshold may be ineffective. It also covers

(8.75) $$\begin{gathered}N_{\mathrm{dom}}=2\left\lceil\frac{1}{2}\exp\!\left(\max\{10^{4},2\log 2/(\ell_{1}\sigma_{*}^{2})\}\right)\right\rceil,\\ N_{z,0}=2\left\lceil\frac{1}{2}\exp\!\left(2\exp((\ell_{1}\sigma_{*}^{2})^{-1/3})\right)\right\rceil,\qquad N_{\mathrm{LEN,app}}(\sigma_{*}).\end{gathered}$$

These structural thresholds alone do not imply (8.74).

Relative to the comparison $-S_{\mathrm{sw}}/W\geq-2\overline{H}$, the constant difference is

(8.76) $$\Delta D_{L3} =\left(2-\frac{c_{K2,G0}}{4}\right)\overline{H}$$

$$=0.022638464724298154340603\ldots.$$

The exact cost of using the common coefficient instead of the optional coefficient is

$$\frac{\overline{H}}{4}(c_{K2,G0}-c)=0.000049489563023753677508253183146688036059673516909.$$

This is scalar arithmetic. Its use in the final count requires (8.72) and the common error bounds. The capped-row application is (8.40)–(8.41). The row-three estimate is (8.47), proved in the row-three subsection. All errors use the common source (8.3). The inputs are Paper I, Theorem 2.8; Fact 3.3; Lemma 6.5(ii) for (3.48); and the prime estimates of Fact 3.1. Hence (8.72) holds at the threshold of (8.74). The constant (8.76) is a gross coefficient difference. It enters the final count only together with the switched error $C_{\mathrm{sw}}\sigma_{*}+P_{\mathrm{sw}}\sigma_{*}^{9}+V_{\mathrm{sw}}$ of (8.72). In the final count this error is bounded by the caps (9.14) ($C_{\mathrm{sw}}\sigma_{*}\leq 10^{-5}$, $P_{\mathrm{sw}}\sigma_{*}^{9}\leq 10^{-5}$) and by (8.74), in total at most $1/40000$.

9. The final bound

We keep the notation of Sections 1 and 2. All error terms below are nonnegative. The analytic inputs are (A1)–(A5) of Table 1.2.

9.1. Addition of the comparisons

Put $\delta_{\mathrm{err}}=10^{-4}$. The constant of (1.14) is

(9.1) $$D_{B}={} D_{\mathrm{II}}+\frac{5719}{10^{6}}+M_{5,\mathrm{internal}}+g_{\mathrm{RA,lo}}$$

$$-\delta_{\mathrm{err}}+\left(2-\frac{c_{K2,G0}}{4}\right)\overline{H},\qquad\overline{H}=\frac{389253}{5000000}.$$

Here $D_{\mathrm{II}}$ is Paper II’s original lower constant [27, (6.3a)]. The six endpoint gains $g_{6}$ of (5.18) are not used. The two profile gains enter separately. In particular,

(9.2) $$D_{\mathrm{II}} =\frac{45849916338009170593}{25000000000000000000},$$

$$M_{5,\mathrm{internal}} =-K_{\mathrm{hi}}\frac{6073714261}{10^{9}}-L_{G1},$$

$$g_{\mathrm{RA,lo}} =\frac{989598353}{10^{10}}=0.0989598353.$$

The last line is the conservative rational floor for the integral gain $\mathfrak{g}_{\mathrm{RA}}$. It is not a central estimate. Section 6 separates it from the integral gain. The certified integral enclosure is $[0.0989598353353,0.0989598353355]$. We take $K_{\mathrm{hi}}$ and $L_{G1}$ as defined in [27, (A.10d), (4.84)]. Their exact values are

(9.3) $$K_{\mathrm{hi}} =\frac{3989480791070578466390443009830933753748442947478110112223}{10^{58}},$$

$$L_{G1} =-\frac{243950428265194656254206921939947762868871361874409552095097}{100433627766186892221372630771322662657637687111424552206336}.$$

Thus $M_{5,\mathrm{internal}}=M_{n}/M_{d}$ exactly, where

(9.4) $$M_{n}=\sum_{i=0}^{5}m_{i}10^{20i},\qquad M_{d}=\sum_{i=0}^{5}d_{i}10^{20i}.$$

The coefficients are printed in Table 9.1.

Table 9.1. Exact shifted-profile gain

$i$	$m_{i}$	$d_{i}$
0	66922148532305907989	0
1	34893223604885136721	0
2	3973350097562772509	0
3	87266878234046404033	48635364229120000000
4	28374567121249055735	73384187692692674921
5	3998	680564

The exact rational $c_{\mathrm{O1}}$ is defined in (8.60) and Table 8.1. It records only the coefficient from the historical, non-certified sharper comparison. The main count uses $c_{K2,G0}$ in both comparisons. The common functional comparison is (8.59). For comparison with the optional sharper coefficient, exact rational arithmetic gives

(9.5) $$0<c-c_{\mathrm{O1}}<4.081\cdot 10^{-46},\qquad 0<\frac{c-c_{\mathrm{O1}}}{4}\overline{H}<7.943\cdot 10^{-48}.$$

This is a rational comparison only. It gives no bound for $2e^{-\gamma}U_{0}$. The further cost of the common coefficient relative to $c$ is exactly

$$\frac{c_{K2,G0}-c}{4}\overline{H}=0.000049489563023753677508253183146688036059673516909.$$

The resulting main switched gain is exactly

$$\Delta D_{L3}=0.02263846472429815434060381043792394540367262995113415.$$

The optional sharper comparisons are not required for (9.1). Each gain is proved on the original arithmetic source of its comparison. The gains $5719/10^{6}$ and $M_{5,\mathrm{internal}}$ come from Proposition 5.10, through Propositions B.1 and B.2. The gain $g_{\mathrm{RA,lo}}$ comes from Proposition 6.13, and $\Delta D_{L3}$ from (8.72). Table 9.4 lists the bindings.

Table 9.2 itemizes the constant. Its decimal column gives approximations only. The defining formulas determine the exact sum.

Table 9.2. The six terms in $D_{B}$

Term	Value
$D_{\mathrm{II}}$	$1.83399665352036682372$
$5719/10^{6}$	$0.005719$
$M_{5,\mathrm{internal}}$	$0.005874949945026357\ldots$
$g_{\mathrm{RA,lo}}$	$0.0989598353$
$-\delta_{\mathrm{err}}$	$-0.0001$
$\Delta D_{L3}=(2-c_{K2,G0}/4)\overline{H}$	$0.022638464724298154\ldots$
$D_{B}$	$1.967088903489691335\ldots$

Adding (1.10) and (1.13) cancels the sixth endpoint:

(9.6) $$\overline{J}_{6}-\overline{J}_{6}=0,\qquad 2\overline{H}-\frac{c_{K2,G0}}{4}\overline{H}=\left(2-\frac{c_{K2,G0}}{4}\right)\overline{H}.$$

Thus their conditional consequence is

(9.7) $$\frac{4G_{b}(N)}{W(N)}\geq D_{B}+\delta_{\mathrm{err}}-c_{K2,G0}I(b)-\mathcal{E}_{B}(N;b,\sigma).$$

This uses the complete $S_{2}$ of Paper II. The finite identities keep their original signs and multiplicities. The fixed errors in Paper II are subtracted once [27, (6.4)–(6.6)], and not again in (9.1).

9.2. Ten error allocations

Use the exterior error from (9.8). For the table below, a conservative majorant is convenient:

(9.8) $$E_{K}(N;b,\sigma) =A_{K}(b)C_{\mathrm{AA}}\sigma+\frac{3884000000000}{9999}\sigma^{9}+r_{K}(N;b,\sigma)$$

$$\leq 4A_{K}(b)C_{\mathrm{AA}}\sigma+\frac{3884000000000}{9999}\sigma^{9}+r_{K}(N;b,\sigma),$$

$$A_{K}(b) =2e^{-\gamma}I(b)W_{\mathrm{AA},K},\qquad P_{K}=\frac{3884000000000}{9999},$$

$$r_{K}(N;b,\sigma) =o_{b,\sigma}(1).$$

Here $W_{\mathrm{AA},K}$ is the exterior remainder-weight bound. It belongs to the source in Section 7. The extra factor four only enlarges this nonnegative error. It leaves the principal constant unchanged. Similarly, the switched comparison requires

(9.9) $$E_{\mathrm{sw}}(N;\sigma) \leq\frac{650000000}{1111}C_{E}\sigma+\frac{48544888063158}{27775}\sigma^{9}+V_{\mathrm{sw}}(N;\sigma),$$

$$V_{\mathrm{sw}}(N;\sigma) =o_{\sigma}(1).$$

Here $V_{\mathrm{sw}}$ is the positive part in (8.71). The conditional limit is (8.73). These two limits concern the actual sums stated there. Both majorants are the error terms of (7.67) under (A3) and of (8.72) under (A4), with $C_{\mathrm{sw}}$ from (8.61). By (A5), $C_{\mathrm{AA}}$ and $C_{E}$ do not depend on $\sigma$. The exterior weight $W_{\mathrm{AA},K}$ is $W_{\mathrm{AA}}$ of (7.38):

(9.10) $$\begin{gathered}A_{K}(b)=2e^{-\gamma}I(b)W_{\mathrm{AA},K},\qquad W_{\mathrm{AA},K}=500\Bigl{(}1+H_{\log}+\frac{H_{\log}^{2}}{2}+\frac{H_{\log}^{3}}{6}\Bigr{)},\\ H_{\log}=\log\frac{\mu_{*}}{u}=\log\frac{3125}{32}>0.\end{gathered}$$

Since $I(b)>0$, this gives $0<A_{K}(b)<\infty$. In particular $A_{K}(b)C_{\mathrm{AA}}\sigma\leq 4A_{K}(b)C_{\mathrm{AA}}\sigma$.

The retained and Root–A comparisons require

(9.11) $$E^{\prime}(N;\sigma)=o_{\sigma}(1),\qquad E_{\mathrm{RA}}(N;\sigma)=o_{\sigma}(1).$$

For Root–A, use $E_{\mathrm{RA}}(N;\sigma)$ from (6.72). The first includes every conversion and the saved-array contribution $\sum_{j\in\mathcal{H}}|\Lambda^{0}_{j,N}|$ in (1.12); that contribution is not charged again. The pointwise error may use the coarser bound

(9.12) $$E_{\mathrm{exc}}(N)=\frac{4E_{\mathrm{pt}}(N)}{W(N)}\leq\frac{2197N^{1-\alpha}}{W(N)}=o(1).$$

Here $E_{\mathrm{pt}}$ is defined in (4.12). The explicit pointwise argument in Section 4 gives $E_{\mathrm{pt}}=0$ for $N\geq N_{\mathrm{pt}}$. The coarser majorant retains the same error bound. The inequality follows from (4.13), using both the pointwise bound and the exceptional-set cardinality. Since $C(N)\geq C_{2}>0$ and $\alpha>0$, the coarser majorant tends to zero: $N^{-\alpha}(\log N)^{2}\to 0$. No sharper exceptional coefficient is needed. Proposition 5.10 gives $E^{\prime}(N;\sigma)\to 0$, and Proposition 6.13 gives $E_{\mathrm{RA}}(N;\sigma)\to 0$ under (A2), each at fixed $\sigma$. Lemma 4.1 gives $E_{\mathrm{exc}}(N)=0$ for $N\geq N_{\mathrm{pt}}$.

Fix $b$, all absolute constants, and the source geometry first. Choose positive absolute constants $C_{E}^{\mathrm{L3}}$ and $C_{E}^{\mathrm{II}}$ for the Section 8 and Paper II sieve bounds. Define

(9.13) $$C_{E}:=\max\{C_{E}^{\mathrm{L3}},C_{E}^{\mathrm{II}}\}.$$

By (A5), these constants are independent of $\sigma$ and $N$. Here $C_{\mathrm{AA}}=4C_{\mathrm{AA37}}$ by (7.39). The constant $C_{E}$ dominates the absolute constant of (3.22) used in (8.25). The underlying sieve bounds come from Paper I, Theorem 2.6, (2.31) [26, Theorem 2.6, (2.31)]. Choose one rational $\sigma_{*}>0$ satisfying

(9.14) $$0<\sigma_{*} \leq 10^{-10},$$

$$\sigma_{*} \leq\frac{713332}{625\cdot 10^{21}C_{E}},$$

$$\sigma_{*} \leq\frac{1}{400000A_{K}(b)C_{\mathrm{AA}}},$$

$$\sigma_{*} \leq\left(\frac{9999}{388400000000000000}\right)^{1/9},$$

$$\sigma_{*} \leq\frac{1111}{65000000000000C_{E}},$$

$$\sigma_{*} \leq\left(\frac{27775}{4854488806315800000}\right)^{1/9},$$

$$\tau_{*} =(\sigma_{*})^{9}.$$

By (9.10) and (A5), every right side in (9.14) is positive and independent of $\sigma$. Hence such a rational $\sigma_{*}$ exists. The second restriction preserves Paper II’s fixed error bound. It is exactly

(9.15) $$\sigma_{*}\leq\frac{4g_{\mathrm{AA}}}{C_{E}W_{*}},\qquad g_{\mathrm{AA}}=\frac{178333}{6250000000},\qquad W_{*}=10^{14}\quad\text{\cite[cite]{[\@@bibref{}{PaperII}{}{}, (1.26), (1.27), (6.6)]}}.$$

The other four restrictions control new fixed errors. Freeze every finite partition and logarithmic precision next. Only then choose the common threshold.

The ten allocations are printed in Table 9.3. The two zero rows are unused capacity.

Table 9.3. The error bound $\delta_{\mathrm{err}}=10^{-4}$

	Error term	Allocation
1	$4A_{K}(b)C_{\mathrm{AA}}\sigma_{*}$	$\delta_{\mathrm{err}}/10$
2	$\tfrac{3884000000000}{9999}\tau_{*}$	$\delta_{\mathrm{err}}/10$
3	$\tfrac{650000000}{1111}C_{E}\sigma_{*}$	$\delta_{\mathrm{err}}/10$
4	$\tfrac{48544888063158}{27775}\tau_{*}$	$\delta_{\mathrm{err}}/10$
5	$E^{\prime}(N;\sigma_{*})+V_{\mathrm{sw}}(N;\sigma_{*})$	$\tfrac{\delta_{\mathrm{err}}}{20}+\tfrac{\delta_{\mathrm{err}}}{20}$
6	$0$	$\delta_{\mathrm{err}}/10$
7	$E_{\mathrm{RA}}(N;\sigma_{*})$	$\delta_{\mathrm{err}}/10$
8	$r_{K}(N;b,\sigma_{*})$	$\delta_{\mathrm{err}}/10$
9	$E_{\mathrm{exc}}(N)$	$\delta_{\mathrm{err}}/10$
10	$0$	$\delta_{\mathrm{err}}/10$

Its capacities sum exactly:

(9.16) $$4\frac{\delta_{\mathrm{err}}}{10}+\left(\frac{\delta_{\mathrm{err}}}{20}+\frac{\delta_{\mathrm{err}}}{20}\right)+5\frac{\delta_{\mathrm{err}}}{10}=\delta_{\mathrm{err}}.$$

The first four bounds follow from (9.14). The five remaining nonzero bounds follow from the stated limits. Each limit is taken at fixed $\sigma_{*}$.

Proposition 9.1 (Error budget).

Assume (A1)–(A5), and let $\sigma_{*}$ be rational with (9.14). Then every even $N\geq N_{0}^{*}(\sigma_{*})$ satisfies

(9.17) $$\mathcal{E}_{B}(N;b,\sigma_{*})\leq\frac{8}{10}\delta_{\mathrm{err}}<\delta_{\mathrm{err}}.$$

Proof.

Rows 1 and 3. The third and fifth lines of (9.14) give

$$4A_{K}(b)C_{\mathrm{AA}}\sigma_{*}\leq\frac{4}{400000}=\frac{\delta_{\mathrm{err}}}{10},\qquad\frac{650000000}{1111}C_{E}\sigma_{*}\leq\frac{650000000}{65000000000000}=\frac{\delta_{\mathrm{err}}}{10}.$$

Rows 2 and 4. The first line gives $\tau_{*}\leq 10^{-90}$. This is below $9999/388400000000000000$ and $27775/4854488806315800000$, so the fourth and sixth lines hold, and

$$\frac{3884000000000}{9999}\tau_{*}\leq\frac{\delta_{\mathrm{err}}}{10},\qquad\frac{48544888063158}{27775}\tau_{*}\leq\frac{\delta_{\mathrm{err}}}{10}.$$

Rows 5, 7 and 8. At the fixed $\sigma_{*}$, $E^{\prime}\to 0$ by (A1), $V_{\mathrm{sw}}\to 0$ by (8.73) under (A4), $E_{\mathrm{RA}}\to 0$ by (A2), and $r_{K}\to 0$ by (A3). By (2.6), the thresholds $T_{E^{\prime}},T_{V},T_{E_{\mathrm{RA}}},T_{r},T_{E_{\mathrm{exc}}}$ of (2.5) exist, and above them $E^{\prime}\leq\delta_{\mathrm{err}}/20$, $V_{\mathrm{sw}}\leq\delta_{\mathrm{err}}/20$, $E_{\mathrm{RA}}\leq\delta_{\mathrm{err}}/10$ and $r_{K}\leq\delta_{\mathrm{err}}/10$. Row 9. Lemma 4.1 gives $E_{\mathrm{exc}}(N)=0$ for $N\geq N_{\mathrm{pt}}$, so $T_{E_{\mathrm{exc}}}\leq N_{\mathrm{pt}}$. Rows 6 and 10 are zero. Every threshold used is bounded by the common maximum in (2.5). Also $N_{\mathrm{pt}}<N_{\mathrm{tr}}$ by Lemma 9.2. Hence all of them hold for $N\geq N_{0}^{*}(\sigma_{*})$. By (1.11), (9.8) and (9.9),

$$\mathcal{E}_{B}\leq{} \bigl{(}E^{\prime}+V_{\mathrm{sw}}\bigr{)}+E_{\mathrm{RA}}+\Bigl{(}\frac{650000000}{1111}C_{E}\sigma_{*}+\frac{48544888063158}{27775}\tau_{*}\Bigr{)}$$

$$+\Bigl{(}4A_{K}C_{\mathrm{AA}}\sigma_{*}+\frac{3884000000000}{9999}\tau_{*}+r_{K}\Bigr{)}+E_{\mathrm{exc}}.$$

The bounds above give at most $\frac{\delta_{\mathrm{err}}}{10}+\frac{\delta_{\mathrm{err}}}{10}+\frac{2\delta_{\mathrm{err}}}{10}+\frac{3\delta_{\mathrm{err}}}{10}+0$, which is below the nonzero allocations $\frac{8\delta_{\mathrm{err}}}{10}$ of Table 9.3. ∎

Take the named source, application and error thresholds listed in Section 2. The Section 7 definitions of $T_{K}$ and $T_{r}$ already include the main-mass threshold $N_{\mathrm{mass}}(b,\sigma_{*})$ and the AA threshold $N_{\mathrm{AA}}(b,\sigma_{*})$. Thus the common maximum in $\mathcal{T}_{*}$ below includes both requirements at the fixed $b,\sigma_{*}$. The common threshold is the single definition

(9.18) $$N_{0}^{*}(\sigma_{*})\quad\text{as defined in \eqref{eq:not-common-threshold}}.$$

Its existence depends on all those estimates. The floor alone supplies none of them. By its definition and Lemma 9.2, $N_{0}^{*}(\sigma_{*})\geq\max\{N_{\mathrm{tr}},N_{6}\}>N_{\mathrm{pt}}$. The Section 2 threshold definitions include $N_{\mathrm{pt}}$ through $T_{\mathrm{src}}$ and $N_{6}$ through $T_{\mathrm{RA}}$. The bound $N_{\mathrm{tr}}$ enters directly in the common maximum. Lemma 4.1 holds for $N\geq N_{\mathrm{pt}}$ and Lemma 4.2 for $N\geq N_{6}$. Thus the order is

(9.19) $$\text{constants}\ \longrightarrow\ \sigma_{*}\ \longrightarrow\ \text{finite choices}\ \longrightarrow\ N_{0}^{*}(\sigma_{*})\ \longrightarrow\ N.$$

No parameter $\sigma_{*}(N)$ is used.

Proposition 9.1 gives

(9.20) $$\mathcal{E}_{B}(N;b,\sigma_{*})\leq\delta_{\mathrm{err}}\qquad(N\geq N_{0}^{*}(\sigma_{*}),\ N\in 2\mathbb{Z}).$$

Combining (9.7) and (9.20) gives

(9.21) $$\boxed{\displaystyle\frac{4G_{b}(N)}{W(N)}\geq D_{B}-c_{K2,G0}I(b).}$$

It holds for every even $N\geq N_{0}^{*}(\sigma_{*})$ under (A1)–(A5).

9.3. The numerical margin

All constants in (9.1) have exact definitions. The exterior price uses the G0 coefficient in (7.36). The same coefficient prices L3. No $G_{*},U_{0}$ sharper domination is used in this arithmetic or in the main-count hypotheses. The constants have the following outward enclosures:

(9.22) $$\begin{gathered}1.967088903489691335582703173494<D_{B},\\ D_{B}<1.967088903489691335582703173495,\\[2.0pt] 1.966859949178424046530804965343<c_{K2,G0}I(b),\\ c_{K2,G0}I(b)<1.966859949178424046530804965344,\\[2.0pt] 0.000228954311267289051898208151<D_{B}-c_{K2,G0}I(b),\\ D_{B}-c_{K2,G0}I(b)<0.000228954311267289051898208152.\end{gathered}$$

Here

(9.23) $$c_{K2,G0}=\frac{6836825163875517756286846321650754372931094689}{10^{45}},\qquad I(b)=\log\frac{4000007}{2999993}.$$

The coefficient is defined in (7.36). The final comparison is (7.67). Hence

(9.24) $$D_{B}-c_{K2,G0}I(b)>0.0002289>\frac{1}{5000}>0.$$

The table below identifies each exact constant by its defining display and lists the comparisons that use it unchanged.

Table 9.4. Constants and the comparisons that use them

Constant	Definition	Comparison
$D_{\mathrm{II}}$	(1.8), [27, (6.3a)]	Proposition 5.10, by (5.13)
$5719/10^{6}$	(5.25)	Proposition 5.10, by Proposition B.1
$M_{5,\mathrm{internal}}$, $K_{\mathrm{hi}}$, $L_{G1}$	(9.3), Table 9.1	Proposition 5.10, by Proposition B.2
$g_{\mathrm{RA,lo}}$	(6.70)	Proposition 6.13, (6.72)
$\overline{H}$, $c_{K2,G0}$	(9.1), (7.36)	(8.72), with (8.61)
$c_{K2,G0}$, $I(b)$	(9.23)	Proposition 7.5, (7.67)
$A_{K}(b)$, $C_{\mathrm{AA}}$, $P_{K}$	(9.10), (7.39), (7.67)	Table 9.3, rows 1 and 2
$C_{\mathrm{sw}}$, $P_{\mathrm{sw}}$	(8.61)	Table 9.3, rows 3 and 4
$\delta_{\mathrm{err}}$	(9.1)	Proposition 9.1

No other constant enters $D_{B}$ or the error budget.

The logarithm enclosure uses rational arithmetic. Put $t_{\log}=500007/3500000$ and

(9.25) $$S_{m}=2\sum_{k=0}^{m}\frac{t_{\log}^{2k+1}}{2k+1},\qquad S_{m}<I(b)<S_{m}+\frac{2t_{\log}^{2m+3}}{(2m+3)(1-t_{\log}^{2})}.$$

Here $I(b)=2\operatorname{arctanh}t_{\log}$ and $0<t_{\log}<1$. Bounding each remaining denominator by $2m+3$ proves the upper bound. Taking $m=24$ gives (9.22). Thus no rounded logarithm enters the margin test.

For reference, put $\zeta=10^{-6}$ and $\mathfrak{E}_{*}=\exp(D_{B}/c_{K2,G0})$. The buffered critical exponent is

(9.26) $$a_{\mathrm{buf}}^{*}=\frac{\mathfrak{E}_{*}+1}{(1-\zeta)\mathfrak{E}_{*}-\zeta}.$$

To derive it, write $b_{*}=1/(1+\mathfrak{E}_{*})+\zeta$. The equation $a(1-b_{*})=1$ gives (9.26). For the present constants,

(9.27) $$1.749974884137892433433207484622<a_{\mathrm{buf}}^{*}<1.749974884137892433433207484623<\frac{7}{4}.$$

This enclosure also uses rational bounds. For $x=D_{B}/c_{K2,G0}$ and $m=36$, put

(9.28) $$T_{m}=\sum_{k=0}^{m}\frac{x^{k}}{k!},\qquad T_{m}<e^{x}<T_{m}+\frac{x^{m+1}}{(m+1)!}\frac{1}{1-x/(m+2)}.$$

Here $0<x<1$. The tail ratio is at most $x/(m+2)$. The function in (9.26) decreases with $\mathfrak{E}_{*}$. Substitution of the two rational bounds proves (9.27).

The endpoint in (9.27) is equivalent to the positive scalar margin at this fixed $b$. It supplies no theorem at a smaller $a$ without new comparisons at the corresponding cutoff. If the optional Lemma 5.3 is used, $D_{B}$ increases by $g_{6}=0.0020549454896034927680\ldots$ of (5.18). The margin then becomes $0.0022838998008707818199\ldots$, and $a_{\mathrm{buf}}^{*}$ becomes $1.7497494980881623276859\ldots$. Theorem 1 does not use this.

For an optional arithmetic comparison, define $D_{B,c}=D_{B}+(c_{K2,G0}-c)\overline{H}/4$. Using $c$ in both K2 and L3 would require two further inputs. This paragraph concerns the optional scalar comparison at the same fixed cutoff; it supplies no new cutoff-dependent estimate. The first is the separate sharper functional domination $G_{*}\leq U_{0}$ from equation B.28. The second is a certified $2e^{-\gamma}U_{0}\leq c$. It does not follow from the coarse interval in equation B.27. The main proof uses neither. Its scalar endpoint is

(9.29) $$E_{c}=\exp(D_{B,c}/c),\qquad a^{*}_{\mathrm{buf},c}=\frac{E_{c}+1}{(1-\zeta)E_{c}-\zeta}.$$

The same rational Taylor enclosure gives

(9.30) $$1.749889173139093190128834678<a^{*}_{\mathrm{buf},c}<1.749889173139093190128834679<\frac{7}{4}.$$

Thus, on the optional sharper branch, $a^{*}_{\mathrm{buf},c}$ is $1.7498891731390\ldots$. The main theorem uses the explicitly certified G0 coefficient defined and bounded in (7.36); the display above is only the optional exponent endpoint.

9.4. The transfer

Lemma 9.2 (Transfer).

The parameters (2.3) satisfy

(9.31) $$\kappa_{\mathrm{tr}}=\frac{3}{4}(1-b)-b=\frac{7}{4000000},\qquad N_{\mathrm{tr}}=10^{12000000/7}>10^{12}+1=N_{\mathrm{pt}}.$$

For even $N\geq N_{\mathrm{tr}}$, every $n=rq\in\mathcal{A}_{N}$ with primes $r\leq q$ and $r<N^{b}$ satisfies

(9.32) $$q>\varepsilon N^{1-b},\qquad q^{3/4}>\varepsilon^{3/4}N^{b+\kappa_{\mathrm{tr}}}\geq N^{b}>r.$$

Proof.

Since $a=7/4$, $a(1-b)-1=\frac{3}{4}(1-b)-b$, and $b=2999993/7000000$ gives $7/4000000$. Also $\varepsilon^{-(a-1)/\kappa_{\mathrm{tr}}}=10^{4\cdot 3000000/7}$, and $12000000/7>13$. For $n=rq\in\mathcal{A}_{N}$ we have $n>\varepsilon N$, so $q=n/r>\varepsilon N^{1-b}$ and $q^{3/4}>\varepsilon^{3/4}N^{3(1-b)/4}=\varepsilon^{3/4}N^{b+\kappa_{\mathrm{tr}}}$. Finally $N^{\kappa_{\mathrm{tr}}}\geq\varepsilon^{-3/4}$ exactly when $N\geq N_{\mathrm{tr}}$. ∎

Proof of Theorem 1

Use the established inputs (A1)–(A5) and fix a rational $\sigma_{*}$ satisfying (9.14). Let $N\geq N_{0}^{*}(\sigma_{*})$ be even. Then $N\geq\max\{N_{\mathrm{tr}},N_{6}\}>N_{\mathrm{pt}}$. Lemma 4.1 gives the first line of (1.10). At $\sigma=\sigma_{*}$, the established comparisons (A1), (A2) and (A4) give its other three lines, and (A3) gives (1.13). The zero-gain combination Lemma 4.3 adds these four comparisons to (9.7). Proposition 9.1 gives (9.20), hence (9.21). Equation (9.24) then gives $G_{b}(N)>0$, since $W(N)>0$ by (2.2). Hence the set in (1.3) is nonempty.

Choose one counted integer $n=N-p$. If $n$ is prime, take $q=n$ and $r=1$. Otherwise $n=rq$ with primes $r\leq q$ and $r<N^{b}$, and Lemma 9.2 gives $r<q^{3/4}$. Therefore $N=p+rq$ and $r\leq q^{3/4}$. $\square$

Appendices

Appendix A proves three auxiliary estimates. Appendix B records finite calculations, numerical certificates, and exact arithmetic. All cross-references to these appendices are internal to this document.

Appendix A Proofs of three estimates of Section 6

This section proves Lemmas 6.5, 6.8 and 6.9. Numbers (6.$k$) and 6.$k$ refer to Section 6. Numbers (A.1) and later refer to this section. The only inputs are Papers I and II and the classical results cited at the point of use. Throughout, $\tau_{j}$ is the $j$-fold divisor function and $\tau=\tau_{2}$. Let $\omega(n)$ be the number of distinct prime divisors of $n$. Let $P^{+}(n)$ and $P^{-}(n)$ be its largest and least prime factors. The sum $\sum^{*}_{b\bmod k}$ runs over reduced residues.

A.1. Elementary inequalities

For integers $j\geq 1$, $k\geq 1$ and $Y\geq 1$,

(A.1) $$\tau(n)^{j}\leq\tau_{2^{j}}(n),\qquad\sum_{n\leq Y}\frac{\tau_{k}(n)}{n}\leq(1+\log Y)^{k},\qquad\sum_{n\leq Y}\tau_{k}(n)\leq Y(1+\log Y)^{k-1}.$$

The first is multiplicative. At a prime power it is $(a+1)^{j}\leq\binom{a+2^{j}-1}{2^{j}-1}$, which follows by induction on $j$ from

$$(a+1)\binom{a+k-1}{k-1}\leq\binom{a+2k-1}{2k-1}.$$

The right side is the coefficient of $p^{a}$ in the convolution of two $k$-fold divisor functions. Each of its $a+1$ terms is at least the left factor $\binom{a+k-1}{k-1}$, since $(j^{\prime}+i)(a-j^{\prime}+i)\geq i(a+i)$ for $0\leq j^{\prime}\leq a$. The two sums in (A.1) follow by writing $n=n_{1}\cdots n_{k}$, enlarging the constraint to $n_{i}\leq Y$, and summing one factor last. Also

(A.2) $$\frac{n}{\varphi(n)}\leq\tau(n),\qquad\varphi(mn)\geq\varphi(m)\varphi(n),\qquad\tau(mn)\leq\tau(m)\tau(n),$$

prime power by prime power, and $\tau(n)\ll_{\epsilon}n^{\epsilon}$.

For a finitely supported sequence $\xi$, a modulus $k$ and an integer $h\geq 1$, put

(A.3) $$\Delta_{\xi}(k,b;h)=\sum_{\begin{subarray}{c}n\equiv b\ (k)\\ (n,h)=1\end{subarray}}\xi_{n}-\frac{1}{\varphi(k)}\sum_{(n,kh)=1}\xi_{n},\qquad\Delta_{\xi}(k,b)=\Delta_{\xi}(k,b;1),$$

(A.4) $$V(k;\xi)=\sum_{b\bmod k}^{*}|\Delta_{\xi}(k,b)|^{2}.$$

If $\xi$ is supported in an interval of length at most $Y$, then

(A.5) $$V(k;\xi)\leq\sum_{b\bmod k}\Bigl{|}\sum_{n\equiv b\ (k)}\xi_{n}\Bigr{|}^{2}\leq\Bigl{(}\frac{Y}{k}+1\Bigr{)}\|\xi\|_{2}^{2}.$$

Here $\Delta_{\xi}(k,b)$ is the deviation of the reduced class sums from their mean. The variance is at most the second moment. Cauchy’s inequality in each class gives the second bound.

The sequence $e_{n}=\mathbf{1}_{H<n\leq 2H}$ satisfies, for all $k,h\geq 1$ and $(b,k)=1$,

(A.6) $$|\Delta_{e}(k,b;h)|\leq 3\tau(h).$$

By inclusion–exclusion over $d\mid h$, $(d,k)=1$, the class count equals $(H/k)\prod_{p\mid h,\,p\nmid k}(1-1/p)+O(\tau(h))$, with error at most $\tau(h)$ in absolute value. The comparator has the same main term and error at most $\tau(h)\tau(k)/\varphi(k)\leq 2\tau(h)$.

Let $\ell=\log\log N$, $z=N^{1/\ell^{3}}$ and $y=N^{1/\ell}$. Here $N$ is the large parameter of the lemma at hand. For each fixed $C\geq 0$, Rankin’s method gives

(A.7) $$\sum_{\begin{subarray}{c}d>y\\ P^{+}(d)\leq z\end{subarray}}\frac{\tau(d)^{C}}{d}\ll_{C}(\log N)^{K_{C}}e^{-\ell^{2}}$$

with a fixed $K_{C}$. With $\lambda=1/\log z$, the left side is at most

$$y^{-\lambda}\prod_{p\leq z}\Bigl{(}1+\sum_{e\geq 1}(e+1)^{C}p^{-e(1-\lambda)}\Bigr{)},\qquad y^{-\lambda}=e^{-\ell^{2}}.$$

For $p\leq z$ one has $p^{\lambda}\leq e$ and $1-\lambda\geq 1/2$, so the factor at $p$ is $1+O_{C}(1/p)$. Mertens’ theorem gives $\sum_{p\leq z}1/p\leq\ell+O(1)$. Hence the product is $\ll(\log N)^{K_{C}}$. Since $e^{-\ell^{2}}=(\log N)^{-\ell}$, the right side of (A.7) is $O_{C,A}((\log N)^{-A})$ for every fixed $A$.

A.2. Lemma 6.5

Recall the setting of Section 6. Let $\xi$ be complex, supported in $[1,Z]$, $Z\geq 3$, and fix $B_{\mathrm{div}}\geq 0$. The hypothesis (6.35) is that, for every $P>0$, uniformly in $h,k\geq 1$ and reduced $b\bmod k$,

(6.35) $$\left|\sum_{\begin{subarray}{c}n\equiv b\ (k)\\ (n,h)=1\end{subarray}}\xi_{n}-\frac{1}{\varphi(k)}\sum_{(n,kh)=1}\xi_{n}\right|\ll_{P}\|\xi\|_{2}Z^{1/2}\tau(h)^{B_{\mathrm{div}}}(\log(2Z))^{-P},$$

with constants common to the family of pairs $(Z,\xi)$. The conclusion (6.36) is

(6.36) $$\sum_{k\leq Z(\log Z)^{-B_{1}(A)}}\sum_{b\bmod k}^{*}|\Delta_{\xi}(k,b)|^{2}\ll_{A}\|\xi\|_{2}^{2}Z(\log Z)^{-A}.$$

In the selected-factor form of Paper I, (3.48), $\xi_{n}=\alpha_{n}1_{(n,r)=1}$, $Z=\lceil 2T\rceil$, and (6.37) reads

(6.37) $$\sum_{q\leq 3Q}\sum_{b\bmod q}^{*}|\Delta_{\xi}(q,b)|^{2}\ll N_{\mathcal{I}}TL^{-A_{3}},\qquad 3Q\leq Z(\log Z)^{-B_{F}},\quad B_{F}=A_{3}+15,$$

where $N_{\mathcal{I}}$ is the interval mass in Paper I, (3.8).

Lemma 6.5 (Transferred mean-square estimate). 

1. If (6.35) holds, then (6.36) holds for every fixed $A>0$. Here $B_{1}(A)$ and the implied constant depend only on $A$, $B_{\mathrm{div}}$ and the common constants.

2. Assume the hypotheses (3.1), (3.2), (3.6), (3.8) with its length condition, (3.11) and (3.25)–(3.27) of Paper I. Then the estimate (6.37) holds uniformly over the parameter boxes and selected factors of Paper I, Theorem 3.2. Hence that theorem holds with this input in place of [3, Theorem 0(a)] at (3.48).

Proof of (i).

Put $E=\|\xi\|_{2}^{2}$, $\lambda=\log(2Z)$ and, for $1\leq Q\leq Z$,

$$S(Q)=\sum_{k\leq Q}V(k;\xi).$$

Since $2Z\leq Z^{2}$, $\log Z\leq\lambda\leq 2\log Z$.

Characters. Orthogonality on the reduced classes gives $\Delta_{\xi}(k,b)=\varphi(k)^{-1}\sum_{\chi\neq\chi_{0}}\overline{\chi}(b)\sum_{n}\xi_{n}\chi(n)$. The centring removes exactly the principal character. Hence

(A.8) $$S(Q)=\sum_{k\leq Q}\frac{1}{\varphi(k)}\sum_{\begin{subarray}{c}\chi\bmod k\\ \chi\neq\chi_{0}\end{subarray}}\Bigl{|}\sum_{n}\xi_{n}\chi(n)\Bigr{|}^{2}.$$

Each $\chi\neq\chi_{0}$ modulo $k$ is induced by a unique primitive $\psi\bmod f$ with $f\mid k$, $f>1$. Write $k=fe$. Then $\chi(n)=\psi(n)\mathbf{1}_{(n,e)=1}$, since $\psi$ already vanishes on primes of $f$. The pair $(e,\psi)$ determines $(k,\chi)$, and $\varphi(fe)\geq\varphi(f)\varphi(e)$ by (A.2). Therefore

(A.9) $$S(Q)\leq\sum_{e\leq Q}\frac{1}{\varphi(e)}\sum_{1<f\leq Q/e}\frac{1}{\varphi(f)}\sum_{\psi\bmod f}^{\rm prim}|C_{e}(\psi)|^{2},\qquad C_{e}(\psi)=\sum_{(n,e)=1}\xi_{n}\psi(n).$$

Small conductors. Let $F\geq 1$ and $1<f\leq F$. Since $\sum_{b}^{*}\psi(b)=0$,

$$C_{e}(\psi)=\sum_{b\bmod f}^{*}\psi(b)\Bigl{\{}\sum_{\begin{subarray}{c}n\equiv b\ (f)\\ (n,e)=1\end{subarray}}\xi_{n}-\frac{1}{\varphi(f)}\sum_{(n,ef)=1}\xi_{n}\Bigr{\}}.$$

The bracket is the left side of (6.35) with $k=f$ and $h=e$. For a fixed $a>0$ it is at most $C_{a}E^{1/2}Z^{1/2}\tau(e)^{B_{\rm div}}\lambda^{-a}$. There are at most $\varphi(f)$ primitive characters modulo $f$, so

(A.10) $$\sum_{1<f\leq F}\frac{1}{\varphi(f)}\sum_{\psi}^{\rm prim}|C_{e}(\psi)|^{2}\leq C_{a}^{2}EZ\tau(e)^{2B_{\rm div}}\lambda^{-2a}\sum_{f\leq F}\varphi(f)^{2}\leq C_{a}^{2}EZ\tau(e)^{2B_{\rm div}}F^{3}\lambda^{-2a}.$$

Large conductors. The multiplicative large sieve, Paper I, (2.19), applied to $c_{n}=\xi_{n}\mathbf{1}_{(n,e)=1}$ with support in $[1,Z]$ and $\|c\|_{2}^{2}\leq E$, gives

$$\sum_{f\leq 2V}\sum_{\psi}^{\rm prim}|C_{e}(\psi)|^{2}\ll(V^{2}+Z)E.$$

If $p_{1}<\cdots<p_{s}$ are the primes of $f$, then $p_{i}\geq i+1$ and $f/\varphi(f)\leq\prod_{i}(i+1)/i=s+1\leq 1+\log f/\log 2$. Thus $1/\varphi(f)\leq 3\lambda/V$ for $V<f\leq 2V$, and

$$\sum_{V<f\leq 2V}\frac{1}{\varphi(f)}\sum_{\psi}^{\rm prim}|C_{e}(\psi)|^{2}\ll\lambda E\Bigl{(}V+\frac{Z}{V}\Bigr{)}.$$

Summing over $V=2^{j}F<Q$ gives, uniformly in $e$,

(A.11) $$\sum_{F<f\leq Q/e}\frac{1}{\varphi(f)}\sum_{\psi}^{\rm prim}|C_{e}(\psi)|^{2}\ll\lambda E\Bigl{(}Q+\frac{Z}{F}\Bigr{)}.$$

If $F\geq Q$, this range is empty.

Summation over $e$. Let $r=\lceil 2B_{\rm div}+1\rceil$ and $m=2^{r}$. By (A.2) and (A.1),

$$\sum_{e\leq Q}\frac{\tau(e)^{2B_{\rm div}}}{\varphi(e)}\leq\sum_{e\leq Q}\frac{\tau(e)^{r}}{e}\leq(1+\log Q)^{m},\qquad\sum_{e\leq Q}\frac{1}{\varphi(e)}\leq(1+\log Q)^{2}.$$

Inserting (A.10) and (A.11) in (A.9) gives

(A.12) $$S(Q)\ll_{a}E\lambda^{m+3}\Bigl{(}ZF^{3}\lambda^{-2a}+\frac{Z}{F}+Q\Bigr{)}.$$

Put

(A.13) $$B_{1}(A)=A+2^{\lceil 2B_{\rm div}+1\rceil}+7,\qquad F=\lambda^{B_{1}},\qquad a=2B_{1}.$$

For $Q\leq Z(\log Z)^{-B_{1}}\leq 2^{B_{1}}Z\lambda^{-B_{1}}$ each term in (A.12) is $\ll EZ\lambda^{m+3-B_{1}}\leq EZ\lambda^{-A}$. This is (6.36). The constants depend only on $A$, $B_{\rm div}$ and the constant $C_{2B_{1}}$ of (6.35). For $B_{\rm div}=1$ one has $r=3$, $m=8$ and $B_{1}(A)=A+15$. ∎

Proof of (ii).

Fix a box and a selected factor $r$ of Paper I, (3.26). Put $\xi_{n}=\alpha_{n}\mathbf{1}_{(n,r)=1}$, with $\alpha$ as in Paper I, (3.8). Then $\xi$ is supported in $I_{T}\subset(T,2T]\subset[1,Z]$, $Z=\lceil 2T\rceil$, $|\xi_{n}|\leq 1$, and $E=\|\xi\|_{2}^{2}\leq N_{\mathcal{I}}+1$. Its discrepancy (A.3) at $h=1$ is $\Delta(\alpha^{(r)};q,b)$ of Paper I, (3.46). The support of $\psi_{0}$ in Paper I, (3.27), gives $q\leq 5Q/2<3Q$.

Low norm. If $E\leq N_{\mathcal{I}}L^{-A_{3}-1}$, then (A.5), $Q<T$ and $\log(3Q)\leq L$ give

$$\sum_{q\leq 3Q}V(q;\xi)\leq(N_{\mathcal{I}}(1+\log 3Q)+3Q)E\ll TLE\leq N_{\mathcal{I}}TL^{-A_{3}}.$$

This is the first case of Paper I, (3.48).

High norm. Let $E>N_{\mathcal{I}}L^{-A_{3}-1}$. For $d,k\geq 1$ and $(l,k)=1$,

$$\Delta_{\xi}(k,l;d)=\sum_{\begin{subarray}{c}n\in I_{T},\ n\equiv l\ (k)\\ (n,dr)=1\end{subarray}}\alpha_{n}-\frac{1}{\varphi(k)}\sum_{\begin{subarray}{c}n\in I_{T}\\ (n,dkr)=1\end{subarray}}\alpha_{n},$$

which is the discrepancy of Paper I, (3.25), with $I=I_{T}$ and auxiliary divisor $dr$. Hence, for each $A_{\rm SW}>0$,

(A.14) $$|\Delta_{\xi}(k,l;d)|\ll_{A_{\rm SW}}T\tau(d)\tau(r)L^{-A_{\rm SW}}.$$

By Paper I, (3.26), $r$ is $u_{0}$-rough and $r\leq X^{0.6}\leq N^{0.6}$. With $\log u_{0}=\ell_{0}\sigma^{2}\log N$ from Paper I, (3.2), the number $\Omega(r)$ of prime factors with multiplicity satisfies $\Omega(r)\leq 0.6/(\ell_{0}\sigma^{2})$, so

$$\tau(r)\leq 2^{\Omega(r)}=O_{\sigma}(1).$$

By Paper I, (3.8), $N_{\mathcal{I}}\asymp TL^{-C_{\rm box}}$, hence

$$E^{1/2}Z^{1/2}\gg TL^{-(C_{\rm box}+A_{3}+1)/2}.$$

By Paper I, (3.11), $T\geq X^{60\sigma}$, so $\log(2Z)\asymp_{\sigma}L$. Take $A_{\rm SW}=P+(C_{\rm box}+A_{3}+1)/2$ in (A.14). This gives (6.35) for $\xi$ with $B_{\rm div}=1$, at every precision $P$. The constants depend only on $\sigma$, $C_{\rm box}$, $A_{3}$, $P$ and the constants of Paper I, (3.25). These are common to all boxes and selected factors.

By part (i) with $B_{\rm div}=1$ and $A=A_{3}$, the range parameter is $B_{1}(A_{3})=A_{3}+15=B_{F}$. Paper I, (3.6), gives $Q<TX^{-29\sigma}$, hence

$$3Q<3TX^{-29\sigma}\leq Z(\log Z)^{-B_{F}}$$

for all sufficiently large $X$. Part (i) yields

$$\sum_{q\leq 3Q}V(q;\xi)\ll EZ(\log Z)^{-A_{3}}\ll N_{\mathcal{I}}TL^{-A_{3}}.$$

This is (6.37), uniformly in the box and the selected factor.

In the proof of Paper I, Theorem 3.2, [3, Theorem 0(a)] is used only to obtain the second case of (3.48). See the list of inputs after the statement of that theorem in Paper I. The two cases above give (3.48) for every box and selected factor, with the same constants and quantifiers. The rest of that proof is unchanged. ∎

A.3. Lemma 6.8

Recall the notation of Section 6. Put $L=\log N$, $\ell=\log\log N$,

$$z=N^{1/\ell^{3}},\qquad y=N^{1/\ell},\qquad v=v_{\mathrm{RA}}=\frac{5227}{10000},\qquad w=\frac{9}{1000},\qquad\sigma_{\mathrm{p}}=10^{-6},\qquad\theta=\frac{7}{32},$$

and, for an integer $q$ and a residue $e$,

(6.21) $$\mathcal{K}_{q}(e)=\mathbf{1}_{e\equiv N\ (q)}-\frac{\mathbf{1}_{(e,q)=1}}{\varphi(q)}$$

(the last line of (6.21)). For a parent $(H,2H]$ with $N^{1/3}/8\leq H\leq 2N^{.46+O(1/L)}$, (6.38) defines

(6.38) $$D_{1}=HN^{-\sigma_{\mathrm{p}}},\qquad D_{2}=QN^{w+\sigma_{\mathrm{p}}}/H,\qquad Q=N^{v}L^{-B}.$$

For a sequence $f$ on $(H,2H]$, (6.32) defines $\Delta_{f}(I;k,b;h)$ as (A.3) with $n$ restricted to the interval $I$. Then (6.34) is the bound

(6.34) $$|\Delta_{f}(I;k,b;h)|\ll_{P}H\tau(h)L^{-P},\qquad(b,k)=1,$$

uniformly for $I\subset(H,2H]$ and $k,h\geq 1$.

Lemma 6.8 (Two-factor convolution estimate).  Let $N^{1/3}/8\leq H\leq 2N^{.46+O(1/L)}$ and $HM\asymp N$. Let $a_{n}$, $n\sim H$, be one-bounded, supported on $P^{-}(n)>z$, and satisfy (6.34) at the original parent $H$, where $z=N^{1/(\log\log N)^{3}}$. Let $b_{m}$, $m\sim M$, be one-bounded. Let $U_{1},V_{1}$ be dyadic modulus scales, allowing the singleton scale $1/2$ for the integer $1$, such that

$$U_{1}\leq D_{1},\qquad V_{1}\leq D_{2},\qquad U_{1}V_{1}\leq N^{v_{\mathrm{RA}}}.$$

Let $f(r)$, $r\sim U_{1}$, and $g(s)$, $s\sim V_{1}$, be arbitrary one-bounded arrays, with the unary restrictions $r<D_{1}$, $s<D_{2}$, $(r,N)=(s,N)=1$ in force. Neither a pointwise condition $rs<Q$ nor a joint condition $(r,s)=1$ is imposed on the pair. Then, for every fixed $A>0$, uniformly in these occupied-parent scales, $f$, $g$, unary interval cuts of $a_{n}$ and $b_{m}$, and intervals $I\subset[N/8,2N]$,

(6.45) $$\sum_{r,s}f(r)g(s)\sum_{\begin{subarray}{c}n,m\\ nm\in I\end{subarray}}a_{n}b_{m}\mathcal{K}_{rs}(nm)\ll_{A}NL^{-A}.$$

Write $H=N^{h}$. Fixed-factor enlargements of the parents change $h$ by $O(1/L)$, so

(A.15) $$\tfrac{1}{3}-o(1)\leq h\leq\tfrac{23}{50}+o(1),\qquad M\gg N^{.54-o(1)}.$$

All implied constants below may depend on $A$, on the fixed comparison constants and on the constants in (6.34). They do not depend on $N$, $H$, $f$, $g$, $a$ or $b$. All fixed precisions are chosen before the final threshold for $N$.

Step 0: reductions

A unary cut $a_{n}\mathbf{1}_{J}(n)$ satisfies (6.34) with the same constants, since $\Delta_{a\mathbf{1}_{J}}(I;k,b;h)=\Delta_{a}(I\cap J;k,b;h)$. The error is still measured at $H$. A unary cut of $b_{m}$ keeps it one-bounded. The product cut $nm\in I$ is removed in Step 9. Until then we prove, for every fixed $A>0$,

(A.16) $$\mathcal{S}:=\sum_{r\sim U_{1},\ s\sim V_{1}}f(r)g(s)\sum_{n\sim H,\ m\sim M}a_{n}b_{m}\mathcal{K}_{rs}(nm)\ll_{A}NL^{-A}.$$

Step 1: a shifted divisor moment and the mean square

For each fixed $j$ and each interval $J$ of length $O(M)$ in $(M/2,4M]$,

(A.17) $$\sum_{\begin{subarray}{c}n\sim H,\ m\in J\\ mn\neq N\end{subarray}}\tau(|mn-N|)^{j}\ll_{j}MHL^{C_{j}}.$$

Let $1\leq u\leq N^{3}$. Its prime factors above $N^{1/8}$ have total multiplicity at most $24$. Group the others greedily into at most $25$ factors, each at most $N^{1/4}$. Submultiplicativity of $\tau$ gives

$$\tau(u)^{j}\leq 2^{24j}\sum_{\begin{subarray}{c}d\mid u\\ d\leq N^{1/4}\end{subarray}}\tau(d)^{25j}.$$

For fixed $n$ and $d$, put $h_{0}=(d,n)$. The congruence $d\mid mn-N$ has no solution unless $h_{0}\mid N$, and otherwise $O(Mh_{0}/d+1)$ solutions $m\in J$. Using $(d,n)\leq\sum_{e\mid(d,n)}e$ and (A.1),

$$\sum_{d\leq N^{1/4}}\tau(d)^{25j}\Bigl{(}\frac{M(d,n)}{d}+1\Bigr{)}\ll L^{C}\bigl{(}M\tau(n)^{25j+1}+N^{1/4}\bigr{)}.$$

Summing over $n$ with (A.1), and using $N^{1/4}\leq M$, proves (A.17).

Put $Z_{H}=\lceil 2H\rceil$. For each fixed $P>0$ and every $K^{\prime}\leq HN^{-\sigma_{\mathrm{p}}/2}$,

(A.18) $$\sum_{k\leq 5K^{\prime}}\tau(k)V(k;a)\ll_{P}H^{2}L^{-P}.$$

First let $R>0$ be fixed. If $\|a\|_{2}^{2}\leq HL^{-R}$, then (A.5) gives $\sum_{k\leq 5K^{\prime}}V(k;a)\ll(H(1+\log 5K^{\prime})+K^{\prime})\|a\|_{2}^{2}\ll H^{2}L^{1-R}$. If $\|a\|_{2}^{2}>HL^{-R}$, then $\|a\|_{2}Z_{H}^{1/2}\geq HL^{-R/2}$. Taking $I=(H,2H]$ in (6.34) at precision $P^{\prime}+R/2$ gives (6.35) for $\xi=a$ with $Z=Z_{H}$, $B_{\rm div}=1$ and common constants, since $\log(2Z_{H})\asymp L$. This argument is uniform at every requested precision $P$ (after choosing $P^{\prime}$ accordingly). Lemma 6.5(i) with $A=P^{\prime}$ and $B_{1}(P^{\prime})=P^{\prime}+15$ gives

$$\sum_{k\leq Z_{H}(\log Z_{H})^{-P^{\prime}-15}}V(k;a)\ll\|a\|_{2}^{2}Z_{H}(\log Z_{H})^{-P^{\prime}}\ll H^{2}L^{-P^{\prime}}.$$

The range contains $k\leq 5K^{\prime}$, because $5K^{\prime}/H\leq 5N^{-\sigma_{\mathrm{p}}/2}$ beats every fixed power of $L$. Thus in both cases $\sum_{k\leq 5K^{\prime}}V(k;a)\ll H^{2}L^{-\min(R-1,P^{\prime})}$. To insert $\tau(k)$, split at $\tau(k)=L^{R^{\prime}}$. Below the split the cost is $L^{R^{\prime}}$. Above it, $\tau(k)\leq L^{-R^{\prime}}\tau(k)^{2}$, and $V(k;a)\leq(2H/k+1)(H+1)$ by (A.5); hence by (A.1)

$$\sum_{\begin{subarray}{c}k\leq 5K^{\prime}\\ \tau(k)>L^{R^{\prime}}\end{subarray}}\tau(k)V(k;a)\ll L^{-R^{\prime}}H^{2}\sum_{k\leq 5K^{\prime}}\frac{\tau(k)^{2}}{k}\ll H^{2}L^{4-R^{\prime}}.$$

Choosing $R^{\prime}=P+4$ and then $R,P^{\prime}\geq P+R^{\prime}+1$ proves (A.18).

For a $z$-rough integer $t\leq N^{C}$ put $a^{(t)}_{n}=a_{n}\mathbf{1}_{(n,t)=1}$. Since $\omega(t)\leq\log t/\log z\ll\ell^{3}$ and each prime $p\mid t$ exceeds $z$,

$$\|a-a^{(t)}\|_{2}^{2}\leq\sum_{p\mid t}\Bigl{(}\frac{2H}{p}+1\Bigr{)}\ll\frac{HL}{z},$$

and by (A.5) and (A.1), uniformly in $t$,

(A.19) $$\sum_{k\leq 5K^{\prime}}\tau(k)V(k;a-a^{(t)})\ll\frac{H^{2}L^{3}}{z}.$$

Step 2: large smooth parts

Write $s=dt$ with $P^{+}(d)\leq z<P^{-}(t)$. Consider the pairs $(r,s)$ with $d>y$. Then the $z$-smooth part of $q=rs$ exceeds $y$. A modulus $q$ has at most $\tau(q)$ factorizations $q=rs$, and $q\leq 4U_{1}V_{1}\leq 4N^{v}$. For $(q,N)=1$, counting $m$ for each $n$ coprime to $q$ gives

$$\Bigl{|}\sum_{n,m}a_{n}b_{m}\mathcal{K}_{q}(nm)\Bigr{|}\ll\frac{N}{q}+H+\frac{N}{\varphi(q)}.$$

By (A.2), the contribution of these pairs to $\mathcal{S}$ is

$$\ll N\sum_{\begin{subarray}{c}q\leq 4N^{v}\\ q=d^{\prime}h^{\prime},\ d^{\prime}>y,\ P^{+}(d^{\prime})\leq z\end{subarray}}\frac{\tau(q)^{2}}{q}+H\sum_{q\leq 4N^{v}}\tau(q)\ll NL^{C}\sum_{\begin{subarray}{c}d^{\prime}>y\\ P^{+}(d^{\prime})\leq z\end{subarray}}\frac{\tau(d^{\prime})^{2}}{d^{\prime}}+HN^{v}L.$$

Here $\tau(q)^{2}\leq\tau(d^{\prime})^{2}\tau(h^{\prime})^{2}$ and the sum over $h^{\prime}$ is bounded by (A.1). By (A.7) the first term is $O_{A}(NL^{-A})$. The second is at most $N^{.9827+o(1)}$, since $h+v\leq.46+.5227+o(1)$. It remains to treat $d\leq y$.

Step 3: the first Cauchy square

Split $d$ and $t$ into dyadic shells $d\sim D$, $t\sim S$, with $D\leq y$, $DS\asymp V_{1}$. There are $O(L)$ choices of $D$ and $O(1)$ of $S$. Put

$$\lambda_{d}(t)=g(dt)\mathbf{1}_{dt\sim V_{1}}\mathbf{1}_{P^{-}(t)>z}\mathbf{1}_{P^{+}(d)\leq z},\qquad K=U_{1}D,\qquad k=rd.$$

Then $|\lambda_{d}(t)|\leq 1$, $rs=kt$, and $(kt,N)=1$ on the supports. One block of $\mathcal{S}$ is

$$\mathcal{S}_{0}=\sum_{r,d,m}f(r)b_{m}\sum_{t}\lambda_{d}(t)\sum_{n}a_{n}\mathcal{K}_{kt}(mn).$$

Fix a smooth $\psi\geq 0$ supported in $[1/2,5/2]$, equal to $1$ on $[1,2]$. Cauchy’s inequality in $(r,d,m)$ gives

(A.20) $$|\mathcal{S}_{0}|^{2}\leq CKM\,\mathcal{S}^{\prime},\qquad\mathcal{S}^{\prime}=\sum_{r,d}\psi\Bigl{(}\frac{r}{U_{1}}\Bigr{)}\mathbf{1}_{(rd,N)=1}\mathbf{1}_{d\sim D}\sum_{m}\psi\Bigl{(}\frac{m}{M}\Bigr{)}\Bigl{|}\sum_{t}\lambda_{d}(t)\sum_{n}a_{n}\mathcal{K}_{kt}(mn)\Bigr{|}^{2}.$$

The indicator $\mathbf{1}_{(rd,N)=1}$ is part of the original support and is kept. The scale conditions and (A.15) give, for large $N$,

(A.21) $$\frac{K}{H}\leq N^{-\sigma_{\mathrm{p}}/2},\qquad\frac{KS}{M}\leq N^{-.0172},\qquad M\gg N^{.54-o(1)},$$

because $K\leq D_{1}y$, $KS\asymp U_{1}V_{1}\leq N^{v}$ and $v-.54=-.0173$. In particular $kt=o(M)$ and $K=o(H)$.

Expand $\mathcal{S}^{\prime}=\mathcal{S}_{3}-2\operatorname{Re}\mathcal{S}_{2}+\mathcal{S}_{1}$ over tuples $\mathbf{t}=(r,d,t_{1},t_{2},n_{1},n_{2})$ with coefficient

$$c(\mathbf{t})=\psi(r/U_{1})\mathbf{1}_{(rd,N)=1}\mathbf{1}_{d\sim D}\lambda_{d}(t_{1})\overline{\lambda_{d}(t_{2})}a_{n_{1}}\overline{a_{n_{2}}},\qquad\omega(\mathbf{t})=\mathbf{1}_{(n_{1},kt_{1})=(n_{2},kt_{2})=1}:$$

$$\mathcal{S}_{3} =\sum_{\mathbf{t}}c\sum_{m}\psi(m/M)\mathbf{1}_{kt_{1}\mid mn_{1}-N}\mathbf{1}_{kt_{2}\mid mn_{2}-N},$$

$$\mathcal{S}_{2} =\sum_{\mathbf{t}}c\sum_{m}\psi(m/M)\mathbf{1}_{kt_{1}\mid mn_{1}-N}\frac{\mathbf{1}_{(mn_{2},kt_{2})=1}}{\varphi(kt_{2})},$$

$$\mathcal{S}_{1} =\sum_{\mathbf{t}}c\sum_{m}\psi(m/M)\frac{\mathbf{1}_{(mn_{1},kt_{1})=1}\mathbf{1}_{(mn_{2},kt_{2})=1}}{\varphi(kt_{1})\varphi(kt_{2})}.$$

In $\mathcal{S}_{3}$ the factor $\omega$ is automatic, since $(kt_{i},N)=1$.

Step 4: restoration before the Chinese remainder theorem

Call a tuple gcd-bad if one of

(A.22) $$(n_{1},n_{2})>1,\quad(t_{1},t_{2})>1,\quad(n_{1},t_{2})>1,\quad(n_{2},t_{1})>1,\quad(k,t_{1}t_{2})>1$$

holds. Call a tuple near-diagonal if $|n_{1}-n_{2}|<HL^{-C_{M}}$; this condition is charged separately below. On the other tuples $k,t_{1},t_{2}$ are pairwise coprime. For a set $\mathcal{B}$ of tuples define the formal terms

(A.23) $$MT(\mathcal{B})=\sum_{\mathcal{B}}\frac{c\,\omega\,M\hat{\psi}(0)}{kt_{1}t_{2}\varphi(k)},\qquad\mathcal{Z}(\mathcal{B})=\sum_{\mathcal{B}}\frac{c\,\omega\,\mathbf{1}_{n_{1}\equiv n_{2}\ (k)}M\hat{\psi}(0)}{kt_{1}t_{2}}.$$

We show

(A.24) $$\begin{gathered}\mathcal{S}^{\prime}=M\mathcal{E}+\mathcal{Z}(\mathrm{all})-MT(\mathrm{all})+O(R_{0}),\\ R_{0}\ll_{P}\frac{MH^{2}}{K}L^{-P}+MHL^{C}\Bigl{[}\frac{H}{zK}+1\Bigr{]}+N^{o(1)}\Bigl{[}\frac{H^{2}}{K}+H^{2}S\Bigr{]}+N^{o(1)}\frac{(H+KS)^{2}}{K},\end{gathered}$$

where $\mathcal{E}$ is the nonzero-frequency sum on good tuples defined in Step 6. The exponent $C$ is fixed after $P$ and the truncation parameters.

Resonant $m$. If $mn_{i}=N$, then $m\mid N$. There are $\tau(N)=N^{o(1)}$ such $m$. For fixed $(r,d)$ and such $m$, the congruence terms over $(t,n)$ number at most $\sum_{t}(2H/(kt)+1)\ll H/K+S$. The comparator terms are $\ll L^{C}H/K$. Hence these $m$ contribute $\ll N^{o(1)}K(H/K+S)^{2}$ to $\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{3}$ and to every subset of tuples below. This is the last term of $R_{0}$.

Divisor truncation. For the other $m$ put $u_{i}=|mn_{i}-N|\geq 1$. Let $\Sigma^{*}$ be the event $\tau(u_{1}),\tau(u_{2})\leq L^{C_{\Sigma}}$. On the part of the complement with $\tau(u_{1})\geq\tau(u_{2})$, the triples $(r,d,t_{1})$ with $kt_{1}\mid u_{1}$ number at most $\tau(u_{1})^{3}$. The $t_{2}\mid u_{2}$ number at most $\tau(u_{1})$, and $n_{2}$ lies in one class modulo $k$, with $O(H/K)$ choices. With $\mathbf{1}_{\tau(u_{1})>L^{C_{\Sigma}}}\leq\tau(u_{1})L^{-C_{\Sigma}}$, (A.17) gives a contribution to $\mathcal{S}_{3}$ of

(A.25) $$\ll L^{-C_{\Sigma}}\frac{H}{K}\sum_{m,n_{1}}\psi(m/M)\tau(u_{1})^{5}\ll\frac{MH^{2}}{K}L^{C_{5}-C_{\Sigma}}.$$

The other part is symmetric. Choose $C_{\Sigma}\geq C_{5}+P$.

Bad tuples in $\mathcal{S}_{3}$. In this paragraph, “bad” means $\Sigma^{*}$ intersected with the five gcd failures in (A.22); the near-diagonal set is not included in this term and is charged separately below. On $\Sigma^{*}$ fix $(m,n_{1})$. There are at most $L^{3C_{\Sigma}}$ triples $(r,d,t_{1})$ and at most $O(L)$ candidate primes $p$ for a failure in (A.22). Once $n_{2}$ is fixed, there are at most $L^{C_{\Sigma}}$ choices of $t_{2}$. Every witnessing prime exceeds $z$: it divides a $z$-rough $n_{i}$ or $t_{i}$. Since $(k,N)=1$ and $k\mid mn_{1}-N$, we have $(m,k)=(n_{1},k)=1$ and $n_{2}\equiv n_{1}\pmod{k}$. Each failure places $n_{2}$ in one class modulo $kp$:

• $p\mid(n_{1},n_{2})$: $p\nmid k$; combine $n_{2}\equiv 0\ (p)$ with $n_{2}\equiv n_{1}\ (k)$.

• $p\mid(t_{1},t_{2})$, $p\nmid k$: $p\nmid m$, since $p\mid mn_{1}-N$ and $p\nmid N$; combine $mn_{2}\equiv N\ (p)$ with $n_{2}\equiv n_{1}\ (k)$. If $p\mid k$, then $kt_{2}\mid mn_{2}-N$ and $p\mid t_{2}$ give $kp\mid mn_{2}-N$.

• $p\mid(n_{1},t_{2})$: $p\nmid k$ and $p\nmid m$; combine $mn_{2}\equiv N\ (p)$ with $n_{2}\equiv n_{1}\ (k)$.

• $p\mid(n_{2},t_{1})$: $p\nmid k$; combine $n_{2}\equiv 0\ (p)$ with $n_{2}\equiv n_{1}\ (k)$.

• $p\mid(k,t_{1}t_{2})$: since $p>z\geq P^{+}(d)$, $p\mid r$. If $p\mid t_{2}$, then $kt_{2}\mid mn_{2}-N$ gives $kp\mid mn_{2}-N$. If $p\mid t_{1}$ only, exchange the two copies first.

The number of $n_{2}$ in one class modulo $kp$ is $O(H/(zK)+1)$. Hence

(A.26) $$\mathcal{S}_{3}(\text{bad})\ll MHL^{4C_{\Sigma}+1}\Bigl{[}\frac{H}{zK}+1\Bigr{]}.$$

The near-diagonal $|n_{1}-n_{2}|<HL^{-C_{M}}$ is counted in the same way, with $H/z$ replaced by $HL^{-C_{M}}$. Choose $C_{M}\geq P+4C_{\Sigma}+1$. This gives the first two terms of $R_{0}$. No joint condition $(r,s)=1$ has been used. A prime $p\leq z$ dividing $(r,s)$ divides $d$ and only enlarges $k$. A prime $p>z$ dividing $(r,s)$ divides $k$ and some $t_{i}$. This is the last case of (A.22).

Bad tuples in the density terms. For $\mathcal{S}_{1}$, $\mathcal{S}_{2}$ and $MT$, dropping coprimality leaves the positive density

$$\ll L^{C_{\varphi}}\frac{M}{k^{2}t_{1}t_{2}},$$

since $kt_{1}=o(M)$ and $q/\varphi(q)\leq L^{C_{\varphi}}$ for $q\leq N^{O(1)}$. (In $\mathcal{S}_{2}$ the $m$-sum is nonzero only if $n_{1}$ is invertible modulo $kt_{1}$, by $(k,N)=(t_{1},N)=1$ and $kt_{1}\mid mn_{1}-N$. Then $m$ lies in one class modulo $kt_{1}$.) For $\mathcal{Z}$ retain $\omega$ and $n_{2}\equiv n_{1}\ (k)$, with weight $M/(kt_{1}t_{2})$. In particular $(n_{1},k)=(n_{2},k)=1$ there. Without the bad condition these weights sum to $\ll L^{C_{\varphi}}MH^{2}/K$ and $\ll MH^{2}/K$, using $\sum_{t\sim S}1/t\leq 1$, $k=rd\asymp K$ and, in $\mathcal{Z}$, $O(H/K)$ values of $n_{2}$ in one class modulo $k$. As in $\mathcal{S}_{3}$ every witnessing prime $p$ exceeds $z$. In each case below the prime $p$ divides $n_{1}$ or $t_{1}$, which is fixed first together with $p$. Since $\omega(n_{1}),\omega(t_{1})\ll L$, there are $O(L)$ choices of $p$. We use

$$\begin{gathered}\#\{n_{2}\sim H:\ p\mid n_{2}\}\leq\frac{H}{p}+1,\qquad\sum_{\begin{subarray}{c}t_{2}\sim S\\ p\mid t_{2}\end{subarray}}\frac{1}{t_{2}}\leq\frac{1}{S}\cdot\frac{2S}{p}=\frac{2}{p},\\ \#\{r\leq\tfrac{5}{2}U_{1}:\ p\mid r\}\leq\frac{5U_{1}}{2p};\end{gathered}$$

the last two have no $+1$, because they count multiples of $p$ in $[1,2S]$ and $[1,\frac{5}{2}U_{1}]$.

• $p\mid(n_{1},n_{2})$: fix $n_{1}$ and $p\mid n_{1}$. In $\mathcal{S}_{1},\mathcal{S}_{2},MT$ there are $\leq H/z+1$ values of $n_{2}$. In $\mathcal{Z}$, $p\nmid k$ since $(n_{1},k)=1$; combining $n_{2}\equiv 0\ (p)$ with $n_{2}\equiv n_{1}\ (k)$ leaves $O(H/(zK)+1)$ values of $n_{2}$.

• $p\mid(t_{1},t_{2})$: fix $t_{1}$ and $p\mid t_{1}$; the $t_{2}$-sum is at most $2/p\leq 2/z$. The $n_{2}$-count is unchanged; $p\mid k$ is allowed.

• $p\mid(n_{1},t_{2})$: fix $n_{1}$ and $p\mid n_{1}$; the $t_{2}$-sum is at most $2/z$, and the $n_{2}$-count is unchanged.

• $p\mid(n_{2},t_{1})$: fix $t_{1}$ and $p\mid t_{1}$. In $\mathcal{S}_{1},\mathcal{S}_{2},MT$ there are $\leq H/z+1$ values of $n_{2}$. In $\mathcal{Z}$, $p\mid k$ would contradict $(n_{2},k)=1$, so $p\nmid k$; combining $n_{2}\equiv 0\ (p)$ with $n_{2}\equiv n_{1}\ (k)$ leaves $O(H/(zK)+1)$ values of $n_{2}$.

• $p\mid(k,t_{1}t_{2})$: the weights are symmetric in the two copies, so let $p\mid t_{1}$; fix $t_{1}$ and $p\mid t_{1}$. Since $p>z\geq P^{+}(d)$, $p\mid r$, and there are at most $5U_{1}/(2p)\leq 5U_{1}/(2z)$ values of $r$. The $d$-, $t_{2}$- and $n$-counts are unchanged.

In every case the bad part of each density term is $\ll MHL^{C_{\varphi}+1}[H/(zK)+1]$. This is part of $R_{0}$. The near-diagonal is counted in the same way, with $H/z$ replaced by $HL^{-C_{M}}$. There are $O(HL^{-C_{M}}+1)$ values of $n_{2}$, and $O(HL^{-C_{M}}/K+1)$ in one class modulo $k$. Hence the near-diagonal is $\ll MH^{2}L^{C_{\varphi}-C_{M}}/K+MHL^{C_{\varphi}}$. This is part of $R_{0}$ once also $C_{M}\geq P+C_{\varphi}$. Thus all bad tuples may be removed from $\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{3},\mathcal{Z},MT$ at the cost $R_{0}$.

Good tuples. On good tuples the two congruences in $\mathcal{S}_{3}$ are compatible exactly on the support $\omega(\mathbf{t})=1$ and $n_{1}\equiv n_{2}\pmod{k}$. Then $m$ lies in one class $\mu$ modulo $kt_{1}t_{2}$, with

(A.27) $$\frac{\mu}{kt_{1}t_{2}}\equiv\frac{N\overline{n_{1}t_{2}}}{kt_{1}}+\frac{N\overline{n_{2}kt_{1}}}{t_{2}}\pmod{1}.$$

Poisson summation gives

(A.28) $$\sum_{m\equiv\mu\ (kt_{1}t_{2})}\psi\Bigl{(}\frac{m}{M}\Bigr{)}=\frac{M}{kt_{1}t_{2}}\sum_{h\in\mathbb{Z}}\hat{\psi}\Bigl{(}\frac{hM}{kt_{1}t_{2}}\Bigr{)}e\Bigl{(}\frac{h\mu}{kt_{1}t_{2}}\Bigr{)}.$$

The term $h=0$ gives $\mathcal{Z}(\mathrm{good})$. All three expanded summands vanish unless $\omega(\mathbf{t})=1$. The following main-term computations are on that support. For $\mathcal{S}_{1}$ the smooth coprime $m$-sum is $M\hat{\psi}(0)\varphi(kt_{1}t_{2})/(kt_{1}t_{2})+O(\tau(kt_{1}t_{2}))$. Division by $\varphi(kt_{1})\varphi(kt_{2})=\varphi(k)^{2}\varphi(t_{1})\varphi(t_{2})$ gives $MT(\mathrm{good})$ with total error $N^{o(1)}H^{2}/K$. For $\mathcal{S}_{2}$ expand $\mathbf{1}_{(m,t_{2})=1}$ by Möbius inversion. The main term is $M\hat{\psi}(0)\varphi(t_{2})/(kt_{1}t_{2})$, with error $O(\tau(t_{2}))$. Division by $\varphi(kt_{2})=\varphi(k)\varphi(t_{2})$ gives $MT(\mathrm{good})$ with total error $N^{o(1)}H^{2}S$. Completing $\mathcal{Z}$ and $MT$ back to all tuples costs $R_{0}$. This proves (A.24).

Step 5: the zero mode

Expanding the centred class counts gives the exact identity, valid for all tuples including those with gcd failures,

(A.29) $$\mathcal{Z}(\mathrm{all})-MT(\mathrm{all})=M\hat{\psi}(0)\sum_{r,d,t_{1},t_{2}}\frac{\psi(r/U_{1})\mathbf{1}_{(rd,N)=1}\lambda_{d}(t_{1})\overline{\lambda_{d}(t_{2})}}{kt_{1}t_{2}}\sum_{b\bmod k}^{*}\Delta_{a^{(t_{1})}}(k,b)\overline{\Delta_{a^{(t_{2})}}(k,b)}.$$

For fixed $(k,t_{1},t_{2})$ both sides equal

$$\sum_{b\bmod k}^{*}A_{1}(b)\overline{A_{2}(b)}-\frac{1}{\varphi(k)}\sum_{b\bmod k}^{*}A_{1}(b)\sum_{b\bmod k}^{*}\overline{A_{2}(b)},\qquad A_{i}(b)=\sum_{\begin{subarray}{c}n\equiv b\ (k)\\ (n,t_{i})=1\end{subarray}}a_{n}.$$

Use $|u\bar{v}|\leq(|u|^{2}+|v|^{2})/2$ and

$$V(k;a^{(t)})\leq 2V(k;a)+2V(k;a-a^{(t)}).$$

The sums $\sum_{t\sim S}1/t$ are $O(1)$, a given $k$ has at most $\tau(k)$ factorizations $k=rd$, and $k\leq 5K$. By (A.18) with $K^{\prime}=K$ and (A.19),

(A.30) $$|\mathcal{Z}(\mathrm{all})-MT(\mathrm{all})|\ll_{P}\frac{MH^{2}}{K}\Bigl{[}L^{-P}+\frac{L^{3}}{z}\Bigr{]}.$$

Only $|\lambda_{d}|\leq 1$ is used, so $\lambda_{d}$ may vary with $d$. The factor $1/K$ is cancelled by $KM$ in (A.20).

By (A.20), (A.24), (A.30) and (A.21), the contributions to $|\mathcal{S}_{0}|$ of all terms except $M\mathcal{E}$ are

(A.31) $$\ll NL^{-P/2}+NL^{C}\Bigl{[}z^{-1/2}+\Bigl{(}\frac{K}{H}\Bigr{)}^{1/2}\Bigr{]}+N^{1+o(1)}\Bigl{[}M^{-1/2}+\Bigl{(}\frac{KS}{M}\Bigr{)}^{1/2}\Bigr{]}+N^{.857}.$$

For the last term, $M^{1/2}(H+KS)\leq N^{(1+h)/2+o(1)}+N^{(1-h)/2+v+o(1)}$, and $(1+h)/2\leq.73+o(1)$, $(1-h)/2+v\leq.856034$.

Step 6: nonzero frequencies

Fix $d\sim D$ and let $\mathcal{E}_{d}$ be the part of $\mathcal{E}$ with this $d$:

$$\mathcal{E}_{d}=\sum_{\begin{subarray}{c}\text{good }\mathbf{t},\ \omega(\mathbf{t})=1\\ n_{1}\equiv n_{2}\ (k)\end{subarray}}\frac{c(\mathbf{t})}{kt_{1}t_{2}}\sum_{h\neq 0}\hat{\psi}\Bigl{(}\frac{hM}{kt_{1}t_{2}}\Bigr{)}e\Bigl{(}\frac{h\mu}{kt_{1}t_{2}}\Bigr{)}.$$

Put $\eta=10^{-9}$ and

(A.32) $$H_{*}=C\frac{KS^{2}N^{\eta}}{M},$$

with $C$ larger than every $kt_{1}t_{2}/(KS^{2})$. Since $|\hat{\psi}(\xi)|\ll_{j}|\xi|^{-j}$, the terms $|h|>H_{*}$ contribute $O(N^{-90})$ in total, taking $j>200/\eta$. If $H_{*}<1$, $\mathcal{E}_{d}$ is negligible. This covers $t=1$, $S=1/2$, where $H_{*}\ll KN^{\eta}/M\leq N^{-.07}$. Hence we may assume $S\geq 1$ and $H_{*}\geq 1$.

Reciprocity. For coprime $u,v$, $\bar{u}/v+\bar{v}/u\equiv 1/(uv)\pmod{1}$. Applied to the first term of (A.27), with $u=n_{1}t_{2}$ and $v=kt_{1}$, it gives

$$e\Bigl{(}\frac{h\mu}{kt_{1}t_{2}}\Bigr{)}=e\Bigl{(}\frac{Nh(n_{1}-n_{2})\overline{n_{2}kt_{1}}}{n_{1}t_{2}}\Bigr{)}e\Bigl{(}\frac{Nh}{kt_{1}t_{2}n_{1}}\Bigr{)}.$$

Put $f_{0}=(n_{1}-n_{2})/k$, a nonzero integer on good tuples. Cancelling $k$,

(A.33) $$e\Bigl{(}\frac{Nhf_{0}\overline{n_{2}t_{1}}}{n_{1}t_{2}}\Bigr{)}.$$

The conditions on $t_{1}$ are

(A.34) $$(t_{1},Nc)=1,\qquad(t_{1},J)=1,\qquad b=n_{2},\quad c=n_{1}t_{2},\quad J=kn_{2}.$$

The first is kept. To separate the second, colour every prime in $[z,N^{3}]$ blue or red independently with probability $1/2$. Since $t_{1}$ is $z$-rough,

(A.35) $$\mathbf{1}_{(t_{1},J)=1}=\mathbb{E}_{\rm col}\Bigl{[}2^{\omega(t_{1})+\omega_{z}(J)}\mathbf{1}_{\text{all primes of }t_{1}\text{ blue}}\,\mathbf{1}_{\text{all primes}\ \geq z\text{ of }J\text{ red}}\Bigr{]},$$

where $\omega_{z}(J)$ counts the primes $\geq z$ of $J$. A shared prime makes the event empty. Otherwise its probability is the reciprocal of the power of $2$. Both counts are $O(\ell^{3})$, so the multipliers are $\exp(O(\ell^{3}))=N^{o(1)}$. Fix one colouring.

Split $f_{0}$ and $h$ into signed dyadic blocks $|f_{0}|\sim F$ and $h=\varepsilon h_{+},\ h_{+}\sim H_{0}>0$, with $F\ll H/K$ and $H_{0}\leq H_{*}$. Below, $h_{+}$ is renamed $h>0$. Fix $d$, this block, a sign $\varepsilon$, and one prime colouring from (A.35). The outer tuple is $\mathbf{o}=(r,n_{1},n_{2},t_{2})$, with $k=rd$, $b=n_{2}$, $c=n_{1}t_{2}$ and $f_{0}=(n_{1}-n_{2})/k$. Its support contains $k\mid n_{1}-n_{2}$, the near-diagonal exclusion, and the conditions on $(r,d,n_{1},n_{2},t_{2})$ inherited from the good tuples and $\omega=1$. It also requires every prime $\geq z$ of $J=kn_{2}$ to be red. The remaining $t_{1}$-conditions are $(t_{1},Nc)=1$, $t_{1}\sim S$ and all primes of $t_{1}$ blue. The colouring enforces $(t_{1},J)=1$. Write

$$u=\frac{hS}{H_{0}t_{1}},\quad Y_{2}(\mathbf{o})=\frac{\varepsilon H_{0}M}{kt_{2}S},\quad Y_{1}(\mathbf{o})=\frac{\varepsilon NH_{0}}{kt_{2}n_{1}S},\quad W_{\mathbf{o}}(u)=w_{0}(u)\hat{\psi}(Y_{2}(\mathbf{o})u)e(Y_{1}(\mathbf{o})u),$$

where $w_{0}$ is a fixed smooth cutoff supported in $(1/4,4)$ and equal to $1$ on $[1/2,2]$. Since $H_{0}\leq H_{*}$, $k\asymp K$, $t_{2}\asymp S$, and $N/(n_{1}M)\asymp 1$, uniformly in all outer tuples, $|Y_{1}(\mathbf{o})|+|Y_{2}(\mathbf{o})|\ll N^{\eta}$. On the original support, the remaining Fourier weight is exactly $W_{\mathbf{o}}(hS/(H_{0}t_{1}))$. Define the Mellin transform and its uniform majorant by

$$\widetilde{W}_{\mathbf{o}}(it)=\int_{0}^{\infty}W_{\mathbf{o}}(u)u^{it}\,\frac{du}{u},\qquad m(t)=C_{\psi,w_{0}}\min\{1,N^{2\eta}(1+|t|)^{-2}\}.$$

The compact support, two integrations by parts, and the bounds on $Y_{1},Y_{2}$ give

$$\sup_{\mathbf{o},t}\frac{|\widetilde{W}_{\mathbf{o}}(it)|}{m(t)}\leq 1,\qquad\int_{\mathbb{R}}m(t)\,dt\ll_{\psi,w_{0}}N^{\eta},$$

with the fixed constant $C_{\psi,w_{0}}$ chosen large enough. Mellin inversion is the outer-dependent identity

$$W_{\mathbf{o}}\Bigl{(}\frac{hS}{H_{0}t_{1}}\Bigr{)}=\frac{1}{2\pi}\int_{\mathbb{R}}m(t)\underbrace{\Bigl{(}\frac{H_{0}}{S}\Bigr{)}^{it}\frac{\widetilde{W}_{\mathbf{o}}(it)}{m(t)}}_{\text{outer coefficient}}\underbrace{h^{-it}t_{1}^{it}}_{\text{inner factors}}\,dt.$$

In particular $\widetilde{W}_{\mathbf{o}}(it)/m(t)$ is not an inner $t_{1}$-weight: it depends on $k,t_{2},n_{1}$ and the fixed sign and block.

The $t$-integral precedes Cauchy. Let $B_{\rm in}\geq\sup_{t_{1}}2^{\omega(t_{1})}$ and $B_{\rm out}\geq\sup_{\mathbf{o}}2^{\omega_{z}(kn_{2})}$ on this block, with $B_{\rm in}B_{\rm out}=\exp(O(\ell^{3}))=N^{o(1)}$. For fixed real Mellin parameter $t$ set

$$\lambda^{\prime}_{v}(t) =\frac{S}{v}\lambda_{d}(v)\frac{2^{\omega(v)}}{B_{\rm in}}\mathbf{1}_{\text{all primes of }v\text{ blue}}\,v^{it},$$

$$\eta_{h}(t) =h^{-it}\mathbf{1}_{h\sim H_{0}}\mathbf{1}_{h\leq H_{*}},$$

$$A_{\mathbf{o}}(t) =\frac{KS}{kt_{2}}\,\psi(r/U_{1})\overline{\lambda_{d}(t_{2})}a_{n_{1}}\overline{a_{n_{2}}}\frac{2^{\omega_{z}(kn_{2})}}{B_{\rm out}}\mathbf{1}_{\text{all primes }\geq z\text{ of }kn_{2}\text{ red}}\,\Bigl{(}\frac{H_{0}}{S}\Bigr{)}^{it}\frac{\widetilde{W}_{\mathbf{o}}(it)}{m(t)}.$$

The factor $\mathbf{1}_{h\leq H_{*}}$ is the one-bounded mask on a terminal frequency block; it is independent of the outer tuple. The sum over $\mathbf{o}$ is restricted to the outer support just specified. Choose the fixed cutoff $\psi$ with $0\leq\psi\leq 1$. On the dyadic support $KS/(kt_{2})\leq 2$. By the definitions of $m$, $B_{\rm in}$ and $B_{\rm out}$, every other factor above has modulus at most one. Hence $|A_{\mathbf{o}}(t)|\leq 2$ and $|\lambda^{\prime}_{v}(t)|,|\eta_{h}(t)|\leq 1$, uniformly in the real $t$ and all original outer parameters. Using $1/(kt_{1}t_{2})=(KS^{2})^{-1}(KS/(kt_{2}))(S/t_{1})$, the contribution of this block and colouring to $\mathcal{E}_{d}$ is

$$\frac{B_{\rm in}B_{\rm out}}{2\pi KS^{2}}\int_{\mathbb{R}}m(t)\sum_{\mathbf{o}}A_{\mathbf{o}}(t)\sum_{\begin{subarray}{c}v\sim S\\ (v,Nc)=1\end{subarray}}\ \sum_{h\sim H_{0}}\lambda^{\prime}_{v}(t)\eta_{h}(t)e\Bigl{(}\frac{\varepsilon Nhf_{0}\overline{bv}}{c}\Bigr{)}\,dt.$$

Here $(b,c)=1$ on the good support. The colour average and the $O(L^{2})$ signed dyadic blocks are taken after the estimate. Their cost is $N^{o(1)}$.

Second square, pointwise in $t$. Group the outer tuples by $(b,c,f_{0})$. For fixed $(b,c,f_{0})$, the factorization $n_{1}t_{2}=c$ has at most $\tau(c)=N^{o(1)}$ choices. Then $k=(n_{1}-b)/f_{0}$ and $r=k/d$ are determined. The number of outer tuples is $\ll N^{o(1)}H^{2}S$. Let $\rho\geq 0$ be a smooth majorant, at least $1$ on their ranges $b\ll H$, $c\ll HS$. Since $b=n_{2}$ is $z$-rough, $(b,p_{0})=1$ for the least prime $p_{0}$ not dividing $N$, as $p_{0}\ll\log N<z$. Define, for each real $t$,

(A.36) $$W_{4}(t) =\sum_{\begin{subarray}{c}(b,c)=1\\ (b,p_{0})=1\end{subarray}}\rho(b,c)\sum_{f_{0}\sim F}\Bigl{|}\sum_{\begin{subarray}{c}v\sim S\\ (v,Nc)=1\end{subarray}}\sum_{h\sim H_{0}}\lambda^{\prime}_{v}(t)\eta_{h}(t)e\Bigl{(}\frac{\varepsilon Nhf_{0}\overline{bv}}{c}\Bigr{)}\Bigr{|}^{2},$$

$$|\mathcal{E}_{d}| \ll N^{o(1)}\frac{H}{KS^{3/2}}\int_{\mathbb{R}}m(t)W_{4}(t)^{1/2}\,dt\leq N^{\eta+o(1)}\frac{H}{KS^{3/2}}(W_{4}^{*})^{1/2},$$

$$W_{4}^{*} =\sup_{\begin{subarray}{c}\mathrm{signed\ blocks},\ \mathrm{colourings}\\ t\in\mathbb{R}\end{subarray}}W_{4}(t).$$

Here $W_{4}(t)$ is formed with a fixed block and colouring; the supremum is taken afterwards. The displayed bound for the full $\mathcal{E}_{d}$ follows from the colour expectation and the sum over the $O(L^{2})$ blocks. Their cost is a uniform $N^{o(1)}$. Thus no outer coefficient is moved through Cauchy or the $t$-integral. Cauchy is applied inside the integral for each $t$. No outer-dependent Mellin factor appears in $\lambda^{\prime}_{v}(t)$.

For every fixed $t$, expand the square with $z_{1}=vv^{\prime}$, $k_{0}=h_{1}v^{\prime}-h_{2}v$ and $y_{1}=f_{0}k_{0}$. Since $|\lambda^{\prime}_{v}(t)|,|\eta_{h}(t)|\leq 1$, the diagonal $k_{0}=0$ obeys, uniformly in $t$,

(A.37) $$W_{\rm diag}(t)\leq N^{o(1)}H_{*}S\cdot H\cdot F\cdot HS\leq N^{\eta+o(1)}\frac{H^{4}S^{4}}{N}.$$

For $k_{0}\neq 0$, write $m=|y_{1}|\ll FH_{*}S$ and separate the two signs of $y_{1}$. The off-diagonal is a sum of expressions

(A.38) $$\sum_{z_{1},m}b^{\pm}_{z_{1},m}(t)\sum_{\begin{subarray}{c}(b,c)=(b,p_{0})=(z_{1},c)=1\end{subarray}}\rho(b,c)e\Bigl{(}\pm\frac{Nm\overline{z_{1}b}}{c}\Bigr{)},$$

where $b^{\pm}_{z_{1},m}(t)$ depends only on $(d,S,F,H_{0},\varepsilon,\text{colouring},t;z_{1},m)$, and not on $(b,c,f_{0})$ or the original outer tuple. Its summands are products of the one-bounded weights $\lambda^{\prime}_{v}(t),\eta_{h}(t)$, and $|b^{\pm}_{z_{1},m}(t)|\leq B_{z_{1},\pm m}$ for the counting majorant defined below. The support has $(z_{1},Np_{0})=1$.

Coefficient norm. Fix $C>0$. For $1\leq S\leq N^{C}$, $1\leq H^{\prime}\leq S$ and $1/2\leq F\leq N^{C}$ let $B_{z_{1},y_{1}}$ count $t,t^{\prime}\sim S$, $(tt^{\prime},N)=1$, $1\leq|h_{1}|,|h_{2}|\leq H^{\prime}$, $f_{0}\sim F$, with $tt^{\prime}=z_{1}$ and $f_{0}(h_{1}t^{\prime}-h_{2}t)=y_{1}\neq 0$. Write $y_{N}$ for the largest divisor of $y$ composed of primes dividing $N$. Then

(A.39) $$\sum_{z_{1},y_{1}\neq 0}B_{z_{1},y_{1}}^{2}(y_{1,N})^{\theta}\leq N^{o(1)}FH^{\prime 2}S^{2}.$$

At fixed $(z_{1},y_{1})$ there are at most $\tau(z_{1})\tau(|y_{1}|)=N^{o(1)}$ choices of $(t,t^{\prime},f_{0})$. By Cauchy’s inequality the left side is at most

$$N^{o(1)}\Bigl{[}\sum_{f\leq 2F}f_{N}^{\theta}\Bigr{]}\mathcal{Q},\qquad\mathcal{Q}=\sum_{t,t^{\prime}}\sum_{k_{0}\neq 0}(k_{0,N})^{\theta}\mathcal{N}_{t,t^{\prime}}(k_{0})^{2},$$

where $\mathcal{N}_{t,t^{\prime}}(k_{0})$ counts $(h_{1},h_{2})$ with $h_{1}t^{\prime}-h_{2}t=k_{0}$. Since $f_{N}^{\theta}\leq\sum_{d\mid f,\ d\mid N^{\infty}}d^{\theta}$, the $f$-sum is at most $2FP_{0}$, with $P_{0}=\prod_{p\mid N}(1-p^{\theta-1})^{-1}$.

Write $g_{0}=(t,t^{\prime})$, $t=g_{0}u$, $t^{\prime}=g_{0}v$ and $S^{\prime}=S/g_{0}$. A fibre of $(h_{1},h_{2})\mapsto k_{0}$ is an arithmetic progression with step $(u,v)$. It has at most $1+2H^{\prime}/S^{\prime}$ points, and at most one if $S^{\prime}\geq 2H^{\prime}$. Also

$$(k_{0,N})^{\theta}=\sum_{d\mid k_{0},\ d\mid N^{\infty}}\phi_{\theta}(d),\qquad 0\leq\phi_{\theta}(d)\leq d^{\theta},$$

by telescoping each prime power. Let $A_{d}(t,t^{\prime})$ count $(h_{1},h_{2})$ with $k_{0}\neq 0$ and $d\mid k_{0}$. Then

(A.40) $$\mathcal{Q}\leq\sum_{\begin{subarray}{c}d\mid N^{\infty}\\ d\leq 4H^{\prime}S\end{subarray}}d^{\theta}\Bigl{[}\sum_{t,t^{\prime}}A_{d}(t,t^{\prime})+\sum_{t,t^{\prime}:\,S^{\prime}<2H^{\prime}}\frac{2H^{\prime}}{S^{\prime}}A_{d}(t,t^{\prime})\Bigr{]}.$$

For the first sum write $(h_{1},h_{2})=e\,\mathbf{w}$ with $\mathbf{w}$ primitive, $\|\mathbf{w}\|_{\infty}=j\leq H^{\prime}$. There are at most $8j$ such $\mathbf{w}$ and $e\leq H^{\prime}/j$. Put $d_{e}=d/(d,e)$. Since $tt^{\prime}$ is prime to $d$, a contributing $\mathbf{w}$ has $(w_{1},d_{e})=1$ (and $w_{1}=0$ forces $d_{e}=1$). Counting $t$ and then $t^{\prime}$ in one class modulo $d_{e}$, and using $\sum_{e\leq E}(d,e)\leq\tau(d)E$,

(A.41) $$\sum_{t,t^{\prime}}A_{d}(t,t^{\prime})\leq C\Bigl{[}\tau(d)\frac{H^{\prime 2}S^{2}}{d}+H^{\prime 2}S\Bigr{]}.$$

For the second sum $(g_{0},d)=1$. There are $O(H^{\prime}S^{\prime}/d)$ nonzero levels of $h_{1}v-h_{2}u$ divisible by $d$. Each has $O(H^{\prime}/S^{\prime})$ points, so $A_{d}\leq CH^{\prime 2}/d$. Summing over $g_{0}$, with $\sum_{g_{0}\leq 2S}(S/g_{0}+1)^{2}(g_{0}/S)\leq S(8+\log 2S)$,

(A.42) $$\sum_{t,t^{\prime}:\,S^{\prime}<2H^{\prime}}\frac{2H^{\prime}}{S^{\prime}}A_{d}\leq C(1+\log 2S)\frac{H^{\prime 3}S}{d}\leq C(1+\log 2S)\frac{H^{\prime 2}S^{2}}{d}.$$

Finally

$$\sum_{d\mid N^{\infty}}\tau(d)d^{\theta-1}=P_{0}^{2},\qquad\sum_{\begin{subarray}{c}d\mid N^{\infty}\\ d\leq 4H^{\prime}S\end{subarray}}d^{\theta}\leq(4H^{\prime}S)^{1/2}P_{2},\qquad P_{2}=\prod_{p\mid N}(1-p^{\theta-1/2})^{-1}.$$

Since $H^{\prime}\leq S$, $H^{\prime 2}S(H^{\prime}S)^{1/2}\leq H^{\prime 2}S^{2}$. Both $P_{0}$ and $P_{2}$ are $N^{o(1)}$: with the primes of $N$ in increasing order, $\log P_{2}\ll\sum_{i\leq\omega(N)}p_{i}^{\theta-1/2}\ll(\log N)^{\theta+1/2}$. This proves (A.39).

With $H^{\prime}=H_{*}$ we have $H_{*}/S\ll KSN^{\eta}/M\leq N^{-.0171}$ by (A.21). For every real Mellin parameter $t$, fixed $d,S,F,H_{0},\varepsilon$ and colouring, the coefficient sequences $b^{\pm}_{z_{1},m}(t)$ in (A.38) satisfy the uniform weighted estimate

$$\sum_{z_{1},m\geq 1}(m_{N})^{\theta}|b^{\pm}_{z_{1},m}(t)|^{2}\leq\sum_{z_{1},y_{1}\neq 0}B_{z_{1},y_{1}}^{2}(y_{1,N})^{\theta}\ll N^{o(1)}FH_{*}^{2}S^{2}.$$

This follows from (A.39), since the inner complex weights have modulus at most one. In particular the implied constant is independent of $t$ and of all original outer data $(r,k,n_{1},n_{2},t_{2},b,c,f_{0})$. The dependence on these data was placed in $A_{\mathbf{o}}(t)$ before Cauchy. Writing $\widetilde{b}_{t}^{\pm}(j)$ for the slices of Paper I, (2.43), with $n_{T}=p_{0}m$ and $a_{T}=\pm N$, gives

(A.43) $$\sup_{t\in\mathbb{R}}\sum_{j\mid N^{\infty}}j^{\theta}\|\widetilde{b}_{t}^{\pm}(j)\|_{2}^{2}\ll N^{o(1)}F(H_{*}S)^{2},\qquad\sup_{t\in\mathbb{R}}\|b_{t}^{\pm}\|_{2}^{2}\ll N^{o(1)}F(H_{*}S)^{2}.$$

Since $(p_{0},N)=1$, the $N$-part of $n_{T}=p_{0}m$ is precisely $m_{N}$. This is the norm required by Paper I, (2.42)–(2.43), for every $t$, without any Mellin factor in the inner coefficients.

Application of Paper I, Theorem 2.8. In (A.38) put

(A.44) $$\begin{gathered}r_{T}=z_{1}\sim Z_{1},\quad s_{T}=p_{0},\quad R_{T}=Z_{1},\quad S_{T}=p_{0}/2,\\ d_{T}=b,\quad c_{T}=c,\quad n_{T}=p_{0}m,\quad Y:=FH_{*}S,\quad N_{T}=Cp_{0}Y\leq N^{o(1)}Y,\quad p_{0}\ll\log N,\quad a_{T}=\pm N,\end{gathered}$$

with $A=N$. Since $t,t^{\prime}\sim S$, $z_{1}=tt^{\prime}\in(S^{2},4S^{2}]$, which meets two dyadic windows. Split the $z_{1}$-sum into $z_{1}\in(S^{2},2S^{2}]$ and $z_{1}\in(2S^{2},4S^{2}]$. Apply the theorem to each, with $Z_{1}=S^{2}$ and $Z_{1}=2S^{2}$ respectively. This costs a factor $2$ in the final bound, and $Z_{1}\asymp S^{2}$ in both windows. Then $s_{T}\in(S_{T},2S_{T}]$ and $S_{T}\geq 3/2$, since $N$ is even and $p_{0}\geq 3$. The support has $(a_{T},r_{T}s_{T})=1$. The domain of the theorem is exactly

$$(z_{1}b,p_{0}c)=1\iff(b,c)=(z_{1},c)=(z_{1},p_{0})=(b,p_{0})=1,$$

and $(z_{1},p_{0})=1$ holds by roughness. Exactly as in Paper I, (3.82)–(3.84), reducing the inverse modulo $c$ gives

$$e\Bigl{(}\frac{a_{T}n_{T}\overline{z_{1}b}}{p_{0}c}\Bigr{)}=e\Bigl{(}\pm\frac{Nm\overline{z_{1}b}}{c}\Bigr{)},$$

also when $p_{0}\mid c$. Split $\rho$ into finitely many smooth pieces on strict dyadic windows in $(b,c)$. Multiply by fixed smooth cutoffs equal to $1$ on the supported $n_{T},z_{1},p_{0}$. At $n_{T}\geq 1$ use a cutoff vanishing below $1/2$. All derivatives satisfy Paper I, (2.39), with uniform seminorms. Take $\epsilon=\epsilon_{0}=10^{-12}$. All parameters lie in $[1,N^{10}]$, so $\mathcal{X}=2+ACDNRS\leq 3N^{60}$ in Paper I, (2.41). The prefactor there satisfies $\mathcal{X}^{\epsilon+18\epsilon_{0}}=\mathcal{X}^{19\cdot 10^{-12}}\leq N^{1.2\cdot 10^{-9}}$ for large $N$. Write $\delta_{X}=1.2\cdot 10^{-9}$ for this fixed loss. This loss and the factors $N^{\eta}=N^{10^{-9}}$ are far below three power savings obtained below. These are $N^{-.00829}$, $N^{-.1317}$ and $N^{-.2914}$, one for each term of the bound. It leaves the final saving $N^{-.001}$ intact. Powers of $p_{0}$ are $N^{o(1)}$.

Put $B=H$ and $C_{\mathrm{aux}}=HS$. Then $Y\ll Z_{1}$, since $Y/Z_{1}\ll N^{\eta}H^{2}S/N\leq N^{-.008}$ (see (A.47)). On a nonempty $f_{0}$-block the near-diagonal removal gives $F\geq cHL^{-C_{M}}/K$, hence $NY\geq cH^{2}S^{3}N^{\eta}L^{-C_{M}}\geq C_{\mathrm{aux}}BZ_{1}$. All exponents of $B,C_{\mathrm{aux}}$ in Paper I, (2.42), are nonnegative. By (2.42) and (A.43),

$$J_{*}^{2}\leq N^{o(1)}\bigl{[}N^{\theta}(C_{\mathrm{aux}}BZ_{1}+NY)+C_{\mathrm{aux}}BZ_{1}(C_{\mathrm{aux}}/Z_{1})^{2\theta}+B^{2}Y\bigr{]}Z_{1}F(H_{*}S)^{2};$$

the regular spectral product is $C_{\mathrm{aux}}(C_{\mathrm{aux}}B\sqrt{Z_{1}})^{2\theta}Z_{1}^{1-\theta}(BZ_{1})^{1-2\theta}=C_{\mathrm{aux}}^{1+2\theta}BZ_{1}^{2-2\theta}$. With $B^{2}<N$, $F\ll H/K$ and (A.32),

(A.45) $$\frac{J_{*}^{2}}{H^{4}S^{6}}\leq N^{3\eta+o(1)}[L_{A}+L_{R}],\qquad L_{A}=\frac{N^{\theta}HKS^{5}}{N^{2}},\qquad L_{R}=\Bigl{(}\frac{H}{S}\Bigr{)}^{2\theta}\frac{HKS^{5}}{N^{2}}.$$

Fix a real $t$. Apply the diagonal bound (A.37), Paper I, Theorem 2.8 with the uniform coefficient norm (A.43), and (A.45). The resulting estimate is the same for every real $t$. Thus their supremum $W_{4}^{*}$ satisfies

(A.46) $$|\mathcal{E}_{d}|\ll N^{\eta+o(1)}\frac{H}{KS^{3/2}}(W_{4}^{*})^{1/2},\qquad\frac{W_{4}^{*}}{H^{2}S^{3}}\ll N^{\eta+o(1)}\frac{H^{2}S}{N}+N^{3\eta/2+\delta_{X}+o(1)}(L_{A}+L_{R})^{1/2}.$$

All Mellin and colouring costs remain $N^{\eta+o(1)}$ as recorded in (A.36). No outer-dependent coefficient enters the norm.

Step 7: exponent reserves

With $S\ll V_{1}/D$, $K=U_{1}D$ and the three scale conditions, $H^{2}S\leq H^{2}V_{1}$, $KS^{5}\leq U_{1}V_{1}^{5}$ and $H^{2\theta}KS^{5-2\theta}\leq H^{2\theta}U_{1}V_{1}^{5-2\theta}$. The smooth part $D$ only helps. Using $U_{1}\leq HN^{-\sigma_{\mathrm{p}}}$, $V_{1}\leq N^{v+w+\sigma_{\mathrm{p}}}/H$ and $U_{1}V_{1}\leq N^{v}$,

(A.47) $$1-v-h-\sigma_{\mathrm{p}}-w \geq\tfrac{8299}{1000000},$$

$$2-\theta-5v+3h-4(\sigma_{\mathrm{p}}+w) \geq\tfrac{65873}{500000},$$

$$2-(5-2\theta)v+(3-4\theta)h-(4-2\theta)(\sigma_{\mathrm{p}}+w) \geq\tfrac{13989529}{48000000},$$

on $1/3\leq h\leq 23/50$. The first minimum is at $h=.46$, the other two at $h=1/3$. These are the savings of $H^{2}S/N$, $L_{A}$ and $L_{R}$. Hence $H^{2}S/N\leq N^{-.00829}$, $L_{A}\leq N^{-.1317}$, $L_{R}\leq N^{-.2914}$, and (A.46) gives

(A.48) $$|\mathcal{E}_{d}|\leq C\frac{H^{2}}{K}N^{-.001}.$$

Summing over the $O(D)$ values of $d$, $|\mathcal{E}|\leq CDH^{2}N^{-.001}/K$. Its contribution to $|\mathcal{S}_{0}|$ through (A.20) is

$$\bigl{[}KM\cdot M\cdot DH^{2}N^{-.001}/K\bigr{]}^{1/2}\leq CND^{1/2}N^{-.0005}\leq N^{1-.0004},$$

since $D\leq y=N^{o(1)}$.

Step 8: parameter order and conclusion of (A.16)

Fix $A$. Choose $P\geq 2A+20$; then $C_{\Sigma}$; then $C_{M}\geq\max(P+4C_{\Sigma}+1,\,P+C_{\varphi})$. Here $C_{\varphi}$ is the absolute exponent with $q/\varphi(q)\leq L^{C_{\varphi}}$. The smooth weights, $\eta$, $\epsilon$ and $\epsilon_{0}$ are fixed numbers. Choose the threshold for $N$ last. By (A.31) and Step 7,

$$|\mathcal{S}_{0}|\ll NL^{-A-3}+N^{1-\sigma_{\mathrm{p}}/8}L^{C},$$

since $(K/H)^{1/2}\leq N^{-\sigma_{\mathrm{p}}/4}$ and $z^{-1/2}$ beats every power of $L$. Sum over the $O(L)$ shells of $D$ and the $O(1)$ shells of $S$, and add Step 2. This proves (A.16), uniformly in $f,g,a,b$ and the scales. Of $f$ and $g$ we used only one-boundedness and the unary support $(rs,N)=1$.

Step 9: the product interval

Let $I\subset[N/8,2N]$ and $C_{\mathrm{grid}}\geq 1$. Divide $(H,2H]$ and $(M,2M]$ into $O(L^{C_{\mathrm{grid}}})$ consecutive intervals of relative length at most $L^{-C_{\mathrm{grid}}}$. Call a rectangle $J_{1}\times J_{2}$ interior if $J_{1}J_{2}\subset I$, exterior if $J_{1}J_{2}\cap I=\emptyset$, and boundary otherwise. On each interior rectangle apply (A.16) to $a_{n}\mathbf{1}_{J_{1}}(n)$ and $b_{m}\mathbf{1}_{J_{2}}(m)$, with saving $A+2C_{\mathrm{grid}}$. By Step 0 the first array satisfies (6.34) at the original parent $H$. For fixed $n$, the $m$ in boundary rectangles lie in $O(1)$ intervals of total length $O(ML^{-C_{\mathrm{grid}}})$. A modulus $q\leq 4N^{v}$ has at most $\tau(q)$ factorizations $q=rs$. Counting $m$ for each $n$,

$$\sum_{q\leq 4N^{v}}\tau(q)\sum_{n\sim H}\Bigl{[}\frac{ML^{-C_{\mathrm{grid}}}}{q}+1+\frac{ML^{-C_{\mathrm{grid}}}}{\varphi(q)}\Bigr{]}\ll NL^{C_{1}-C_{\mathrm{grid}}}+HN^{v}L,$$

with a fixed $C_{1}$. The last term saves a fixed power, since $h+v\leq.9827+o(1)$. Taking $C_{\mathrm{grid}}=A+C_{1}$ proves (6.45). ∎

A.4. Lemma 6.9

Recall (6.47):

(6.47) $$H_{\mathrm{c}}M_{\mathrm{c}}\asymp N,\qquad N^{.50}\ll H_{\mathrm{c}}\ll N^{.54},\qquad Q_{1}\ll N^{.4545},\qquad Q_{2}\ll N^{.0682}.$$

Lemma 6.9 (Central rectangular distribution).  Let $x$ be even and $H_{\mathrm{c}}M_{\mathrm{c}}\asymp x$. Let the bounds in (6.47) hold with $N$ replaced by $x$, with fixed comparison constants. Let $a_{n},b_{m}$ be one-bounded on the complete dyadic rectangle $n\sim H_{\mathrm{c}},m\sim M_{\mathrm{c}}$. Let the tested $a_{n}$ satisfy (6.34) at its original parent, with one fixed divisor exponent and uniform constants at every requested precision. Write

$$\mathcal{D}(q,a)=\sum_{n\sim H_{\mathrm{c}},m\sim M_{\mathrm{c}}}a_{n}b_{m}\left(1_{nm\equiv a\ (q)}-\frac{1_{(nm,q)=1}}{\varphi(q)}\right).$$

Then, for every fixed $A>0$,

(6.48) $$\sum_{r\sim Q_{1}}\max_{(b,r)=1}\sum_{\begin{subarray}{c}s\sim Q_{2}\\ \mu(rs)^{2}=1,\ (rs,x)=1\end{subarray}}\max_{\begin{subarray}{c}a\bmod rs,\ (a,rs)=1\\ a\equiv b\ (r)\end{subarray}}|\mathcal{D}(rs,a)|\ll_{A}x(\log x)^{-A}.$$

The constant is uniform over these families, rectangles, squarefree modulus supports and unary interval restrictions of $a_{n}$ and $b_{m}$. All errors are measured at the original parents. One outer residue $b$ is chosen before $s$ varies. In the application it is $N\bmod r$.

Write $H=H_{\mathrm{c}}$, $M=M_{\mathrm{c}}$, $L=\log x$, $\ell=\log\log x$, $z=x^{1/\ell^{3}}$, $y=x^{1/\ell}$. Let $c_{0}\geq 0$ be the fixed divisor exponent in the hypothesis, so that for every $D>0$, uniformly in intervals $I\subset(H,2H]$, $k,h\geq 1$ and $(u,k)=1$,

(A.49) $$|\Delta_{a}(I;k,u;h)|\leq C_{D}H\tau(h)^{c_{0}}L^{-D}.$$

Since $\Delta_{a\mathbf{1}_{J}}(I;k,u;h)=\Delta_{a}(I\cap J;k,u;h)$, a unary restriction of $a$ satisfies (A.49) with the same constants. A unary restriction of $b$ keeps it one-bounded. We may therefore assume that $a$ and $b$ are the restricted arrays. We keep $(rs,x)=1$ in (6.48) and throughout the signed residue-difference argument. The individual maxima in (6.48) are nonnegative, but the signed differences in Step 2 are not. Enlarging their support does not preserve order. Every selected coefficient below is extended by zero outside $\mu(rs)^{2}=1$ and $(rs,x)=1$. On this support $(r,s)=1$ and $\varphi(rs)=\varphi(r)\varphi(s)$. Throughout, $rs\leq 4Q_{1}Q_{2}\leq Cx^{v}$ with $v=.5227$, and $M\gg x^{.46}$.

Step 1: the projected comparator

For $t=rs$ and reduced $a\equiv b\pmod{r}$ put $S_{t}(a)=\sum_{n,m}a_{n}b_{m}\mathbf{1}_{nm\equiv a\ (t)}$ and

$$\Delta(a,b;r,s) =S_{t}(a)-\frac{\varphi(r)}{\varphi(t)}\sum_{\begin{subarray}{c}(nm,t)=1\\ nm\equiv b\ (r)\end{subarray}}a_{n}b_{m},$$

$$\mathcal{F}(b;r,s) =\frac{\varphi(r)}{\varphi(t)}\sum_{(nm,t)=1}a_{n}b_{m}\Bigl{[}\mathbf{1}_{nm\equiv b\ (r)}-\frac{1}{\varphi(r)}\Bigr{]}.$$

Then exactly

(A.50) $$\mathcal{D}(t,a)=\Delta(a,b;r,s)+\mathcal{F}(b;r,s).$$

With $a^{(s)}_{n}=a_{n}\mathbf{1}_{(n,s)=1}$,

(A.51) $$\mathcal{F}(b;r,s)=\frac{1}{\varphi(s)}\sum_{\begin{subarray}{c}m\sim M\\ (m,rs)=1\end{subarray}}b_{m}\Delta_{a^{(s)}}(r,b\overline{m}).$$

Mean square for $a^{(s)}$. The identity $\Delta_{a^{(s)}}(I;k,u;h)=\Delta_{a}(I;k,u;sh)$ and (A.2) give

$$|\Delta_{a^{(s)}}(I;k,u;h)|\leq C_{D}H\tau(s)^{c_{0}}\tau(h)^{c_{0}}L^{-D}.$$

Put $h_{s}=a^{(s)}/\tau(s)^{c_{0}}$, which is one-bounded, and $e_{n}=\mathbf{1}_{H<n\leq 2H}$. By the parallelogram identity $\|h_{s}+e\|^{2}+\|h_{s}-e\|^{2}\geq 2\|e\|^{2}$, there is $\varepsilon_{s}\in\{\pm 1\}$ with

$$g_{s}=h_{s}+\varepsilon_{s}e,\qquad\|g_{s}\|_{2}^{2}\geq\|e\|_{2}^{2}\geq H/2,\qquad\|g_{s}\|_{2}^{2}\leq 4\|e\|_{2}^{2}\leq 8H.$$

Let $Z=\lceil 2H\rceil$ and $c_{1}=\max(c_{0},1)$. By (A.49) for $h_{s}$, by (A.6) for $e$, and since $H\leq 2^{1/2}\|g_{s}\|_{2}Z^{1/2}$ and $\|e\|_{2}Z^{1/2}\geq H/2$, both $g_{s}$ and $e$ satisfy (6.35) with $B_{\rm div}=c_{1}$. The constants are independent of $s$, the sign and the restriction. Here $\log(2Z)\asymp L$, as $H\gg x^{.50}$. Lemma 6.5(i) applies to both. Since $\Delta_{a^{(s)}}=\tau(s)^{c_{0}}(\Delta_{g_{s}}-\varepsilon_{s}\Delta_{e})$ and $|u-w|^{2}\leq 2|u|^{2}+2|w|^{2}$, for every fixed $S_{0}>0$,

(A.52) $$\sum_{k\leq 2Q_{1}}V(k;a^{(s)})\ll_{S_{0}}\tau(s)^{2c_{0}}H^{2}L^{-S_{0}}.$$

The range condition $2Q_{1}\leq Z(\log Z)^{-B_{1}(S_{0})}$ holds for large $x$, because $Q_{1}/H\ll x^{.4545-.50}$.

Bound for $\mathcal{F}$. Each reduced class modulo $r$ contains at most $M/r+1$ values $m\sim M$. The map $m\mapsto b\overline{m}$ permutes the reduced classes. Cauchy’s inequality in (A.51) gives, uniformly in $b$,

$$\Bigl{|}\sum_{m}b_{m}\Delta_{a^{(s)}}(r,b\overline{m})\Bigr{|}^{2}\leq 2M\Bigl{(}\frac{M}{r}+1\Bigr{)}V(r;a^{(s)}).$$

Cauchy’s inequality over $r\sim Q_{1}$ and (A.52) then give

$$\sum_{r\sim Q_{1}}\max_{b}\Bigl{|}\sum_{m}b_{m}\Delta_{a^{(s)}}(r,b\overline{m})\Bigr{|}\ll\bigl{[}Q_{1}M(M/Q_{1}+1)\tau(s)^{2c_{0}}H^{2}L^{-S_{0}}\bigr{]}^{1/2}\ll\tau(s)^{c_{0}}HML^{-S_{0}/2},$$

since $Q_{1}\leq M$ by a fixed power. Summing over $s$ with (A.2) and (A.1),

(A.53) $$\sum_{r\sim Q_{1}}\max_{(b,r)=1}\sum_{s\sim Q_{2}}|\mathcal{F}(b;r,s)|\ll xL^{-S_{0}/2}\sum_{s\leq 2Q_{2}}\frac{\tau(s)^{c_{0}}}{\varphi(s)}\ll_{A}xL^{-A},$$

choosing $S_{0}$ after $A$ (the full order is fixed in Step 8).

Step 2: residue differences

For coprime $r,s$ there are exactly $\varphi(s)$ reduced lifts $a^{\prime}\bmod t$ of $b\bmod r$. Every $nm$ coprime to $t$ with $nm\equiv b\ (r)$ lies in exactly one of them. Hence

(A.54) $$\Delta(a,b;r,s)=\frac{1}{\varphi(s)}\sum_{\begin{subarray}{c}a^{\prime}\bmod t,\ (a^{\prime},t)=1\\ a^{\prime}\equiv b\ (r)\end{subarray}}\bigl{[}S_{t}(a)-S_{t}(a^{\prime})\bigr{]}.$$

For each $r$, choose one reduced class $b_{r}\pmod{r}$ maximizing the masked sum of the projected differences, before $s$ varies. For each $s$ with $\mu(rs)^{2}=1$ and $(rs,x)=1$, choose a reduced lift $a_{r,s}$ maximizing $|\Delta(a,b_{r};r,s)|$. By (A.54) and the triangle inequality, some reduced lift $a^{\prime}_{r,s}\equiv b_{r}\pmod{r}$ satisfies

$$|S_{rs}(a_{r,s})-S_{rs}(a^{\prime}_{r,s})|\geq|\Delta(a_{r,s},b_{r};r,s)|.$$

Choose $|\gamma_{r,s}|=1$ to make this difference nonnegative real, and put $\gamma_{r,s}=0$ outside the masked squarefree support. By (A.50), (A.53) and (A.54), it suffices to prove, uniformly for all these residues and phases,

(A.55) $$\mathcal{Z}_{0}^{(x)}=\sum_{r\sim Q_{1}}\sum_{\begin{subarray}{c}s\sim Q_{2}\\ \mu(rs)^{2}=1\\ (rs,x)=1\end{subarray}}\gamma_{r,s}\bigl{[}S_{rs}(a_{r,s})-S_{rs}(a^{\prime}_{r,s})\bigr{]}\ll_{A}xL^{-A}.$$

This signed sum keeps the original mask.

Step 3: large smooth parts and large divisor functions

We quote the completed-hyperbola lemma of Paper I:

Let $c\bmod q$ be a unit. For integer intervals $J_{1},J_{2}$, put

(3.107) $$\mathcal{D}=\#\{(u,v)\in J_{1}\times J_{2}:uv\equiv c(q)\}-\frac{\#J_{1}^{*}\#J_{2}^{*}}{\varphi(q)}.$$

The Weil–Estermann bound [11, 12]

(3.108) $$|S(h,k;q)|\leq\tau(q)q^{1/2}(h,k,q)^{1/2}.$$

implies, uniformly in interval lengths and positions,

(3.109) $$|\mathcal{D}|\ll\tau(q)q^{1/2}\log^{2}(2q)+\tau(q)^{2}\log^{2}(2q)/\varphi(q).$$

For (A.56), set the lemma’s $q=t=rs$, $J_{1}=(H,2H]\cap\mathbb{Z}$, $J_{2}=(M,2M]\cap\mathbb{Z}$, and take $c=a$ or $c=a^{\prime}$. Both are units modulo $t$ by the reduced-lift condition. Here $J_{i}^{*}=\{u\in J_{i}:(u,t)=1\}$. The source’s $\mathcal{D}$ is an unweighted counting discrepancy. It differs from our weighted $\mathcal{D}(t,a)$. Define

$$C_{t}(c)=\#\{(n,m)\in J_{1}\times J_{2}:nm\equiv c\pmod{t}\}.$$

The quoted lemma gives $C_{t}(c)\leq\#J_{1}^{*}\#J_{2}^{*}/\varphi(t)+O(E_{t})$, where $E_{t}=\tau(t)t^{1/2}\log^{2}(2t)+\tau(t)^{2}\log^{2}(2t)/\varphi(t)$. Since $|a_{n}|,|b_{m}|\leq 1$, one has $|S_{t}(c)|\leq C_{t}(c)$. This is the only passage from unweighted counts to the arbitrary complex arrays. Every pair counted has $(nm,t)=1$ automatically because $c$ is a unit. No coprimality hypothesis on the arrays is inferred. Consequently,

(A.56) $$|S_{t}(a)-S_{t}(a^{\prime})|\leq C_{t}(a)+C_{t}(a^{\prime})\leq\frac{2\#J_{1}^{*}\#J_{2}^{*}}{\varphi(t)}+O(E_{t})\leq\frac{8Cx}{\varphi(t)}+O(E_{t}),$$

using $HM\leq Cx$. The mask $(t,x)=1$ remains on the summation support but is not a hypothesis of Lemma 3.4. In the smooth-part and divisor-function eliminations below, we first take absolute values termwise. Only then are the nonnegative majorants enlarged. Write $s=k\varrho$ with $P^{+}(k)\leq z<P^{-}(\varrho)$. If $k\geq y$, then the $z$-smooth part of $t$ is at least $y$. A given $t$ has at most $\tau(t)$ factorizations $t=rs$. By (A.56), (A.2), (A.1) and (A.7), the sum of the absolute values of these masked terms in $\mathcal{Z}_{0}^{(x)}$ is at most

$$\ll x\sum_{h^{\prime}\leq Cx^{v}}\frac{\tau(h^{\prime})^{2}}{h^{\prime}}\sum_{\begin{subarray}{c}d^{\prime}\geq y\\ P^{+}(d^{\prime})\leq z\end{subarray}}\frac{\tau(d^{\prime})^{2}}{d^{\prime}}+\sum_{t\leq Cx^{v}}\tau(t)E_{t}\ll_{A}xL^{-A}+x^{3v/2}L^{K},$$

and $3v/2=.78405<1$. In the same way the pairs with $\tau(rs)>L^{D_{\tau}}$ contribute

$$\ll xL^{-D_{\tau}}\sum_{t\leq Cx^{v}}\frac{\tau(t)^{3}}{t}+x^{3v/2}L^{K}\ll xL^{K^{\prime}-D_{\tau}}+x^{3v/2}L^{K}.$$

Choose $D_{\tau}=D_{\tau}(A)\geq K^{\prime}+A$ after the requested precision $A$. From now on $k<y$, and every remaining modulus and every divisor of it has $\tau\leq L^{D_{\tau}}$.

Step 4: the first Cauchy square

Rename the outer modulus $r$ as $q\sim Q:=Q_{1}$ and write the inner one as $s=kr$, with $k\sim K\leq y$ smooth and $r\sim R$ squarefree and $z$-rough. Then $KR\asymp Q_{2}$. On the support of (A.55), $q,k,r$ are pairwise coprime and squarefree, $(qkr,x)=1$, and $QKR\ll x^{v}$. There are $O(L^{2})$ choices of $(K,R)$. Extend the selected coefficients by zero and put

$$\Gamma_{qkr}=\gamma_{q,kr}\,\mathbf{1}_{\mu(qkr)^{2}=1}\mathbf{1}_{(qkr,x)=1},\qquad|\Gamma_{qkr}|\leq 1.$$

The $\Gamma_{qkr}$ are identical in both choices $a,a^{\prime}$ of residue. For one shell the masked signed sum is

$$\mathcal{Z}^{(x)}=\sum_{\begin{subarray}{c}q\sim Q,\ k\sim K\\ (qk,x)=1\end{subarray}}\sum_{m}b_{m}\sum_{\begin{subarray}{c}r\sim R,\ n\\ (r,x)=1\end{subarray}}\Gamma_{qkr}a_{n}\bigl{[}\mathbf{1}_{mn\equiv a_{qkr}\ (qkr)}-\mathbf{1}_{mn\equiv a^{\prime}_{qkr}\ (qkr)}\bigr{]},$$

where $a_{qkr}\equiv a^{\prime}_{qkr}\equiv b_{q}\pmod{q}$ and $b_{q}$ does not depend on $k,r$. Put

$$\mathcal{F}^{(x)}_{qk}(m)=\sum_{\begin{subarray}{c}r\sim R,\ n\\ (r,x)=1\end{subarray}}\Gamma_{qkr}a_{n}\bigl{[}\mathbf{1}_{mn\equiv a_{qkr}\ (qkr)}-\mathbf{1}_{mn\equiv a^{\prime}_{qkr}\ (qkr)}\bigr{]}.$$

With $\psi$ as in Step 3 of the proof of Lemma 6.8,

(A.57) $$|\mathcal{Z}^{(x)}|^{2}\ll QKM\sum_{\begin{subarray}{c}q\sim Q,\ k\sim K\\ (qk,x)=1\end{subarray}}\sum_{m}\psi(m/M)|\mathcal{F}^{(x)}_{qk}(m)|^{2}.$$

For $A_{i}=A_{i}(q,k,r_{i})$ chosen from $a_{qkr_{i}},a^{\prime}_{qkr_{i}}$, the four correlations, with signs $+,-,-,+$, are precisely

$$\mathcal{C}_{x}(A_{1},A_{2})={} \sum_{\begin{subarray}{c}q\sim Q,\ k\sim K\\ (qk,x)=1\end{subarray}}\sum_{m}\psi(m/M)\sum_{\begin{subarray}{c}r_{1},r_{2}\sim R\\ (r_{1}r_{2},x)=1\end{subarray}}\Gamma_{qkr_{1}}\overline{\Gamma_{qkr_{2}}}$$

$$\qquad\times\sum_{n_{1},n_{2}}a_{n_{1}}\overline{a_{n_{2}}}\mathbf{1}_{mn_{1}\equiv A_{1}\ (qkr_{1})}\mathbf{1}_{mn_{2}\equiv A_{2}\ (qkr_{2})}.$$

We prove at arbitrary fixed precision $P>0$ that

(A.58) $$\mathcal{C}_{x}(A_{1},A_{2})=MT_{x}+O_{P}\Bigl{(}\frac{MH^{2}}{QKL^{P}}\Bigr{)},$$

where $MT_{x}$ contains the same mask and selected coefficients. It is independent of the choices of $A_{1},A_{2}$. Since the coefficients and support are the same in all four correlations, their main terms cancel. Equation (A.57) then gives $|\mathcal{Z}^{(x)}|\ll xL^{-P/2}$ before the $O(L^{2})$ shell sum.

Write $g=(r_{1},r_{2})$, $r_{1}=gu$, $r_{2}=gv$, $(u,v)=1$, with $g\sim G$ dyadic. Put $d_{0}=qkg$ and $U=R/G$. Then $q,k,g,u,v$ are pairwise coprime. The product of the two $\Gamma$-supports imposes $(qkguv,x)=1$. This modulus-only restriction remains in the correlation and its zero mode. It may be omitted only from a termwise absolute-value majorant. It may not be omitted from a signed expression or from an inner sum before squaring. The two congruences $mn_{1}\equiv A_{1}\ (d_{0}u)$, $mn_{2}\equiv A_{2}\ (d_{0}v)$ are soluble exactly when

(A.59) $$n_{2}\equiv A_{2}\overline{A_{1}}\,n_{1}\pmod{d_{0}},\qquad(n_{1},d_{0}u)=(n_{2},d_{0}v)=1,$$

and then $m$ lies in one reduced class modulo $l=d_{0}uv$. Modulo $q$ both residues equal $b_{q}$, so $n_{2}\equiv n_{1}\pmod{q}$.

Step 5: large common factor

Let $G\geq H/(QKL^{C_{g}})$, with $C_{g}$ fixed later. The target is $\mathcal{B}=MH^{2}/(QK)\asymp xH/(QK)$. In the large-$G$ estimate we first take absolute values of each weighted correlation term. Hence its nonnegative counting majorant may be enlarged to all $(q,k,g,u,v)$, while $\mathcal{C}_{x}$ itself stays masked. For fixed $(q,k,g,u,v)$ the pairs $(n_{1},n_{2})$ in (A.59) lie in a congruence lattice of determinant $d_{0}$. Let $\mathbf{w}$ be a shortest nonzero vector. If the line $\mathbb{R}\mathbf{w}$ meets the positive rectangle, choose its sign so its coordinates are positive; otherwise the $\beta=0$ contribution is empty. Complete it to a reduced basis $(\mathbf{w},\mathbf{w}^{\prime})$. The component of $\mathbf{w}^{\prime}$ orthogonal to $\mathbf{w}$ has length $h_{\perp}=d_{0}/|\mathbf{w}|\geq\sqrt{3}|\mathbf{w}|/2$. In a disc of radius $CH$ the points $\alpha\mathbf{w}+\beta\mathbf{w}^{\prime}$ with $\beta\neq 0$ have $|\beta|\ll H/h_{\perp}$ and, for fixed $\beta$, $O(H/|\mathbf{w}|+1)$ values of $\alpha$. Such points exist only if $h_{\perp}\ll H$. Then $|\mathbf{w}|\ll h_{\perp}\ll H$ and $(H/h_{\perp})^{2}\ll H^{2}/d_{0}$, so they number $O(H^{2}/d_{0})$. Each such pair has $O(M/l+1)$ values of $m$. Hence these points contribute

$$\ll QKGU^{2}\cdot\frac{H^{2}}{QKG}\Bigl{(}1+\frac{M}{QKGU^{2}}\Bigr{)}=H^{2}U^{2}+\frac{MH^{2}}{QKG}.$$

For $\beta=0$, $(n_{1},n_{2})=\alpha\mathbf{w}$ with $\alpha\geq 1$, and the two congruences fix $\alpha m\leq Cx$ in one unit class $c\pmod{l}$. Here $l=d_{0}uv=qkguv\geq 1$, and $c$ is a unit. Each original congruence has a unit right-hand side $A_{i}$. Hence $m$ and $\alpha$ are units modulo every factor of $l$ on this line. For $X\leq Cx$, cover the positive integer hyperbola $\alpha m\leq X$ by the $O(L)$ integer rectangles $J_{1,0}=\{1\}$, $J_{2,0}=[1,\lfloor X\rfloor]$, and, for $j\geq 1$ with $2^{j-1}<X$,

$$J_{1,j}=(2^{j-1},2^{j}]\cap\mathbb{Z},\qquad J_{2,j}=[1,\lfloor X/2^{j-1}\rfloor]\cap\mathbb{Z}.$$

These rectangles may overcount. This is harmless for the positive quantity $\sum_{e\leq X,e\equiv c\ (l)}\tau(e)$. Each has $\#J_{1,j}\#J_{2,j}\ll X$, and their number is $O(L)$. Apply the quoted Paper I, Lemma 3.4 with its modulus $q=l$, unit $c$, and these two integer intervals. Empty intervals contribute zero. The positive counts give, uniformly in $c$ and $X$,

(A.60) $$\sum_{\begin{subarray}{c}e\leq X\\ e\equiv c\ (l)\end{subarray}}\tau(e)\ll\frac{XL}{\varphi(l)}+\Bigl{[}\tau(l)l^{1/2}+\frac{\tau(l)^{2}}{\varphi(l)}\Bigr{]}L^{3}$$

uniformly in the unit $c$ and $X\leq Cx$. A modulus $l$ has at most $\tau_{5}(l)\leq\tau(l)^{4}$ representations $l=qkguv$. Summing (A.60) with (A.1),

$$\ll xL^{C}+(QKGU^{2})^{3/2}L^{C}.$$

With $W=QK\leq x^{.4545+o(1)}$, $R\ll x^{.0682}$ and $WR\ll x^{.5227}$, on this range of $G$,

$$\frac{MH^{2}/(QKG)}{\mathcal{B}}=\frac{1}{G}\leq x^{-.0455+o(1)},\quad\frac{H^{2}U^{2}}{\mathcal{B}}\ll\frac{W^{3}R^{2}L^{2C_{g}}}{xH}\leq x^{-.0001+o(1)},\quad\frac{x}{\mathcal{B}}\leq x^{-.0455+o(1)},$$

and $(QKGU^{2})^{3/2}\leq(W^{2}R^{2}L^{C_{g}}/H)^{3/2}\leq x^{.8181+o(1)}$, while $\mathcal{B}\geq x^{1.0455-o(1)}$. So the large-$G$ part of $\mathcal{C}_{x}$ is $O(\mathcal{B}L^{-P})$ at any requested fixed precision $P$. The mask was removed only after taking absolute values.

Step 6: small common factor, Poisson and the zero mode

Let $G<H/(QKL^{C_{g}})$. For one class $c\bmod l$,

(A.61) $$\sum_{m\equiv c\ (l)}\psi\Bigl{(}\frac{m}{M}\Bigr{)}=\frac{M}{l}\sum_{h\in\mathbb{Z}}\hat{\psi}\Bigl{(}\frac{hM}{l}\Bigr{)}e\Bigl{(}\frac{hc}{l}\Bigr{)},$$

and the terms $|h|>x^{\eta}l/M$ contribute $O(x^{-100})$ per tuple. We take $\eta=10^{-8}$.

The zero mode is $M\hat{\psi}(0)/l$ times the weighted number of compatible $(n_{1},n_{2})$. It still carries the product of the two $\Gamma$-supports, and thus $(qkguv,x)=1$. After taking absolute values of the error, its positive majorant can be enlarged. This is the only use of the unmasked range in (A.62). Put $f_{v}=a\mathbf{1}_{(n,v)=1}$. For fixed $n_{1}$, the $n_{2}$-sum in (A.59) is a sum of $\overline{f_{v}}$ over one reduced class $c(n_{1})$ modulo $d_{0}$. Replace it by $\varphi(d_{0})^{-1}\sum_{(n_{2},d_{0}v)=1}\overline{a_{n_{2}}}$. The error is $\overline{\Delta_{f_{v}}(d_{0},c(n_{1}))}$. The map $n_{1}\mapsto c(n_{1})$ hits each reduced class $O(H/d_{0})$ times, so Cauchy’s inequality gives

$$\sum_{n_{1}}|\Delta_{f_{v}}(d_{0},c(n_{1}))|\ll Hd_{0}^{-1/2}V(d_{0};f_{v})^{1/2}.$$

A given $d_{0}$ has at most $\tau_{3}(d_{0})\leq L^{2D_{\tau}}$ representations, and $d_{0}\asymp D_{0}=QKG$. Multiplying by $M/(d_{0}uv)$ and summing over $u,v$, the zero-mode error is

(A.62) $$\ll\frac{MH}{D_{0}}L^{2D_{\tau}}\sup_{v}\Bigl{[}\sum_{d_{0}\leq C^{\prime}D_{0}}V(d_{0};f_{v})\Bigr{]}^{1/2}.$$

Since $v$ divides a remaining modulus, $\tau(v)\leq L^{D_{\tau}}$. The identity $\Delta_{f_{v}}(I;k,u;h)=\Delta_{a}(I;k,u;vh)$ and (A.49) show that $f_{v}$ satisfies (A.49) at precision $D-c_{0}D_{\tau}$. As in Step 1, add $\varepsilon_{v}e$ to make the norm at least $(H/2)^{1/2}$. Both sequences satisfy (6.35) with $B_{\rm div}=c_{1}$ and uniform constants. Lemma 6.5(i) and linearity give, uniformly in $v$,

$$\sum_{d_{0}\leq C^{\prime}D_{0}}V(d_{0};f_{v})\ll_{S_{0}}H^{2}L^{-S_{0}},$$

because $C^{\prime}D_{0}\leq C^{\prime}H/L^{C_{g}}\leq Z(\log Z)^{-B_{1}(S_{0})}$ once $C_{g}>B_{1}(S_{0})+1$. With $D_{\tau}=D_{\tau}(A)$ fixed in Step 3, choose first $S_{0}>4D_{\tau}+2A$, then $C_{g}>B_{1}(S_{0})+1$, then the threshold for $x$. Since $D_{0}\geq cQK$, (A.62) is $O(\mathcal{B}L^{2D_{\tau}-S_{0}/2})=O(\mathcal{B}L^{-A})$.

What remains of the zero mode is the masked main term

(A.63) $$MT_{x,G}=M\hat{\psi}(0)\sum_{\begin{subarray}{c}q,k,g,u,v\\ (qkguv,x)=1\end{subarray}}\Gamma_{qkgu}\overline{\Gamma_{qkgv}}\frac{\bigl{[}\sum_{(n_{1},d_{0}u)=1}a_{n_{1}}\bigr{]}\bigl{[}\sum_{(n_{2},d_{0}v)=1}\overline{a_{n_{2}}}\bigr{]}}{d_{0}uv\varphi(d_{0})},$$

with the previously imposed coprime/squarefree and dyadic conditions understood. This expression depends on the modulus support and $\Gamma$, but not on $A_{1},A_{2}$. Put $MT_{x}=\sum_{G\ \rm small}MT_{x,G}$.

Step 7: the second square

It remains to bound the frequencies $h\neq 0$ in the masked correlation. Write

$$\frac{M}{l}=\frac{M}{QKGU^{2}}\cdot\frac{QKGU^{2}}{qkguv};$$

the second factor is a product of bounded weights in $(q,k,g)$, in $u$ and in $v$. The mask factors exactly as

$$\mathbf{1}_{(qkguv,x)=1}=\mathbf{1}_{(qkg,x)=1}\mathbf{1}_{(u,x)=1}\mathbf{1}_{(v,x)=1}.$$

Keep the first factor with the outer $(q,k,g,n_{1},n_{2})$ data and the last two with the $(u,v,h)$-sum, together with $\Gamma_{qkgu}\overline{\Gamma_{qkgv}}$. No masked term is reintroduced into a signed sum.

Every $z$-rough $u,v\leq x^{2}$ has $O(\ell^{3})$ prime factors. Colour each prime below $x^{2}$ red or blue independently. A coprime pair $(u,v)$ is covered if all primes of $u$ are red and all of $v$ blue. This has probability at least $2^{-O(\ell^{3})}$. With $J_{0}=\lceil 2^{O(\ell^{3})}(10\log x+1)\rceil$ independent colourings, the probability that one of the at most $x^{4}$ pairs is never covered is $<1$. Fix such a list and assign each pair to its first covering colouring. This partitions the coprime pairs into $J\leq J_{0}=x^{o(1)}$ classes, and $(uu^{\prime},vv^{\prime})=1$ for any two pairs in one class.

On one class let $\mathcal{Y}$ be the sum over compatible $(n_{1},n_{2})$, $(u,v)$ and $0<|h|\leq\mathsf{F}$, after removing $M/(QKGU^{2})$, where

(A.64) $$\mathsf{F}=Cx^{\eta}QKGU^{2}/M.$$

If $\mathsf{F}<1$ there is nothing to bound. By CRT the class of $m$ in the phase is

$$\frac{c}{l}\equiv\frac{A_{1}\overline{n_{1}}\,\overline{v}}{d_{0}u}+\frac{A_{2}\overline{n_{2}}\,\overline{d_{0}u}}{v}\pmod{1}.$$

Cauchy’s inequality over the masked outer data $(q,k,g,n_{1},n_{2})$, $(qkg,x)=1$, with $n_{2}\equiv n_{1}\ (q)$ removes $a$ and gives

(A.65) $$|\mathcal{Y}|^{2}\ll H^{2}KG\,\mathcal{D}_{3},$$

since there are at most $QKG\cdot O(H(H/Q+1))\ll H^{2}KG$ outer tuples ($Q<H$). Here $\mathcal{Y}$ and $\mathcal{D}_{3}$ keep the $x$-mask. The sum $\mathcal{D}_{3}$ runs over $(qkg,x)=1$ and $n_{2}\equiv n_{1}\ (q)$. Its terms are a smooth majorant times the square of the $(u,v,h)$-sum. This inner sum contains $\mathbf{1}_{(u,x)=1}\mathbf{1}_{(v,x)=1}$ and the selected $\Gamma$ coefficients. The congruence modulo $kg$ stays inside the square. In the expansion of this square each pair of copies keeps its zero–one masks. The later diagonal, Weil and Ramanujan bounds are termwise bounds after taking absolute values of the expanded contributions. Hence they may discard zero–one factors. We do not use the false inequality $|\sum_{(uv,x)=1}z_{uv}|^{2}\leq|\sum_{u,v}z_{uv}|^{2}$.

Expanding gives $(u,v,h)$ and $(u^{\prime},v^{\prime},h^{\prime})$ with $0<|h|,|h^{\prime}|\leq\mathsf{F}$. Put $A=A_{1}(u)$, $A^{\prime}=A_{1}(u^{\prime})$, $B=A_{2}(v)$, $B^{\prime}=A_{2}(v^{\prime})$ and

$$\Delta=hu^{\prime}v^{\prime}-h^{\prime}uv.$$

The two congruences modulo $kg$ must agree, or the term vanishes. Let $t\equiv B\overline{A}\pmod{d_{0}}$. The $(n_{1},n_{2})$-sum is

$$\mathcal{K}=\sum_{\begin{subarray}{c}n_{2}\equiv tn_{1}\ (d_{0})\\ (n_{1},d_{0}uu^{\prime})=1,\ (n_{2},vv^{\prime})=1\end{subarray}}\psi\Bigl{(}\frac{n_{1}}{H}\Bigr{)}\psi\Bigl{(}\frac{n_{2}}{H}\Bigr{)}e\Bigl{(}\frac{c_{1}\overline{n_{1}}}{d_{0}uu^{\prime}}\Bigr{)}e\Bigl{(}\frac{c_{2}\overline{n_{2}}}{vv^{\prime}}\Bigr{)},$$

$$c_{1}\equiv(Ahu^{\prime}v^{\prime}-A^{\prime}h^{\prime}uv)\overline{vv^{\prime}}\pmod{d_{0}uu^{\prime}},\qquad c_{2}\equiv(Bhu^{\prime}v^{\prime}-B^{\prime}h^{\prime}uv)\overline{d_{0}uu^{\prime}}\pmod{vv^{\prime}}.$$

All inverses exist, since $(uu^{\prime},vv^{\prime})=1$ and both are prime to $d_{0}$. Completing first in $n_{2}$ and then in $n_{1}$,

(A.66) $$\mathcal{K}=\frac{H^{2}}{d_{0}^{2}uu^{\prime}vv^{\prime}}\sum_{\ell_{1},\ell_{2}\in\mathbb{Z}}\hat{\psi}\Bigl{(}\frac{\ell_{2}H}{d_{0}vv^{\prime}}\Bigr{)}\hat{\psi}\Bigl{(}\frac{\ell_{1}H}{d_{0}uu^{\prime}}\Bigr{)}S(c_{2},\ell_{2}\overline{d_{0}};vv^{\prime})\,S(c_{1},\ell_{1}+\ell_{2}c_{3};d_{0}uu^{\prime}),$$

with $c_{3}\equiv uu^{\prime}t\,\overline{vv^{\prime}}\pmod{d_{0}uu^{\prime}}$ and the Kloosterman sum $S(a,b;m)$ of Paper I, (2.35). The first completion produces the phase $e(\ell_{2}tn_{1}\overline{vv^{\prime}}/d_{0})$, which combines with the $n_{1}$-phase to give $c_{3}$. Let

(A.67) $$\Lambda=C_{1}x^{\eta}QKGU^{2}/H.$$

Frequencies with $|\ell_{i}|>\Lambda$ contribute $O(x^{-100})$ per tuple. If $\Lambda<1$, only $(\ell_{1},\ell_{2})=(0,0)$ remains.

The target for $\mathcal{D}_{3}$ is $T_{3}=H^{2}U^{4}G/K$. By (A.65), a bound $\mathcal{D}_{3}\ll T_{3}x^{-\rho}$ gives $|\mathcal{Y}|\ll H^{2}U^{2}Gx^{-\rho/2}$.

Diagonal $\Delta=0$. Then $uv\mid hu^{\prime}v^{\prime}$, so for fixed $(h,u^{\prime},v^{\prime})$ there are $x^{o(1)}$ choices of $(h^{\prime},u,v)$. Bounding $\mathcal{K}$ trivially by $O(H^{2}/d_{0})$, as $d_{0}<H$,

$$\mathcal{D}_{3}^{\rm diag}\ll x^{o(1)}H^{2}\mathsf{F}U^{2},\qquad\frac{\mathcal{D}_{3}^{\rm diag}}{T_{3}}\ll x^{\eta+o(1)}\frac{QK^{2}H}{x}\leq x^{-.0055+\eta+o(1)}.$$

The divisor-counting bound $x^{o(1)}$ for the tuple count immediately above is loose. Only this $x^{o(1)}$ factor is used in the estimate for $\mathcal{D}_{3}^{\rm diag}$.

Gcd bound. Put $e=(u,u^{\prime})$, $f=(v,v^{\prime})$, $u_{0}=u/e$, $u_{1}=u^{\prime}/e$, $v_{0}=v/f$, $v_{1}=v^{\prime}/f$. Then

(A.68) $$(c_{1},uu^{\prime})(c_{2},vv^{\prime})\leq(h,u_{0}v_{0})(h^{\prime},u_{1}v_{1})(c_{1},e^{2})(c_{2},f^{2}).$$

The four factors are supported on disjoint sets of primes. If $p\mid u$, $p\nmid u^{\prime}$, then $c_{1}\equiv Ahu^{\prime}v^{\prime}\overline{vv^{\prime}}\pmod{p}$ with a unit multiplier, so $(c_{1},p)=(h,p)$. The cases $p\mid u^{\prime}$ only, $p\mid v$ only and $p\mid v^{\prime}$ only are identical, with $(h^{\prime},p)$, $(h,p)$ and $(h^{\prime},p)$. If $p\mid e$, the numerator of $c_{1}$ is $p(Aha^{\prime}v^{\prime}-A^{\prime}h^{\prime}a^{\prime\prime}v)$ with $u=pa^{\prime\prime}$, $u^{\prime}=pa^{\prime}$. Thus the $p$-part of $(c_{1},e^{2})$ is at most $p^{2}$. The case $p\mid f$ is similar. Modulo $q$ all four residues equal $b_{q}$, so $c_{1}\equiv b_{q}\overline{vv^{\prime}}\Delta\pmod{q}$ and $(c_{1},q)=(\Delta,q)$. For $p\mid kg$ we use only $(c_{1},kg)\leq kg$. Hence

(A.69) $$(c_{1},d_{0}uu^{\prime})(c_{2},vv^{\prime})\leq kg\,(\Delta,q)\,(h,uv)(h^{\prime},u^{\prime}v^{\prime})(u,u^{\prime})^{2}(v,v^{\prime})^{2}.$$

Long dual range, $\Lambda\geq 1$, $\Delta\neq 0$. There are $O(\Lambda^{2})$ pairs $(\ell_{1},\ell_{2})$. Paper I, (2.35), gives

$$|S(a,b;m)|\leq\tau(m)m^{1/2}(a,m)^{1/2}.$$

Use the square root of (A.69) and $(kg)^{1/2}(QKG)^{1/2}\ll Q^{1/2}KG$. Then

$$\begin{multlined}|\mathcal{K}|\ll x^{2\eta+o(1)}Q^{1/2}KGU^{2}\\ \times(h,uv)^{1/2}(h^{\prime},u^{\prime}v^{\prime})^{1/2}(u,u^{\prime})(v,v^{\prime})(\Delta,q)^{1/2}.\end{multlined}|\mathcal{K}|\ll x^{2\eta+o(1)}Q^{1/2}KGU^{2}\\ \times(h,uv)^{1/2}(h^{\prime},u^{\prime}v^{\prime})^{1/2}(u,u^{\prime})(v,v^{\prime})(\Delta,q)^{1/2}.$$

The averages

$$\sum_{1\leq|h|\leq\mathsf{F}}(h,a)\leq 2\mathsf{F}\tau(a),\qquad\sum_{q\sim Q}(\Delta,q)\leq CQ\tau(|\Delta|),\qquad\sum_{u,u^{\prime}\sim U}(u,u^{\prime})\ll U^{2}\log(2U)$$

follow from $(h,a)=\sum_{d\mid(h,a)}\varphi(d)$ and counting multiples. Summing over the $O(KG)$ choices of $(k,g)$,

$$\mathcal{D}_{3}^{\rm long}\ll x^{2\eta+o(1)}Q^{3/2}K^{2}G^{2}U^{6}\mathsf{F}^{2},\qquad\frac{\mathcal{D}_{3}^{\rm long}}{T_{3}}\ll x^{4\eta+o(1)}\frac{Q^{7/2}K^{5}R^{6}}{G^{3}x^{2}}\leq x^{-.00005+4\eta+o(1)},$$

using $U=R/G$, $G\geq\tfrac{1}{2}$, $K\leq y$; the possible absolute factor $G^{-3}\leq 8$ is absorbed in the implied constant and in $x^{o(1)}$, and then

$$2-\tfrac{7}{2}(.4545)-6(.0682)=.00005.$$

Short dual range, $\Lambda<1$, $\Delta\neq 0$. Now

$$\mathcal{K}=\frac{H^{2}\hat{\psi}(0)^{2}}{d_{0}^{2}uu^{\prime}vv^{\prime}}\,c_{vv^{\prime}}(c_{2})\,c_{d_{0}uu^{\prime}}(c_{1})+O(x^{-100}),$$

with Ramanujan sums $c_{m}(a)=S(a,0;m)$, and $|c_{m}(a)|\leq(a,m)$. By (A.69) each term is

$$\ll\frac{H^{2}}{Q^{2}KGU^{4}}(\Delta,q)(h,uv)(h^{\prime},u^{\prime}v^{\prime})(u,u^{\prime})^{2}(v,v^{\prime})^{2}.$$

With $\sum_{u,u^{\prime}\sim U}(u,u^{\prime})^{2}\ll U^{3}$ and the averages above,

$$\mathcal{D}_{3}^{\rm short}\ll x^{o(1)}\frac{H^{2}\mathsf{F}^{2}U^{2}}{Q},\qquad\frac{\mathcal{D}_{3}^{\rm short}}{T_{3}}\ll x^{2\eta+o(1)}\frac{H^{2}QK^{3}R^{2}}{Gx^{2}}\leq x^{-.3291+2\eta+o(1)},$$

since $2(.54)+.4545+2(.0682)-2=-.3291$.

Step 8: conclusion

After the $x^{o(1)}$ factors the three ratios are at most $x^{-.00005+5\eta}$. These factors include $J$, the divisor bounds, $K\leq y$ and fixed powers of $L$. Take $\rho=10^{-6}$. Then $|\mathcal{Y}|\ll H^{2}U^{2}Gx^{-\rho/2}$ on each masked colouring class. Restoring $M/(QKGU^{2})$ and the $J=x^{o(1)}$ classes, the nonzero frequencies contribute $\ll\mathcal{B}x^{-\rho/3}$ to each small-$G$ block of $\mathcal{C}_{x}$. The modulus mask was preserved through both Cauchy squares. Only termwise absolute-value majorants were enlarged. There are $O(L)$ common-factor blocks. Together with Steps 5 and 6 this proves (A.58) with the same masked $MT_{x}$ in each of its four choices of residues. Their signed combination is $MT_{x}-MT_{x}-MT_{x}+MT_{x}=0$. Fix a requested final exponent $A$. First choose the correlation precision $P>2A+10$. Then choose $D_{\tau}=D_{\tau}(P)\geq K^{\prime}+P$ (Step 3). Then choose $S_{0}>4D_{\tau}+2P$, also large enough to make (A.53) smaller than $xL^{-A-10}$. Next take $C_{g}>B_{1}(S_{0})+1$, and finally the threshold for $x$. Equation (A.57) gives $\mathcal{Z}^{(x)}\ll xL^{-P/2}$ on each shell. Sum over its $O(L^{2})$ choices of $(K,R)$ and add the masked Step 1 and Step 3 errors. This proves exactly (6.48), with $(rs,x)=1$ as stated. Every constant depends only on $A$, the comparison constants in (6.47) and the constants in (A.49). ∎

Appendix B Finite calculations

This appendix gives the finite quantities used above. Each terminating decimal denotes an exact rational number. All decimal interval endpoints are rounded outwards; decimal approximations in explanatory prose are not additional certified bounds. The comparisons with prime sums occur in the main text.

B.1. Series bounds

For $|z|<1$,

(B.1) $$\left|\operatorname{arctanh}z-\sum_{k=0}^{m-1}\frac{z^{2k+1}}{2k+1}\right|\leq\frac{|z|^{2m+1}}{(2m+1)(1-z^{2})}.$$

Write any positive rational as $t=2^{a}v$, where $2/3\leq v<4/3$. Then use

(B.2) $$\log t=a\log 2+2\operatorname{arctanh}\frac{v-1}{v+1},\qquad\log 2=2\operatorname{arctanh}(1/3).$$

Also, for $0\leq z<1$,

(B.3) $$0\leq\operatorname{Li}_{2}(z)-\sum_{n=1}^{m}\frac{z^{n}}{n^{2}}\leq\frac{z^{m+1}}{(m+1)^{2}(1-z)}.$$

These inequalities enclose the elementary functions below. Integer comparisons then give the printed rational bounds.

B.2. Four profile differences

Put

$$\alpha=\frac{100}{1149},\quad\beta=\frac{50}{309},\quad\mu_{*}=\frac{25}{128},\quad\lambda=\frac{93}{400},\quad h_{G}=\frac{9}{5000},\quad K=\frac{57}{64}-\frac{1}{500},$$

(B.4) $$\delta_{h}=\frac{h_{G}}{\alpha},\qquad C=K-\frac{11}{5}\alpha,\qquad B(x)=\min\{\beta,\max\{\alpha,C-5x/2\}\}.$$

Define the elementary functions

$$\widehat{f}(s) =\frac{\log(s-1)}{s} (2\leq s\leq 4),$$

(B.5) $$\widehat{F}(s) =\frac{1}{s} (1<s\leq 3),$$

(B.6) $$s\widehat{F}(s) =1+\log(s-2)\log(s-1)-\operatorname{Li}_{2}\!\left(\frac{s-2}{s-1}\right)$$

$$\quad-\frac{\log^{2}(s-1)}{2}+\operatorname{Li}_{2}(1/2)+\frac{\log^{2}2}{2} (3<s<5),$$

The two sieve normalizations satisfy

$$\widehat{F}=F/(2e^{\gamma}),\qquad\widehat{f}=f/(2e^{\gamma}).$$

The raw multiplier $2/(e^{\gamma}\alpha)$, a signed row coefficient $\pm 1/4$, and $F=2e^{\gamma}\widehat{F}$ or $f=2e^{\gamma}\widehat{f}$ combine to the displayed $1/\alpha$ factor. The directed integrals below use these hatted functions. No further $1/4$ or $2e^{\gamma}$ factor is applied. Also $\operatorname{Li}_{2}(t)=\sum_{n\geq 1}t^{n}/n^{2}$ for $0\leq t<1$. Write

$$D_{F}(s)=\widehat{F}(s)-\widehat{F}(s+\delta_{h}),\qquad D_{f}(s)=\widehat{f}(s+\delta_{h})-\widehat{f}(s).$$

For the three intervals in Table B.1, put

(B.7) $$\Delta_{j} =\frac{1}{\alpha}\int_{I_{j}}D_{F}\!\left(\frac{K-5x/2}{\alpha}\right)\frac{dx}{x}\quad(j=4,5,7),$$

(B.8) $$\Delta_{10} =\frac{1}{\alpha}\int_{\mu_{*}}^{\lambda}\int_{\alpha}^{B(x)}D_{f}\!\left(\frac{K-5x/2-y}{\alpha}\right)\frac{dy\,dx}{xy}.$$

These $\Delta_{j}$ are those of equation (5.22). Closed or half-open interval endpoints have measure zero. The curved domain is that of [27, (2.38)–(2.40)].

Proposition B.1.

A directed interval computation certifies the following outward enclosure of the four elementary integrals:

(B.9) $$\frac{572004}{10^{8}}<\Delta_{4}+\Delta_{5}+\Delta_{7}+\Delta_{10}<\frac{606486}{10^{8}}.$$

In particular, their sum minus $10^{-20}$ exceeds $5719/10^{6}$. Also,

(B.10) $$\frac{\log 3}{4\alpha^{2}}\left(\frac{5}{2}\lambda+\beta\right)\frac{\lambda-\mu_{*}}{\mu_{*}}<\frac{21137254516197687}{4120000000000000}<\frac{5137}{1000}<\frac{21}{4}<\frac{179}{25}.$$

Table B.1. Outward decimal bounds from the directed computation. Every endpoint is an exact terminating rational.

$j$	$I_{j}$	lower bound	upper bound
$4$	$[25/128,1/4]$	$0.00162068$	$0.00169539$
$5$	$[1/4,7109/28000]$	$0.00040327$	$0.00041325$
$7$	$[25/128,93/400]$	$0.00056463$	$0.00059051$
$10$	curved domain	$0.00313145$	$0.00336572$
sum		$0.00572003$	$0.00606487$

Proof.

The profiles are evaluated with directed interval arithmetic. An independent verifier rechecks the directed output. Put $h_{G}=9/5000$, $p(s)=-\widehat{F}^{\prime}(s)$, and let $d$ be the right-branch derivative profile for $\widehat{f}$, so $d(s)=\widehat{f}^{\prime}(s)$ for $2<s<4$ and $d(2)=1/2$ by continuous extension. The three one-dimensional rows each use 2048 equal cells:

$$x_{k}=a_{j}+\frac{k(b_{j}-a_{j})}{2048}\qquad(0\leq k\leq 2048).$$

For a cell $[x_{0},x_{1}]=[x_{k},x_{k+1}]$, define

$$s_{-}=\frac{K-5x_{1}/2}{\alpha},\qquad s_{+}=\frac{K-5x_{0}/2+h_{G}}{\alpha}.$$

Since $D_{F}(s)=\int_{0}^{h_{G}/\alpha}p(s+t)\,dt$, the code’s lower and upper cell contributions are exactly

(B.11) $$L=\frac{h_{G}}{\alpha^{2}}\frac{x_{1}-x_{0}}{x_{1}}p(s_{+}),\qquad U=\frac{h_{G}}{\alpha^{2}}\frac{x_{1}-x_{0}}{x_{0}}p(s_{-}).$$

These hold because $s_{-}\leq(K-5x/2)/\alpha+t\leq s_{+}$ for every $x\in[x_{0},x_{1}]$ and $0\leq t\leq h_{G}/\alpha$; also $x_{0}\leq x\leq x_{1}$.

For the fourth integral, split at

$$x_{B}=\frac{506848823}{2366940000}.$$

On each piece $[a,b]$ of $[\mu_{*},x_{B}]\cup[x_{B},\lambda]$, the S10 routine uses $x_{i}=a+i(b-a)/96$ and $z_{j}=j/96$ for $0\leq i,j\leq 96$. For a cell $[x_{i},x_{i+1}]\times[z_{j},z_{j+1}]$, with $0\leq i,j<96$, put

$$b_{0}=B(x_{i}),\quad b_{1}=B(x_{i+1}),\quad y_{-}=\alpha+(b_{1}-\alpha)z_{j},\quad y_{+}=\alpha+(b_{0}-\alpha)z_{j+1},$$

$$q_{-}=\frac{K-5x_{i+1}/2-y_{+}}{\alpha},\qquad q_{+}=\frac{K-5x_{i}/2-y_{-}+h_{G}}{\alpha}.$$

The substitution is $y=\alpha+(B(x)-\alpha)z$. Its executed lower and upper contributions are

(B.12) $$L=\frac{h_{G}}{\alpha^{2}}(x_{i+1}-x_{i})(z_{j+1}-z_{j})\frac{b_{1}-\alpha}{x_{i+1}y_{+}}d(q_{+}),\qquad U=\frac{h_{G}}{\alpha^{2}}(x_{i+1}-x_{i})(z_{j+1}-z_{j})\frac{b_{0}-\alpha}{x_{i}y_{-}}d(q_{-}).$$

For every point in this cell and $0\leq t\leq h_{G}/\alpha$, $q_{-}\leq(K-5x/2-y)/\alpha+t\leq q_{+}$, and

$$\frac{b_{1}-\alpha}{x_{i+1}y_{+}}\leq\frac{B(x)-\alpha}{xy}\leq\frac{b_{0}-\alpha}{x_{i}y_{-}}.$$

Hence the two positive factors are bounded separately, as in the code. The two rectangular grids contain $2\cdot 96^{2}$ cells. The total executed cell count is

$$3\cdot 2048+2\cdot 96\cdot 96=24576.$$

The endpoint bounds use the following branch formulas. For $2<s<4$,

$$d(s)=\frac{s/(s-1)-\log(s-1)}{s^{2}}>0,\qquad d^{\prime}(s)=-\frac{1}{s(s-1)^{2}}-\frac{2\{s/(s-1)-\log(s-1)\}}{s^{3}}<0.$$

They extend continuously to $s=2$ with $d(2)=1/2$ and right derivative $d^{\prime}_{+}(2)=-1$. Thus the branch profile is positive and decreasing on $[2,4]$. For positivity: the numerator decreases, and at $s=4$ it is $4/3-\log 3>0$, since $e^{4/3}>1+4/3+(4/3)^{2}/2+(4/3)^{3}/6>3$. For $s>3$, put $A(s)=\widehat{F}(s)-\widehat{f}(s-1)$. Then

$$(sA(s))^{\prime}=-s\,\widehat{f}^{\prime}(s-1)<0,\qquad p(s)=A(s)/s,\qquad p^{\prime}(s)=-2A(s)/s^{2}-\widehat{f}^{\prime}(s-1)/s.$$

At $s=3$ the two pieces give $p(3)=1/9$, so $p$ is continuous there; the displayed differential identities are used on $s>3$. The exact directed evaluation at $s_{\max}=73930107/16000000$ gives $A(s_{\max})>1597/100000>0$. Hence $A(s)>0$ throughout the required range $3<s\leq s_{\max}$, and there $p>0$ and $p^{\prime}<0$. For $s\leq 3$, $p(s)=1/s^{2}$ is also positive and decreasing. The S10 mixed-corner range is

$$\frac{17359415273}{7910400000}\leq q_{-}\leq q_{+}\leq\frac{57930107}{16000000}<4,\qquad\frac{17359415273}{7910400000}>\frac{219}{100}.$$

These inequalities justify using the decreasing positive $d$ in (B.12). The lower bound refers only to the executed mixed-corner interval.

The code evaluates the formulas by rational geometry and 256-bit Arb ball arithmetic. In profile_direct.py, logarithms use Arb’s log() operation. Put $w=s-2$. The dilogarithm uses Arb’s polylog(2) operation at the negative argument $-w$, together with $\pi^{2}/12$. These are the code’s evaluations of (B.5). By the Landen identity

$$\operatorname{Li}_{2}(-w)=-\operatorname{Li}_{2}\!\left(\frac{w}{1+w}\right)-\frac{1}{2}\log^{2}(1+w),\qquad\operatorname{Li}_{2}(1/2)+\frac{1}{2}\log^{2}2=\frac{\pi^{2}}{12}$$

the Arb expression equals the displayed formula. Arb supplies outward ball enclosures. The code does not use the atanh and dilogarithm tail inequalities above. The exact directed rational endpoints are written to the result record. The terminating decimals in the table are outward projections of each row’s directed endpoint. Their displayed column sums are $0.00572003$ and $0.00606487$. The sharper bound (B.9) adds the unrounded rational endpoints in RESULT.json before decimal projection. Summing before rounding proves the stated integer comparisons. Finally, 64 terms of $\log 3=2\sum_{n\geq 0}2^{-(2n+1)}/(2n+1)$ and its geometric remainder give $\log 3<10987/10000$. This proves (B.10). ∎

Proposition B.1 concerns elementary integrals and one strip coefficient. Its application to the prime sums requires the source comparisons, uniform errors and a single charge of the strip.

B.3. The shifted three-variable integral

Here $F,f$ are the linear-sieve functions. Their normalization is

(B.13) $$\begin{gathered}F(s)=2e^{\gamma}/s\quad(1\leq s\leq 3),\qquad f(s)=0\quad(s\leq 2),\\ (sF(s))^{\prime}=f(s-1)\quad(s>3),\qquad(sf(s))^{\prime}=F(s-1)\quad(s>2),\\ 0\leq f\leq 1\leq F,\qquad f\text{ is nondecreasing}.\end{gathered}$$

These classical properties are the inputs to this calculation. Put $u=1/500$, $\mu_{*}=25/128$ and $h_{G}=9/5000$. Define

(B.14) $$v_{1}(s) =\min\left(\frac{1+s}{2},\frac{32-34s}{39},\frac{57}{64}-\frac{3s}{2}\right), v(s) =v_{1}(s)-u,$$

$$\psi(x,y) =\begin{cases}x,&x\geq y,\\ -\infty,&x<y,\end{cases}$$

$$\Theta_{*}(x,y,z) =\max\{v(x),v(y),v(x+y),v(x+y+z),$$

$$\hskip 27.0pt\psi(v(x+z),x+2y+z),\psi(v(y+z),2x+y+z)\}.$$

The unguarded branches include $v(x)>0$ for $u\leq x\leq\mu_{*}$. Thus replacing the inactive value $-\infty$ by $0$ leaves this maximum unchanged. For $d=0,h_{G}$, put

(B.15) $$J_{d} =\int_{u\leq z\leq y\leq x\leq\mu_{*}}\frac{1-f((\Theta_{*}(x,y,z)+d-x-y-z)/z)}{xyz^{2}}\,dx\,dy\,dz,$$

$$H_{\log} =\log(\mu_{*}/u),$$

$$R_{\mathrm{up}} =500(F(30)-1)+500H_{\log}(1-f(30))+250H_{\log}^{2}(F(30)-1),$$

$$\mathcal{G}_{d} =1/\mu_{*}+R_{\mathrm{up}}+J_{d}.$$

The unshifted $d=0$ profile and functional come from [27, (4.70)–(4.75)]. The parameter $d$ and the shifted functional $\mathcal{G}_{d}$ are defined in this paper. The retained-side argument uses the instance $d=h_{G}$, constructed in (5.52). Paper II’s (5.4)–(5.5) are not the source of this shifted functional.

Proposition B.2.

The shifted scalar satisfies

(B.16) $$0\leq\mathcal{G}_{h_{G}}\leq U_{h}<\frac{6073714261}{10^{9}}.$$

Let $F_{h}^{+}\geq F(30)$, $f_{h}^{-}\leq f(30)$, and $H_{h}^{+}\geq H_{\log}$ be the directed table endpoints. Define

(B.17) $$\mathcal{R}(X,Y,Z) =500(X-1)+500Z(1-Y)+250Z^{2}(X-1),$$

$$R_{h} =\mathcal{R}(F_{h}^{+},f_{h}^{-},H_{h}^{+}),$$

$$U_{h} =1/\mu_{*}+R_{h}+J_{h}^{+}.$$

The construction has $F_{h}^{+}\geq 1$, $f_{h}^{-}\leq 1$. Thus $R_{\mathrm{up}}\leq R_{h}$. This directed endpoint comparison is the monotonicity step of [27, §4.6, (4.73)]. Table B.2 encloses these finite upper bounds. The constants $K_{\mathrm{hi}}$ and $L_{G1}$ are printed exactly in (5.48) and (5.49), with their Paper II locators. The normalization identities are

(B.18) $$M_{5,\mathrm{internal}} =-K_{\mathrm{hi}}\frac{6073714261}{10^{9}}-L_{G1},$$

$$M_{5,\mathrm{internal}}+K_{\mathrm{hi}}\left(\frac{6073714261}{10^{9}}-U_{h}\right) =-K_{\mathrm{hi}}U_{h}-L_{G1}.$$

Table B.2. Directed bounds for the shifted profile with $n_{\mathrm{mesh}}=512$.

Quantity	lower endpoint	upper endpoint
$R_{h}$	$0.000004373236955$	$0.000004373236956$
$J_{h}^{+}$	$0.953709886776891$	$0.953709886776892$
$U_{h}$	$6.073714260013847$	$6.073714260013848$
$6073714261/10^{9}-U_{h}$	$0.000000000986152$	$0.000000000986153$

Proof.

For the outer partition put $n=512$, $x_{0}=u$, $x_{n}=\mu_{*}$, and

(B.19) $$x_{i}=2^{-48}\left\lfloor 2^{48}\left(u+(\mu_{*}-u)\sqrt{i/n}\right)\right\rfloor\quad(1\leq i<n).$$

It is strictly increasing. For the inner integral write

(B.20) $$g_{d}(x,y)=\int_{u}^{y}\frac{1-f((\Theta_{*}(x,y,z)+d-x-y-z)/z)}{z^{2}}\,dz.$$

Each branch residual decreases in $x,y$. Each guard’s active set is a down-set. Thus $g_{d}$ is nondecreasing in $x,y$, guard jumps included.

Let $w_{i}$ be an outward upper bound for $\log(x_{i}/x_{i-1})$, on the $2^{-128}$ lattice. Let $v_{ij}$ enclose $g_{h_{G}}(x_{i},x_{j})$ upwards on that lattice. The diagonal cell has half the logarithmic square’s area. Hence

(B.21) $$J_{h_{G}}\leq J_{h}^{+}=\sum_{i=1}^{n}w_{i}\left(\sum_{j<i}w_{j}v_{ij}+\frac{1}{2}w_{i}v_{ii}\right).$$

Split the inner interval at every affine crossing. Include every guard crossing and each branch kink. On each resulting interval the argument is $A/z+B$. If $\Phi(t)=\int_{1}^{t}(1-f(s))\,ds$, then

(B.22) $$\int_{p}^{q}\frac{1-f(A/z+B)}{z^{2}}\,dz=\frac{\Phi(A/p+B)-\Phi(A/q+B)}{A}\quad(A\neq 0).$$

The trivial bound is $1/p-1/q$. The smaller of the two upper bounds may be used. For $A=0$, the argument is constant.

The delay equations give the primitive explicitly:

(B.23) $$P(t)=\frac{tF(t)}{2e^{\gamma}},\qquad\Phi(t)=t-1-2e^{\gamma}(P(t+1)-1)\quad(t\geq 0).$$

For $t\geq 30$, monotonicity gives

(B.24) $$\Phi(30)\leq\Phi(t)\leq\Phi(30)+(t-30)(1-f(30)).$$

We bound the trapezoid error. If $g$ is either $F$ or $f$ and $[a,a+h_{0}]\subset[1,30]$ is one mesh panel, then

(B.25) $$\int_{a}^{a+h_{0}}g(s)\,ds-{h_{0}\over 2}\bigl{(}g(a)+g(a+h_{0})\bigr{)}=-{h_{0}^{3}\over 12}g^{\prime\prime}(\xi),\qquad a<\xi<a+h_{0},$$

with one-sided interpretations at the integer breakpoints. The mesh has $h_{0}=1/65536$ and every breakpoint is a mesh node. We prove the derivative bound independently of the quadrature table. Write $C=2e^{\gamma}$. The elementary inequality $\gamma\leq H_{6}-\log 6$ and the first three positive terms of the atanh series give

$$\log 2>\frac{842}{1215},\qquad\log 3>\frac{263}{240},\qquad\gamma<\frac{12853}{19440}<\frac{2}{3}<\log 2.$$

Thus $0<C<4$ and $\log 2<1$. The defining functions satisfy

$$F(s)=C/s\ (1\leq s\leq 3),\quad f(s)=0\ (1\leq s\leq 2),\quad f(s)=C\log(s-1)/s\ (2<s\leq 3),$$

$$sF(s)=C+\int_{1}^{s-1}f(t)\,dt,\qquad sf(s)=\int_{1}^{s-1}F(t)\,dt\quad(s>3).$$

Induction on unit intervals gives $0\leq F,f\leq C$. If the delayed functions are in this range, the integrals have length $s-2$, so $F(s)\leq C(s-1)/s\leq C$ and $f(s)\leq C(s-2)/s\leq C$. For either companion pair $(g,h)=(F,f)$ or $(f,F)$, differentiating on open unit intervals gives

$$g^{\prime}(s)=\frac{h(s-1)-g(s)}{s},\qquad g^{\prime\prime}(s)=\frac{h^{\prime}(s-1)}{s}+\frac{2(g(s)-h(s-1))}{s^{2}}\quad(s>3).$$

For $f$, the first-derivative delay identity holds already for $t>2$. For $F$ on $2<t<3$, use $F^{\prime}(t)=-C/t^{2}$. Hence $|F^{\prime}(t)|,|f^{\prime}(t)|\leq C/t$ for $t>2$. Since $g(s),h(s-1)\in[0,C]$, for $s>3$

$$|g^{\prime\prime}(s)|\leq\frac{C}{s(s-1)}+\frac{2C}{s^{2}}\leq\frac{7C}{18}<\frac{14}{9}<12.$$

On the initial intervals,

$$F^{\prime\prime}(s)=2C/s^{3}\ (1<s<3),\quad f^{\prime\prime}(s)=0\ (1<s<2),$$

$$f^{\prime\prime}(s)=C\left(\frac{2\log(s-1)}{s^{3}}-\frac{1}{s(s-1)^{2}}-\frac{2}{s^{2}(s-1)}\right)\ (2<s<3).$$

The formulas imply $0\leq F^{\prime\prime}\leq 8$ on $[1,2]$, $0\leq F^{\prime\prime}\leq 1$ on $[2,3]$, and $-4\leq f^{\prime\prime}\leq\log 2<1$ on $[2,3]$. Inside the last parentheses each negative term has magnitude at most $1/2$. The positive term is at most $\log 2/4$. Therefore $|g^{\prime\prime}|\leq 8<12$ on all smooth panel interiors of $[1,30]$, with one-sided limits at the integer breakpoints.

The archived outward intervals and the independently computed exact rational intervals are recorded with the reproducibility materials. The derivative check starts from the proved invariant and then rebuilds the delay table. It checks every computed unit-interval enclosure against the archived coarse intervals. It confirms that the computed global bound is $\leq 12$, the recorded value:

$$\begin{array}[]{c|cc}\text{range}&F^{\prime\prime}&f^{\prime\prime}\\ \hline\cr[1,2]&[0,8]&[0,0]\\ {}[2,3]&[-8,8]&[-4,4]\\ {}[3,30]&[-12,12]&[-12,12]\end{array}$$

The computed bound is 8. Hence the absolute local remainder is at most $8h_{0}^{3}/12=(2/3)h_{0}^{3}\leq h_{0}^{3}$. The verifier checks this radius against the $h_{0}^{3}$ panel allowance before it accepts the regenerated table. The M5 computation builds the delay table with DEN=65536, S=128, 160-bit Arb interval arithmetic, and 1,900,545 generated nodes. The shifted computation then evaluates the exact $n_{\mathrm{mesh}}=512$ triangular grid, with $512\cdot 513/2=131,328$ inner-grid evaluations. It stores one aggregated row per outer index. Hence SHIFT-N512-fresh.json contains 512 rows for the 131,328 evaluations. The M5 computation uses no Taylor model and no width-$1/32$ step. The historical Taylor certificate provides background information and is not an input to this calculation. The table and shifted rows are regenerated before the directed inclusion check. The verifier compares recorded JSON fields and exact row values. It also checks SHA-256 hashes of the canonical compact-JSON integer arrays used by the real-inclusion verifier. These canonical-array hashes differ from hashes of regenerated JSON files. File formatting is not used as a certificate. Directed summation gives Table B.2. The positive last row proves the strict cap. The 512 aggregated rows give the exact 131328-term sum used for $J_{h}^{+}$. The directed real-inclusion result records the directed tail and the cap check. The reproducibility report documents the execution boundary. The two normalization identities follow by expansion. ∎

The proposition concerns the scalar functional only. The prime-sum comparison is required separately in Section 5.

B.4. Unshifted comparison (historical and not used)

Status: historical and non-certified.

This historical unshifted comparison records an earlier $n=8192$ calculation for context only. No current computation reproduces it. None of its values, table, constants or comparison is an input to the main proof. The comparison concerns the unshifted functional $G_{*}:=\mathcal{G}_{0}$. It is distinct from the main-line coefficient bound $G_{0}=6088440363/10^{9}$. The main K2 and L3 comparisons use the latter via [27, (4.73), (4.75), (4.79), (A.10a)]. They do not use $U_{0}$. The cited bound $G_{*}\leq G_{0}$ is the coefficient used in the main argument. It does not assert the separate, sharper $G_{*}\leq U_{0}$ comparison. This appendix proves that comparison only conditionally, as an optional one. Its historical delay table was constructed separately, but no such table or producer is part of the current proof package. The panel width is $h_{0}=1/65536$. At 192-bit precision, each trapezoid receives radius $h_{0}^{3}$. On smooth panel interiors, $|F^{\prime\prime}|,|f^{\prime\prime}|<12$. The panels meet the integer breakpoints. This bounds each local truncation remainder. The endpoint enclosures pass through the delay recursion in Arb ball arithmetic. Accumulated input radii are thus kept before outward rounding. All table endpoints are then rounded outwards to $2^{-128}$. The primitive and true tail use (B.24). This table gives a historical directed bound for $R_{\mathrm{up}}$. It is kept only to reproduce the earlier non-certified calculation. The separate earlier Taylor models are not part of the M5 computation. Let $F_{0}^{+}\geq F(30)$, $f_{0}^{-}\leq f(30)$, and $H_{0}^{+}\geq H_{\log}$ be these directed endpoints. Define

(B.26) $$R_{0}=\mathcal{R}(F_{0}^{+},f_{0}^{-},H_{0}^{+})\geq R_{\mathrm{up}}.$$

The earlier finite calculation used $n=8192$ in (B.19). The outer sum has $n(n+1)/2=33558528$ nodes. Write $J_{0}^{+}$ for that upper-corner sum. Table B.3 groups its eight consecutive row ranges. Each row lists a lower and upper bound for the sum. The diagonal half-weights are already included.

Table B.3. Historical non-certified contributions to the unshifted sum.

Rows	nodes	lower endpoint	upper endpoint
$1$–$1024$	524800	0.000002727813799350	0.000002727813799351
$1025$–$2048$	1573376	0.001226034866061399	0.001226034866061400
$2049$–$3072$	2621952	0.019651743682986805	0.019651743682986806
$3073$–$4096$	3670528	0.080548405705743750	0.080548405705743751
$4097$–$5120$	4719104	0.146720281588383935	0.146720281588383936
$5121$–$6144$	5767680	0.200161682376550457	0.200161682376550458
$6145$–$7168$	6816256	0.242188572133550572	0.242188572133550573
$7169$–$8192$	7864832	0.275672089024516687	0.275672089024516688

The earlier, non-certified calculation reported

(B.27) $$J_{0}^{+} <0.966171537191592960294580903427,$$

$$R_{0} <0.000004373236943357822206406687,$$

$$U_{0} =1/\mu_{*}+R_{0}+J_{0}^{+}+\frac{1}{2\cdot 10^{85}}$$

$$\in[6.086175910428536318116787310112545656706640625,$$

$$\hskip 23.99997pt6.086175910428536318116787310112545656706640626].$$

The term $1/(2\cdot 10^{85})$ bounds directed outer-sum rounding. The shifted computation does not use this interval. The comparison with the continuous functional is one-sided:

(B.28) $$\text{Historical comparison (not certified): }G_{*}=\mathcal{G}_{0}\leq U_{0}.$$

No claim $2e^{-\gamma}U_{0}\leq c$ follows from the printed interval. The finite row sums alone prove no prime-sum estimate. The source comparison and its error terms remain required.

B.5. The sixth-sum constant

The certified finite enclosure is

(B.29) $$\mathfrak{g}_{\mathrm{RA}}\in[0.0989598353353,\ 0.0989598353355].$$

Here $\mathfrak{g}_{\mathrm{RA}}$ denotes the integral quantity in Section 6. The constant used in the final bound is

(B.30) $$g_{\mathrm{RA,lo}}=\frac{989598353}{10^{10}}=0.0989598353.$$

Thus the certified enclosure gives

(B.31) $$\frac{7499}{9999}\left(J_{J}(1/2)-J_{J}(5227/10000)\right)-\frac{5106819}{31557955000}=\mathfrak{g}_{\mathrm{RA}}>g_{\mathrm{RA,lo}}.$$

Only this conservative lower constant enters the final bound.

B.6. Computer checks

Table B.6 lists the supplementary scripts. The second column states what each one checks. The partitions and defining formulas are printed above. The exact-arithmetic checks use Python’s integer and fraction types. Environment and timing details are recorded in the reproducibility materials. They are not theorem inputs. A full numerical regeneration is distinguished from a compact check. The certificate bundle is available at the location given in the Data availability statement; its SHA-256 manifest hash is 81fc00441c3a3f0924995cd41263feddece827d9b40376dae531191c86f90ed3. The recorded M5 regeneration agrees with the exact 512-row certificate. The reproducibility certificate set is independent. It consists of the H6 RESULT.json, the 512-row M5 output, the Root–A enclosure, and the verifier scripts for those records. Each record is checked against the source hashes and the manifest. None replaces an analytic estimate.

Table B.4. Scripts and finite checks.

Script	Statements checked	Time
Reproducibility replay verifier	manifests, source hashes, TeX structure, finite consistency checks, and the producer verification run	replay
H6/M5 producer verification	executes the H6 computation and M5 $n_{\mathrm{mesh}}=512$ table, shifted probe and real-inclusion verifier, computing the derivative enclosure and comparing regenerated outputs	replayed
H6 directed verifier	regenerates the directed H6 rows from the source inputs and Arb runtime	replayed
M5 derivative verifier	exact rational unit-interval bounds for $F^{\prime\prime}$ and $f^{\prime\prime}$ and the panel radius	replayed
M5 table, shift and inclusion verifiers	regenerates the $DEN=65536$ delay table, $n_{\mathrm{mesh}}=512$ rows, and real inclusion	replayed
Finite consistency verifier	exact finite arithmetic, orbit, and sign checks for the patches	replayed

The reproducibility verifier checks the recorded inputs and then runs the producer verification. That run regenerates the H6 output, the M5 delay table, the shifted rows and the real inclusion on the recorded platform. The earlier unshifted calculation used Python 3.12.3 and python-flint 0.9.0, with four workers. Its slowest worker took $7639$ seconds, rounded upwards. The historical unshifted comparison is optional and not certified. The main proof does not use it, and it does not affect the shifted computation. Neither these scripts nor their agreement establishes a uniform estimate. The final proof requires the analytic comparisons in Sections 5–8.

B.7. Rows of the retained side

The index set $\mathcal{I}$ of Section 5 consists of Paper II’s nineteen original row chains and one outer chain. The outer chain $\mathrm{T5}$ covers the upper part of $S_{5}$, Lemma 5.1. Sixteen of the row chains are listed below, each with its source parent and raw coefficient. The remaining three are Paper II’s single-prefix comparisons for $S_{3}$, $S_{8}$ and $S_{9}$. They keep Paper II’s own sources, restorations and error entries unchanged. The four rows of $\mathcal{H}$ are restored to their parents $S_{4}$, $S_{5}$, $S_{7}$ and $S_{10}$. The lower part of $S_{5}$ has no separate restoration address. The four joint rows are restored once to the joint source of $G_{1}$ with weights $(2,1,1)$.

Row	Parent	Coefficient	Row	Parent	Coefficient
S1-0	S1	$3$	S4	S4	$-1$
S1-1	S1	$-3$	S5-low	S5	$-1$
S1-2	S1	$3$	S7	S7	$-1$
S1-3	S1	$-3$	S10-Ret	S10	$1$
S2-0	S2	$1$	G1-0	G1 $(2,1,1)$	$-1$
S2-1	S2	$-1$	G1-1	G1 $(2,1,1)$	$1$
S2-2	S2	$1$	G1-2	G1 $(2,1,1)$	$-1$
S2-3	S2	$-1$	G1-3	G1 $(2,1,1)$	$1$

These sixteen rows, their seven source-parent restorations and four outer terms carry $91$ error addresses. Each address occurs exactly once. One outer term is the chain $\mathrm{T5}$. The others are the strict BV piece of $S_{2}$, inside its joint group, and two $G_{1}$ comparator entries. The same table has seven further rows with $38$ addresses. They treat $S_{3}$, $S_{8}$ and $S_{9}$ another way, and this paper does not use them.

B.8. Statements for the sixth sum

The sixth-sum argument uses the following statements.

Statement
Lemma 6.2 (rough-factor distribution)
Lemma 6.4 (four-family Siegel–Walfisz); range $N^{1/3}/8\leq H$
Lemmas 6.6 and 6.7 (selector and tensor, applied per shell pair)
Lemma 6.5 (transferred mean square)
Lemma 6.8 (two-factor convolution)
Lemma 6.9 (central rectangles)
Proper powers, (6.26)
Fact 3.2 (maximal prime estimates)
Paper I, (2.21), as used in Paper I, (3.25)

Use of AI tools

The author devised the method of proof. Codex (OpenAI, Astra 6) derived and checked the estimates, wrote the Python code for the certificates in Appendix B and ran blind reviews of the manuscript. Claude Code (Anthropic, Claude Fable 5) checked the proofs, the numerical certificates and the references. The author checked every step of the proof, every table and every reference, and takes full responsibility for the content of the paper.

Statements and declarations

Competing interests. The author has no competing interests.

Funding. No funding was received for this work.

Data availability. The supplementary scripts, exact inputs and compact outputs are archived as paper-iii-public-certificates-v1.tar.gz with release v1 at https://github.com/ryzfsu/goldbach-1plus1.75-theorem/releases/download/v1/paper-iii-public-certificates-v1.tar.gz.

The SHA-256 hash of the archive’s internal SHA256SUMS manifest is 81fc00441c3a3f0924995cd41263feddece827d9b40376dae531191c86f90ed3.