Irrationality of the dilogarithm at $1/3$

Abstract.

We prove that $\operatorname{Li}_{2}(1/3)$ is irrational. Irrationality of $\operatorname{Li}_{2}(1/s)$ was known for integers $s\leq-5$ and $s\geq 6$ by work of Hata and of Rhin and Viola, and Prellberg recently proved it for $s=-4,5,-3$. We define polynomials $G_{m}(T)$ with rational coefficients by an interpolation determinant. The rank of this determinant modulo primes $p$ with $m/8<p<5m/4$ bounds the denominators of the coefficients. A concavity inequality for a logarithmic energy, combined with the inversion $v\mapsto 3/v$, bounds $G_{m}(\operatorname{Li}_{2}(1/3))$. Interval arithmetic is used only for the energy and the potentials of a printed reference measure, on a uniform dyadic subdivision. The integral of the local rank function is a rational sum over $76$ printed affine pieces. The coefficients of $G_{4}$ and the $9\times 9$ determinant modulo $17$ are printed and can be checked by hand. A congruence modulo $17$ shows that $G_{m}(\xi)\neq 0$ for every rational $\xi$ and $m=4\cdot 17^{t}$, $t$ large in terms of $\xi$.

1. Introduction

For $|z|\leq 1$ let $\operatorname{Li}_{2}(z)=\sum_{k\geq 1}z^{k}/k^{2}$. Hata [5, Theorem 1.1] proved that $\operatorname{Li}_{2}(1/s)$ is irrational for every integer $s\leq-5$ and every $s\geq 7$. Rhin and Viola settled $s=6$ by their permutation group method [10, Section 6], and later treated negative arguments [11, Theorem 2.1 and Section 4]. Viola and Zudilin [13] introduced a six-parameter family of double integrals for dilogarithmic values. In 2024 Calegari, Dimitrov and Tang [2, Remark 2.8.2 and Section 15.2] listed the integers still open: $s\in\{-4,-3,-2,2,3,4,5\}$. Prellberg’s 2026 preprint [9, Theorem 1.1] proves the cases $s=-4,5,-3$. We prove the case $s=3$. Our linear forms are values of an interpolation determinant of size $9n$, in place of the double integrals of [5, 9, 10, 13]. The rank of this determinant modulo primes controls the denominators of its coefficients.

Theorem 1.

The number $\operatorname{Li}_{2}(1/3)$ is irrational.

By Landen’s identity [8, (25.12.3)], $\operatorname{Li}_{2}(1/3)+\operatorname{Li}_{2}(-1/2)=-\frac{1}{2}\log^{2}(3/2)$, and $\log(3/2)$ is transcendental by the Hermite–Lindemann theorem, a special case of [1, Theorem 1.4]. So Landen’s identity does not reduce the case $s=3$ to the case $s=-2$.

Put $\theta=\operatorname{Li}_{2}(1/3)$ and $m=4n$. Section 2 defines $G_{m}(T)\in\mathbb{Q}[T]$ by an interpolation determinant. Sections 3 and 4 give

$$\deg G_{m}\leq m\ \text{by Lemma~{}\ref{lem:moment-identity}},\qquad K_{m}G_{m}(T)\in\mathbb{Z}[T],\qquad K_{m}\in\mathbb{Q}_{>0},$$

$$\limsup_{\begin{subarray}{c}m\to\infty\\ 4\mid m\end{subarray}}\frac{\log K_{m}}{m^{2}}\leq\tau<3.87273,$$

(1.1) $$\log|G_{m}(\theta)|\leq-3.8757\,m^{2}+C\,m\log m\quad(m=4n\geq 4).$$

The exact value of $\tau$ is in (3.12), and $3.8757$ is a lower bound for the constant of (4.5). The finite constant $C$ of Lemma 4.4 depends only on the reference measure. Since $3.87273<3.8757$, the quadratic exponent is negative. If $\theta=u/q_{0}$ with $q_{0}>0$, then $q_{0}^{m}K_{m}G_{m}(\theta)\in\mathbb{Z}$. Section 5 shows that it is nonzero for $m=4\cdot 17^{t}$ and $t$ large, and (1.1) shows that it tends to zero.

The only numerical margin in the proof is $\alpha_{0}-\tau>0$. The size of $C$ is not used. Lemma 4.2 is computer-assisted. It is proved by interval arithmetic on the reference measure of Tables C.1 and C.2, over the uniform subdivision of Lemma C.1. The integral (3.8) is an exact sum over the $76$ affine pieces of Table A.2, each determined by the explicit lines of Table A. Tables 2.1, A and A.2 and (5.2) can be checked by hand. Python scripts recheck Lemma 4.2 and part of the remaining proof (Appendix D). Apart from this lemma, the proof is complete without them.

The proof uses three classical inputs.

Fact 1 (prime number theorem [4, Theorems 6 and 420, §§1.8 and 22.4]).

$\theta_{\rm Ch}(y)=\sum_{p\leq y}\log p$ satisfies $\theta_{\rm Ch}(y)/y\to 1$ as $y\to\infty$.

Only this qualitative form is used.

Fact 2 ([12, Lemma I.1.8]).

If $\sigma$ is a real compactly supported signed measure of total mass zero with $\iint|\log|u-w||\,d|\sigma|(u)\,d|\sigma|(w)<\infty$, then $\iint\log|u-w|\,d\sigma(u)\,d\sigma(w)\leq 0$.

A short proof is given in Appendix E.

Fact 3 (Cauchy–Vandermonde [7, Theorem 25], [3, Section 2]).

For $s=(s_{1},\ldots,s_{m})$, $v=(v_{1},\ldots,v_{L})$ and $h=L-m\geq 0$,

$$\det\begin{pmatrix}((s_{i^{\prime}}+v_{b})^{-1})_{1\leq i^{\prime}\leq m,\,1\leq b\leq L}\\ (v_{b}^{\ell-1})_{1\leq\ell\leq h,\,1\leq b\leq L}\end{pmatrix}=\frac{\Delta(s)\Delta(v)}{\prod_{i^{\prime},b}(s_{i^{\prime}}+v_{b})},\qquad\Delta(u)=\prod_{i<j}(u_{j}-u_{i}).$$

A proof, including the sign, is given in the proof of Lemma 2.2.

The use of AI tools is described at the end of the paper.

2. The determinant

Put $a=3$. For $n\geq 1$ let

$$\begin{gathered}m=4n,\qquad\sigma=k=h=n,\qquad L=m+h,\\ I=\{\sigma+k,\ldots,\sigma+k+m-1\},\quad J=\{\sigma,\ldots,\sigma+2m+h-1\},\quad\Lambda=\{1,\ldots,h\},\end{gathered}$$

all ordered increasingly. For $j\geq 0$ put

$$t_{j}=\sum_{u=1}^{j}\frac{a^{j-u}}{u},\qquad b_{j}=\sum_{u=1}^{j}\frac{a^{j-u}}{u^{2}};\qquad t_{0}=b_{0}=0.$$

For $i\in I$, $j\in J$ and $\ell\in\Lambda$ let

$$\begin{gathered}C_{ij}=\begin{cases}(a^{i}-a^{j})/(j-i),&i\neq j,\\ 0,&i=j,\end{cases}\quad B_{ij}=\begin{cases}(t_{i}-t_{j})/(j-i),&i\neq j,\\ b_{i},&i=j,\end{cases}\\ A_{ij}=a^{i}\delta_{ij},\qquad V_{\ell j}=\frac{1}{j+\ell},\qquad M_{ij}=\int_{0}^{1}\!\int_{0}^{1}\frac{x^{i}y^{j}}{xy-3}\,dx\,dy,\end{gathered}$$

and put

(2.1) $$G_{m}(T)=\det\begin{pmatrix}C\\ B-TA\\ V\end{pmatrix},$$

a determinant of size $2m+h=9n$. For an indeterminate $a$ the same formulas define

$$G_{m}(T;a)=\det\begin{pmatrix}C(a)\\ B(a)-TA(a)\\ V\end{pmatrix}\in\mathbb{Q}[a,T];\qquad G_{m}(T)=G_{m}(T;3).$$

Lemma 2.1.

One has $G_{m}\in\mathbb{Q}[T]$, $\deg G_{m}\leq m$, and

(2.2) $$M=B+\log(2/3)\,C-\theta A,\qquad G_{m}(\theta)=\det\begin{pmatrix}C\\ M\\ V\end{pmatrix}.$$

Proof.

Write $\lambda=\log(2/3)$ and

$$\mathfrak{m}_{j}=\int_{0}^{1}\frac{u^{j}}{u-3}\,du,\qquad\mathfrak{n}_{j}=\int_{0}^{1}\frac{u^{j}\log(1/u)}{u-3}\,du.$$

Polynomial division and the geometric series give

$$\begin{gathered}\mathfrak{m}_{0}=\lambda,\quad\mathfrak{n}_{0}=-\theta,\qquad\mathfrak{m}_{j}=a\mathfrak{m}_{j-1}+\frac{1}{j},\quad\mathfrak{n}_{j}=a\mathfrak{n}_{j-1}+\frac{1}{j^{2}}\quad(j\geq 1),\\ \mathfrak{m}_{j}=a^{j}\lambda+t_{j},\qquad\mathfrak{n}_{j}=b_{j}-a^{j}\theta.\end{gathered}$$

The substitution $u=xy$ gives

$$M_{ij}=\frac{\mathfrak{m}_{i}-\mathfrak{m}_{j}}{j-i}\quad(i\neq j),\qquad M_{ii}=\mathfrak{n}_{i}.$$

Hence the first identity in (2.2). Add $\lambda$ times each $C$ row to the corresponding row of $B-\theta A$. Only the $m$ middle rows depend on $T$, and their coefficients are rational. ∎

Put $r=2/5$ and $\gamma(\vartheta)=re^{i\vartheta}$ $(0\leq\vartheta\leq 2\pi)$. Let $[3,1]$ be the segment from $3$ to $1$, and let

$$\Gamma=[3,1]+c_{\Gamma}\gamma,\qquad c_{\Gamma}=\frac{\log 3}{2\pi i},$$

a linear combination of paths with complex coefficients. For every integer $d$,

(2.3) $$\int_{\Gamma}z^{d-1}\,dz=\begin{cases}(1-a^{d})/d,&d\neq 0,\\ 0,&d=0.\end{cases}$$

For $d\neq 0$ integrate $z^{d}/d$ on each path. For $d=0$ the segment gives $-\log 3$ and the circle gives $2\pi ic_{\Gamma}=\log 3$. The total variation is

(2.4) $$T_{r}=2+r\log 3=2.43944\ldots<3.$$

With $\Delta$ as in Fact 3, use variables $x=(x_{1},\ldots,x_{m})$, $v=(v_{1},\ldots,v_{L})$ and $z=(z_{1},\ldots,z_{m})$.

Lemma 2.2.

With product differentials $dx\,dv\,dz$,

(2.5) $$G_{m}(\theta)=\frac{3^{3m^{2}/2-m}}{(m!)^{2}L!}\int_{[0,1]^{m}}\int_{[0,1]^{L}}\int_{\Gamma^{m}}\Phi(x,v,z)\,dz\,dv\,dx,$$

where

$$\Phi(x,v,z)=\frac{\Delta(x)^{2}\Delta(v)^{2}\Delta(z)^{2}\displaystyle\prod_{d=1}^{m}\prod_{b=1}^{L}(z_{d}-v_{b})\displaystyle\prod_{i^{\prime}=1}^{m}x_{i^{\prime}}^{\sigma+k+h}\prod_{b=1}^{L}v_{b}^{\sigma}\prod_{d=1}^{m}z_{d}^{-m-k}}{\displaystyle\prod_{i^{\prime}=1}^{m}\prod_{b=1}^{L}(3-x_{i^{\prime}}v_{b})}.$$

Proof.

By (2.3), $C_{ij}=a^{i}\int_{\Gamma}z^{j-i-1}\,dz$. Put $f_{i}(v)=\int_{0}^{1}x^{i}/(xv-3)\,dx$. Then

$$M_{ij}=\int_{0}^{1}f_{i}(v)v^{j}\,dv,\qquad V_{\ell j}=\int_{0}^{1}v^{\ell-1}v^{j}\,dv.$$

Set $e_{\Delta}=m(m-1)/2$ and $\mathbf{w}=(z_{1},\ldots,z_{m},v_{1},\ldots,v_{L})$. Multilinearity in the rows and antisymmetrization, applied to the $L$ rows $[M;V]$, then to the $m$ rows $C$, and then to the $m$ functions $f_{i}$, give

$$G_{m}(\theta)=\frac{1}{(m!)^{2}L!}\int_{[0,1]^{m}\times[0,1]^{L}\times\Gamma^{m}}D_{z}D_{zv}D_{x}D_{xv}\,dx\,dv\,dz,$$

where

$$D_{z} =\det(a^{i}z_{d}^{-i-1})_{i\in I,\,1\leq d\leq m}=(-1)^{e_{\Delta}}a^{m(\sigma+k)+e_{\Delta}}\Delta(z)\prod_{d}z_{d}^{-\sigma-k-m},$$

$$D_{zv} =\det(\mathbf{w}_{b}^{j})_{1\leq b\leq 2m+h,\,j\in J}=(-1)^{mL}\Delta(z)\Delta(v)\prod_{d}z_{d}^{\sigma}\prod_{b}v_{b}^{\sigma}\prod_{d,b}(z_{d}-v_{b}),$$

$$D_{x} =\det(x_{i^{\prime}}^{i})_{i\in I,\,1\leq i^{\prime}\leq m}=\Delta(x)\prod_{i^{\prime}}x_{i^{\prime}}^{\sigma+k},$$

$$D_{xv} =\det\begin{pmatrix}((x_{i^{\prime}}v_{b}-3)^{-1})_{1\leq i^{\prime}\leq m,\,1\leq b\leq L}\\ (v_{b}^{\ell-1})_{1\leq\ell\leq h,\,1\leq b\leq L}\end{pmatrix}=\frac{(-1)^{mL}3^{e_{\Delta}}\Delta(x)\Delta(v)\prod_{i^{\prime}}x_{i^{\prime}}^{h}}{\prod_{i^{\prime},b}(3-x_{i^{\prime}}v_{b})}.$$

The last equality follows from the Cauchy–Vandermonde identity. Explicitly, for $s_{i^{\prime}}=-3/x_{i^{\prime}}$,

$$\det\begin{pmatrix}((x_{i^{\prime}}v_{b}-3)^{-1})_{i^{\prime},b}\\ (v_{b}^{\ell-1})_{\ell,b}\end{pmatrix}=\prod_{i^{\prime}}x_{i^{\prime}}^{-1}\det\begin{pmatrix}((s_{i^{\prime}}+v_{b})^{-1})_{i^{\prime},b}\\ (v_{b}^{\ell-1})_{\ell,b}\end{pmatrix},$$

and $\Delta(s)=3^{e_{\Delta}}\Delta(x)\prod_{i^{\prime}}x_{i^{\prime}}^{1-m}$, $\prod_{i^{\prime},b}(s_{i^{\prime}}+v_{b})=(-1)^{mL}\prod_{i^{\prime},b}(3-x_{i^{\prime}}v_{b})\prod_{i^{\prime}}x_{i^{\prime}}^{-L}$. Fact 3 and this row scaling give the formula for $D_{xv}$. Continuity covers $x_{i^{\prime}}=0$.

We prove Fact 3 as an identity of rational functions, with $s_{i^{\prime}},v_{b}$ independent indeterminates. We use induction on $m$, for all $h\geq 0$ at once. The case $m=0$ is the Vandermonde identity. Let $m\geq 1$, let $v^{\prime}=(v_{1},\ldots,v_{L-1})$, and let $\widehat{s}_{i^{\prime}}$ denote $s$ with $s_{i^{\prime}}$ omitted. As a function of $v_{L}$, expansion along the last column writes the determinant as $\sum_{i^{\prime}}(-1)^{i^{\prime}+L}M_{i^{\prime}}/(s_{i^{\prime}}+v_{L})$ plus a polynomial of degree at most $h-1$, where $M_{i^{\prime}}$ is the minor without row $i^{\prime}$ and column $L$. The right side is a polynomial of degree $L-1=m+h-1$ in $v_{L}$ divided by $\prod_{i^{\prime}}(s_{i^{\prime}}+v_{L})$, so it has the same form. By the induction hypothesis for $(\widehat{s}_{i^{\prime}},v^{\prime})$, $M_{i^{\prime}}$ is the right side of the fact for $(\widehat{s}_{i^{\prime}},v^{\prime})$. We have $\Delta(s)=(-1)^{i^{\prime}-1}\Delta(\widehat{s}_{i^{\prime}})\prod_{j\neq i^{\prime}}(s_{j}-s_{i^{\prime}})$, and $\Delta(v)=(-1)^{L-1}\Delta(v^{\prime})\prod_{b<L}(s_{i^{\prime}}+v_{b})$ at $v_{L}=-s_{i^{\prime}}$. Hence, after the factors $s_{j}+v_{L}=s_{j}-s_{i^{\prime}}$ $(j\neq i^{\prime})$ and $s_{i^{\prime}}+v_{b}$ $(b<L)$ cancel, the residue of the right side at $v_{L}=-s_{i^{\prime}}$ is

$$(-1)^{i^{\prime}+L}\frac{\Delta(\widehat{s}_{i^{\prime}})\Delta(v^{\prime})}{\prod_{j\neq i^{\prime}}\prod_{b<L}(s_{j}+v_{b})}=(-1)^{i^{\prime}+L}M_{i^{\prime}}.$$

So the residues agree, and the difference of the two sides is a polynomial in $v_{L}$ of degree at most $h-1$, zero if $h=0$. It vanishes at the $L-1\geq h$ points $v_{L}=v_{b}$, $b<L$, where the determinant has two equal columns and $\Delta(v)=0$. Hence it is zero.

The product of the four factors is $(-1)^{e_{\Delta}}a^{m(\sigma+k)+2e_{\Delta}}\Phi$. Since $m=4n$, $e_{\Delta}=2n(4n-1)$ is even and $m(\sigma+k)+2e_{\Delta}=3m^{2}/2-m$. All integrals converge absolutely: $3-xv\geq 2$ on $[0,1]^{2}$, $|z|\geq r$ on $\Gamma$, and $\Gamma$ has finite variation. ∎

An exact calculation gives

(2.6) $$G_{4}(T)=\frac{9}{67228000000000}\sum_{j=0}^{4}q_{j}T^{j},\qquad 67228000000000=2^{11}5^{9}7^{5},$$

with the coefficients in Table 2.1.

Table 2.1. The integer coefficients in (2.6).

$j$	$q_{j}$
$0$	$11390586457461030954563$
$1$	$-124400937938521703661570$
$2$	$509486272807705849983600$
$3$	$-927382286191365446400000$
$4$	$633018449291268096000000$

3. The divisor

Put

$$N_{\mathrm{den}}=10n-1,\qquad U=11n-1,\qquad d_{\ell}=\operatorname{lcm}(1,\ldots,\ell),\qquad\mathcal{D}_{m}=d_{N_{\mathrm{den}}}^{3m}d_{U}^{n}.$$

For $F\in\mathbb{Q}[T]$ let $v_{p}(F)=\min_{j}v_{p}([T^{j}]F)$. From now on $k$ is a running index, and the parameter $k$ of Section 2 equals $n$. Fix the seam $R=8$. Primes $p\leq m/R$ are cleared by $\mathcal{D}_{m}$ alone. For primes $m/R<p<5m/4$ we use the rank bound of Lemma 3.2.

Lemma 3.1.

For an indeterminate $a$,

(3.1) $$P_{m}(a,T):=\mathcal{D}_{m}G_{m}(T;a)\in\mathbb{Z}[a,T].$$

Proof.

The denominators of a $C$ entry divide $d_{N_{\mathrm{den}}}$. An off-diagonal $B$ entry is a difference of harmonic sums divided by $j-i$, so its denominator divides $d_{N_{\mathrm{den}}}^{2}$, and so does that of $b_{i}$. The entries $Ta^{i}$ are integral. The denominators of a $V$ entry divide $d_{U}$. ∎

3.1. Residue classes

For $\rho>0$ and $0\leq x<\rho$ let

(3.2) $$K(x,\rho) =\{k\geq 0:1/2\leq x+k\rho<3/2\},$$

$$\mathcal{L}(x,\rho) =\{l\geq 0:1/4\leq x+l\rho<5/2\},$$

$$W(x,\rho) =\{w\geq 1:0<w\rho-x\leq 1/4\},$$

and

(3.3) $$\mathfrak{r}(x,\rho)=\min(2|K|+|W|,|\mathcal{L}|)-\mathbf{1}_{\{K=\{0\},\,\mathcal{L}=\{0,1\},\,W=\varnothing\}},\qquad\mathfrak{c}(\rho)=1+\int_{0}^{\rho}\mathfrak{r}(x,\rho)\,dx.$$

Lemma 3.2.

Let $p$ be a prime with $p^{2}>U$, so that $p\geq 5$, and let $a=3$. Then

(3.4) $$v_{p}(G_{m})\geq-m-\sum_{u=0}^{p-1}\mathfrak{r}(u/m,p/m)=-m\mathfrak{c}(p/m).$$

Proof.

Multiply the whole rows of $C$, $B-TA$ and $V$ by $p$, $p^{2}$ and $p$. The resulting matrix $S_{p}(T)$ has $\det S_{p}(T)=p^{3m+n}G_{m}(T)$ and entries in $\mathbb{Z}_{(p)}[T]$, and $S_{p}(T)\equiv S_{p}(0)\pmod{p}$ because $p^{2}TA\equiv 0$. Write $i=u+kp$, $j=u+lp$ with $0\leq u<p$. Only the harmonic terms with $p$ in the denominator survive modulo $p$. Their quotients are $p$-units because all indices are less than $p^{2}$, and $a^{p}\equiv a\pmod{p}$, so

$$pt_{u+kp}\equiv a^{u}t_{k},\qquad p^{2}b_{u+kp}\equiv a^{u}b_{k}\pmod{p}.$$

The scaled $C$ and $B$ entries vanish modulo $p$ between distinct residue classes modulo $p$. For a cross-class $B$ entry, the numerator is $p$-integral after the row scaling and the remaining factor $p/(j-i)$ is zero modulo $p$. Within a class, for $j\neq i$, $j-i=p(l-k)$ with $0<|l-k|<p$, so the quotient denominator is a unit. Diagonal entries are given by the separate cases $i=j$ in the definitions of $C$ and $B$. Since $a^{u+kp}\equiv a^{u+k}\pmod{p}$, for $i\neq j$

$$pC_{ij}=\frac{a^{i}-a^{j}}{l-k}\equiv a^{u}\frac{a^{k}-a^{l}}{l-k}=a^{u}C_{kl},\qquad p^{2}B_{ij}=\frac{pt_{i}-pt_{j}}{l-k}\equiv a^{u}B_{kl}\pmod{p}.$$

A $V$ row reduces to zero unless $\ell+u=pw$. Then its entry is $1/(l+w)$, without the factor $a^{u}$, and $l+w=(j+\ell)/p<p$. So, after permutations, the reduction has blocks with $2|K|+|W|$ rows and $|\mathcal{L}|$ columns, where $(x,\rho)=(u/m,p/m)$.

For $K=\{0\}$, $\mathcal{L}=\{0,1\}$, $W=\varnothing$ the block, with its rows divided by the unit $a^{u}$, is

(3.5) $$\begin{pmatrix}0&1-a\\ 0&-1\end{pmatrix}=\begin{pmatrix}0&-2\\ 0&-1\end{pmatrix},$$

of rank one for every $p$. Every other block has rank at most the minimum of its dimensions. So the rank of $S_{p}(0)$ modulo $p$ is at most $\mathrm{rk}_{p}=\sum_{u}\mathfrak{r}(u/m,p/m)$. The matrix has $2m+n$ rows. Row operations over $\mathbb{Z}_{(p)}$, independent of $T$, make its last $2m+n-\mathrm{rk}_{p}$ rows divisible by $p$ coefficientwise. Subtracting the scaling exponent $3m+n$ gives the first inequality in (3.4).

Every boundary in (3.2) has the form $x=c_{0}+b\rho$ with $4c_{0},b\in\mathbb{Z}$. At $\rho=p/m$, $m=4n$, it lies in $m^{-1}\mathbb{Z}$. The $K,\mathcal{L}$ conditions are closed on the left and open on the right in $x$, and the $W$ condition is $w\rho-1/4\leq x<w\rho$. So $\mathfrak{r}$ is constant on each $[u/m,(u+1)/m)$, and

(3.6) $$\sum_{u=0}^{p-1}\mathfrak{r}(u/m,p/m)=m\int_{0}^{p/m}\mathfrak{r}(x,p/m)\,dx.\qed$$

The lines that bound the sets (3.2) are $x=z_{0}-k\rho$ $(z_{0}\in\{1/4,1/2,3/2,5/2\},\ k\geq 0)$ and $x=w\rho-1/4$ $(w\geq 1)$. The first meets the strip $0<x<\rho$ exactly for $z_{0}/(k+1)<\rho<z_{0}/k$, the second exactly for $1/(4w)<\rho<1/(4(w-1))$. For $k=0$ and for $w=1$ the upper endpoint is $+\infty$. So $x=z_{0}-k\rho$ meets the strip for some $\rho\in(1/R,5/4)$ if and only if $4z_{0}/5-1<k<z_{0}R$, and $x=w\rho-1/4$ if and only if $1\leq w<R/4+1$. Hence, for $1/R<\rho<5/4$, the sets (3.2) change only on the $39$ lines

(3.7) $$\begin{gathered}x=0,\quad x=\rho,\quad x=w\rho-1/4\quad(1\leq w\leq 2),\\ x=1/4-k\rho\quad(0\leq k\leq 1),\quad x=1/2-k\rho\quad(0\leq k\leq 3),\\ x=3/2-k\rho\quad(1\leq k\leq 11),\quad x=5/2-k\rho\quad(2\leq k\leq 19).\end{gathered}$$

Cut $(1/R,5/4)$ at the finitely many values of $\rho$ where two of these lines meet in $0\leq x\leq\rho$, and let $\mathcal{C}$ be the set of these cut points. For $\rho\in(1/R,5/4)\setminus\mathcal{C}$ and each $z_{0}$ exactly one line $x=z_{0}-k\rho$ meets $0<x<\rho$, namely $k=\lfloor z_{0}/\rho\rfloor$, and exactly one line $x=w\rho-1/4$ does, namely $w=\lfloor 1/(4\rho)\rfloor+1$. These indices lie in the ranges of (3.7). So on each open interval of $(1/R,5/4)\setminus\mathcal{C}$ the order of the lines in $0<x<\rho$ is fixed, exactly five lines lie there, the sets (3.2) are constant on each of the six $x$-intervals between consecutive lines, and $\mathfrak{c}$ is affine. A change of $\rho$ by $\delta$ moves each of these lines by $O(|\delta|)$ and the endpoint of the integral by $\delta$. So the sets (3.2) change only on strips of total length $O(|\delta|)$. As $\mathfrak{r}$ is bounded, $|\mathfrak{c}(\rho)-\mathfrak{c}(\rho^{\prime})|=O(|\rho-\rho^{\prime}|)$, and $\mathfrak{c}$ is continuous. Table A.2 (Appendix A) lists $76$ consecutive intervals with union $(1/R,5/4)$ up to finitely many points, and the affine function $\mathfrak{c}$ on each. Each row follows from (3.2)–(3.3) and (3.7). The affine functions of adjacent rows agree at all $75$ internal joins. Summation of the rows gives

(3.8) $$\int_{1/R}^{5/4}\mathfrak{c}(\rho)\,d\rho=\frac{580566971}{177365760}.$$

The configuration $K=\{0\}$, $\mathcal{L}=\{0,1\}$, $W=\varnothing$ of (3.5) means $\max(1/2,3/2-\rho)\leq x<3/2$, $\max(1/4,5/2-2\rho)\leq x<5/2-\rho$ and $x<\rho-1/4$. For $\rho<5/4$ this is $\max(1/2,3/2-\rho,5/2-2\rho)\leq x<\rho-1/4$, which is nonempty only for $\rho>11/12$. So it occupies $5/2-2\rho\leq x<\rho-1/4$ for $11/12<\rho<1$, and $1/2\leq x<\rho-1/4$ for $1\leq\rho<5/4$. As $1/R<11/12$, its contribution to (3.8) is

$$-\int_{11/12}^{1}(3\rho-11/4)\,d\rho-\int_{1}^{5/4}(\rho-3/4)\,d\rho=-\frac{5}{48}.$$

3.2. Larger primes

For $p>5n$ define row potentials $\lambda_{C}=0$, $\lambda_{B}(i)$, $\lambda_{V}(\ell)$ and column potentials $\chi(j)$ by

$$5n<p<6n: \quad(\lambda_{B},\lambda_{V},\chi)=(1+\mathbf{1}_{i\geq p},\,1,\,\mathbf{1}_{j\geq p+2n}),$$

$$6n\leq p<8n: \quad(\lambda_{B},\lambda_{V},\chi)=(\mathbf{1}_{i+p<10n},\,1,\,\mathbf{1}_{j\geq p}),$$

$$8n\leq p<10n: \quad(\lambda_{B},\lambda_{V},\chi)=(0,\,1,\,\mathbf{1}_{j\geq p}),$$

$$10n\leq p\leq 11n-1: \quad(\lambda_{B},\lambda_{V},\chi)=(0,\,\mathbf{1}_{p-\ell\in J},\,0),$$

and put

$$E_{p}^{\rm out}=\begin{cases}19n-2p,&5n<p<8n,\\ 11n-p,&8n\leq p\leq U.\end{cases}$$

Lemma 3.3.

For a prime $5n<p\leq U$ one has $v_{p}(G_{m})\geq-E_{p}^{\rm out}$.

Proof.

A $C$ entry has a pole only at $j=i+p$, of order at most one. A $B$ entry has pole order at most

$$\mathbf{1}_{\max(i,j)\geq p}+\mathbf{1}_{\{\max(i,j)\geq p,\ i\equiv j\pmod{p}\}},$$

including the double pole of $b_{i}$ for $i\geq p$. A $V$ entry has a simple pole when $p\mid j+\ell$. As $p^{2}>U$, there are no higher powers.

In the first range every $B$ row has potential at least one, and two if $i\geq p$. A collision $j=i+p$ has $j\geq p+2n$, and its column gives the second unit. In the second range all $i<p$. The columns $j\geq p$ cover the simple $B$ poles and the $C$ poles. A collision needs $i+p<10n$, and its row gives the other unit. In the third range $i<p$ and $i+p\geq 10n$, so only simple $B$ poles remain. In these three ranges every $V$ row has potential one, which covers $j+\ell=p$ and $j+\ell=2p$. In the fourth range $i,j<p$, only $V$ poles remain, and the rows with $\lambda_{V}=1$ are those containing one. So $\lambda_{\rm row}+\chi_{j}$ dominates every pole order. Multiply each row and column by $p$ to the power of its potential. The potentials add up to $(10n-p)+n+(8n-p)$, $(8n-p)+n+(10n-p)$, $n+(10n-p)$ and $11n-p$ in the four ranges. ∎

3.3. The factor at three

Lemma 3.4.

Let $N_{m}=\lceil 53m^{2}/48\rceil-m$. Then

(3.9) $$P_{m}(a,T)\in a^{N_{m}}\mathbb{Z}[a,T].$$

In particular $v_{3}(G_{m})\geq N_{m}-v_{3}(\mathcal{D}_{m})$.

Proof.

Work in $\mathbb{Q}[[a]][T]$. Write

$$S_{j}(a)=\sum_{u\geq 1}\frac{a^{u}}{u^{j}},\qquad\mathsf{v}_{k}(i)=\begin{cases}(i-k)^{-1},&i\neq k,\\ 0,&i=k,\end{cases}\quad\mathsf{z}_{k}(i)=\mathsf{v}_{k}(i)^{2},\quad\mathsf{e}_{k}(i)=\mathbf{1}_{i=k}.$$

Subtract $S_{1}(a)$ times each $C$ row from the paired row of $B-TA$. Comparison of coefficients, including the diagonal, gives

(3.10) $$\begin{pmatrix}C\\ B-TA-S_{1}C\end{pmatrix} =\sum_{k\geq 0}a^{k}\begin{pmatrix}\mathsf{v}_{k}(I)\\ -\mathsf{z}_{k}(I)-(T+S_{2})\mathsf{e}_{k}(I)\end{pmatrix}\mathsf{e}_{k}(J)^{t}$$

$$\quad+\sum_{k\geq 0}a^{k}\begin{pmatrix}\mathsf{e}_{k}(I)\\ \mathsf{v}_{k}(I)\end{pmatrix}\mathsf{v}_{k}(J)^{t}+\sum_{k\geq 0}a^{k}\begin{pmatrix}0\\ -\mathsf{e}_{k}(I)\end{pmatrix}\mathsf{z}_{k}(J)^{t}.$$

For example, the off-diagonal lower entry is $\sum_{k\geq 0}a^{k}\mathsf{v}_{k}(i)\mathsf{v}_{k}(j)-(a^{i}+a^{j})/(i-j)^{2}$, and the diagonal entry is $\sum_{k\geq 0}a^{k}\mathsf{v}_{k}(i)^{2}-a^{i}(T+S_{2})=b_{i}-Ta^{i}$. Each coefficient of $a$ involves finitely many terms.

Expand along the $n$ rows of $V$, and expand each remaining $2m$-row minor by the rank-one summands in (3.10). A nonzero term selects $2m=8n$ distinct rank-one summands. At weight $k$ at most

$$c_{k}=1+\mathbf{1}_{k\in J}+\mathbf{1}_{k\in I}$$

occur: the first summand needs $k\in J$, the third needs $k\in I$. All entries, including $T+S_{2}$, have nonnegative order in $a$. So every term has order at least the sum of the $8n$ smallest available weights. The infinite expansion is used coefficientwise. Modulo $a^{N_{m}+1}$ only finitely many $k$ occur, so Cauchy–Binet applies to a finite truncation.

There is one weight at each $0\leq k<n$, two at each $n\leq k<2n$, and three at each $2n\leq k<6n$. Write $5n=3d+\epsilon^{\prime}$, $0\leq\epsilon^{\prime}<3$. The minimum available-weight sum is

$$\sum_{k=0}^{n-1}k+2\sum_{k=n}^{2n-1}k+3\sum_{k=2n}^{2n+d-1}k+\epsilon^{\prime}(2n+d) =\frac{53n^{2}}{3}-4n+\frac{\epsilon^{\prime}(3-\epsilon^{\prime})}{6}$$

$$=\left\lceil\frac{53n^{2}}{3}\right\rceil-4n=N_{m}.$$

The minors of $V$ do not depend on $a,T$, so $a^{N_{m}}\mid G_{m}(T;a)$ in $\mathbb{Q}[a,T]$. By (3.1), division by the monic $a^{N_{m}}$ gives (3.9). Specialization at $a=3$ gives the last statement. ∎

3.4. Summation over primes

Put $e_{0}(p)=3m\lfloor\log_{p}N_{\mathrm{den}}\rfloor+n\lfloor\log_{p}U\rfloor=v_{p}(\mathcal{D}_{m})$. With the first applicable line taking precedence, let

(3.11) $$E_{p}=\begin{cases}e_{0}(3)-N_{m},&p=3,\\ e_{0}(p),&p\leq m/R\ \text{or}\ p^{2}\leq U\ \text{or}\ p=5m/4,\\ m\mathfrak{c}(p/m),&m/R<p<5m/4,\\ E_{p}^{\rm out},&5m/4<p\leq U,\\ 0,&p>U,\end{cases}\qquad K_{m}=\prod_{p}p^{E_{p}}\in\mathbb{Q}_{>0}.$$

Lemma 3.5.

One has $K_{m}G_{m}(T)\in\mathbb{Z}[T]$ for every $m=4n$, and $\log K_{m}\leq\tau m^{2}+o(m^{2})$ as $m\to\infty$, where

(3.12) $$\begin{gathered}A=\frac{13}{4R}+\int_{1/R}^{5/4}\mathfrak{c}(\rho)\,d\rho+\frac{45}{32}=\frac{902042411}{177365760},\\ \tau=A-\frac{53}{48}\log 3=3.872724286099666640\ldots,\end{gathered}$$

so that $\alpha_{0}-\tau>0.00305$ for the constant $\alpha_{0}$ defined after (4.5) in Section 4.

Proof.

By (3.6), $m\mathfrak{c}(p/m)\in\mathbb{Z}$. Lemmas 3.1, 3.2, 3.3 and 3.4 give $v_{p}(K_{m}G_{m})\geq 0$ for every $p$. The exponent at three may be negative. Since $\mathfrak{r}\leq|\mathcal{L}|$ and $\sum_{u}|\mathcal{L}(u/m,p/m)|=|J|=9n$, every inner exponent is at most $13n$.

If $m\geq 11R^{2}/4$, then $p>m/R$ implies $p^{2}>m^{2}/R^{2}\geq 11m/4>U$. Put

$$\eta_{\rm PNT}(X)=\sup_{y\geq X}|\theta_{\rm Ch}(y)/y-1|.$$

By Fact 1, $\eta_{\rm PNT}(X)\to 0$. For an affine $f$ on $[t_{0},t_{1}]\subset[1/R,11/4]$, Stieltjes integration by parts gives

$$m\sum_{t_{0}m<p<t_{1}m}f(p/m)\log p=m^{2}\int_{t_{0}}^{t_{1}}f(\rho)\,d\rho+O_{f}\bigl{(}\eta_{\rm PNT}(m/R)m^{2}+m\log m\bigr{)}.$$

There are finitely many pieces, and the primes at the finitely many points of $\mathcal{C}\cup\{1/R,5/4,2,11/4\}$, at most one for each point and each $m$, contribute $O(m\log m)$. For $p\leq m/R$ the first powers contribute $(13m/4)\theta_{\rm Ch}(m/R)=\frac{13}{4R}m^{2}+o(m^{2})$, and the higher powers, including those of two and three,

$$O\Bigl{(}m\sum_{j=2}^{\lfloor\log_{2}U\rfloor}\theta_{\rm Ch}(U^{1/j})\Bigr{)}=O(m^{3/2}\log(3m)).$$

Subtract $N_{m}\log 3$ once. For $p>5m/4$ the exponent is $m\mathfrak{c}_{\rm out}(p/m)$ with

$$\mathfrak{c}_{\rm out}(\rho)=\begin{cases}19/4-2\rho,&5/4<\rho<2,\\ 11/4-\rho,&2\leq\rho\leq 11/4,\end{cases}\qquad\int_{5/4}^{11/4}\mathfrak{c}_{\rm out}(\rho)\,d\rho=\frac{9}{8}+\frac{9}{32}=\frac{45}{32}.$$

With (3.8) this gives $A-\frac{53}{48}\log 3=\tau$. The total error is $O(\eta_{\rm PNT}(m/R)m^{2}+m^{3/2}\log(3m))=o(m^{2})$. The bounds $\log 3>2\sum_{k=0}^{99}((2k+1)2^{2k+1})^{-1}$ and $\log 3<2\sum_{k=0}^{99}((2k+1)2^{2k+1})^{-1}+8/(3\cdot 201\cdot 2^{201})$ give the digits of $\tau$ and of $\alpha_{0}-\tau$. ∎

4. Analytic estimate

In (2.5) there are $m$, $L=5m/4$ and $m$ variables $x$, $v$, $z$. Let

$$\mathcal{K}_{x}=\mathcal{K}_{v}=[0,1],\qquad\mathcal{K}_{z}=[1,3]\cup\{z:|z|=r\}.$$

For finite positive measures put

$$I(\sigma,\sigma^{\prime}) =\iint\log|u-w|\,d\sigma(u)\,d\sigma^{\prime}(w),\qquad D(\sigma,\sigma^{\prime})=\iint\log|3-uw|\,d\sigma(u)\,d\sigma^{\prime}(w),$$

$$\mathcal{Q}(\mu) =I(\mu_{x},\mu_{x})+I(\mu_{v},\mu_{v})+I(\mu_{z},\mu_{z})+I(\mu_{v},\mu_{z})-D(\mu_{x},\mu_{v}).$$

For a fixed $\nu=(\nu_{x},\nu_{v},\nu_{z})$ let $\mathcal{U}_{i}(q)=\int\log|q-t|\,d\nu_{i}(t)$, $D_{i}(q)=\int\log|3-qt|\,d\nu_{i}(t)$ $(i=x,v)$,

$$\psi_{x} =2\mathcal{U}_{x}-D_{v},\qquad\psi_{v}=2\mathcal{U}_{v}+\mathcal{U}_{z}-D_{x},\qquad\psi_{z}=2\mathcal{U}_{z}+\mathcal{U}_{v},$$

$$F_{x}(q) =\psi_{x}(q)+\tfrac{3}{4}\log q,\quad F_{v}(q)=\psi_{v}(q)+\tfrac{1}{4}\log q\quad(0<q\leq 1),$$

$$F_{z}(q) =\psi_{z}(q)-\tfrac{5}{4}\log|q|\quad(q\in\mathcal{K}_{z}).$$

For a compact line segment or circle $\Gamma\subset\mathbb{C}$ let $\ell_{\Gamma}$ be arclength on $\Gamma$. A segment meets the open disk $B(q,t)=\{w:|w-q|<t\}$ in length at most $2t$. A circle meets it in arclength at most $2\pi t$. If $t$ is at least the radius, this follows from the length of the circle. Otherwise the intersection is an arc of at most a semicircle, of length at most $\pi/2$ times its chord. So $\ell_{\Gamma}(B(q,t))\leq 2\pi t$ for all $q\in\mathbb{C}$, $t>0$. A positive measure $\eta$ is tame if $\eta\leq H(\ell_{\Gamma_{1}}+\dots+\ell_{\Gamma_{N}})$ for some $H\geq 0$ and compact segments or circles $\Gamma_{1},\ldots,\Gamma_{N}$, that is, if $\eta$ is a finite sum of measures with arclength density at most $H$ on finitely many segments and circles. Tame measures are finite and compactly supported, and sums of tame measures are tame. Let $\mathfrak{M}$ be the set of triples $\mu=(\mu_{x},\mu_{v},\mu_{z})$ of tame measures of masses $(1,5/4,1)$ with $\operatorname{supp}\mu_{x}\subseteq\{|q|\leq 19/16\}$ and $\operatorname{supp}\mu_{v}\subseteq\{\zeta\leq|q|\leq 19/16\}$ for some $\zeta>0$. On these supports $|3-uw|\geq 3-(19/16)^{2}>1$.

Lemma 4.1.

For $\mu,\nu\in\mathfrak{M}$, with $\psi_{x},\psi_{v},\psi_{z}$ formed from $\nu$,

(4.1) $$\mathcal{Q}(\mu)\leq-\mathcal{Q}(\nu)+\int\psi_{x}\,d\mu_{x}+\int\psi_{v}\,d\mu_{v}+\int\psi_{z}\,d\mu_{z}.$$

Proof.

We first show that for tame $\eta,\eta^{\prime}$

$$\iint|\log|u-w||\,d\eta(u)\,d\eta^{\prime}(w)<\infty.$$

By the definition of tame it suffices to take $\eta=\ell_{\Gamma}$, $\eta^{\prime}=\ell_{\Gamma^{\prime}}$. Let $\Lambda\geq 1$ bound $|u-w|$ on $\Gamma\times\Gamma^{\prime}$. For $u\in\Gamma$ the layer-cake formula and the bound on $\ell_{\Gamma^{\prime}}(B(u,t))$ give

$$\int|\log|u-w||\,d\ell_{\Gamma^{\prime}}(w)\leq\int_{0}^{\infty}\ell_{\Gamma^{\prime}}(B(u,e^{-s}))\,ds+\ell_{\Gamma^{\prime}}(\mathbb{C})\log\Lambda\leq 2\pi+\ell_{\Gamma^{\prime}}(\mathbb{C})\log\Lambda,$$

and integration over $u$ gives the bound.

Let $V$ be the space of real signed measures $\eta-\eta^{\prime}$ with $\eta,\eta^{\prime}$ tame and $\eta(\mathbb{C})=\eta^{\prime}(\mathbb{C})$. By this bound $I$ is finite on $V\times V$ and on pairs of tame measures, so all pairings below are finite. For $\sigma=\eta-\eta^{\prime}\in V$ one has $|\sigma|\leq\eta+\eta^{\prime}$, and Fact 2 gives

$$\|\sigma\|^{2}=-I(\sigma,\sigma)\geq 0.$$

So $-I$ is a positive semidefinite symmetric form on $V$, and $|I(\sigma,\sigma^{\prime})|\leq\|\sigma\|\,\|\sigma^{\prime}\|$.

Put $\sigma_{i}=\mu_{i}-\nu_{i}\in V$, $\iota(v)=3/v$ and $\beta=\iota_{*}\sigma_{v}$. The $v$-parts are supported in $A=\{\zeta\leq|q|\leq 19/16\}$ (the smaller $\zeta$). The map $\iota$ sends $\Gamma\setminus\{0\}$ into a line or circle and $A$ into $\{|q|\leq 3/\zeta\}$, and $|\iota^{\prime}(v)|=3/|v|^{2}>1$ on $A$. So $\iota_{*}(\ell_{\Gamma}|_{A})\leq\ell_{\Gamma^{\prime}}$ for a compact segment or circle $\Gamma^{\prime}$. Hence $\iota_{*}\mu_{v}$ and $\iota_{*}\nu_{v}$ are tame, $\beta\in V$, and $\log|v|$ is bounded on $A$. From

$$\log|\iota(v)-\iota(w)| =\log 3+\log|v-w|-\log|v|-\log|w|,$$

$$\log|3-xv| =\log|x-\iota(v)|+\log|v|,$$

and the zero masses,

$$I(\beta,\beta)=I(\sigma_{v},\sigma_{v}),\qquad D(\sigma_{x},\sigma_{v})=I(\sigma_{x},\beta).$$

With $s_{x}=\|\sigma_{x}\|$, $s_{v}=\|\sigma_{v}\|=\|\beta\|$ and $s_{z}=\|\sigma_{z}\|$,

$$\mathcal{Q}(\sigma) =-s_{x}^{2}-s_{v}^{2}-s_{z}^{2}+I(\sigma_{v},\sigma_{z})-I(\sigma_{x},\beta)\leq-s_{x}^{2}-s_{v}^{2}-s_{z}^{2}+s_{v}s_{z}+s_{x}s_{v}$$

$$=-(s_{x}-s_{v}/2)^{2}-(s_{z}-s_{v}/2)^{2}-s_{v}^{2}/2\leq 0.$$

Let $\mathcal{Q}(\cdot,\cdot)$ be the symmetric bilinear form with $\mathcal{Q}(\mu,\mu)=\mathcal{Q}(\mu)$. Then $2\mathcal{Q}(\nu,\mu)=\int\psi_{x}\,d\mu_{x}+\int\psi_{v}\,d\mu_{v}+\int\psi_{z}\,d\mu_{z}$ and

$$\mathcal{Q}(\mu)=\mathcal{Q}(\nu)+2\mathcal{Q}(\nu,\sigma)+\mathcal{Q}(\sigma)\leq 2\mathcal{Q}(\nu,\mu)-\mathcal{Q}(\nu).\qed$$

Lemma 4.2 (computer-assisted).

The measure $\nu$ of Tables C.1 and C.2 lies in $\mathfrak{M}$, $\operatorname{supp}\nu_{i}\subseteq\mathcal{K}_{i}$, and

(4.2) $$\mathcal{Q}(\nu)\geq q_{-}:=-5.193572,$$

(4.3) $$F_{x}\leq K_{x}:=-4.83148,\qquad F_{v}\leq K_{v}:=-4.74318,\qquad F_{z}\leq K_{z}:=0.04318$$

on $(0,1]$, $(0,1]$ and $\mathcal{K}_{z}$. On $|z|=r$ even $F_{z}\leq 0.04283$.

Proof.

By Appendix C, $\nu_{x}$, $\nu_{v}$ and the interval part $\nu_{z}^{\rm int}$ of $\nu_{z}$ are sums of $131$, $143$ and $105$ intervals $[t_{0},t_{1}]$ of positive mass $w$, each carrying $w\,dt/(t_{1}-t_{0})$. The intervals of each part form a chain of adjacent intervals, contained in $[v_{0},1]$ for $\nu_{v}$, $v_{0}=0.00960735979838$. The circular part $\nu_{z}^{\rm Poi}$ is a sum of $167$ Poisson measures $(c,w)$ with $w>0$ and $|c|<r$,

$$\frac{w(r^{2}-|c|^{2})}{2\pi|re^{i\vartheta}-c|^{2}}\,d\vartheta\qquad(0\leq\vartheta\leq 2\pi).$$

Its arclength density is at most $H_{c}$ in (4.7). The masses are $1$, $5/4$ and $1$, and $\nu_{z}$ is invariant under conjugation. Let $H_{x},H_{v},H_{z}$ be the largest values of $w/(t_{1}-t_{0})$ over the intervals of the three parts. So the parts of $\nu$ are tame, $\operatorname{supp}\nu_{v}\subseteq[v_{0},1]$, and $\nu\in\mathfrak{M}$. An interval has potential

$$\frac{w}{t_{1}-t_{0}}\operatorname{Re}\{f(q-t_{0})-f(q-t_{1})\},\qquad f(u)=u\log u-u,$$

with $u\log|u|-u$ on the real line and $0\log 0=0$. A Poisson measure has potential

(4.4) $$w\begin{cases}\log|r^{2}-q\overline{c}|-\log r,&|q|\leq r,\\ \log|q-c|,&|q|\geq r.\end{cases}$$

Energy. Interval pairs use $g(u)=\tfrac{1}{2}u^{2}\log|u|-\tfrac{3}{4}u^{2}$, with $g^{\prime\prime}(u)=\log|u|$. A pair of Poisson measures $(c,w)$, $(c^{\prime},w^{\prime})$ gives $ww^{\prime}(\log|r^{2}-c\overline{c^{\prime}}|-\log r)$, by (4.4) and harmonicity. Interval–Poisson pairs integrate (4.4) with the primitives of $\log|t-c|$ and $\log|r^{2}-t\overline{c}|$, split at $t=r$. For the denominator, with $\mathsf{M}_{j}^{i}=\int t^{j}\,d\nu_{i}$ and $N=150$,

$$D(\nu_{x},\nu_{v})=\frac{5}{4}\log 3-\sum_{j=1}^{N}\frac{\mathsf{M}_{j}^{x}\mathsf{M}_{j}^{v}}{j3^{j}}-\mathcal{E}_{N},\qquad 0\leq\mathcal{E}_{N}\leq\frac{5/4}{2(N+1)3^{N}}.$$

Table B.1 gives the outward enclosures, at $320$-bit precision. The enclosure of $D(\nu_{x},\nu_{v})$ contains the partial sum minus $[0,(5/4)/(2(N+1)3^{N})]$. The interval sum of the rows gives $\mathcal{Q}(\nu)>q_{-}+5\cdot 10^{-7}$.

Small arguments. Put $\varepsilon_{0}=10^{-12}$. For $0<q\leq\varepsilon_{0}$, $\mathcal{U}_{x}(q),\mathcal{U}_{v}(q)\leq 0$, $\mathcal{U}_{z}(q)\leq\log(3+\varepsilon_{0})$, $D_{v}(q)\geq\frac{5}{4}\log(3-\varepsilon_{0})$ and $D_{x}(q)\geq\log(3-\varepsilon_{0})$. Hence

$$F_{x}(q) \leq-\tfrac{5}{4}\log(3-\varepsilon_{0})+\tfrac{3}{4}\log\varepsilon_{0}<-22.09,$$

$$F_{v}(q) \leq\log(3+\varepsilon_{0})-\log(3-\varepsilon_{0})+\tfrac{1}{4}\log\varepsilon_{0}<-6.90.$$

Both are below the caps.

Subdivision. The bounds on $[\varepsilon_{0},1]$, $[\varepsilon_{0},1]$, $[1,3]$ and the upper half of $|z|=r$ are proved on the $126464$ closed pieces (leaves) of Lemma C.1. The lower half of the circle follows by conjugation. All bounds are computed in Arb ball arithmetic [6] at $192$-bit precision.

Let $[l,u]$ be a leaf on a real support, with centre $c_{0}$ and half-width $h_{0}$. The interval components of $F$ come from the chains of $\nu_{x}$ on $x$-leaves and of $\nu_{v}$, $\nu_{z}^{\rm int}$ on $v$- and $z$-leaves. The endpoints of the chains of $\nu_{x}$, $\nu_{v}$, $\nu_{z}^{\rm int}$ cut the initial pieces of the $x$-, $v$-, $z$-support. The chain of $\nu_{z}^{\rm int}$ misses the $v$-support, and that of $\nu_{v}$ meets the $z$-support only at $1$. So the open leaf meets at most one interval $[t_{0},t_{1}]$ of each of these chains, and then lies in it. A $v$-leaf lies in $[0,r]$ or in $[r,1]$. Write $F=F_{C}+F_{N}$, where $F_{C}$ is the sum of the interval components of $F$, with their coefficient $2$, whose interval contains $[l,u]$. Inside its interval a component is convex, since $\mathcal{U}^{\prime\prime}=w/((q-t_{0})(t_{1}-q))>0$ there, so $F_{C}$ lies below its chord on $[l,u]$. With $B\geq\max\{0,\sup_{(l,u)}F_{N}^{\prime\prime}\}$, Taylor’s formula with integral remainder for $F_{N}$ at $c_{0}$ gives

$$\max_{[l,u]}F\leq\max_{q\in\{l,u\}}\bigl{(}F_{C}(q)+F_{N}(c_{0})+F_{N}^{\prime}(c_{0})(q-c_{0})\bigr{)}+\tfrac{1}{2}h_{0}^{2}B,$$

because the chord plus an affine function is affine. The other interval components have $\mathcal{U}^{\prime\prime}=-w/((q-t_{0})(q-t_{1}))\leq 0$ on $(l,u)$. The terms $\frac{3}{4}\log q$ and $\frac{1}{4}\log q$ are concave, and $(-\frac{5}{4}\log q)^{\prime\prime}\leq\frac{5}{4}l^{-2}$ on $[l,u]$. The terms $-D_{v}$, $-D_{x}$ are expanded as $-M\log 3+\sum_{j\geq 1}\mathsf{M}_{j}q^{j}/(j3^{j})$ with the first $N^{\prime}=90$ moments. The omitted value is at most $M/(2(N^{\prime}+1)3^{N^{\prime}})$, the omitted derivative lies in $[0,M/(2\cdot 3^{N^{\prime}})]$, and the second derivative is at most $M/4$. On the real line a Poisson measure has second derivative $w\operatorname{Re}(-\overline{c}^{2}/(r^{2}-q\overline{c})^{2})\leq w|c|^{2}/|r^{2}-q\overline{c}|^{2}$ for $|q|\leq r$ and $w\operatorname{Re}(-(q-c)^{-2})\leq w/|q-c|^{2}$ for $|q|\geq r$; $B$ uses enclosures of these second derivatives over the leaf.

On an arc $q=re^{i\vartheta}$, $|\vartheta-\vartheta_{0}|\leq h_{0}$, of the circle we take the smaller of two bounds. The first is

$$F(\vartheta_{0})+h_{0}|F^{\prime}(\vartheta_{0})|+\tfrac{1}{2}h_{0}^{2}B,\qquad B\geq\max\{0,\sup F^{\prime\prime}\},$$

with derivatives in $\vartheta$ and

$$\frac{d}{d\vartheta}\log|q-s|=\operatorname{Re}\frac{iq}{q-s},\qquad\frac{d^{2}}{d\vartheta^{2}}\log|q-s|=\operatorname{Re}\frac{qs}{(q-s)^{2}}\leq\frac{r|s|}{|q-s|^{2}}.$$

Here $B$ uses enclosures over the arc of these second derivatives for the Poisson terms ($s=c$). For an interval component of mass $w$ and density $H$ on $[t_{0},t_{1}]$ it uses the smaller upper end of two enclosures: of the exact second derivative $H\operatorname{Re}\bigl{(}q\log\frac{q-t_{1}}{q-t_{0}}+\frac{q^{2}}{q-t_{1}}-\frac{q^{2}}{q-t_{0}}\bigr{)}$, and of $w\operatorname{Re}(qs/(q-s)^{2})$ with $s$ ranging over $[t_{0},t_{1}]$. The second bound uses that the distance from $re^{i\vartheta}$ to a point $s\geq 0$ increases with $\vartheta\in[0,\pi]$. It bounds the real-interval terms by their value at the right end of the arc and applies the first bound to the Poisson terms alone. On the arc starting at $\vartheta=0$ only the second bound is used.

On every leaf the upper end of the resulting bound is below the cap of its support in Table B.2: $K_{x}$, $K_{v}$, $K_{z}$ on $[1,3]$, and $0.04283<K_{z}$ on the circle. Table B.2 gives the largest upper end on each support. The least margin, attained on $[1,3]$, is at least $2\cdot 10^{-6}$. ∎

For this $\nu$ put

(4.5) $$\alpha=q_{-}-\tfrac{3}{2}\log 3-K_{x}-\tfrac{5}{4}K_{v}-K_{z}.$$

Since $\frac{3}{2}\log 3<1647919/10^{6}$,

$$\alpha>\alpha_{0}:=q_{-}-\frac{1647919}{10^{6}}-K_{x}-\frac{5}{4}K_{v}-K_{z}=3.875784.$$

Lemma 4.3.

Let $\eta$ be a compactly supported positive measure of mass $M$ with $\eta(B(q,t))\leq At$ for all $q\in\mathbb{C}$, $t>0$. If $0<d=|q-q^{\prime}|\leq 1/3$, then

$$\Bigl{|}\int\log|q-t|\,d\eta(t)-\int\log|q^{\prime}-t|\,d\eta(t)\Bigr{|}\leq(6A+2M)\,d\,(1+\log(1/d)).$$

The components of the reference of Lemma 4.2 are compactly supported, and for them one may take

(4.6) $$A_{x}=2H_{x},\qquad A_{v}=2H_{v},\qquad A_{z}=2H_{z}+\sum_{c}\frac{w_{c}(r+|c|)}{r(r-|c|)},$$

the sum over the $167$ Poisson measures $(c,w_{c})$. It is finite, since $|c|<r$ for every centre.

Proof.

The closed disk $\{w:|w-q|\leq s\}$ is the intersection of the $B(q,s^{\prime})$, $s^{\prime}>s$, so its $\eta$-measure is also at most $As$. Let $N=\{t:|q-t|\leq 2d\}$. On $N$ both distances are at most $3d\leq 1$, and the layer-cake formula gives

$$\int_{N}\log\frac{1}{|q-t|}\,d\eta \leq 2Ad\log\frac{1}{2d}+2Ad,$$

$$\int_{N}\log\frac{1}{|q^{\prime}-t|}\,d\eta \leq 2Ad\log\frac{1}{3d}+3Ad.$$

Off $N$, $|q^{\prime}-t|\geq|q-t|/2$, the logarithmic difference is at most $2d/|q-t|$, and

$$\int_{|q-t|>2d}\frac{d\eta(t)}{|q-t|}\leq\int_{0}^{1}M\,ds+\int_{1}^{1/(2d)}\frac{A}{s}\,ds=M+A\log\frac{1}{2d}.$$

The sum is at most $6Ad\log(1/d)+5Ad+2Md$.

By the bounds on $\ell_{\Gamma}(B(q,t))$ before Lemma 4.1, an interval of density at most $H_{i}$ gives $2H_{i}$, and a Poisson measure gives $2\pi H_{c}$: since $|re^{i\vartheta}-c|\geq r-|c|$ and arclength is $r\,d\vartheta$, its arclength density is at most

(4.7) $$H_{c}=\frac{w(r+|c|)}{2\pi r(r-|c|)}.\qed$$

Put $M_{x}=1$, $M_{v}=5/4$, $M_{z}=1$, and

(4.8) $$B_{i}=6A_{i}+2M_{i},\qquad S=2B_{x}+\tfrac{7}{2}B_{v}+\tfrac{13}{4}B_{z},\qquad C=10+3S.$$

$S$ and $C$ are finite and depend only on $\nu$.

Lemma 4.4.

For every $m=4n\geq 4$,

(4.9) $$\begin{split}\log|G_{m}(\theta)|\leq{}&-\alpha m^{2}+\tfrac{13}{2}m\log m+3S\bigl{(}1+\log(m^{2}/3)\bigr{)}+\tfrac{45}{4}\\ &+m\log(T_{r}/3)-\log\bigl{(}(m!)^{2}L!\bigr{)},\end{split}$$

and hence $\log|G_{m}(\theta)|\leq-\alpha_{0}m^{2}+Cm\log m$ with $C$ as in (4.8).

Proof.

Expand $\Gamma^{m}$ in (2.5) into products of its pieces and take absolute values. By (2.4) the paths contribute a factor at most $T_{r}^{m}$, and $[0,1]^{m}\times[0,1]^{L}$ has volume one. So it suffices to bound $\log|\Phi|$ for $x\in[0,1]^{m}$, $v\in[0,1]^{L}$, $z\in\mathcal{K}_{z}^{m}$. If $\Phi=0$ there is nothing to prove. Otherwise $x_{i^{\prime}},v_{b}>0$ and the points of each species are distinct. Let $\mu_{x}$, $\mu_{v}$, $\mu_{z}$ put mass $1/m$ at each $x_{i^{\prime}}$, $v_{b}$, $z_{d}$. Since $\sigma+k+h=3m/4$, $\sigma=m/4$ and $m+k=5m/4$,

(4.10) $$\frac{\log|\Phi|}{m^{2}}=\mathcal{Q}_{\rm disc}(\mu)+\frac{3}{4}\int\log x\,d\mu_{x}+\frac{1}{4}\int\log v\,d\mu_{v}-\frac{5}{4}\int\log|z|\,d\mu_{z},$$

where $\mathcal{Q}_{\rm disc}$ omits the diagonal in the three self-energies.

Put $\epsilon=m^{-2}\leq 1/16$. Replace every atom $q$ by the uniform measure of mass $1/m$ on the circle $q+2i\epsilon+\epsilon e^{i\vartheta}$, and let $\mu^{\epsilon}$ be the result. Since $\frac{1}{2\pi}\int_{0}^{2\pi}\log|u+\epsilon e^{i\vartheta}|\,d\vartheta=\log\max(|u|,\epsilon)$ and the common translation cancels in differences, averaging does not decrease any off-diagonal positive pair. Each circle has energy $\log\epsilon$, and there are $13m/4$ atoms. The function $\log|3-XV|$ is harmonic in each variable on the product of two smearing disks, since $|X|,|V|\leq 1+3\epsilon\leq 19/16$. Its mean is $\log|3-(x+2i\epsilon)(v+2i\epsilon)|\leq\log(3-xv)+3\epsilon$, because $|(x+2i\epsilon)(v+2i\epsilon)-xv|\leq 4\epsilon+4\epsilon^{2}$ and $3-xv\geq 2$. With the masses $1\cdot 5/4$,

(4.11) $$\mathcal{Q}_{\rm disc}(\mu)\leq\mathcal{Q}(\mu^{\epsilon})+\frac{13}{4m}\log\frac{1}{\epsilon}+\frac{15}{4}\epsilon.$$

The measures $\mu^{\epsilon}$ and $\nu$ lie in $\mathfrak{M}$: $\mu^{\epsilon}$ is a sum of uniform measures on circles, inside $\{|q|\leq 19/16\}$ for $x$ and $v$, with $\operatorname{Im}q\geq\epsilon$ on the $v$-circles. For $\nu$ this is part of Lemma 4.2. By Lemmas 4.1 and 4.2,

$$\mathcal{Q}(\mu^{\epsilon})\leq-q_{-}+\sum_{i}\int\psi_{i}\,d\mu_{i}^{\epsilon}.$$

Each point of a smearing circle lies within $d=3\epsilon\leq 3/16$ of the atom $q$ it replaces. Since $s\mapsto s(1+\log(1/s))$ increases on $(0,1]$, Lemma 4.3 bounds the change of $\mathcal{U}_{i}$ by $B_{i}d(1+\log(1/d))$. For $q\in[0,1]$, $|p-q|\leq d$ and $0\leq t\leq 1$, $|\nabla_{p}\log|3-pt||\leq t/|3-pt|\leq 1$, so $D_{i}$ changes by at most $M_{i}d$. Counting masses, the coefficients of $B_{x},B_{v},B_{z}$ are $2$, $\frac{5}{4}\cdot 2+1=\frac{7}{2}$ and $\frac{5}{4}+2=\frac{13}{4}$, and the $D$ terms give $\frac{5}{4}+\frac{5}{4}=\frac{5}{2}$. Hence

$$\sum_{i}\int\psi_{i}\,d\mu_{i}^{\epsilon}\leq\sum_{i}\int\psi_{i}\,d\mu_{i}+Sd(1+\log(1/d))+\tfrac{5}{2}d.$$

The atoms lie in $(0,1]$, $(0,1]$ and $\mathcal{K}_{z}$. Adding the original external terms of (4.10) turns $\psi_{i}$ into $F_{i}$, and (4.3) gives

$$\frac{\log|\Phi|}{m^{2}}\leq-q_{-}+K_{x}+\tfrac{5}{4}K_{v}+K_{z}+\frac{13}{4m}\log\frac{1}{\epsilon}+3S\epsilon\Bigl{(}1+\log\frac{1}{3\epsilon}\Bigr{)}+\frac{45}{4}\epsilon,$$

with $45/4=15/4+3\cdot 5/2$. Multiply by $m^{2}$, put $\epsilon=m^{-2}$, add the logarithm $(3m^{2}/2-m)\log 3-\log((m!)^{2}L!)$ of the prefactor in (2.5) and $m\log T_{r}$. By (4.5) this is (4.9).

For the last statement drop $-\log((m!)^{2}L!)$ and $m\log(T_{r}/3)<0$. As $m\geq 4$, $3S(1+2\log m-\log 3)\leq 6S\log m\leq 3Sm\log m$ and $45/4\leq 3m\log m$. So the right-hand side is at most $-\alpha m^{2}+(10+3S)m\log m=-\alpha m^{2}+Cm\log m$. ∎

5. Nonvanishing

Let $\mathbb{Z}_{(17)}=\{u/v\in\mathbb{Q}:17\nmid v\}$, and let $v_{17}$ be the $17$-adic valuation, $v_{17}(0)=+\infty$.

Lemma 5.1.

For $m=4$, coefficientwise in $\mathbb{Z}_{(17)}[T]$,

(5.1) $$G_{4}(T)\equiv g(T):=11+5T^{2}+4T^{4}\pmod{17}.$$

Proof.

Here $I=\{2,\ldots,5\}$, $J=\{1,\ldots,9\}$, $\Lambda=\{1\}$, and all denominators divide $6350400=2520^{2}$, prime to $17$. Modulo $17$ the matrix (2.1) is

(5.2) $$\begin{pmatrix}11&0&16&15&7&7&3&13&9\\ 5&16&0&14&11&4&4&9&5\\ 8&15&14&0&8&16&12&12&10\\ 8&7&11&8&0&7&14&2&2\\ 6&16+8T&4&13&10&3&15&14&8\\ 5&4&16+7T&5&13&14&5&16&3\\ 5&13&5&13+4T&4&10&5&6&6\\ 9&10&13&4&3+12T&16&14&1&15\\ 9&6&13&7&3&5&15&2&12\end{pmatrix}.$$

Its determinant is $g(T)$. Reduction of (2.6), where $9/67228000000000\equiv 13\pmod{17}$ and $(q_{0},q_{1},q_{2},q_{3},q_{4})\equiv(10,0,3,0,16)\pmod{17}$, gives the same polynomial. ∎

Lemma 5.2.

Let $\xi\in\mathbb{Q}$. For every integer $t\geq 1$ with $2t+v_{17}(\xi)>0$, $G_{4\cdot 17^{t}}(\xi)\neq 0$.

Proof.

Put $Q_{t}=17^{t}$, $m_{t}=4Q_{t}$ and $h=\sigma=k=Q_{t}$, and

(5.3) $$P_{t}(W)=Q_{t}^{3m_{t}+h}G_{m_{t}}(W/Q_{t}^{2}).$$

This is the determinant obtained after $T=W/Q_{t}^{2}$ by multiplying the $C$, $B-TA$ and $V$ rows by $Q_{t}$, $Q_{t}^{2}$ and $Q_{t}$. The middle rows become $Q_{t}^{2}B-WA$. Every nonzero index, difference and Cauchy denominator has absolute value less than $17Q_{t}$, hence $17$-adic valuation at most $t$. So $Q_{t}t_{i}$ and $Q_{t}^{2}b_{i}$ lie in $\mathbb{Z}_{(17)}$, and for $i\neq j$

$$Q_{t}^{2}B_{ij}=\frac{Q_{t}}{j-i}\,(Q_{t}t_{i}-Q_{t}t_{j}),\qquad Q_{t}C_{ij}=\frac{Q_{t}}{j-i}\,(a^{i}-a^{j}),\qquad Q_{t}V_{\ell j}=\frac{Q_{t}}{j+\ell},$$

with $v_{17}(j-i),v_{17}(j+\ell)\leq t$. So all entries of the scaled matrix lie in $\mathbb{Z}_{(17)}[W]$. Hence $P_{t}\in\mathbb{Z}_{(17)}[W]$.

Let $0\leq u<Q_{t}$. Only the summation indices divisible by $Q_{t}$ survive, and $a^{Q_{t}}\equiv a\pmod{17}$, so

$$Q_{t}t_{Q_{t}j+u}\equiv a^{u}t_{j},\qquad Q_{t}^{2}b_{Q_{t}j+u}\equiv a^{u}b_{j}\pmod{17}.$$

The scaled $C$ and $B$ entries vanish modulo $17$ between distinct residue classes modulo $Q_{t}$. In the class $u$ write $i=Q_{t}i_{0}+u$, $j=Q_{t}j_{0}+u$. Then $2\leq i_{0}\leq 5$, $1\leq j_{0}\leq 9$, and $A_{ii}\equiv a^{u}A_{i_{0}i_{0}}$. The only $V$ row in this class is $\ell=Q_{t}-u$, with

$$\frac{Q_{t}}{j+\ell}=\frac{1}{j_{0}+1}.$$

Cross-class $C$ and $B$ entries have $17$-adic valuation at least $1$. Within a class the nonzero differences $j_{0}-i_{0}$ satisfy $|j_{0}-i_{0}|\leq 7<17$ and are units. All other scaled $V$ entries vanish modulo $17$. So, after permutations, the matrix has $Q_{t}$ diagonal blocks

$$\begin{bmatrix}a^{u}C_{4}\\ a^{u}(B_{4}-WA_{4})\\ V_{4}\end{bmatrix},\qquad 0\leq u<Q_{t},$$

where the subscript $4$ refers to $G_{4}$. Each block has eight rows with the factor $a^{u}$. Hence

(5.4) $$P_{t}(W)\equiv\varepsilon_{t}\,3^{4Q_{t}(Q_{t}-1)}g(W)^{Q_{t}}\pmod{17},\qquad\varepsilon_{t}\in\{1,-1\},$$

coefficientwise. In particular $P_{t}(0)\equiv\pm 3^{4Q_{t}(Q_{t}-1)}11^{Q_{t}}\not\equiv 0\pmod{17}$.

Take $W=Q_{t}^{2}\xi$. Then $v_{17}(W)=2t+v_{17}(\xi)>0$, and since $P_{t}\in\mathbb{Z}_{(17)}[W]$, $P_{t}(W)\equiv P_{t}(0)\not\equiv 0\pmod{17}$. By (5.3), $G_{m_{t}}(\xi)\neq 0$. ∎

The restriction $2t+v_{17}(\xi)>0$ is needed in this argument: $g$ has the roots $7$ and $10$ in $\mathbb{F}_{17}$.

6. Proof of Theorem 1

Suppose $\theta=u/q_{0}$ with $u\in\mathbb{Z}$ and $q_{0}\in\mathbb{Z}_{>0}$. By Lemmas 2.1 and 3.5,

$$Z_{m}:=q_{0}^{m}K_{m}G_{m}(u/q_{0})\in\mathbb{Z}.$$

By Lemma 5.2 with $\xi=\theta$, $Z_{m}\neq 0$ for $m=4\cdot 17^{t}$ and all large $t$. By Lemmas 3.5 and 4.4,

$$\log|Z_{m}|\leq-(\alpha_{0}-\tau)m^{2}+Cm\log m+o(m^{2})+m\log q_{0}.$$

As $\alpha_{0}-\tau>0.00305>0$, the right-hand side tends to $-\infty$ along this sequence. So $0<|Z_{m}|<1$ for large $t$, which is impossible. ∎

Appendix A Arithmetic integration

Table A lists the lines (3.7) for the seam $R=8$. The index ranges are the integer solutions of $4z_{0}/5-1<k<z_{0}R$ and $1\leq w<R/4+1$ (Section 3).

Table A.2 gives, for $\rho_{0}<\rho<\rho_{1}$, the function $\mathfrak{c}(\rho)=c_{0}+c_{1}\rho$. Consecutive rows share an endpoint. The first row starts at $1/R$ and the last ends at $5/4$. The rows are the $76$ open intervals of $(1/R,5/4)\setminus\mathcal{C}$, so a row can be checked at one interior point $\rho$: the six cells of (3.2) at $\rho$ give an affine function of $\rho$, which must equal $c_{0}+c_{1}\rho$. Then

$$\int_{1/R}^{5/4}\mathfrak{c}(\rho)\,d\rho=\sum_{\rm rows}(\rho_{1}-\rho_{0})\Bigl{(}c_{0}+c_{1}\frac{\rho_{0}+\rho_{1}}{2}\Bigr{)}.$$

Table A.1. Line families of (3.7).

family	lines	count
strip	$x=0$, $x=\rho$	$2$
$\mathcal{L}$ lower	$x=1/4-k\rho$, $0\leq k\leq 1$	$2$
$K$ lower	$x=1/2-k\rho$, $0\leq k\leq 3$	$4$
$K$ upper	$x=3/2-k\rho$, $1\leq k\leq 11$	$11$
$\mathcal{L}$ upper	$x=5/2-k\rho$, $2\leq k\leq 19$	$18$
$W$	$x=w\rho-1/4$, $1\leq w\leq 2$	$2$
total		$39$

Table A.2. The affine pieces of $\mathfrak{c}$ on $(1/R,5/4)$, $R=8$: $\mathfrak{c}(\rho)=c_{0}+c_{1}\rho$ for $\rho_{0}<\rho<\rho_{1}$.

$\rho_{0}$	$\rho_{1}$	$c_{0}$	$c_{1}$	$\rho_{0}$	$\rho_{1}$	$c_{0}$	$c_{1}$	$\rho_{0}$	$\rho_{1}$	$c_{0}$	$c_{1}$
$1/8$	$11/84$	$9/2$	$-10$	$2/11$	$11/60$	$11/4$	$2$	$1/3$	$11/32$	$9/2$	$-4$
$11/84$	$5/38$	$7/4$	$11$	$11/60$	$3/16$	$11/4$	$2$	$11/32$	$7/20$	$7/4$	$4$
$5/38$	$9/68$	$17/4$	$-8$	$3/16$	$5/26$	$1/2$	$14$	$7/20$	$5/14$	$7/2$	$-1$
$9/68$	$2/15$	$17/4$	$-8$	$5/26$	$7/36$	$3$	$1$	$5/14$	$3/8$	$6$	$-8$
$2/15$	$7/52$	$9/4$	$7$	$7/36$	$11/56$	$19/4$	$-8$	$3/8$	$11/28$	$15/4$	$-2$
$7/52$	$3/22$	$4$	$-6$	$11/56$	$1/5$	$2$	$6$	$11/28$	$2/5$	$15/4$	$-2$
$3/22$	$11/80$	$5/2$	$5$	$1/5$	$9/44$	$3$	$1$	$2/5$	$5/12$	$7/4$	$3$
$11/80$	$5/36$	$5/2$	$5$	$9/44$	$5/24$	$21/4$	$-10$	$5/12$	$7/16$	$1/2$	$6$
$5/36$	$9/64$	$5/4$	$14$	$5/24$	$11/52$	$4$	$-4$	$7/16$	$9/20$	$9/4$	$2$
$9/64$	$1/7$	$7/2$	$-2$	$11/52$	$3/14$	$4$	$-4$	$9/20$	$11/24$	$9/2$	$-3$
$1/7$	$11/76$	$9/2$	$-9$	$3/14$	$7/32$	$5/2$	$3$	$11/24$	$1/2$	$7/4$	$3$
$11/76$	$7/48$	$7/4$	$10$	$7/32$	$2/9$	$17/4$	$-5$	$1/2$	$11/20$	$19/4$	$-3$
$7/48$	$5/34$	$7/2$	$-2$	$2/9$	$9/40$	$9/4$	$4$	$11/20$	$9/16$	$2$	$2$
$5/34$	$3/20$	$6$	$-19$	$9/40$	$5/22$	$9/4$	$4$	$9/16$	$7/12$	$17/4$	$-2$
$3/20$	$11/72$	$15/4$	$-4$	$5/22$	$11/48$	$19/4$	$-7$	$7/12$	$5/8$	$6$	$-5$
$11/72$	$2/13$	$15/4$	$-4$	$11/48$	$1/4$	$2$	$5$	$5/8$	$2/3$	$19/4$	$-3$
$2/13$	$5/32$	$7/4$	$9$	$1/4$	$11/40$	$9/2$	$-5$	$2/3$	$11/16$	$11/4$	$0$
$5/32$	$7/44$	$1/2$	$17$	$11/40$	$5/18$	$7/4$	$5$	$11/16$	$3/4$	$11/4$	$0$
$7/44$	$9/56$	$9/4$	$6$	$5/18$	$9/32$	$17/4$	$-4$	$3/4$	$5/6$	$1/2$	$3$
$9/56$	$11/68$	$9/2$	$-8$	$9/32$	$2/7$	$17/4$	$-4$	$5/6$	$7/8$	$3$	$0$
$11/68$	$1/6$	$7/4$	$9$	$2/7$	$7/24$	$9/4$	$3$	$7/8$	$11/12$	$19/4$	$-2$
$1/6$	$11/64$	$19/4$	$-9$	$7/24$	$3/10$	$4$	$-3$	$11/12$	$1$	$19/4$	$-2$
$11/64$	$9/52$	$2$	$7$	$3/10$	$11/36$	$5/2$	$2$	$1$	$9/8$	$15/4$	$-1$
$9/52$	$7/40$	$17/4$	$-6$	$11/36$	$5/16$	$5/2$	$2$	$9/8$	$5/4$	$6$	$-3$
$7/40$	$5/28$	$6$	$-16$	$5/16$	$9/28$	$5/4$	$6$
$5/28$	$2/11$	$19/4$	$-9$	$9/28$	$1/3$	$7/2$	$-1$

With (3.8),

$$A=\frac{13}{32}+\int_{1/R}^{5/4}\mathfrak{c}(\rho)\,d\rho+\frac{45}{32},$$

the constant of (3.12). The coefficient of $\log 3$ is $53/48$, and the block (3.5) contributes $-5/48$ to (3.8).

Appendix B Certified values

Table B.1 gives the energy of $\nu$. Here $\nu_{z}^{\rm int}$ and $\nu_{z}^{\rm Poi}$ are the interval and Poisson parts of $\nu_{z}$, $I_{\rm int}$ is the energy of $\nu_{z}^{\rm int}$, $I_{\rm mix}=I(\nu_{z}^{\rm int},\nu_{z}^{\rm Poi})$, $I_{\rm Poi}$ is the energy of $\nu_{z}^{\rm Poi}$, and

$$\mathcal{Q}(\nu)=I(\nu_{x},\nu_{x})+I(\nu_{v},\nu_{v})+I_{\rm int}+2I_{\rm mix}+I_{\rm Poi}+I(\nu_{v},\nu_{z}^{\rm int})+I(\nu_{v},\nu_{z}^{\rm Poi})-D(\nu_{x},\nu_{v}).$$

Every displayed interval is outward rounded, and the row of $D(\nu_{x},\nu_{v})$ contains the tail $\mathcal{E}_{N}$. The interval sum of the first eight rows, with these coefficients, exceeds $q_{-}$, which is (4.2).

Table B.1. The pair decomposition of $\mathcal{Q}(\nu)$ and its enclosures.

term	outward enclosure
$I(\nu_{x},\nu_{x})$	[-1.604087582665193,-1.604087582665192]
$I(\nu_{v},\nu_{v})$	[-2.279694146532524,-2.279694146532523]
$I_{\rm int}$	[-0.159655877431972,-0.159655877431971]
$I_{\rm mix}$	[0.226965417324502,0.226965417324503]
$I_{\rm Poi}$	[-0.495113146744583,-0.495113146744582]
$I(\nu_{v},\nu_{z}^{\rm int})$	[0.278231287125003,0.278231287125004]
$I(\nu_{v},\nu_{z}^{\rm Poi})$	[-0.191796788958401,-0.191796788958400]
$D(\nu_{x},\nu_{v})$	[1.195386024949843,1.195386024949844]
$\mathcal{Q}(\nu)$	[-5.193571445508505,-5.193571445508504]

Table B.2. The caps (4.3): leaves of Lemma C.1, the largest upper bound over the leaves (rounded upward at $10^{-6}$) and the cap. On $\mathcal{K}_{z}$ the cap is $K_{z}=\max(0.04318,0.04283)=0.04318$.

support	leaves	upper bound	cap
$x\in[\varepsilon_{0},1]$	$67584$	$-4.831488$	$-4.83148$
$v\in[\varepsilon_{0},1]$	$37120$	$-4.743190$	$-4.74318$
$z\in[1,3]$	$13568$	$0.043178$	$0.04318$
$|z|=r$, $\operatorname{Im}z\geq 0$	$8192$	$0.042823$	$0.04283$

Appendix C Reference measure and subdivision

Terminating decimals are exact. Other values are printed as fractions $p/q$.

C.1. Intervals

Table C.1 gives $\nu_{x}$, $\nu_{v}$ and $\nu_{z}^{\rm int}$. Each is a chain of adjacent intervals. Its heading gives the start point $b_{0}$, and its row $(k,b_{k},w_{k})$ is the interval $[b_{k-1},b_{k}]$ carrying $w_{k}\,dt/(b_{k}-b_{k-1})$. The three chains have $131$, $143$ and $105$ rows and end at $1$, $1$ and $3$. Outside its chain a part vanishes. The masses are $1$, $5/4$ and $170340748021/500000000000$.

Table C.1. Interval chains of the reference measure. A line gives the rows $(k,b_{k},w_{k})$, $(k+1,b_{k+1},w_{k+1})$ and $(k+2,b_{k+2},w_{k+2})$.

$k$	$b_{k}$	$w_{k}$	$b_{k+1}$	$w_{k+1}$	$b_{k+2}$	$w_{k+2}$
$\nu_{x}$: $b_{0}=0.08426519384873$
1	$0.09379426$	$0.004878668$	$0.1033233$	$0.005388618$	$0.1137017$	$0.006994053$
4	$0.1240801$	$0.007569004$	$0.1352634$	$0.008564016$	$0.1464466$	$0.008818097$
7	$0.1583869$	$0.009590098$	$0.1703271$	$0.009692372$	$0.1829732$	$0.0103232$
10	$0.1956193$	$0.01034485$	$0.2089171$	$0.01087642$	$0.2222149$	$0.0108577$
13	$0.2361074$	$0.01131139$	$0.25$	$0.01127106$	$0.2644278$	$0.01166085$
16	$0.2788557$	$0.01161365$	$0.293757$	$0.01194875$	$0.3086583$	$0.01190387$
19	$0.3239693$	$0.01219079$	$0.3392803$	$0.01215494$	$0.3549354$	$0.01239796$
22	$0.3705905$	$0.01237491$	$0.3865227$	$0.01257814$	$0.4024548$	$0.01257039$
25	$0.4185959$	$0.01273679$	$0.4347369$	$0.01274778$	$0.4510177$	$0.01287879$
28	$0.4672984$	$0.01290965$	$0.4836492$	$0.01300665$	$0.5$	$0.01305858$
31	$0.5163508$	$0.01312226$	$0.5327016$	$0.01319818$	$0.5489823$	$0.01322769$
34	$0.5652631$	$0.01332781$	$0.5814041$	$0.01332593$	$0.5975452$	$0.01345034$
37	$0.6134773$	$0.01341514$	$0.6294095$	$0.01356788$	$0.6450646$	$0.01349774$
40	$0.6607197$	$0.01367922$	$0.6760307$	$0.01357396$	$0.6913417$	$0.01378771$
43	$0.706243$	$0.01364148$	$0.7211443$	$0.01389263$	$0.7355722$	$0.01370571$
46	$0.75$	$0.01399346$	$0.7638926$	$0.01376403$	$0.7777851$	$0.01408874$
49	$0.7910829$	$0.01381373$	$0.8043807$	$0.01418846$	$0.8170268$	$0.01385703$
52	$0.8296729$	$0.01427975$	$0.8416131$	$0.01389719$	$0.8535534$	$0.01437142$
55	$0.8647366$	$0.01392421$	$0.8759199$	$0.0144671$	$0.8862983$	$0.01394331$
58	$0.8966767$	$0.0145547$	$0.9062057$	$0.01394895$	$0.9157348$	$0.0146563$
61	$0.9243738$	$0.01394465$	$0.9330127$	$0.01475476$	$0.9407245$	$0.01392136$
64	$0.9484364$	$0.01485533$	$0.9551881$	$0.01387691$	$0.9619398$	$0.01497938$
67	$0.9677024$	$0.01379006$	$0.9734651$	$0.01513013$	$0.978214$	$0.01364607$
70	$0.9829629$	$0.01534824607$	$0.9864815$	$0.01264469$	$0.99$	$0.01468005$
73	$0.9901963$	$0.0009304798$	$0.9903926$	$0.0008176946$	$0.9923846$	$0.009526088$
76	$0.9943766$	$0.01098753$	$0.9950495$	$0.004282778$	$0.9957224$	$0.004357794$
79	$0.9962801$	$0.003907897$	$0.9968377$	$0.004198966$	$0.9975297$	$0.005760908$
82	$0.9982217$	$0.006780532$	$0.9985756$	$0.003909506$	$0.9989295$	$0.004519428$
85	$0.9989498$	$0.0002978646$	$0.9989702$	$0.0002759106$	$0.9989851$	$0.0002057553$
88	$0.999$	$0.0002063948$	$0.9990431$	$0.0006106305$	$0.9990862$	$0.0006201$
91	$0.9991731$	$0.001304877$	$0.9992599$	$0.00137548$	$0.9993488$	$0.001492587$
94	$0.9994377$	$0.001607745$	$0.9994512$	$0.0002561717$	$0.9994647$	$0.0002462361$
97	$0.9995672$	$0.002058841$	$0.9996696$	$0.002355955$	$0.9996767$	$0.0001814995$
100	$0.9996838$	$0.0001633233$	$0.999753$	$0.00181967$	$0.9998222$	$0.002143887$
103	$0.9998327$	$0.0003687629$	$0.9998432$	$0.0003554893$	$0.9998716$	$0.001051055$
106	$0.9999$	$0.001186225$	$0.9999219$	$0.001022622$	$0.9999438$	$0.001204602$
109	$0.9999515$	$0.0004787153$	$0.9999593$	$0.0005156937$	$0.9999638$	$0.0003258994$
112	$0.9999684$	$0.000343669$	$0.9999753$	$0.000575319$	$0.9999822$	$0.000676004$
115	$0.9999861$	$0.0004312648$	$0.99999$	$0.0005066644$	$0.9999922$	$0.0003232857$
118	$0.9999944$	$0.0003798382$	$0.9999956$	$0.0002423703$	$0.9999968$	$0.0002847743$
121	$0.9999975$	$0.000181713$	$0.9999982$	$0.0002135055$	$0.9999986$	$0.0001362365$
124	$0.999999$	$0.0001600714$	$0.9999992$	$0.0001021401$	$0.9999994$	$0.0001200076$
127	$0.9999996$	$0.00007657429$	$0.9999997$	$0.00008996647$	$0.9999998$	$0.00005740427$
130	$0.9999999$	$0.0001104681$	$1$	$0.0003313966$
$\nu_{v}$: $b_{0}=0.00960735979838$
1	$0.01332222$	$0.001656734$	$0.01703709$	$0.00542487$	$0.02178601$	$0.008072268$
4	$0.02653494$	$0.008280399$	$0.03229758$	$0.009887851$	$0.03806023$	$0.009561701$
7	$0.04481193$	$0.01077175$	$0.05156363$	$0.01030386$	$0.05927546$	$0.01128366$
10	$0.0669873$	$0.01079812$	$0.07562625$	$0.01163129$	$0.08426519$	$0.01116664$
13	$0.09379426$	$0.01188784$	$0.1033233$	$0.01146128$	$0.1137017$	$0.01209265$
16	$0.1240801$	$0.01171582$	$0.1352634$	$0.01227961$	$0.1464466$	$0.0119444$
19	$0.1583869$	$0.01244886$	$0.1703271$	$0.01216045$	$0.1829732$	$0.01261611$
22	$0.1956193$	$0.01236727$	$0.2089171$	$0.0127843$	$0.2222149$	$0.0125745$
25	$0.2361074$	$0.01295137$	$0.25$	$0.01278215$	$0.2644278$	$0.01312539$
28	$0.2788557$	$0.01298995$	$0.293757$	$0.01330075$	$0.3086583$	$0.01319874$
31	$0.3239693$	$0.01347729$	$0.3392803$	$0.01340707$	$0.3549354$	$0.01365223$
34	$0.3705905$	$0.01360987$	$0.3865227$	$0.01381909$	$0.4024548$	$0.01378893$
37	$0.4185959$	$0.01398487$	$0.4347369$	$0.01401875$	$0.4510177$	$0.01417834$
40	$0.4672984$	$0.014227$	$0.4836492$	$0.01434849$	$0.5$	$0.01442008$
43	$0.5163508$	$0.01450478$	$0.5327016$	$0.01460288$	$0.5489823$	$0.01464888$
46	$0.5652631$	$0.01477283$	$0.5814041$	$0.01478305$	$0.5975452$	$0.01493237$
49	$0.6134773$	$0.01490385$	$0.6294095$	$0.01508317$	$0.6450646$	$0.01501368$
52	$0.6607197$	$0.01522317$	$0.6760307$	$0.01511253$	$0.6913417$	$0.01535607$
55	$0.706243$	$0.01519763$	$0.7211443$	$0.01548108$	$0.7355722$	$0.01527532$
58	$0.75$	$0.01559771$	$0.7638926$	$0.01534276$	$0.7777851$	$0.0157047$
61	$0.7910829$	$0.01539736$	$0.8043807$	$0.01581345$	$0.8170268$	$0.01544195$
64	$0.8296729$	$0.01591008$	$0.8416131$	$0.0154806$	$0.8535534$	$0.01600481$
67	$0.8647366$	$0.01550264$	$0.8759199$	$0.01610211$	$0.8862983$	$0.01551437$
70	$0.8966767$	$0.01618908$	$0.9062057$	$0.01551011$	$0.9157348$	$0.01629058$
73	$0.9243738$	$0.01549424$	$0.9330127$	$0.01638817$	$0.9407245$	$0.01545718$
76	$0.9484364$	$0.0164881$	$0.9551881$	$0.01539698$	$0.9619398$	$0.0166144$
79	$0.9677024$	$0.01529057$	$0.9734651$	$0.01677105$	$0.978214$	$0.01512197$
82	$0.9829629$	$0.01700337185$	$0.9864815$	$0.01400531$	$0.99$	$0.01625595$
85	$0.9901963$	$0.001030202$	$0.9903926$	$0.0009054735$	$0.9923846$	$0.01054699$
88	$0.9943766$	$0.0121635$	$0.9950495$	$0.004740679$	$0.9957224$	$0.004823687$
91	$0.9962801$	$0.004325506$	$0.9968377$	$0.004647527$	$0.9975297$	$0.006376076$
94	$0.9982217$	$0.007504222$	$0.9985756$	$0.004326653$	$0.9989295$	$0.005001531$
97	$0.9989498$	$0.0003296327$	$0.9989702$	$0.0003053404$	$0.9989851$	$0.0002277013$
100	$0.999$	$0.0002284092$	$0.9990431$	$0.000675759$	$0.9990862$	$0.0006862373$
103	$0.9991731$	$0.001444043$	$0.9992599$	$0.001522167$	$0.9993488$	$0.001651755$
106	$0.9994377$	$0.001779182$	$0.9994512$	$0.0002834868$	$0.9994647$	$0.0002724926$
109	$0.9995672$	$0.002278363$	$0.9996696$	$0.002607138$	$0.9996767$	$0.0002008494$
112	$0.9996838$	$0.0001807361$	$0.999753$	$0.002013666$	$0.9998222$	$0.002372435$
115	$0.9998327$	$0.0004080737$	$0.9998432$	$0.0003933854$	$0.9998716$	$0.001163099$
118	$0.9999$	$0.001312675$	$0.9999219$	$0.001131631$	$0.9999438$	$0.001333007$
121	$0.9999515$	$0.0005297437$	$0.9999593$	$0.0005706635$	$0.9999638$	$0.0003606381$
124	$0.9999684$	$0.0003803018$	$0.9999753$	$0.0006366439$	$0.9999822$	$0.0007480609$
127	$0.9999861$	$0.0004772341$	$0.99999$	$0.0005606707$	$0.9999922$	$0.0003577452$
130	$0.9999944$	$0.0004203256$	$0.9999956$	$0.000268205$	$0.9999968$	$0.0003151286$
133	$0.9999975$	$0.000201082$	$0.9999982$	$0.0002362633$	$0.9999986$	$0.0001507581$
136	$0.999999$	$0.0001771335$	$0.9999992$	$0.0001130273$	$0.9999994$	$0.0001327994$
139	$0.9999996$	$0.00008473636$	$0.9999997$	$0.00009955619$	$0.9999998$	$0.0000635229$
142	$0.9999999$	$0.000122243$	$1$	$0.0003667206$
$\nu_{z}^{\rm int}$: $b_{0}=28303039933611/17500000000000$
1	$1.647939$	$0.000800039$	$1.678561$	$0.001634018$	$1.709871$	$0.002195722$
4	$1.741181$	$0.002569887$	$1.773045$	$0.002941359$	$1.80491$	$0.003206511$
7	$1.837192$	$0.003492292$	$1.869474$	$0.003707447$	$1.902035$	$0.003935108$
10	$1.934597$	$0.004118914$	$1.967298$	$0.00430909$	$2$	$0.004469524$
13	$2.032702$	$0.00462588$	$2.065403$	$0.004777443$	$2.097965$	$0.004901502$
16	$2.130526$	$0.005045769$	$2.162808$	$0.005145023$	$2.19509$	$0.005284974$
19	$2.226955$	$0.005358049$	$2.258819$	$0.005500629$	$2.290129$	$0.005547422$
22	$2.321439$	$0.005693562$	$2.352061$	$0.005716104$	$2.382683$	$0.005869112$
25	$2.412486$	$0.0058643$	$2.442289$	$0.006028879$	$2.471144$	$0.00599805$
28	$2.5$	$0.00617374$	$2.527785$	$0.0061167$	$2.55557$	$0.006304648$
31	$2.582166$	$0.00622033$	$2.608761$	$0.00642816$	$2.634054$	$0.006311412$
34	$2.659346$	$0.006538253$	$2.683226$	$0.006392705$	$2.707107$	$0.006640591$
37	$2.729473$	$0.006459666$	$2.75184$	$0.006737854$	$2.772597$	$0.006515946$
40	$2.793353$	$0.006824422$	$2.812411$	$0.006559483$	$2.83147$	$0.006912101$
43	$2.848748$	$0.006592625$	$2.866025$	$0.006992728$	$2.881449$	$0.006611407$
46	$2.896873$	$0.007069417$	$2.910376$	$0.006615346$	$2.92388$	$0.007152991$
49	$2.935405$	$0.006594439$	$2.94693$	$0.00724478$	$2.956428$	$0.006543148$
52	$2.965926$	$0.007358282$	$2.973356$	$0.00643243$	$2.980785$	$0.007538103528$
55	$2.985393$	$0.00532701$	$2.99$	$0.006421216$	$2.990722$	$0.001161462$
58	$2.991445$	$0.001127666$	$2.992911$	$0.002498185$	$2.994377$	$0.002795714$
61	$2.995607$	$0.00261479$	$2.996838$	$0.003080948$	$2.997348$	$0.001436316$
64	$2.997859$	$0.001592602$	$2.9979$	$0.0001394852$	$2.99794$	$0.000132549$
67	$2.998056$	$0.0003909291$	$2.998172$	$0.0004039291$	$2.998197$	$0.00008733582$
70	$2.998222$	$0.00008626531$	$2.998371$	$0.0005458644$	$2.99852$	$0.0005686528$
73	$2.998725$	$0.0008328935$	$2.998929$	$0.0009109751$	$2.998965$	$0.0001665063$
76	$2.999$	$0.0001617129$	$2.99917$	$0.0008451466$	$2.999339$	$0.0009489754$
79	$2.999388$	$0.0002982447$	$2.999438$	$0.0003002494$	$2.999561$	$0.0008270791$
82	$2.999684$	$0.0009730542$	$2.999685$	$0.00001266974$	$2.999686$	$0.000009434384$
85	$2.999754$	$0.0006105242$	$2.999822$	$0.0007153116$	$2.999861$	$0.0004647103$
88	$2.9999$	$0.0005473779$	$2.999909$	$0.0001464355$	$2.999919$	$0.0001487465$
91	$2.999931$	$0.0002195401$	$2.999944$	$0.000241353$	$2.999956$	$0.0002613112$
94	$2.999968$	$0.0003071303$	$2.999975$	$0.0001959502$	$2.999982$	$0.0002302158$
97	$2.999986$	$0.0001468951$	$2.99999$	$0.0001725934$	$2.999992$	$0.0001101306$
100	$2.999994$	$0.0001293995$	$2.999996$	$0.00008256952$	$2.999997$	$0.00009701663$
103	$2.999998$	$0.00006190621$	$2.999999$	$0.0001191514$	$3$	$0.0003570495$

C.2. Poisson measures

Table C.2 gives $\nu_{z}^{\rm Poi}$. A row $(c,w)$ with $\operatorname{Im}c>0$ stands for the two Poisson measures on $|q|=r=2/5$ with centres $c$ and $\overline{c}$, each of mass $w$. The row $k=0$, with $c$ real, stands for one Poisson measure. The measures are those of the proof of Lemma 4.2, and their logarithmic potential is (4.4). This gives $2\cdot 83+1=167$ measures of total mass $329659251979/500000000000$.

Table C.2. Poisson centres $c$ and masses $w$: rows $k\geq 1$ have $\operatorname{Im}c>0$, in increasing order of $\arg c$, and stand also for $\overline{c}$; row $k=0$ is the real centre.

$k$	$\operatorname{Re}c$	$\operatorname{Im}c$	$w$	$k$	$\operatorname{Re}c$	$\operatorname{Im}c$	$w$
0	$-0.394$	$0$	$0.006687968$	42	$-0.2046129$	$0.341376$	$0.0002611643$
1	$0.2535036$	$0.308895$	$0.0002724889$	43	$-0.2080587$	$0.3345857$	$0.004943139$
2	$0.2370883$	$0.3196766$	$0.001734768001$	44	$-0.2188947$	$0.327599$	$0.006091431004$
3	$0.221117$	$0.3309249$	$0.003160258$	45	$-0.2294962$	$0.3202616$	$0.004978119$
4	$0.2046129$	$0.341376$	$0.002355698$	46	$-0.2370883$	$0.3196766$	$0.0003172227$
5	$0.197$	$0.341214$	$0.00203316$	47	$-0.239852$	$0.3125812$	$0.004953756$
6	$0.1876159$	$0.3510047$	$0.001945593$	48	$-0.249951$	$0.3045661$	$0.006293054005$
7	$0.1742617$	$0.3533679$	$0.00333003$	49	$-0.2597823$	$0.2962249$	$0.00498701$
8	$0.1626066$	$0.3588803$	$0.003486625$	50	$-0.2672805$	$0.2948985$	$0.0003407953$
9	$0.152308$	$0.3677041$	$0.002375737$	51	$-0.2693354$	$0.2875664$	$0.005025951$
10	$0.1387865$	$0.3687469$	$0.003817593$	52	$-0.2786001$	$0.2786001$	$0.006379229002$
11	$0.1266471$	$0.3730905$	$0.003942579$	53	$-0.2875664$	$0.2693354$	$0.005074617$
12	$0.1155333$	$0.3808623$	$0.002775625$	54	$-0.2948985$	$0.2672805$	$0.0003451508$
13	$0.1019747$	$0.3805748$	$0.004181110002$	55	$-0.2962249$	$0.2597823$	$0.005068562$
14	$0.08946805$	$0.3837075$	$0.004331115$	56	$-0.3045661$	$0.249951$	$0.006456653$
15	$0.07764595$	$0.3903525$	$0.003082157$	57	$-0.3125812$	$0.239852$	$0.005132843$
16	$0.06418082$	$0.3887375$	$0.004515421$	58	$-0.3196766$	$0.2370883$	$0.0003275158$
17	$0.05142732$	$0.3906293$	$0.004618404$	59	$-0.3202616$	$0.2294962$	$0.00520169$
18	$0.03901082$	$0.3960835$	$0.003346546$	60	$-0.327599$	$0.2188947$	$0.006408025$
19	$0.02576883$	$0.3931564$	$0.004623723$	61	$-0.3345857$	$0.2080587$	$0.005265605$
20	$0.01289132$	$0.393789$	$0.005370324$	62	$-0.341376$	$0.2046129$	$0.0002748567$
21	$0$	$0.398$	$0.002954771$	63	$-0.341214$	$0.197$	$0.005351403$
22	$-0.01289132$	$0.393789$	$0.00558512$	64	$-0.347477$	$0.1857303$	$0.006322745001$
23	$-0.02576883$	$0.3931564$	$0.004147672$	65	$-0.3533679$	$0.1742617$	$0.00539497$
24	$-0.03861875$	$0.3921028$	$0.005377501$	66	$-0.3597877$	$0.1701669$	$0.0002230895$
25	$-0.05142732$	$0.3906293$	$0.004364202$	67	$-0.3588803$	$0.1626066$	$0.005494093$
26	$-0.05839873$	$0.3936923$	$0.0002681214$	68	$-0.3640085$	$0.1507773$	$0.006236832$
27	$-0.06418082$	$0.3887375$	$0.004435699$	69	$-0.3687469$	$0.1387865$	$0.005515612$
28	$-0.07686559$	$0.3864294$	$0.00549873$	70	$-0.3747345$	$0.1340822$	$0.0002252512$
29	$-0.08946805$	$0.3837075$	$0.004569928$	71	$-0.3730905$	$0.1266471$	$0.005431967$
30	$-0.09670611$	$0.3860724$	$0.0002359134$	72	$-0.3770345$	$0.1143722$	$0.006398725$
31	$-0.1019747$	$0.3805748$	$0.004673523$	73	$-0.3805748$	$0.1019747$	$0.005431584$
32	$-0.1143722$	$0.3770345$	$0.005552904$	74	$-0.3860724$	$0.09670611$	$0.0002813461$
33	$-0.1266471$	$0.3730905$	$0.004783363$	75	$-0.3837075$	$0.08946805$	$0.005361819$
34	$-0.1340822$	$0.3747345$	$0.0001976397$	76	$-0.3864294$	$0.07686559$	$0.00656059$
35	$-0.1387865$	$0.3687469$	$0.004891999$	77	$-0.3887375$	$0.06418082$	$0.005341411$
36	$-0.1507773$	$0.3640085$	$0.005599664001$	78	$-0.3936923$	$0.05839873$	$0.000338135$
37	$-0.1626066$	$0.3588803$	$0.00497547$	79	$-0.3906293$	$0.05142732$	$0.005290732$
38	$-0.1701669$	$0.3597877$	$0.0002060286$	80	$-0.3921028$	$0.03861875$	$0.00668558776$
39	$-0.1742617$	$0.3533679$	$0.004926876$	81	$-0.3931564$	$0.02576883$	$0.005271869$
40	$-0.1857303$	$0.347477$	$0.005847720003$	82	$-0.3975206$	$0.01952893$	$0.0003596048$
41	$-0.197$	$0.341214$	$0.004984704$	83	$-0.393789$	$0.01289132$	$0.005295008$

C.3. Dyadic subdivision

The four supports are

$$s=x,v:\ [\varepsilon_{0},1],\qquad s=z:\ [1,3],\qquad s=\circ:\ \{re^{i\pi t}:0\leq t\leq 1\},$$

the last parametrized by $t\in[0,1]$. The initial pieces of a support are the closed intervals between consecutive points of the following sets, in increasing order:

• $\varepsilon_{0}$, $1$, and the endpoints of the $\nu_{x}$-chain in $(\varepsilon_{0},1)$;

• $\varepsilon_{0}$, $1$, $r=2/5$, and the endpoints of the $\nu_{v}$-chain in $(\varepsilon_{0},1)$;

• $1$, $3$, and the endpoints of the $\nu_{z}^{\rm int}$-chain in $(1,3)$;

• $t=k/32$, $0\leq k\leq 32$.

Since the chains start at $0.0842\ldots$, $v_{0}$ and $1.617\ldots$ and end at $1$, $1$ and $3$, and $r$ is not an endpoint of the $\nu_{v}$-chain, there are $132$, $145$, $106$ and $32$ initial pieces.

Lemma C.1.

Let $d_{x}=9$, $d_{v}=8$, $d_{z}=7$, $d_{\circ}=8$. Bisect every initial piece of the support $s$ to depth $d_{s}$: a piece $[A,B]$ gives the $2^{d_{s}}$ leaves

$$[A+(B-A)(j-1)2^{-d_{s}},\,A+(B-A)j2^{-d_{s}}],\qquad 1\leq j\leq 2^{d_{s}},$$

in $q$ on the real supports and in $t$ on the circle. The leaves of each support have rational endpoints, disjoint interiors and union equal to the support. There are $132\cdot 2^{9}$, $145\cdot 2^{8}$, $106\cdot 2^{7}$ and $32\cdot 2^{8}$ leaves, $126464$ in all (Table C.3).

Proof.

Within a piece the leaves are consecutive and their lengths add up to $B-A$. The pieces of a support are consecutive and cover it. The counts are products. ∎

The leaves are the closed pieces of the proof of Lemma 4.2. On each of them the bound of that proof is below the cap, so no piece needs a finer subdivision.

Table C.3. Uniform bisection depths.

support	initial pieces	depth $d_{s}$	leaves
$x$	$132$	$9$	$67584$
$v$	$145$	$8$	$37120$
$z$ real	$106$	$7$	$13568$
$\circ$	$32$	$8$	$8192$
total	$415$		$126464$

Appendix D Computer checks

The ancillary scripts recheck printed data and run finite tests (Table D). They use exact rational arithmetic and, for the analytic bounds, Arb ball arithmetic [6] through python-flint 0.9.0 under Python 3.12.3.

Table D.1. Computer checks.

check	statements checked
determinant	the exact seed determinant: (2.6) and Table 2.1. Finite tests: $C_{ij}=3^{i}$ times the right side of (2.3) with $d=j-i$, for $0\leq i\leq 16$ and $|j-i|\leq 16$; the four determinant evaluations in Lemma 2.2 and their product for $m\leq 3$ and three choices of $(s,k,h)$.
arithmetic	recomputes the arrangement of (3.7) for $R=8$ in exact rational arithmetic; checks the line families of Table A, every row of Table A.2 (by evaluating $\mathfrak{c}$ from (3.2)–(3.3) at two interior points), (3.8), $A$ in (3.12), $45/32$ and $-5/48$; rational bounds for $\tau$ and $\alpha_{0}-\tau$; integrality of $K_{m}G_{m}$ for small $n$.
analytic	Lemmas 4.2 and C.1: the chains, masses, conjugation symmetry and $|c|<r$ of Tables C.1 and C.2; Tables B.1, B.2 and C.3, (4.2), (4.3) and $\alpha_{0}$, with the bound of the proof of Lemma 4.2 on each of the $126464$ leaves.
mod $17$	(5.2), (5.1), and (5.4) for $t=1$ at $W=0,1,2$.

Appendix E Proof of Fact 2

For $t>0$ let $g_{t}(y)=(\pi t)^{-1}e^{-|y|^{2}/t}$ on $\mathbb{C}$, a centred Gaussian density. Then $g_{s}*g_{t}=g_{s+t}$ and $g_{t}(-y)=g_{t}(y)$. For $a,b>0$, $\int_{0}^{\infty}(e^{-a/t}-e^{-b/t})\,dt/t=\log(b/a)$, and the integrand has constant sign. With $a=|y|^{2}$ and $b=1$ this gives, for $y\neq 0$,

$$-\log|y|=\frac{\pi}{2}\int_{0}^{\infty}\bigl{(}g_{t}(y)-g_{t}(1)\bigr{)}\,dt,\qquad\int_{0}^{\infty}|g_{t}(y)-g_{t}(1)|\,dt=\frac{2}{\pi}\bigl{|}\log|y|\bigr{|}.$$

The hypothesis of Fact 2 makes the diagonal $|\sigma|\otimes|\sigma|$-null and, by the second identity, $g_{t}(u-w)-g_{t}(1)$ integrable for $|\sigma|\otimes|\sigma|\otimes dt$. By Fubini’s theorem and $\sigma(\mathbb{C})=0$ (for each $t>0$, $g_{t}$ is bounded, so $\iint g_{t}(1)\,d\sigma\,d\sigma=g_{t}(1)\sigma(\mathbb{C})^{2}=0$),

$$-\iint\log|u-w|\,d\sigma(u)\,d\sigma(w)=\frac{\pi}{2}\int_{0}^{\infty}\iint g_{t}(u-w)\,d\sigma(u)\,d\sigma(w)\,dt.$$

For fixed $t$, $g_{t}(u-w)=\int g_{t/2}(x-u)g_{t/2}(x-w)\,dx$ ($dx$ the Lebesgue measure on $\mathbb{C}$), and the corresponding triple integral against $|\sigma|\otimes|\sigma|$ is at most $|\sigma|(\mathbb{C})^{2}/(\pi t)$. So, by Fubini again,

$$\iint g_{t}(u-w)\,d\sigma(u)\,d\sigma(w)=\int_{\mathbb{C}}\Bigl{(}\int g_{t/2}(x-u)\,d\sigma(u)\Bigr{)}^{2}dx\geq 0.$$

Hence $\iint\log|u-w|\,d\sigma(u)\,d\sigma(w)\leq 0$. Compact support is not used.

Use of AI tools

The author devised the method of proof. Codex (OpenAI, GPT-6.1 Sol) computed the reference measure $\nu$ and the other printed data, and wrote the Python scripts in anc/. Claude Code (Anthropic, Claude Opus 5.5) checked the scripts, the data and the tables. 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. All data used in the proof are printed in Appendices A–C. The anc/ attachment holds the scripts of Appendix D and a README.md. The entry points are check_determinant.py, check_arith.py, check_analytic.py and check_mod17.py. The helpers are arith.py, caps.py, combine.py, energy.py, indep_caps_params.py, mod17_replay.py, parse_tex.py, refmeasure.py, tex2refjson.py and tex_tables.py. The scripts read the tables and constants from the TeX source of this manuscript. The analytic replay writes to a temporary directory. The reports in anc/verification/ record the SHA-256 of the TeX source they checked. The file anc/SCRIPTS_SHA256SUMS lists the SHA-256 of the scripts and the README.md. Its own SHA-256 is

890192149e125730cc0b9b0f85ddd7dab8e9ac391a6d14703849fb74d54f8cd7.