Irrationality of the dilogarithm at 1/4

Abstract.

We prove that $\operatorname{Li}_{2}(1/4)$ 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 $s=-4,5,-3$. The linear forms in $1$ and $\operatorname{Li}_{2}(1/4)$ come from one double integral of Viola and Zudilin at $z=4$. We expand it into Rhin–Viola integrals in three ways and apply their factorial transformations to each term. This gives a common factor of the coefficients of the $N$th form. The third expansion adds the primes between $265N$ and $322N$ to this factor, which is too small without them. We bound the integral by the weighted arithmetic-geometric mean inequality and by concavity on circular arcs.

1. Introduction

For $|w|\leq 1$ let $\operatorname{Li}_{2}(w)=\sum_{n\geq 1}w^{n}/n^{2}$. Hata [3] proved that $\operatorname{Li}_{2}(1/s)$ is irrational for every integer $s\leq-5$ and every $s\geq 7$. Rhin and Viola adapted their permutation group method [5] to the dilogarithm and settled $s=6$ [6]. They later treated negative arguments [7]. Viola and Zudilin [8] introduced a six-parameter family of double integrals. In 2024 Calegari, Dimitrov and Tang [1, 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 [4] treats $s=-4,5,-3$. We prove the case $s=4$.

Theorem 1.

The number $\operatorname{Li}_{2}(1/4)$ is irrational.

By Landen’s identity [2, (25.12.3)], $\operatorname{Li}_{2}(1/4)+\operatorname{Li}_{2}(-1/3)=-\frac{1}{2}\log^{2}(3/4)$, so Theorem 1 does not follow directly from the case $s=-3$.

Put $\theta=\operatorname{Li}_{2}(1/4)$, $d_{m}=\operatorname{lcm}(1,\dots,m)$, $d_{0}=1$, and for $N\geq 1$

(1.1) $$G_{N} =J_{4}(240N,240N,190N,157N,215N,165N),$$

$$D_{N} =d_{323N}\,d_{322N}\,4^{132N}\,3^{107N},$$

with $J_{4}$ as in Section 2. By [8, Lemma 2.1], $D_{N}G_{N}\in\mathbb{Z}+\mathbb{Z}\theta$. Sections 3 and 4 give $\Delta_{N}\mid D_{N}$ with

$$(D_{N}/\Delta_{N})\,G_{N}\in\mathbb{Z}+\mathbb{Z}\theta,$$

$$\lim_{N\to\infty}\frac{\log\Delta_{N}}{40N}=\delta>4.7670,\qquad\lim_{N\to\infty}\frac{\log D_{N}}{40N}<23.63856,$$

and Sections 5 and 6 give

$$G_{N}>0\ \text{for large }N,\qquad\limsup_{N\to\infty}\frac{\log G_{N}}{40N}\leq-18.87333.$$

Since $23.63856-4.7670-18.87333<0$, Theorem 1 follows.

The arithmetic follows [4, Section 3.2]. Identity (2.1) for $G_{N}$ and for its image under (F2) gives the analogues of [4, (3.11)–(3.12)]. The factorial transformations of [6] act on each term. For $265N<p<322N$ these expansions give only $M=0$ in Lemma 3.2. These primes account for $57/40$ of $\delta$, and the rest of $\delta$ is below $3.35<23.63856-18.87333$. Expansion (2.2) gives one power of each such $p$. The bounds of Lemma 3.2 depend only on $N/p$, so we need not exclude the exceptional primes of [4, (3.6)–(3.7)]. One linear form suffices, since $G_{N}>0$ for large $N$.

Two Python scripts recheck only part of the proof (Appendix C). The proof is complete without them. The use of AI tools is described at the end of the paper.

2. Integrals and known results

Let $z>1$. For integers $h,j,k,l,m,q\geq 0$ with $j+k-m,\ j+q-m,\ h+m-k,\ h+q-k\geq 0$ put

$$R_{z}(x,y)=z^{k-l-q}\,\frac{x^{j}(1-x)^{h}\,y^{k}(1-y)^{l}\,(1-y+zy)^{j+q-m}}{\bigl{(}x(1-y)+zy\bigr{)}^{j+k-m+1}},$$

and, with small positively oriented circles,

$$J_{z}^{(0)} =\int_{0}^{1}\!\!\int_{0}^{1}R_{z}\,dy\,dx, J_{z}^{(1)} =\int_{0}^{1}\frac{1}{2\pi i}\oint_{|y-\frac{x}{x-z}|=\varepsilon}R_{z}\,dy\,dx,$$

$$J_{z}^{(2)} =\frac{1}{2\pi i}\oint_{|x-z|=\varepsilon^{\prime}}\frac{1}{2\pi i}\oint_{|y-\frac{x}{x-z}|=\varepsilon}R_{z}\,dy\,dx, J_{z} =J_{z}^{(0)}-(\log z)J_{z}^{(1)}.$$

For $t=(h,j,k,l,m)\in\mathbb{Z}_{\geq 0}^{5}$ define $I_{z}^{(0)}(t)$, $I_{z}^{(1)}(t)$, $I_{z}^{(2)}(t)$ and $I_{z}(t)$ in the same way from

$$z^{-l-m}\,\frac{x^{j}(1-x)^{h}\,y^{k}(1-y)^{l}}{\bigl{(}x(1-y)+zy\bigr{)}^{j+k-m+1}}\qquad\text{\cite[cite]{[\@@bibref{}{RV05}{}{}, (2.1)--(2.4)]}}.$$

For $\mathcal{J}=J_{z}$ or $\mathcal{J}=I_{z}(t)$, with components $\mathcal{J}^{(0)},\mathcal{J}^{(1)},\mathcal{J}^{(2)}$, put

$$c(\mathcal{J})=\bigl{(}c_{1}(\mathcal{J}),c_{2}(\mathcal{J})\bigr{)}=\bigl{(}\mathcal{J}+\operatorname{Li}_{2}(1/z)\,\mathcal{J}^{(2)},\ \mathcal{J}^{(2)}\bigr{)}.$$

For $t=(h,j,k,l,m)$ let

$$u(t)=(l+m-j,\ h+m-k,\ h+j-l,\ j+k-m),\qquad T(t)=h!\,j!\,k!\,l!\,m!,$$

$$H(t)=\max_{i}u_{i},\qquad K(t)=\max\{u_{1},\min(u_{2},u_{3}),u_{4}\},$$

and call $t$ admissible if $u(t)\in\mathbb{Z}_{\geq 0}^{4}$. Put

$$\rho(t)=(h,m,l,k,j),\qquad\sigma(t)=(h+m-k,\ j+k-m,\ m,\ l,\ k),$$

and let $O(t)$ be the orbit of $t$ under the group generated by $\rho$ and $\sigma$. Both maps permute the multiset $\{h,j,k,l,m,u_{1},u_{2},u_{3},u_{4}\}$ [6, (3.2), (3.15)], so every $s\in O(t)$ is admissible when $t$ is.

1. [8, Lemma 2.1] Let $H_{1}=\max\{l+q,\,h+m-k,\,h+j-l,\,j+k-m\}$, $H_{2}=\max\{l+q,\,\min(h+m-k,h+j-l),\,j+k-m\}$, $\alpha=\max\{0,l+q-k,l+q-m\}$ and $\beta=\max\{0,k+l-h\}$. There are $P,Q\in\mathbb{Z}[z]$ with

   $$d_{H_{1}}d_{H_{2}}\,z^{\alpha}(1-z)^{\beta}\,c(J_{z})=(P(z),Q(z)).$$

2. [8, Lemma 3.1] $c(J_{z}(h,j,k,l,m,q))=c(J_{z}(j,h,m,l,k,q))$.

3. [6, Theorem 2.1 and (2.9)] Let $\alpha(t)=\max\{j+k,k+l,l+m\}$ and $\beta(t)=\max\{0,k+l-h\}$. There are $P_{t},Q_{t}\in\mathbb{Z}[z]$ with

   $$d_{H(t)}d_{K(t)}\,z^{\alpha(t)}(z-1)^{\beta(t)}\,c(I_{z}(t))=(P_{t}(z),Q_{t}(z)).$$

4. [6, (3.10)–(3.16)] If $t$ is admissible and $s\in O(t)$, then $c(I_{z}(t))/T(t)=c(I_{z}(s))/T(s)$.

In the sources, (F1) and (F3) read $\varphi\mathcal{J}=P-Q\operatorname{Li}_{2}(1/z)$ and $\varphi\mathcal{J}^{(2)}=Q$, with $\varphi$ the displayed multiplier; (F2) and (F4) are stated there for each component. Numbering in [8] refers to the preprint version.

Lemma 2.1.

Let $a=j+q-m$. Then

(2.1) $$c(J_{z}(h,j,k,l,m,q)) =\sum_{\lambda=0}^{a}\binom{a}{\lambda}z^{j+k+\lambda}\,c(I_{z}(h,\,j,\,k+\lambda,\,l+a-\lambda,\,m+\lambda)),$$

(2.2) $$c(J_{z}(h,j,k,l,m,q)) =\sum_{\lambda=0}^{a}\binom{a}{\lambda}z^{j+k}(1-z)^{\lambda}\,c(I_{z}(h,\,j,\,k,\,l+\lambda,\,m)).$$

Proof.

Expand $(1-y+zy)^{a}$ as $((1-y)+zy)^{a}$ and as $(z+(1-z)(1-y))^{a}$. Each term has the denominator and the pole $y=x/(x-z)$ of the corresponding $I_{z}$-integrand. The powers of $z$ are $z^{k-l-q+\lambda}\cdot z^{(l+a-\lambda)+(m+\lambda)}=z^{j+k+\lambda}$ and $z^{k-l-q+a-\lambda}\cdot z^{(l+\lambda)+m}=z^{j+k}$. The identities hold for each component, hence for $c$. ∎

Identity (2.1) is used in the proof of [8, Lemma 2.1] and in [4, (3.11)–(3.12)].

3. The divisor

3.1. Local bounds

For the tuple in (1.1), (F1) gives $H_{1}=323N$, $H_{2}=322N$, $\alpha=132N$, $\beta=107N$. Since $(1-z)^{\beta}=\pm 3^{107N}$ at $z=4$,

(3.1) $$D_{N}\,c(G_{N})\in\mathbb{Z}^{2},\qquad G_{N}=c_{1}(G_{N})-\theta\,c_{2}(G_{N}).$$

Fix $N\geq 1$ and a prime $p>3$ with $p^{2}>430N$. Put $x=N/p$ and $F_{a}=\lfloor ax\rfloor$, and let $[\,\cdot\,]$ be the Iverson bracket. For $0\leq n\leq 430N$, $v_{p}(n!)=\lfloor n/p\rfloor$ and $v_{p}(d_{n})=[\,n\geq p\,]$.

Lemma 3.1.

Let $t$ be admissible with all entries of $t$ and $u(t)$ in $[0,430N]$, and let $s\in O(t)$. Then, for $i=1,2$,

$$v_{p}\bigl{(}c_{i}(I_{4}(t))\bigr{)}\ \geq\ \sum_{\nu=1}^{5}\Bigl{(}\lfloor\frac{t_{\nu}}{p}\rfloor-\lfloor\frac{s_{\nu}}{p}\rfloor\Bigr{)}-[\,H(s)\geq p\,]-[\,K(s)\geq p\,].$$

Proof.

By (F4), $c(I_{4}(t))=T(t)T(s)^{-1}c(I_{4}(s))$. The entries of $s$ and $u(s)$ are those of $t$ and $u(t)$ in some order. Apply (F3) to $s$, where $z=4$ and $z-1=3$ are units at $p$. ∎

Lemma 3.2.

Suppose $p<322N$. For $b\in\{190,215\}$ put $e_{b}=2b-165$ and

$$C =2F_{240}-F_{265}-F_{215},$$

$$U_{b} =2F_{240}+2F_{b}-F_{e_{b}}-F_{323}-F_{322},$$

$$V_{b} =2F_{b}+F_{157}-F_{e_{b}}-F_{322}.$$

Then, for $i=1,2$,

(3.2) $$v_{p}\bigl{(}D_{N}\,c_{i}(G_{N})\bigr{)}\ \geq\ M:=\max\{0,\,C,\,U_{190},\,U_{215},\,V_{190},\,V_{215}\}.$$

If moreover $p>265N$, then $v_{p}(D_{N}\,c_{i}(G_{N}))\geq 1$.

Proof.

Here $v_{p}(D_{N})=2$. By (F2), $c(G_{N})$ is unchanged if (1.1) is replaced by $(240N,240N,215N,157N,190N,165N)$. So (2.1), applied to (1.1) for $b=190$ and to this tuple for $b=215$, gives

$$c(G_{N})=\sum_{\lambda=0}^{bN}\binom{bN}{\lambda}4^{(240+b)N+\lambda}\,c(I_{4}(t)),$$

where $t=(240,240,b,157+b,405-b)N+(0,0,1,-1,1)\lambda$. Here $u(t)=(322,\,645-2b,\,323-b,\,e_{b})N+(0,0,1,0)\lambda$; all entries of $t$ and $u(t)$ lie in $[0,430N]$. The tuples

$$s_{1}=\sigma t =(645-2b,\ e_{b},\ 405-b,\ 157+b,\ b)N+(0,0,1,-1,1)\lambda,$$

$$s_{2}=\sigma\rho\sigma t =(323-b,\ 322,\ e_{b},\ 405-b,\ 157+b)N+(1,0,0,1,-1)\lambda,$$

$$s_{3}=\sigma\rho\sigma\rho\sigma t =(240,\ 240,\ 322,\ e_{b},\ 405-b)N+(0,0,0,0,1)\lambda$$

lie in $O(t)$. Put $\tau=\lambda/p$. Then $v_{p}\binom{bN}{\lambda}=F_{b}-\lfloor\tau\rfloor-\lfloor bx-\tau\rfloor$. Also $[\,H(s)\geq p\,]+[\,K(s)\geq p\,]\leq 2=v_{p}(D_{N})$ for each $s\in O(t)$, so Lemma 3.1 shows that the $\lambda$-th term of $D_{N}c_{i}(G_{N})$ has valuation at least

$$g(s)=\sum_{\nu}\bigl{(}\lfloor t_{\nu}/p\rfloor-\lfloor s_{\nu}/p\rfloor\bigr{)}+F_{b}-\lfloor\tau\rfloor-\lfloor bx-\tau\rfloor.$$

Direct subtraction gives

$$g(s_{1}) =C+\bigl{(}F_{b}-\lfloor\tau\rfloor-\lfloor bx-\tau\rfloor\bigr{)},$$

$$g(s_{2}) =2F_{240}+F_{b}-F_{e_{b}}-F_{322}+\bigl{(}\lfloor bx+\tau\rfloor-\lfloor\tau\rfloor\bigr{)}$$

$$\qquad-\bigl{(}\lfloor bx-\tau\rfloor+\lfloor(323-b)x+\tau\rfloor\bigr{)},$$

$$g(s_{3}) =F_{b}-F_{e_{b}}-F_{322}+\bigl{(}\lfloor bx+\tau\rfloor-\lfloor\tau\rfloor\bigr{)}$$

$$\qquad+\bigl{(}\lfloor(157+b)x-\tau\rfloor-\lfloor bx-\tau\rfloor\bigr{)}.$$

By $\lfloor y\rfloor+\lfloor w\rfloor\leq\lfloor y+w\rfloor$,

$$F_{b}-\lfloor\tau\rfloor-\lfloor bx-\tau\rfloor\geq 0,\qquad\lfloor bx+\tau\rfloor-\lfloor\tau\rfloor\geq F_{b},$$

$$\lfloor(157+b)x-\tau\rfloor-\lfloor bx-\tau\rfloor\geq F_{157},\qquad\lfloor bx-\tau\rfloor+\lfloor(323-b)x+\tau\rfloor\leq F_{323}.$$

Hence $g(s_{1})\geq C$, $g(s_{2})\geq U_{b}$ and $g(s_{3})\geq V_{b}$ for every $\lambda$. The ultrametric inequality and (3.1) give (3.2).

Now let $p>265N$. By (2.2),

$$c(G_{N})=\sum_{\lambda=0}^{190N}\binom{190N}{\lambda}4^{430N}(-3)^{\lambda}\,c(I_{4}(t^{\prime})),$$

where

$$t^{\prime} =(240,240,190,157,215)N+(0,0,0,1,0)\lambda,$$

$$u(t^{\prime}) =(132,265,323,215)N+(1,0,-1,0)\lambda,$$

$$s^{\prime}=\sigma\rho\sigma t^{\prime} =(323,132,215,215,157)N+(-1,1,0,0,1)\lambda,$$

$$u(s^{\prime}) =(240,265,240,190)N.$$

So $[\,H(s^{\prime})\geq p\,]=[\,K(s^{\prime})\geq p\,]=0$. Also $v_{p}\binom{190N}{\lambda}=0$ and $F_{240}=F_{215}=F_{190}=0$. By Lemma 3.1 and $v_{p}(D_{N})=2$, each term of $D_{N}c_{i}(G_{N})$ has valuation at least $2-\lfloor 323x-\tau\rfloor-\lfloor 132x+\tau\rfloor\geq 2-\lfloor 455x\rfloor\geq 1$, as $455x<2$. ∎

3.2. Periodic intervals

Since $5$ divides $240$, $265$ and $215$, and $2\cdot 240=265+215$, we get $C(x+1/5)=C(x)$. For $1\leq k\leq 47$ put $j_{k}=\lfloor 5k/48\rfloor$, $\varrho_{k}=5k-48j_{k}$, $a_{k}=k/240$ and

$$b_{k}=\begin{cases}(k-j_{k})/215,&\varrho_{k}\leq 21,\\ (k+j_{k}+1)/265,&\varrho_{k}\geq 22.\end{cases}$$

Lemma 3.3.

For $1\leq k\leq 47$ one has $a_{k}<b_{k}\leq(k+1)/240$ and $C=1$ on $(a_{k},b_{k})$.

Proof.

Here $0<\varrho_{k}<48$, $(k-j_{k})/215-a_{k}=\varrho_{k}/10320$, $(k+j_{k}+1)/265-a_{k}=(48-\varrho_{k})/12720$ and $265(k-j_{k})-215(k+j_{k}+1)=10\varrho_{k}-215$. So $b_{k}=\min\{(k-j_{k})/215,\,(k+j_{k}+1)/265\}>a_{k}$, and $48j_{k}\leq 5k$ gives $b_{k}\leq(k+j_{k}+1)/265\leq(k+1)/240$. For $a_{k}<x<b_{k}$,

$$k-j_{k}-1<k-\tfrac{5k}{48}<215x<k-j_{k},\qquad k+j_{k}\leq k+\tfrac{5k}{48}<265x<k+j_{k}+1,$$

and $F_{240}=k$. Hence $C=2k-(k-j_{k}-1)-(k+j_{k})=1$. ∎

Put $\mathcal{C}=\bigcup_{r=0}^{9}\bigcup_{k=1}^{47}\,(r/5+a_{k},\ r/5+b_{k})\subset(1/240,2)$; by Lemma 3.3 these $470$ intervals are pairwise disjoint. Let $\mathcal{E}_{1},\dots,\mathcal{E}_{18}$ be the open intervals

$$\begin{array}[]{llll}(1/322,\,1/265)&(1/215,\,2/323)&(2/215,\,3/322)&(3/215,\,4/265),\\ (4/190,\,7/323)&(7/265,\,9/322)&(5/157,\,11/323)&(7/190,\,12/322),\\ (9/215,\,14/323)&(11/215,\,17/322)&(11/190,\,19/323)&(17/265,\,21/322),\\ (11/157,\,23/323)&(14/190,\,24/322)&(15/190,\,26/323)&(19/215,\,29/322),\\ (21/190,\,36/322)&(22/190,\,38/323).\end{array}$$

They are pairwise disjoint and lie in $(1/322,1/5)$. Define

$$\mu(x)=\mathbf{1}_{\mathcal{C}}(x)+\sum_{i=1}^{18}\mathbf{1}_{\mathcal{E}_{i}}(x).$$

Proposition 3.4.

Let $N\geq 1$ and let $p>3$ be a prime with $p^{2}>430N$. Then $\mu(N/p)\leq v_{p}(D_{N})$ and $v_{p}(D_{N}\,c_{i}(G_{N}))\geq\mu(N/p)$ for $i=1,2$.

Proof.

Let $x=N/p$ with $\mu(x)>0$. Then $1/322<x<2$, so $p<322N$, $v_{p}(D_{N})=2$ and $\mu(x)\leq 2$. If $x\notin\bigcup_{i}\mathcal{E}_{i}$, then $\mu(x)=\mathbf{1}_{\mathcal{C}}(x)\leq[\,C=1\,]\leq M$ by Lemmas 3.3 and 3.2. If $x\in\mathcal{E}_{1}$, then $x\notin\mathcal{C}$ and $p>265N$, so Lemma 3.2 gives $v_{p}(D_{N}\,c_{i}(G_{N}))\geq 1=\mu(x)$. If $x\in\mathcal{E}_{2}\cup\dots\cup\mathcal{E}_{18}$, then $M\geq 1+[\,C=1\,]\geq\mu(x)$ by Table A.1, which lists the floors $F_{a}$ for $a\in\{157,190,215,240,265,322,323\}$ on pieces with no jump inside. ∎

4. Density

Lemma 4.1.

Let $\Delta_{N}=\prod p^{\mu(N/p)}$ over primes $p>3$ with $p^{2}>430N$. Then $\Delta_{N}\mid D_{N}$, $(D_{N}/\Delta_{N})\,c(G_{N})\in\mathbb{Z}^{2}$, and

$$\lim_{N\to\infty}\frac{\log\Delta_{N}}{40N} =\delta:=\frac{1}{40}\int_{0}^{2}\mu(x)\,\frac{dx}{x^{2}}>4.7670,$$

$$\lim_{N\to\infty}\frac{\log D_{N}}{40N} =c_{D}:=\frac{645+132\log 4+107\log 3}{40}<23.63856.$$

Proof.

The first two claims follow from Proposition 3.4 and (3.1). If $\mu(N/p)>0$, then $p>N/2$, so $p>3$ and $p^{2}>430N$ for $N>1720$. For each of the $488$ intervals $I$ of $\mathcal{C}$ and $\mathcal{E}_{1},\dots,\mathcal{E}_{18}$, the prime number theorem gives

$$\sum_{N/p\in I}\log p=N\int_{I}x^{-2}\,dx+o(N).$$

Summing over $I$ gives $\delta$. Likewise $\log d_{n}=n+o(n)$ gives $c_{D}$, and $c_{D}<23.63856$ by Table B.1.

The intervals of $\mathcal{C}$ with $r=0$ contribute

$$\delta_{0}=\frac{1}{40}\sum_{k=1}^{47}\Bigl{(}\frac{1}{a_{k}}-\frac{1}{b_{k}}\Bigr{)}=\frac{1}{40}\Bigl{(}\sum_{k=1}^{47}\frac{240}{k}-\sum_{\varrho_{k}\leq 21}\frac{215}{k-j_{k}}-\sum_{\varrho_{k}\geq 22}\frac{265}{k+j_{k}+1}\Bigr{)}>1.6187,$$

with the two cases listed in Table A.2. Let $\ell=\sum_{k}(b_{k}-a_{k})=1139/22790$ and $\bar{y}=\ell^{-1}\sum_{k}\int_{a_{k}}^{b_{k}}y\,dy=164041/1640160$. For $r\geq 1$, Jensen’s inequality for $y\mapsto(r/5+y)^{-2}$ gives

$$\sum_{k}\Bigl{(}\frac{1}{r/5+a_{k}}-\frac{1}{r/5+b_{k}}\Bigr{)}=\sum_{k}\int_{a_{k}}^{b_{k}}\frac{dy}{(r/5+y)^{2}}\geq\frac{\ell}{(r/5+\bar{y})^{2}}.$$

So the intervals with $1\leq r\leq 9$ contribute at least $\frac{\ell}{40}\sum_{r}(r/5+\bar{y})^{-2}>0.0260$ (Table B.1). The intervals $\mathcal{E}_{i}$ contribute $\frac{202007398913}{64696932300}>3.1223$. Thus $\delta>1.6187+0.0260+3.1223=4.7670$. ∎

5. Analytic estimates

5.1. The double integral

Write $J_{N}^{(0)}$, $J_{N}^{(1)}$ for $J_{4}^{(0)}$, $J_{4}^{(1)}$ at the tuple (1.1).

Lemma 5.1.

Let $y=\eta/(\eta-4)$. Then $R_{4}(x,y)\,dy=-f_{N}(x,\eta)\,d\eta$, where

$$f_{N}(x,\eta)=\frac{\Psi(x,\eta)^{N}}{(x-\eta)(4-\eta)},\qquad\Psi(x,\eta)=\frac{x^{240}(1-x)^{240}\,\eta^{190}(1-\eta)^{190}}{(x-\eta)^{215}(4-\eta)^{322}}.$$

In particular, with

$$\Phi(x,\tau)=\Psi(x,-\tau)=\frac{x^{240}(1-x)^{240}\,\tau^{190}(1+\tau)^{190}}{(x+\tau)^{215}(4+\tau)^{322}},$$

we have

$$J_{N}^{(0)}=\int_{0}^{1}\!\!\int_{0}^{\infty}\frac{\Phi(x,\tau)^{N}}{(x+\tau)(4+\tau)}\,d\tau\,dx.$$

Proof.

We have

$$y=\frac{-\eta}{4-\eta},\qquad 1-y=\frac{4}{4-\eta},\qquad 1+3y=\frac{4(1-\eta)}{4-\eta},$$

$$x(1-y)+4y=\frac{4(x-\eta)}{4-\eta},\qquad dy=\frac{-4\,d\eta}{(4-\eta)^{2}}.$$

The exponents are $j=h=240N$, $k=190N$, $l=157N$, $j+q-m=190N$, $j+k-m+1=215N+1$ and $k-l-q=-132N$. The powers of $4$ cancel, since $-132+157+190-215=0$ and $4^{-1}\cdot 4=1$. Those of $4-\eta$ add to $-(322N+1)$, and $(-\eta)^{190N}=\eta^{190N}$. For $J_{N}^{(0)}$ put $\tau=-\eta$. Then $\tau$ runs over $[0,\infty)$ as $y$ runs over $[0,1)$, and $-f_{N}\,d\eta=f_{N}(x,-\tau)\,d\tau$. ∎

Lemma 5.2.

For $N\geq 1$, $10^{-9}e^{-40(18.88)N}<J_{N}^{(0)}<e^{-40(18.87333)N}$.

Proof.

Weighted AM–GM, $x+\tau\geq 215\,(x/25)^{25/215}(\tau/190)^{190/215}$, gives

$$\Phi \leq\Bigl{(}\frac{5}{43}\Bigr{)}^{25}\Bigl{(}\frac{38}{43}\Bigr{)}^{190}x^{215}(1-x)^{240}\,\frac{(1+\tau)^{190}}{(4+\tau)^{322}}$$

$$\leq\Phi_{+}:=\Bigl{(}\frac{5}{43}\Bigr{)}^{25}\Bigl{(}\frac{38}{43}\Bigr{)}^{190}\Bigl{(}\frac{43}{91}\Bigr{)}^{215}\Bigl{(}\frac{48}{91}\Bigr{)}^{240}\frac{(95/22)^{190}}{(161/22)^{322}},$$

with maxima at $x=43/91$ and $\tau=73/22$. By Table B.1, $-\log\Phi_{+}>40(18.87333)$. Also

$$\int_{0}^{1}\!\!\int_{0}^{\infty}\frac{d\tau\,dx}{(x+\tau)(4+\tau)}=\int_{0}^{1}\frac{\log(4/x)}{4-x}\,dx<\frac{1+\log 4}{3}<1,$$

so $J_{N}^{(0)}<\Phi_{+}^{N}$. For the lower bound let $\mathcal{B}=[0.4711,0.4713]\times[3.3999,3.4001]$. Bounding each factor of $\Phi$ by its least value on $\mathcal{B}$ gives

$$\Phi\geq\Phi_{-}:=\frac{0.4711^{240}\,0.5287^{240}\,3.3999^{190}\,4.3999^{190}}{3.8714^{215}\,7.4001^{322}}\qquad\text{on }\mathcal{B},$$

and $(x+\tau)(4+\tau)<29$ there. By Table B.1, $-\log\Phi_{-}<40(18.88)$. Hence $J_{N}^{(0)}\geq\operatorname{area}(\mathcal{B})\,\Phi_{-}^{N}/29=4\cdot 10^{-8}\,\Phi_{-}^{N}/29>10^{-9}e^{-40(18.88)N}$. ∎

5.2. The residue integral

Put $A=(3+4i)/5$, $B=(5-12i)/13$ and, for $0\leq t,u\leq 1$,

$$X(t)=\frac{t}{t+A(1-t)},\qquad Y(u)=\frac{u}{u+B(1-u)},\qquad\omega=t(1-u)+u(1-t).$$

Lemma 5.3.

For $0\leq t,u\leq 1$ and $x\in[0,1]$:

1. $|X^{\prime}|\leq 5/4$, $|Y^{\prime}|\leq 13/9$, $|X(1-X)|\leq\frac{5}{4}t(1-t)\leq\frac{5}{4}\omega$, $|Y(1-Y)|\leq\frac{13}{9}u(1-u)\leq\frac{13}{9}\omega$, $|Y|\leq 1$ and $|4-Y|\geq 3$;

2. $|X-Y|\geq 7\omega/\sqrt{65}$, and $|x-Y|\geq 2\omega_{x}/\sqrt{13}$ with $\omega_{x}=x(1-u)+u(1-x)$;

3. $\operatorname{Im}X(t)<0<\operatorname{Im}Y(u)$ for $0<t,u<1$.

Proof.

One has $|t+A(1-t)|^{2}=1-\frac{4}{5}t(1-t)\geq\frac{4}{5}$, $|u+B(1-u)|^{2}=1-\frac{16}{13}u(1-u)\geq\frac{9}{13}$, $X^{\prime}=A/(t+A(1-t))^{2}$ and $1-X=A(1-t)/(t+A(1-t))$; similarly for $Y$. Also $|u+B(1-u)|^{2}-u^{2}=(1-u)(13-3u)/13\geq 0$, so $|Y|\leq 1$. Next $t(1-t)\leq\omega$, since $\omega-t(1-t)$ equals $u(1-2t)+t^{2}\geq 0$ for $t\leq\frac{1}{2}$ and is at least $(1-t)^{2}$ for $t\geq\frac{1}{2}$; likewise $u(1-u)\leq\omega$. For (2), $X-Y=\frac{tB(1-u)-uA(1-t)}{(t+A(1-t))(u+B(1-u))}$ and the denominator has modulus at most $1$. With $\xi=t(1-u)$, $\zeta=u(1-t)$ and $\gamma=\operatorname{Re}(B\bar{A})=-33/65$,

$$|tB(1-u)-uA(1-t)|^{2}=\xi^{2}+\zeta^{2}-2\gamma\xi\zeta=\tfrac{1-\gamma}{2}\,\omega^{2}+\tfrac{1+\gamma}{2}(\xi-\zeta)^{2}\geq\tfrac{49}{65}\,\omega^{2}.$$

For real $x$ take $A=1$, $t=x$ and $\gamma=\operatorname{Re}B=5/13$. Part (3) follows from $\operatorname{Im}X=-\frac{4}{5}t(1-t)/|t+A(1-t)|^{2}$ and its analogue. ∎

Lemma 5.4.

For $N\geq 1$, $J_{N}^{(1)}=\pi^{-1}\operatorname{Im}Z_{N}$, where

$$Z_{N}=\int_{0}^{1}\!\!\int_{0}^{1}f_{N}\bigl{(}X(t),Y(u)\bigr{)}\,X^{\prime}(t)\,Y^{\prime}(u)\,dt\,du.$$

Proof.

Residues are invariant under the change of variables of Lemma 5.1, and $y=x/(x-4)$ corresponds to $\eta=x$. Hence $J_{N}^{(1)}=-\int_{0}^{1}\operatorname*{Res}_{\eta=x}f_{N}(x,\eta)\,dx$. Fix $x\in(0,1)$. By Lemma 5.3, $Y$ runs from $0$ to $1$ in the upper half-plane and $\bar{Y}$ in the lower one, both in the closed unit disc. So $\bar{Y}$ followed by $Y$ reversed is a positively oriented Jordan curve around $x$, and the other pole $\eta=4$ lies outside. As $f_{N}$ has real coefficients, the residue theorem gives

$$\operatorname*{Res}_{\eta=x}f_{N}=\frac{\overline{W(x)}-W(x)}{2\pi i}=-\frac{\operatorname{Im}W(x)}{\pi},\qquad W(x)=\int_{0}^{1}f_{N}(x,Y(u))\,Y^{\prime}(u)\,du.$$

By Lemma 5.3, also with $t$ replaced by $x$,

$$|f_{N}(x,Y)\,Y^{\prime}|\ll_{N}\omega_{x}^{215N-1},\qquad|f_{N}(X,Y)\,X^{\prime}Y^{\prime}|\ll_{N}\omega^{215N-1},$$

so both are bounded on $[0,1]^{2}$. For fixed $u\in(0,1)$ the function $w\mapsto f_{N}(w,Y(u))$ is holomorphic off $w=Y(u)$, and $\operatorname{Im}Y(u)>0\geq\operatorname{Im}X(t)$. Cauchy’s theorem moves $\int_{0}^{1}dx$ to the arc $X$, and Fubini’s theorem gives $\int_{0}^{1}W(x)\,dx=Z_{N}$. ∎

Lemma 5.5.

For $N\geq 1$, $|J_{N}^{(1)}|<e^{-40(18.8995)N}$.

Proof.

Put $r=t/(1-t)=e^{R}$ and $s=u/(1-u)=e^{S}$, so $X=r/(r+A)$ and $Y=s/(s+B)$. With

$$Q_{1}=r^{2}+\tfrac{6}{5}r+1,\quad Q_{2}=s^{2}+\tfrac{10}{13}s+1,\quad Q_{3}=r^{2}+\tfrac{66}{65}rs+s^{2},\quad Q_{4}=9s^{2}+\tfrac{120}{13}s+16,$$

one has $|X(1-X)|=r/Q_{1}$, $|Y(1-Y)|=s/Q_{2}$, $|X-Y|^{2}=Q_{3}/(Q_{1}Q_{2})$ and $|4-Y|^{2}=Q_{4}/Q_{2}$. So $\log|\Psi(X,Y)|=L$, where

$$L=240\log r+190\log s-\tfrac{265}{2}\log Q_{1}+\tfrac{157}{2}\log Q_{2}-\tfrac{215}{2}\log Q_{3}-161\log Q_{4}.$$

Let $\ell_{\gamma}(v)=\log(\cosh v+\gamma)$ and $\kappa=\log(4/3)$. Since $Q_{1}=2r(\cosh R+\frac{3}{5})$, $Q_{2}=2s(\cosh S+\frac{5}{13})$, $Q_{3}=2rs(\cosh(R-S)+\frac{33}{65})$ and $Q_{4}=24s(\cosh(S-\kappa)+\frac{5}{13})$,

$$L={} -\tfrac{323}{2}\log 2-161\log 24-\tfrac{265}{2}\ell_{3/5}(R)+\tfrac{157}{2}\ell_{5/13}(S)$$

$$-\tfrac{215}{2}\ell_{33/65}(R-S)-161\,\ell_{5/13}(S-\kappa).$$

Here $\ell_{\gamma}^{\prime\prime}=(1+\gamma\cosh v)/(\cosh v+\gamma)^{2}>0$ and, for $0<\gamma<1$,

$$(\log\ell_{\gamma}^{\prime\prime})^{\prime}=-\sinh v\,\frac{2+\gamma\cosh v-\gamma^{2}}{(1+\gamma\cosh v)(\cosh v+\gamma)},\qquad|(\log\ell_{\gamma}^{\prime\prime})^{\prime}|\leq\frac{2|\sinh v|}{\cosh v+\gamma}<2,$$

so $\ell_{\gamma}^{\prime\prime}(v-\kappa)\geq\frac{9}{16}\ell_{\gamma}^{\prime\prime}(v)$. Therefore

$$-d^{2}L={} \tfrac{265}{2}\ell_{3/5}^{\prime\prime}(R)\,dR^{2}+\tfrac{215}{2}\ell_{33/65}^{\prime\prime}(R-S)(dR-dS)^{2}$$

$$+\bigl{(}161\,\ell_{5/13}^{\prime\prime}(S-\kappa)-\tfrac{157}{2}\ell_{5/13}^{\prime\prime}(S)\bigr{)}dS^{2}$$

is positive definite, since $161\cdot\frac{9}{16}>\frac{157}{2}$. So $L$ is strictly concave and $L_{RS}>0$. The partial derivatives are

$$L_{R} =240-265\,\frac{r(r+3/5)}{Q_{1}}-215\,\frac{r(r+33s/65)}{Q_{3}},$$

$$L_{S} =190+157\,\frac{s(s+5/13)}{Q_{2}}-215\,\frac{s(s+33r/65)}{Q_{3}}-322\,\frac{s(9s+60/13)}{Q_{4}}.$$

Table B.2 gives $L_{R},L_{S}>0$ at $(r,s)=(\frac{29}{25},\frac{139}{100})$ and $L_{R},L_{S}<0$ at $(\frac{59}{50},\frac{71}{50})$. Consider the box $\frac{29}{25}\leq r\leq\frac{59}{50}$, $\frac{139}{100}\leq s\leq\frac{71}{50}$. Since $L_{RS}>0$, we get $L_{R}>0$ on its face $r=\frac{29}{25}$, $L_{R}<0$ on $r=\frac{59}{50}$, $L_{S}>0$ on $s=\frac{139}{100}$ and $L_{S}<0$ on $s=\frac{71}{50}$. So the maximum of $L$ on the box is an interior critical point, hence the global maximum. At $(r,s)=(\frac{117}{100},\frac{7}{5})$, Table B.2 gives $-\frac{1}{4}<L_{R}<0<L_{S}<\frac{3}{10}$ and $L<-755.99$. Concavity gives

$$\sup L<-755.99+\tfrac{1}{4}\log\tfrac{117}{116}+\tfrac{3}{10}\log\tfrac{71}{70}<-755.99+\tfrac{1}{464}+\tfrac{3}{700}<-755.98.$$

By Lemma 5.3, $|X^{\prime}Y^{\prime}|<2$, $|X-Y|\geq\frac{4}{5}\omega$ and $|4-Y|\geq 3$. With $w=1-2u$,

$$\int_{0}^{1}\!\!\int_{0}^{1}\frac{dt\,du}{\omega}=\int_{0}^{1}\frac{\log\frac{1-u}{u}}{1-2u}\,du=\frac{1}{2}\int_{-1}^{1}\log\frac{1+w}{1-w}\,\frac{dw}{w}=2\sum_{n\geq 0}\frac{1}{(2n+1)^{2}}=\frac{\pi^{2}}{4}.$$

By Lemma 5.4,

$$|J_{N}^{(1)}|\leq\frac{1}{\pi}\cdot 2\cdot\frac{5}{4}\cdot\frac{1}{3}\cdot\frac{\pi^{2}}{4}\,e^{-40(18.8995)N}=\frac{5\pi}{24}\,e^{-40(18.8995)N}<e^{-40(18.8995)N}.\qed$$

6. Proof of Theorem 1

By Lemmas 5.2 and 5.5, $J_{N}^{(0)}>10^{-9}e^{-40(18.88)N}>(\log 4)\,|J_{N}^{(1)}|$ for large $N$. Hence $G_{N}=J_{N}^{(0)}-(\log 4)J_{N}^{(1)}>0$ and $G_{N}<3e^{-40(18.87333)N}$. Put

$$\Lambda_{N}=\frac{D_{N}}{\Delta_{N}}\,G_{N}=\frac{D_{N}}{\Delta_{N}}\,c_{1}(G_{N})-\frac{D_{N}}{\Delta_{N}}\,c_{2}(G_{N})\,\theta.$$

By (3.1) and Lemma 4.1, $\Lambda_{N}\in\mathbb{Z}+\mathbb{Z}\theta$. Also $\Lambda_{N}>0$ for large $N$, and

$$\limsup_{N\to\infty}\frac{\log\Lambda_{N}}{40N}\leq c_{D}-\delta-18.87333<23.63856-4.7670-18.87333=-0.00177.$$

If $q_{0}\theta\in\mathbb{Z}$ for an integer $q_{0}\geq 1$, then $q_{0}\Lambda_{N}$ is a positive integer for large $N$ and $q_{0}\Lambda_{N}\to 0$, which is impossible. ∎

Appendix A Floor data

Table A.1 cuts each $\mathcal{E}_{i}$ at the jumps of the $F_{a}$. On each piece, $F_{a}=\lfloor ax_{0}\rfloor$ with $x_{0}$ the left end. The columns $C$, $U_{b}$, $V_{b}$ follow from Lemma 3.2, and the last column is $1+[\,C=1\,]$. The first row uses (2.2) instead of $C$, $U_{b}$, $V_{b}$.

Table A.1. Floors on the pieces of $\mathcal{E}_{1},\dots,\mathcal{E}_{18}$.

piece	$F_{157}$	$F_{190}$	$F_{215}$	$F_{240}$	$F_{265}$	$F_{322}$	$F_{323}$	$C$	$U_{190}$	$U_{215}$	$V_{190}$	$V_{215}$	$1{+}[\,C{=}1\,]$
$(1/322,1/265)$	$0$	$0$	$0$	$0$	$0$	$1$	$1$	by (2.2)					$1$
$(1/215,1/190)$	$0$	$0$	$1$	$1$	$1$	$1$	$1$	$0$	$-1$	$1$	$-2$	$0$	$1$
$[1/190,2/323)$	$0$	$1$	$1$	$1$	$1$	$1$	$1$	$0$	$1$	$1$	$0$	$0$	$1$
$(2/215,3/322)$	$1$	$1$	$2$	$2$	$2$	$2$	$3$	$0$	$-1$	$1$	$-1$	$1$	$1$
$(3/215,4/265)$	$2$	$2$	$3$	$3$	$3$	$4$	$4$	$0$	$-1$	$1$	$-1$	$1$	$1$
$(4/190,7/323)$	$3$	$4$	$4$	$5$	$5$	$6$	$6$	$1$	$2$	$1$	$1$	$0$	$2$
$(7/265,9/323)$	$4$	$5$	$5$	$6$	$7$	$8$	$8$	$0$	$1$	$-1$	$1$	$-1$	$1$
$[9/323,6/215)$	$4$	$5$	$5$	$6$	$7$	$8$	$9$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[6/215,9/322)$	$4$	$5$	$6$	$6$	$7$	$8$	$9$	$-1$	$-1$	$0$	$0$	$1$	$1$
$(5/157,7/215)$	$5$	$6$	$6$	$7$	$8$	$10$	$10$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[7/215,8/240)$	$5$	$6$	$7$	$7$	$8$	$10$	$10$	$-1$	$-1$	$0$	$0$	$1$	$1$
$[8/240,9/265)$	$5$	$6$	$7$	$8$	$8$	$10$	$10$	$1$	$1$	$2$	$0$	$1$	$2$
$[9/265,11/323)$	$5$	$6$	$7$	$8$	$9$	$10$	$10$	$0$	$1$	$1$	$0$	$0$	$1$
$(7/190,12/323)$	$5$	$7$	$7$	$8$	$9$	$11$	$11$	$0$	$1$	$-1$	$1$	$-1$	$1$
$[12/323,8/215)$	$5$	$7$	$7$	$8$	$9$	$11$	$12$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[8/215,12/322)$	$5$	$7$	$8$	$8$	$9$	$11$	$12$	$-1$	$-1$	$0$	$0$	$1$	$1$
$(9/215,8/190)$	$6$	$7$	$9$	$10$	$11$	$13$	$13$	$0$	$-1$	$1$	$-2$	$0$	$1$
$[8/190,14/323)$	$6$	$8$	$9$	$10$	$11$	$13$	$13$	$0$	$1$	$1$	$0$	$0$	$1$
$(11/215,10/190)$	$8$	$9$	$11$	$12$	$13$	$16$	$16$	$0$	$-1$	$1$	$-1$	$1$	$1$
$[10/190,17/322)$	$8$	$10$	$11$	$12$	$13$	$16$	$17$	$0$	$0$	$0$	$1$	$1$	$1$
$(11/190,14/240)$	$9$	$11$	$12$	$13$	$15$	$18$	$18$	$-1$	$0$	$-1$	$1$	$0$	$1$
$[14/240,19/323)$	$9$	$11$	$12$	$14$	$15$	$18$	$18$	$1$	$2$	$1$	$1$	$0$	$2$
$(17/265,21/323)$	$10$	$12$	$13$	$15$	$17$	$20$	$20$	$0$	$1$	$-1$	$1$	$-1$	$1$
$[21/323,14/215)$	$10$	$12$	$13$	$15$	$17$	$20$	$21$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[14/215,21/322)$	$10$	$12$	$14$	$15$	$17$	$20$	$21$	$-1$	$-1$	$0$	$0$	$1$	$1$
$(11/157,17/240)$	$11$	$13$	$15$	$16$	$18$	$22$	$22$	$-1$	$-1$	$0$	$0$	$1$	$1$
$[17/240,23/323)$	$11$	$13$	$15$	$17$	$18$	$22$	$22$	$1$	$1$	$2$	$0$	$1$	$2$
$(14/190,24/323)$	$11$	$14$	$15$	$17$	$19$	$23$	$23$	$0$	$1$	$-1$	$1$	$-1$	$1$
$[24/323,16/215)$	$11$	$14$	$15$	$17$	$19$	$23$	$24$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[16/215,24/322)$	$11$	$14$	$16$	$17$	$19$	$23$	$24$	$-1$	$-1$	$0$	$0$	$1$	$1$
$(15/190,17/215)$	$12$	$15$	$16$	$18$	$20$	$25$	$25$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[17/215,19/240)$	$12$	$15$	$17$	$18$	$20$	$25$	$25$	$-1$	$-1$	$0$	$0$	$1$	$1$
$[19/240,21/265)$	$12$	$15$	$17$	$19$	$20$	$25$	$25$	$1$	$1$	$2$	$0$	$1$	$2$
$[21/265,26/323)$	$12$	$15$	$17$	$19$	$21$	$25$	$25$	$0$	$1$	$1$	$0$	$0$	$1$
$(19/215,14/157)$	$13$	$16$	$19$	$21$	$23$	$28$	$28$	$0$	$-1$	$1$	$-2$	$0$	$1$
$[14/157,17/190)$	$14$	$16$	$19$	$21$	$23$	$28$	$28$	$0$	$-1$	$1$	$-1$	$1$	$1$
$[17/190,29/323)$	$14$	$17$	$19$	$21$	$23$	$28$	$28$	$0$	$1$	$1$	$1$	$1$	$1$
$[29/323,29/322)$	$14$	$17$	$19$	$21$	$23$	$28$	$29$	$0$	$0$	$0$	$1$	$1$	$1$
$(21/190,36/323)$	$17$	$21$	$23$	$26$	$29$	$35$	$35$	$0$	$1$	$-1$	$1$	$-1$	$1$
$[36/323,24/215)$	$17$	$21$	$23$	$26$	$29$	$35$	$36$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[24/215,36/322)$	$17$	$21$	$24$	$26$	$29$	$35$	$36$	$-1$	$-1$	$0$	$0$	$1$	$1$
$(22/190,25/215)$	$18$	$22$	$24$	$27$	$30$	$37$	$37$	$0$	$0$	$-2$	$1$	$-1$	$1$
$[25/215,28/240)$	$18$	$22$	$25$	$27$	$30$	$37$	$37$	$-1$	$-1$	$0$	$0$	$1$	$1$
$[28/240,31/265)$	$18$	$22$	$25$	$28$	$30$	$37$	$37$	$1$	$1$	$2$	$0$	$1$	$2$
$[31/265,38/323)$	$18$	$22$	$25$	$28$	$31$	$37$	$37$	$0$	$1$	$1$	$0$	$0$	$1$

Table A.2. The two cases in the definition of $b_{k}$.

$j_{k}$	$\varrho_{k}\leq 21$	$k-j_{k}$	$\varrho_{k}\geq 22$	$k+j_{k}+1$
0	$1\leq k\leq 4$	$1,\dots,4$	$5\leq k\leq 9$	$6,\dots,10$
1	$10\leq k\leq 13$	$9,\dots,12$	$14\leq k\leq 19$	$16,\dots,21$
2	$20\leq k\leq 23$	$18,\dots,21$	$24\leq k\leq 28$	$27,\dots,31$
3	$29\leq k\leq 33$	$26,\dots,30$	$34\leq k\leq 38$	$38,\dots,42$
4	$39\leq k\leq 42$	$35,\dots,38$	$43\leq k\leq 47$	$48,\dots,52$

Appendix B Numerical values

Values are truncated to the digits shown.

Table B.1. Constants.

quantity	value
$-\log\Phi_{+}/40$	$18.8733376741\ldots$
$-\log\Phi_{-}/40$	$18.8775167011\ldots$
$c_{D}$	$23.6385592638\ldots$
$\delta_{0}$	$1.6187800851\ldots$
$\frac{\ell}{40}\sum_{r=1}^{9}(r/5+\bar{y})^{-2}$	$0.0260767666\ldots$
$\frac{1}{40}\sum_{i}\int_{\mathcal{E}_{i}}x^{-2}\,dx$	$3.1223644109\ldots$

Table B.2. Values of $L_{R}$, $L_{S}$ and $L$.

$(r,s)$	$L_{R}$	$L_{S}$	$L$
$(29/25,\,139/100)$	$0.57635\ldots$	$0.62026\ldots$
$(59/50,\,71/50)$	$-0.52544\ldots$	$-1.02457\ldots$
$(117/100,\,7/5)$	$-0.22659\ldots$	$0.26833\ldots$	$-755.99795\ldots$

Appendix C Computer checks

The two scripts in Table C.1 recheck only the statements listed there, in exact rational arithmetic. They are included as ancillary files. The proof does not use them. For $\log v$ the analytic script writes $v=2^{e}q$ with $e\in\mathbb{Z}$, $1\leq q\leq 2$, and uses $30$ terms of $\log q=2\sum_{n\geq 0}w^{2n+1}/(2n+1)$, $w=(q-1)/(q+1)$, with tail at most $2w^{61}/(61(1-w^{2}))$.

Table C.1. Scripts.

script	statements checked	time
check_arith.py	the intervals $\mathcal{E}_{i}$ and Tables A.1, A.2 as printed; Lemma 3.3; the tuples $s_{1},s_{2},s_{3},s^{\prime}$ and the floor sums in Lemma 3.2; Proposition 3.4 for $1/265\leq x<2$ at each jump of the $F_{a}$ and between consecutive jumps, and $M=0$, $\mu=1$ on $\mathcal{E}_{1}$; the numbers in Lemma 4.1; $57/40$ and $3.35$ in Section 1; rows 4–6 of Table B.1	10 s
check_arith.py --brute 1,2,3,4,5,6,7,8	for $N\leq 8$ and all primes $p$ with $3<p\leq 323N$ and $p^{2}>430N$ (for larger $p$, $\mu(N/p)=0$), the best bound from Lemma 3.1 over (2.1) with $b=190,215$ and (2.2), all $\lambda$ and all $s\in O(t)$, is at least $\mu(N/p)$	3 min
check_analytic.py	the exponents in Lemma 5.1; the constants in Lemma 5.2; the identities in Lemmas 5.3 and 5.5 at rational points; rows 1–3 of Table B.1 and Table B.2 as printed; $23.63856-4.7670-18.87333=-0.00177$	1 s

Use of AI tools

Codex (OpenAI, GPT-5.6 Sol) searched for the parameters in (1.1) and wrote the Python code and the interval lists. Claude Code (Anthropic, Claude Fable 5) checked the code, the parameters 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.