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\hypersetup{pdftitle={The book Ramsey number R(B57,B58)=231},pdfauthor={Deep Bhattacharjee}}

\journal{Discrete Mathematics}

\begin{document}

\begin{frontmatter}

\title{The book Ramsey number $R(B_{57},B_{58})=231$}

\author{Deep Bhattacharjee}
\ead{d.bhattacharjee@erl-forschung.de}
\ead{itsdeep@live.com}
\ead[url]{https://orcid.org/0000-0003-0466-750X}
\affiliation{organization={Formerly, Electro-Gravitational Space Propulsion Laboratory (EGSPL)},
  city={Bhubaneswar}, state={Odisha}, postcode={751030}, country={India}}

\begin{abstract}
We construct a graph on $230$ vertices with no book $B_{57}$ whose complement contains no $B_{58}$. With the upper bound of Rousseau and Sheehan this gives $R(B_{57},B_{58})=231$, the smallest case of Wesley's conjecture $R(B_{n-1},B_n)=4n-1$ that was open. The graph is a two-block circulant graph over $\Z_{115}$, and the proof reduces to $57$ inequalities.
\end{abstract}

\begin{keyword}
Ramsey number \sep book graph \sep block-circulant graph \sep local search \sep formal verification
\MSC[2020] 05C55 \sep 05C25 \sep 05B10
\end{keyword}

\end{frontmatter}


\section{Introduction}

The book $B_k$ is the graph $K_2+\overline{K_k}$: $k$ triangles sharing one edge, its spine. For graphs $G$ and $H$, the Ramsey number $R(G,H)$ is the least $N$ such that every graph on $N$ vertices contains $G$ or has a complement containing $H$. Rousseau and Sheehan~\cite{RS} proved, from Goodman's count of monochromatic triangles~\cite{Go}, that
\begin{equation}\label{eq:RS}
R(B_{n-1},B_n)\le 4n-1\qquad(n\ge2).
\end{equation}
Wesley~\cite{We} proved equality for $n\le20$ and whenever $2n-1$ is a prime power congruent to $1$ modulo $4$, and conjectured equality for every $n$; the conjecture is one of the open problems of Epoch AI's FrontierMath collection~\cite{Ep}. Turturean~\cite{Tu} constructed extremal graphs for every $n\le57$ and for fourteen further values below $100$, partly through a cyclotomic construction of Szekeres~\cite{Sz}; Chung, Du and Wesley~\cite{CDW} found three further infinite families and settled $n=67$ by a finite certificate~\cite{CDWd}, for which~\cite{Bh} gives a proof checkable by hand; and Price obtained $n=70$~\cite{Ep}. Before the present note the smallest open case was $n=58$.\footnote{The values $n\le100$ that none of the constructions cited here covers were $58$, $59$, $60$, $64$, $65$, $68$, $72$, $73$, $76$, $78$, $80$, $81$, $84$, $88$, $89$, $92$, $93$, $94$, $96$ and $100$. This list is our reading of~\cite{CDW,CDWd,Ep,Tu,We} as of 15 October 2026; it agrees with the ledger in~\cite[Section~6.1]{CDWd}.}

\begin{maintheorem}
$R(B_{57},B_{58})=231$.
\end{maintheorem}

The upper bound is~\eqref{eq:RS}, and the lower bound is a graph on $230$ vertices with no $B_{57}$ whose complement has no $B_{58}$. For odd $n$ the graph of~\cite{Bh} and the Paley graphs of~\cite{We} are regular two-block circulant graphs over $\Z_{2n-1}$ whose two blocks are complementary circulants of degree $n-1$. When $n$ is even such a graph does not exist, because a symmetric subset of $\Z_{2n-1}\setminus\{0\}$ has even size. Turturean's examples for even $n$~\cite{Tu} enlarge the first block by one element and shrink the second by one, and our graph has that shape. Section~\ref{sec:blocks} reduces the two book conditions for it to one inequality~\eqref{eq:star} on the connection sets, a reduction that underlies the search of~\cite{Tu} and that we prove here in full. Section~\ref{sec:58} gives the connection sets for $n=58$, and Table~\ref{tab:main} verifies~\eqref{eq:star} at the $57$ values of $d$ that matter. The graph has also been checked by two independent programs and in Lean~4 (see the end of the paper).

\section{Two-block circulant graphs of even order}\label{sec:blocks}

Throughout, $n\ge4$ is even, $v=2n-1$ and $G=\Z_v$. For $X,Y\subseteq G$ and $d\in G$ put
\[
\NN_{X,Y}(d)=|X\cap(Y+d)|=\#\{y\in Y:\ y+d\in X\}.
\]
Translating by $-d$ shows $\NN_{X,X}(-d)=\NN_{X,X}(d)$, and $\sum_{d\in G}\NN_{X,X}(d)=|X|^2$.

Let $A\subseteq G\setminus\{0\}$ with $A=-A$ and $|A|=n$, let $C=G\setminus(A\cup\{0\})$, so that $C=-C$ and $|C|=n-2$, and let $B\subseteq G$ with $|B|=n-1$. The graph $\Gamma=\Gamma(A,B)$ has vertex set $G\times\{0,1\}$, and
\[
(x,0)\sim(y,0)\iff y-x\in A,\qquad (x,1)\sim(y,1)\iff y-x\in C,
\]
and $(x,0)\sim(y,1)$ if and only if $y-x\in B$. Since $A$ and $C$ are symmetric and avoid $0$, $\Gamma$ is a simple graph on $4n-2$ vertices. A vertex $(x,0)$ has $|A|+|B|=2n-1$ neighbours and a vertex $(x,1)$ has $|C|+|B|=2n-3$; the translations $(x,i)\mapsto(x+t,i)$ are automorphisms. Write $\lambda(p,q)$ for the number of common neighbours of two vertices, $\mu(p,q)$ for the number of vertices other than $p$ and $q$ adjacent to neither, $\ind{P}$ for the indicator of a statement $P$, and
\[
f(d)=\NN_{A,A}(d)+\NN_{B,B}(d)\qquad(d\in G).
\]

\begin{lemma}\label{lem:lambda}
For $d\ne0$,
\[
\lambda\big((0,0),(d,0)\big)=f(d),\qquad \lambda\big((0,1),(d,1)\big)=f(d)-3+2\ind{d\in A},
\]
and for every $d\in G$, $\lambda\big((0,0),(d,1)\big)=n-1-\ind{d\in B}$.
\end{lemma}

\begin{proof}
A vertex $(z,0)$ is a common neighbour of $(0,0)$ and $(d,0)$ when $z\in A$ and $z-d\in A$, and $(z,1)$ is one when $z\in B$ and $z-d\in B$; this gives $\NN_{A,A}(d)+\NN_{B,B}(d)$.

For $(0,1)$ and $(d,1)$, the common neighbours $(z,1)$ are counted by $\NN_{C,C}(d)$. A vertex $(z,0)$ is adjacent to $(0,1)$ when $-z\in B$ and to $(d,1)$ when $d-z\in B$; with $w=-z$ these say $w\in B$ and $w+d\in B$, so there are $\NN_{B,B}(-d)=\NN_{B,B}(d)$ of them. To compute $\NN_{C,C}(d)$, let $S=A\cup\{0\}$, so that $|S|=n+1$ and $C=G\setminus S$. Then $|C\cap(C+d)|=v-|S\cup(S+d)|=v-2(n+1)+|S\cap(S+d)|$. The set $S\cap(S+d)$ is the disjoint union of $A\cap(A+d)$, of $\{0\}$ when $-d\in A$, and of $\{d\}$ when $d\in A$; neither $0$ nor $d$ lies in $A\cap(A+d)$, because $0\notin A$. As $A=-A$ and $v-2(n+1)=-3$, we get $\NN_{C,C}(d)=\NN_{A,A}(d)-3+2\ind{d\in A}$.

For $(0,0)$ and $(d,1)$, a vertex $(z,0)$ is a common neighbour when $z\in A$ and $d-z\in B$, that is, $z=d-b$ with $b\in B$ and $d-b\in A$; since $A=-A$ the last condition reads $b-d\in A$. A vertex $(z,1)$ is a common neighbour when $z\in B$ and $z-d\in C$. Together, $\lambda=\#\{b\in B:\ b-d\in A\cup C\}=\#\{b\in B:\ b\ne d\}=n-1-\ind{d\in B}$.
\end{proof}

\begin{lemma}\label{lem:mu}
Let $p,q$ be distinct non-adjacent vertices of $\Gamma$. Then $\mu(p,q)=\lambda(p,q)-2$ if both lie in $G\times\{0\}$, $\mu(p,q)=\lambda(p,q)+2$ if both lie in $G\times\{1\}$, and $\mu(p,q)=\lambda(p,q)$ otherwise.
\end{lemma}

\begin{proof}
The neighbourhoods of $p$ and $q$ lie among the other $4n-4$ vertices, and their union has $\deg p+\deg q-\lambda(p,q)$ elements, so $\mu(p,q)=4n-4-\deg p-\deg q+\lambda(p,q)$. Insert the degrees $2n-1$ and $2n-3$.
\end{proof}

\begin{proposition}\label{prop:star}
$\Gamma(A,B)$ contains no $B_{n-1}$ and its complement contains no $B_n$ if and only if
\begin{equation}\label{eq:star}
f(d)\le n+1-3\,\ind{d\in A}\qquad\text{for every } d\in G\setminus\{0\}.
\end{equation}
In that case $R(B_{n-1},B_n)=4n-1$.
\end{proposition}

\begin{proof}
An edge lying in $k$ triangles is the spine of a $B_k$, and a non-edge $pq$ is the spine of a $B_k$ in the complement exactly when $\mu(p,q)\ge k$. So the two conditions say that $\lambda(p,q)\le n-2$ for every edge and $\mu(p,q)\le n-1$ for every pair of distinct non-adjacent vertices. By translation it suffices to take $p=(0,i)$, and we use Lemmas~\ref{lem:lambda} and~\ref{lem:mu}.

For $p=(0,0)$ and $q=(d,0)$ with $d\ne0$, the pair is an edge if and only if $d\in A$. For $d\in A$ the condition is $f(d)\le n-2$, and for $d\notin A$ it is $f(d)-2\le n-1$; together they are~\eqref{eq:star}. For $p=(0,1)$ and $q=(d,1)$, the pair is an edge if and only if $d\in C$, that is, $d\notin A$. For $d\notin A$ the condition is $f(d)-3\le n-2$, and for $d\in A$ it is $(f(d)-1)+2\le n-1$; together they are again~\eqref{eq:star}. For $p=(0,0)$ and $q=(d,1)$ the pair is an edge if and only if $d\in B$; then $\lambda=n-2$, and otherwise $\mu=\lambda=n-1$, so the condition holds with equality in both cases. The pairs $(0,1),(d,0)$ are translates of these. The last sentence follows from~\eqref{eq:RS}, since $\Gamma$ has $4n-2$ vertices.
\end{proof}

\begin{remark}\label{rem:slack}
Summing over $d\ne0$ gives $\sum f(d)=n(n-1)+(n-1)(n-2)=2(n-1)^2$ and $\sum\big(n+1-3\ind{d\in A}\big)=(n+1)(2n-2)-3n=(n-2)(2n+1)$. So when~\eqref{eq:star} holds, the slacks $s(d)=n+1-3\ind{d\in A}-f(d)$ are non-negative integers with sum $n-4$, and~\eqref{eq:star} is an equality for at least $n+2$ of the $2n-2$ values of $d$. For the regular graphs used when $n$ is odd~\cite{Bh,We} the corresponding sum is $n-1$, so the even case is tighter. The cross pairs never matter: $B$ enters only through its autocorrelation $\NN_{B,B}$. Since $f(-d)=f(d)$ and $\ind{-d\in A}=\ind{d\in A}$, it suffices to check~\eqref{eq:star} for $1\le d\le n-1$.
\end{remark}

\section{The case \texorpdfstring{$n=58$}{n = 58}}\label{sec:58}

Let $n=58$ and $v=115=5\cdot23$. Let $A=\pm A_+$ and let $B$ be the sets
\begin{align*}
A_+=\{&3, 10, 11, 14, 16, 18, 21, 23, 24, 25, 26, 27, 28, 29,\\
&30, 32, 36, 37, 38, 43, 44, 46, 47, 48, 49, 52, 54, 55,\\
&56\},\\
B=\{&0, 2, 5, 6, 7, 9, 10, 11, 15, 17, 20, 21, 22, 23,\\
&24, 26, 30, 34, 38, 43, 44, 46, 47, 50, 52, 53, 54, 56,\\
&59, 65, 66, 67, 69, 71, 72, 73, 74, 75, 80, 81, 83, 84,\\
&85, 88, 90, 92, 93, 97, 100, 102, 105, 106, 107, 109, 110, 111,\\
&114\}.
\end{align*}
The $29$ elements of $A_+$ lie in $\{1,\dots,57\}$, so $A=A_+\cup(-A_+)$ has $58$ elements, avoids $0$ and satisfies $A=-A$; and $|B|=57$. Written out,
\begin{align*}
A=\{&3, 10, 11, 14, 16, 18, 21, 23, 24, 25, 26, 27, 28, 29,\\
&30, 32, 36, 37, 38, 43, 44, 46, 47, 48, 49, 52, 54, 55,\\
&56, 59, 60, 61, 63, 66, 67, 68, 69, 71, 72, 77, 78, 79,\\
&83, 85, 86, 87, 88, 89, 90, 91, 92, 94, 97, 99, 101, 104,\\
&105, 112\}.
\end{align*}
Figure~\ref{fig:torus} draws both sets on the torus $\Z_5\times\Z_{23}$.

\begin{figure}[t]
\centering
\includegraphics[width=\textwidth]{figures/torus.png}
\caption{The sets $A$ and $B$ for $n=58$. The residue $x\in\Z_{115}$ sits at angle $2\pi x/5$ around the tube and $2\pi x/23$ around the axis.}
\label{fig:torus}
\end{figure}

\begin{table}[p]
\centering
\footnotesize
\setlength{\tabcolsep}{4.2pt}
\caption{The inequality~\eqref{eq:star} for $n=58$ at $1\le d\le57$, with $b(d)=59-3\ind{d\in A}$ and the slack $s(d)=b(d)-f(d)$ of Remark~\ref{rem:slack}. The slacks add up to $27$, half of $n-4=54$.}
\label{tab:main}
\begin{tabular}{r c r r r r r c r c r r r r r}
\toprule
$d$ & $\ind{d\in A}$ & $\NN_{A,A}$ & $\NN_{B,B}$ & $f$ & $b$ & $s$ & & $d$ & $\ind{d\in A}$ & $\NN_{A,A}$ & $\NN_{B,B}$ & $f$ & $b$ & $s$\\
\cmidrule{1-7}\cmidrule{9-15}
1 & 0 & 32 & 27 & 59 & 59 & 0 && 30 & 1 & 26 & 28 & 54 & 56 & 2\\
2 & 0 & 32 & 27 & 59 & 59 & 0 && 31 & 0 & 30 & 29 & 59 & 59 & 0\\
3 & 1 & 29 & 27 & 56 & 56 & 0 && 32 & 1 & 27 & 28 & 55 & 56 & 1\\
4 & 0 & 28 & 30 & 58 & 59 & 1 && 33 & 0 & 30 & 28 & 58 & 59 & 1\\
5 & 0 & 31 & 27 & 58 & 59 & 1 && 34 & 0 & 30 & 27 & 57 & 59 & 2\\
6 & 0 & 31 & 27 & 58 & 59 & 1 && 35 & 0 & 30 & 27 & 57 & 59 & 2\\
7 & 0 & 33 & 26 & 59 & 59 & 0 && 36 & 1 & 27 & 29 & 56 & 56 & 0\\
8 & 0 & 32 & 27 & 59 & 59 & 0 && 37 & 1 & 26 & 29 & 55 & 56 & 1\\
9 & 0 & 28 & 31 & 59 & 59 & 0 && 38 & 1 & 28 & 27 & 55 & 56 & 1\\
10 & 1 & 26 & 29 & 55 & 56 & 1 && 39 & 0 & 31 & 28 & 59 & 59 & 0\\
11 & 1 & 35 & 21 & 56 & 56 & 0 && 40 & 0 & 32 & 26 & 58 & 59 & 1\\
12 & 0 & 30 & 29 & 59 & 59 & 0 && 41 & 0 & 31 & 28 & 59 & 59 & 0\\
13 & 0 & 30 & 29 & 59 & 59 & 0 && 42 & 0 & 31 & 27 & 58 & 59 & 1\\
14 & 1 & 30 & 26 & 56 & 56 & 0 && 43 & 1 & 29 & 27 & 56 & 56 & 0\\
15 & 0 & 26 & 32 & 58 & 59 & 1 && 44 & 1 & 26 & 28 & 54 & 56 & 2\\
16 & 1 & 28 & 27 & 55 & 56 & 1 && 45 & 0 & 30 & 28 & 58 & 59 & 1\\
17 & 0 & 29 & 30 & 59 & 59 & 0 && 46 & 1 & 29 & 27 & 56 & 56 & 0\\
18 & 1 & 30 & 26 & 56 & 56 & 0 && 47 & 1 & 26 & 29 & 55 & 56 & 1\\
19 & 0 & 29 & 30 & 59 & 59 & 0 && 48 & 1 & 27 & 28 & 55 & 56 & 1\\
20 & 0 & 31 & 26 & 57 & 59 & 2 && 49 & 1 & 28 & 28 & 56 & 56 & 0\\
21 & 1 & 25 & 31 & 56 & 56 & 0 && 50 & 0 & 29 & 30 & 59 & 59 & 0\\
22 & 0 & 31 & 28 & 59 & 59 & 0 && 51 & 0 & 29 & 30 & 59 & 59 & 0\\
23 & 1 & 29 & 27 & 56 & 56 & 0 && 52 & 1 & 27 & 29 & 56 & 56 & 0\\
24 & 1 & 28 & 28 & 56 & 56 & 0 && 53 & 0 & 32 & 27 & 59 & 59 & 0\\
25 & 1 & 28 & 28 & 56 & 56 & 0 && 54 & 1 & 27 & 29 & 56 & 56 & 0\\
26 & 1 & 28 & 27 & 55 & 56 & 1 && 55 & 1 & 27 & 28 & 55 & 56 & 1\\
27 & 1 & 27 & 29 & 56 & 56 & 0 && 56 & 1 & 27 & 29 & 56 & 56 & 0\\
28 & 1 & 27 & 29 & 56 & 56 & 0 && 57 & 0 & 31 & 28 & 59 & 59 & 0\\
29 & 1 & 27 & 29 & 56 & 56 & 0 &&  & & & & & & \\
\bottomrule
\end{tabular}
\end{table}

\begin{theorem}\label{thm:main}
The sets $A$ and $B$ above satisfy~\eqref{eq:star} for $n=58$. Consequently $\Gamma(A,B)$ is a graph on $230$ vertices, with degrees $115$ and $113$, that contains no $B_{57}$ and whose complement contains no $B_{58}$, and $R(B_{57},B_{58})=231$.
\end{theorem}

\begin{proof}
By Remark~\ref{rem:slack} it suffices to check~\eqref{eq:star} for $1\le d\le57$, and Table~\ref{tab:main} lists $\NN_{A,A}(d)$ and $\NN_{B,B}(d)$ there; each entry is the number of elements of $A$, or of $B$, whose translate by $d$ stays in the set.\footnote{For instance $\NN_{A,A}(1)=32$ is the number of pairs $a,a+1$ of consecutive residues in $A$: the runs $10,11$; $23,\dots,30$; $36,37,38$; $43,44$; $46,\dots,49$; $54,55,56$; $59,60,61$; $66,\dots,69$; $71,72$; $77,78,79$; $85,\dots,92$; $104,105$ contribute $1+7+2+1+3+2+2+3+1+2+7+1=32$, and neither $0$ nor $114$ lies in $A$.} In every row $f(d)\le59-3\ind{d\in A}$. Proposition~\ref{prop:star} gives the rest.
\end{proof}

\begin{figure}[t]
\centering
\includegraphics[width=0.92\textwidth]{figures/slack.png}
\caption{The slack $s(d)=59-3\ind{d\in A}-f(d)$ for $n=58$ at every $d\ne0$, with $d$ placed by its residues modulo $5$ and $23$; the black cell is $d=0$, and a dot marks $d\in A$. Equality holds for $70$ values of $d$, the slack is $1$ for $34$ and $2$ for $10$.}
\label{fig:slack}
\end{figure}

\section{Remarks}\label{sec:remarks}

\begin{remark}
The function $f$ is not constant on $A$ or on $C$ (Figure~\ref{fig:slack}): on the $58$ elements of $A$ it takes the values $54$, $55$ and $56$, for $4$, $18$ and $36$ of them, and on the $56$ elements of $C$ the values $57$, $58$ and $59$, for $6$, $16$ and $34$ of them. So the example is not a difference-set construction of the kind behind the infinite families in~\cite{Sz,Tu,We}.
\end{remark}

\begin{remark}
Unlike the graph for $n=67$ in~\cite{Bh}, this one has no multiplier symmetry beyond $d\mapsto-d$: the only units $u$ of $\Z_{115}$ with $uA=A$ are $\pm1$, the only one with $uB=B$ is $1$, and $B$ meets $-B$ in $29$ elements. So the $57$ rows of Table~\ref{tab:main} cannot be grouped into orbits as in~\cite{Bh}.
\end{remark}

\begin{remark}
We found the example by a computer search following the two-phase scheme of Turturean~\cite{Tu}. The first phase chooses $A$ so that the bounds $\NN_{B,B}(d)\le n+1-3\ind{d\in A}-\NN_{A,A}(d)$ leave a large margin in Fourier space, where they become linear conditions on the power spectrum of $B$. The second phase fixes $A$ and searches for $B$ by local search over exchanges of one element of $B$ with one outside it, guided by weights on the inequalities~\eqref{eq:star}; when no exchange lowers the weighted excess, the weights of the violated inequalities are raised, as in the breakout method of Morris~\cite{Mo}. No step of the proof above depends on the search.
\end{remark}

\begin{question}
The smallest open case of $R(B_{n-1},B_n)=4n-1$ is now $n=59$, with $v=117$; there $n$ is odd, and the regular two-block circulant graphs of~\cite{Bh,We} are available. Is there a two-block circulant graph over $\Z_{117}$, or over $\Z_3^2\times\Z_{13}$, that settles it?
\end{question}

\section*{Formal verification}
Theorem~\ref{thm:main} and the equality $R(B_{57},B_{58})=231$ are also proved in Lean~4~\cite{MU}, without any library beyond the Lean core. For the lower bound the development defines the graph $\Gamma(A,B)$ on the vertices $0,\dots,229$ by the two bit masks of $A$ and $B$, and the kernel evaluates the common neighbours and common non-neighbours of all $26{,}335$ pairs of vertices and checks that the adjacency is symmetric: every edge has at most $56$ common neighbours and every pair of distinct non-adjacent vertices has at most $57$ common non-neighbours. For the upper bound~\eqref{eq:RS} the development proves Goodman's count in the form used by Rousseau and Sheehan: if every edge of a graph on $N$ vertices lies in at most $a$ triangles and every non-edge has at most $b$ common non-neighbours, then counting the ordered triples of distinct vertices by the colours of their three pairs gives
\[
N(N-1)(N-2)\le\sum_{w}\bigl(3\,d(w)\,\bar d(w)+a\,d(w)+b\,\bar d(w)\bigr),
\]
where $d(w)$ is the degree of $w$ and $\bar d(w)=N-1-d(w)$. For $N=231$, $a=56$ and $b=57$ each term on the right is at most $230\cdot229$, with equality only when $d(w)=115$, and then the degrees would add up to the odd number $231\cdot115$. The final theorem states that $231$ is the least $N$ such that every graph on $N$ vertices contains $B_{57}$ or has a complement containing $B_{58}$; it depends only on the standard axioms \texttt{propext}, \texttt{Classical.choice} and \texttt{Quot.sound}. The graph is also checked by two independent programs, in Python and in C, which read the sets $A$ and $B$ and test all pairs directly. The development, the programs and the search are archived in~\cite{Ar}.

\section*{Declaration of competing interest}
The author has no competing interests to declare.

\section*{Funding}
This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

\section*{Data availability}
The data, the Lean~4 development, the C and Python checks and the search are release v2.1.0 of the repository \url{https://github.com/creelie/Breakthrough-problems-2}, \url{https://github.com/creelie/Breakthrough-problems-2/releases/tag/v2.1.0}.

\section*{Declaration of generative AI and AI-assisted technologies in the manuscript preparation process}
During the preparation of this work the author used Claude (Anthropic) to search for the example, to run the computer checks and the Lean formalization, and to draft the text. After using this tool, the author reviewed and edited the content as needed and takes full responsibility for the content of the published article.

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