\documentclass[11pt]{amsart} \usepackage[margin=1.1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs} \usepackage[hidelinks]{hyperref} \usepackage{xcolor} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\cR}{\mathcal{R}} \newcommand{\eps}{\varepsilon} \newcommand{\AUTHOR}{Ella Agnes Puerto} \title[Iterating the GNNW bound]{Iterating the Gupta--Ndiaye--Norin--Wei method:\\ a computer-verified bound $R(k)\le 3.7698^{\,k+o(k)}$} \author{\AUTHOR} \address{Independent researcher} \email{eagnespuerto@gmail.com} \date{October 2026. Preprint DOI: \href{https://doi.org/10.5281/zenodo.23885218}{10.5281/zenodo.23885218}. This preprint is licensed under CC-BY-4.0; the accompanying code under the MIT license} \begin{document} \begin{abstract} Gupta, Ndiaye, Norin and Wei (GNNW) recast the Campos--Griffiths--Morris--Sahasrabudhe method as an inductive theorem that turns a known bound on off-diagonal Ramsey numbers into a better one. With two applications they obtained $R(k)\le (3.7992\ldots)^{k+o(k)}$. They also reported an unverified, AI-suggested third application giving $3.78233\ldots$. We make three observations about their framework. (i) The free function $M$ can be optimised pointwise. (ii) For a fixed input, the best output exponent is the least solution of an explicit first-order differential inequality. (iii) GNNW's Theorem~14 needs only $F\in C^1$ with locally Lipschitz derivative. We then construct an explicit chain of eight applications and verify every hypothesis by interval arithmetic on the whole range $0<\lambda\le 1$, with a hand-proved estimate covering $\lambda\to0$. The verification was done twice, with independent code, libraries and enclosure strategies. This proves $R(k)\le 3.76973^{\,k+o(k)}$, and in particular confirms GNNW's $3.78233$. Numerically, iterating the optimal step converges to about $3.7690$, which suggests this is close to the limit of the GNNW method in its present form. \end{abstract} \maketitle \section{Introduction} The diagonal Ramsey number $R(k)$ is the least $n$ such that every red/blue colouring of the edges of $K_n$ contains a monochromatic $K_k$. Erd\H{o}s and Szekeres~\cite{ES35} proved $R(k)\le 4^{k}$, and Erd\H{o}s~\cite{Erd47} proved $R(k)\ge \sqrt2^{\,k}$. Whether $\lim_k R(k)^{1/k}$ exists, and what it is, is Erd\H{o}s Problem~\#77~\cite{EP77}. Campos, Griffiths, Morris and Sahasrabudhe~\cite{CGMS} gave the first exponential improvement of the upper bound, $R(k)\le(4-\eps)^k$. Gupta, Ndiaye, Norin and Wei~\cite{GNNW} simplified the argument to a single inductive statement (their Theorem~14) and optimised it, obtaining \[ R(k)\le \bigl(4e^{-0.14/e}\bigr)^{k+o(k)}=(3.7992\ldots)^{k+o(k)}. \] In their Remark~17 they record a further, unverified, AI-generated iteration that would give $(3.78233\ldots)^{k+o(k)}$. Balister et al.~\cite{BBCGHMST} gave another short proof of an exponential improvement, extended to many colours. \begin{theorem}\label{thm:main} $R(k)\le 3.76973^{\,k+o(k)}$. More precisely, there is an explicit function $F_8$, given by the node data in~\S\ref{sec:data}, with $e^{F_8(1)}\le 3.7697215$ and $R(k,\ell)\le e^{F_8(\ell/k)k+o(k)}$ for all $k\ge \ell$. \end{theorem} \begin{theorem}\label{thm:ai} The bound of GNNW's Remark~17 holds: $R(k)\le(3.782329)^{k+o(k)}$. \end{theorem} Theorem~\ref{thm:ai} is superseded by Theorem~\ref{thm:main}, but it settles the status of the bound recorded in~\cite[Remark~17]{GNNW}. \smallskip \noindent\textbf{Nature of the proof.} Both theorems follow from GNNW's Theorem~14 and Lemma~15, which we use as stated in~\cite{GNNW} (see \S\ref{sec:gnnw} for the precise hypotheses and two small clarifications). The new input is a choice of functions, together with a verification that they satisfy the hypotheses. The verification consists of one-variable inequalities between elementary functions on $(0,1]$. We prove them on $[10^{-7},1]$ by interval arithmetic, and on $(0,10^{-7}]$ by Lemma~\ref{lem:tail}, a hand-proved estimate whose constants are again evaluated in interval arithmetic. The verification was carried out by two programs that share no code (\S\ref{sec:verify}). \smallskip \noindent\textbf{The limit of the method.} In \S\ref{sec:limit} we explain why the outcome of one GNNW step, for a given input, is governed by an ODE. We report that iterating the optimal step from the Erd\H{o}s--Szekeres bound converges numerically to about $3.76897$. This part is numerical evidence, not a theorem. It is consistent with GNNW's expectation that going below $3.75$ needs new ideas. \section{The GNNW framework}\label{sec:gnnw} Write $\lambda=\ell/k\in(0,1]$ and $h(\lambda)=(1+\lambda)\log(1+\lambda)-\lambda\log\lambda$, so that $\log\binom{k+\ell}{\ell}=h(\ell/k)k+o(k)$. Following~\cite{GNNW}, let $\cR_0$ be the set of $(x,y)\in(0,1)^2$ such that $R(k,\ell)\le x^{-k}y^{-\ell}$ whenever $k+\ell$ is large. Let $\cR$ be its closure and $\cR_*$ its interior. We use the following facts from \cite[Observation~10]{GNNW}: $(x,1-x)\in\cR$ for $00}$ be smooth, and let $M,X,Y:(0,1]\to(0,1)$ with $M$ continuous. Suppose that for all $0<\lambda\le1$ we have $F'(\lambda)>0$, \[ X(\lambda)=\bigl(1-e^{-F'(\lambda)}\bigr)^{\frac1{1-M(\lambda)}}\bigl(1-M(\lambda)\bigr),\qquad (X(\lambda),Y(\lambda))\in\cR, \] \begin{equation}\label{eq:phi} \Phi(\lambda):=F(\lambda)+\tfrac12\bigl(\log X(\lambda)+\lambda\log M(\lambda) +\lambda\log Y(\lambda)\bigr)>0 . \end{equation} Then $R(k,\ell)\le e^{F(\ell/k)k+o(k)}$ for all $k\ge\ell$. \end{theorem} \begin{lemma}[{\cite[Lemma~15]{GNNW}}]\label{lem:15} Let $f:(0,1]\to\mathbb{R}$ be differentiable and strictly concave with $f'>0$, and suppose $R(k,\ell)\le e^{f(\ell/k)k+o(k)}$ for $1\le\ell\le k$. Put $A(t)=e^{-f'(t)}$ and $B(t)=e^{tf'(t)-f(t)}$. Suppose $Ab$). Then $(x,Y_f(x))\in\cR$ for all $00$. \end{proposition} \begin{proof} In the proof in~\cite{GNNW}, the smoothness of $F$ is used in three places, and nowhere else; Theorem~13 and Lemma~12 there do not involve $F$. \begin{enumerate} \item Claim~14.1 uses the uniform continuity of $F$ and $F'$ on $[\eps',1]$, which holds for $F\in C^1$. \item The induction step uses the mean value theorem to write $k\bigl(F(\ell/k)-F((\ell-1)/k)\bigr)=F'(\xi)$ with $\xi\in[(\ell-1)/k,\ell/k]$. This needs only differentiability. \item It then bounds $|F'(\ell/k)-F'(\xi)|\le \frac1k\max_{[\eps'/2,1]}|F''|$. The bound $|F'(\ell/k)-F'(\xi)|\le \mathrm{Lip}(F'|_{[\eps'/2,1]})/k$ works equally well. \qedhere \end{enumerate} \end{proof} \begin{remark}[On $\cR$ versus $\cR_*$]\label{rem:interior} Claim~14.1 of~\cite{GNNW} applies their Theorem~13, which requires $(x_\lambda,y_\lambda)\in\cR_*$, to $x_\lambda=e^{-\delta}p_\lambda^{1/(1-M(\lambda))}(1-M(\lambda))$ and $y_\lambda=e^{-\delta}Y(\lambda)$, where $p_\lambda=1-e^{-F'(\lambda)+2\delta}$. This membership is not spelled out there, but it is immediate. Since $\delta>0$ we have $p_\lambda<1-e^{-F'(\lambda)}$, so $x_\lambda0$, and $(1+t)f'(t)-f(t)>0$ on $(0,1]$. Then $f$ satisfies the analytic hypotheses of Lemma~\ref{lem:15}. \end{lemma} \begin{proof} $f'$ is continuous and strictly decreasing on each piece, hence on $(0,1]$. So $f$ is strictly concave, and $A=e^{-f'}$ is strictly increasing. The function $\log B=tf'-f$ is continuous and piecewise $C^1$ with derivative $tf''<0$, so $B$ is strictly decreasing. $A0$. As $t\to0$ we have $f'(t)=\log\frac{1+t}{t}+g'(t)\to\infty$, so $A\to0$. Also $tf'-f=-\log(1+t)+tg'(t)-g(t)\to0$, because $g'$ is bounded and $g(0)=0$; so $B\to1$. \end{proof} \section{Choosing the functions}\label{sec:choose} \subsection{$M$ is pointwise} In~\eqref{eq:phi}, $\Phi(\lambda)$ depends on $M$ only through the single number $M(\lambda)$. So for given $F$ and $Y$-curve, the best $M$ is found by maximising, for each $\lambda$, the one-variable function \[ m\ \longmapsto\ \log X+\lambda\log m+\lambda\log Y(X),\qquad X=(1-e^{-F'(\lambda)})^{1/(1-m)}(1-m). \] Theorem~\ref{thm:14} only needs $M$ to be continuous, so we take $M$ to be a shape-preserving interpolant of the pointwise maximisers (\S\ref{sec:data}). \subsection{The best exponent solves an ODE}\label{sec:ode} Fix the $Y$-curve. Let $Q(\lambda,s)$ be the maximum above with $F'(\lambda)$ replaced by $s$, and let $S(\lambda,v)=\inf\{s:-\tfrac12 Q(\lambda,s)0$ on $(0,1]$; that $00$; that $F>0$, which follows from $\Phi>0$ because every logarithm in~\eqref{eq:phi} is negative; and that the predecessor's hypotheses for Lemma~\ref{lem:15pw} hold on $(0,1]$. \subsection{Enclosures} All quantities are enclosed in intervals with outward rounding. The $Y$-curve of the predecessor $f$ is decreasing in $x$, so on a cell it suffices to bound $Y$ below at $\sup X$. For a point $x$ in the branch $xb$), we isolate the parameter $t$ with $A(t)=x$ (respectively $B(t)=x$) by certified bisection, using the monotonicity from Lemma~\ref{lem:15pw}. We then evaluate $B$ (respectively $A$) at the appropriate endpoint. At a junction we move to a certified point of the neighbouring branch, which is valid because $Y_f$ is decreasing. Concretely, near the junction at $b$ the hyperbola $e^{-f(1)}/x$ is an upper bound for $Y_f$ when $x>b$, so it must not be used there. \subsection{Two independent programs} \begin{enumerate} \item \emph{Mean-value certifier} (\texttt{certify.py}, \texttt{mpmath.iv}). On each cell $I$ with midpoint $c$ it proves $\Phi(I)\subseteq \Phi(c)+\Phi'(I)(I-c)$, using closed forms for $\Phi'$. These include $d\log Y/d\log X\in\{-t,-1,-1/t\}$ on the three branches. The bound is valid because $\Phi$ is continuous and piecewise $C^1$. \item \emph{Direct auditor} (\texttt{audit\_arb.py}, Arb~\cite{Arb} via python-flint). It shares no code with the first program, and evaluates $\Phi/\lambda$ directly, with no derivative, in the exactly equivalent form \[ \frac{\Phi}{\lambda}=\Bigl[\tfrac{(1+\lambda)\log(1+\lambda)}{\lambda}+\tfrac{g}{\lambda}\Bigr] +\tfrac12\Bigl[r(M)\tfrac{M}{\lambda}+\tfrac{r(z)\,z/\lambda}{1-M}\Bigr] +\tfrac12\log\tfrac{M}{\lambda}+\tfrac12\log\tfrac{Y}{\lambda}, \] where $z=e^{-F'}=\frac{\lambda}{1+\lambda}e^{-g'}$ and $r(y)=\log(1-y)/y$. In this form the $\log(1/\lambda)$ terms cancel algebraically rather than numerically. \end{enumerate} Both programs bisect cells adaptively on $[10^{-7},1]$, starting from geometric grids. Both also verify the hypotheses of Lemma~\ref{lem:15pw} for every predecessor on all of $(0,1]$: by cells on $[t_0,1]$, and on $(0,t_0]$ by the explicit bounds $f''\le 1-1/t_0+\sup g''$, $f'\ge\log(1/t_0)+\inf g'$ and $(1+t)f'-f\ge f'-f\ge \log(1/t_0)+\inf g'-h(t_0)-\sup g$. Here $t_0=10^{-6}$ for the first program and $10^{-7}$ for the second. As a control, both programs reject several known-false steps. \subsection{Small $\lambda$} \begin{lemma}\label{lem:tail} Let $0<\lambda_t\le10^{-3}$. Suppose that on $[0,\lambda_t]$ we have $F=h+g$ with $g(0)=0$, $a=g'(0)$ and $|g''|\le G_2$, and that $M(\lambda)=c(\lambda)\lambda$ with $c(\lambda)\in[c_-,c_+]$. Put $E=e^{-a}$, $m_+=c_+\lambda_t$, $\bar z=Ee^{G_2\lambda_t}$, $k_-=c_-+Ee^{-G_2\lambda_t}/(1+\lambda_t)$ and $u_{\max}=\lambda_t(c_++\bar z)$. Assume $\lambda_t\bar z<1$ and $m_+<1$. Then for all $\lambda\in(0,\lambda_t]$, \[ \frac{\Phi(\lambda)}{\lambda}\ \ge\ 1+a-\frac{G_2\lambda_t}{2} +\frac12\Bigl[-\frac{c_+}{1-m_+}-\frac{\bar z}{(1-\lambda_t\bar z)(1-m_+)}+\log c_-+\log k_-+T_Y\Bigr]. \] Here $T_Y=\log(1-u_{\max}/2)$ when $Y=1-X$. When $Y=Y_f(X)$ for a predecessor $f=h+g_p$ with $g_p(0)=0$, $b=g_p'(0)$, $|g_p''|\le B_2$ on $[0,u_{\max}]$ and $u_{\max}<-\log B(1)$, we have $T_Y=-b-(1+\tfrac32B_2)u_{\max}$. \end{lemma} \begin{proof} We bound the terms of $\Phi/\lambda$ one at a time. \begin{itemize} \item \emph{The $F$ term.} Since $(1+\lambda)\log(1+\lambda)\ge\lambda$ and $g(\lambda)\ge a\lambda-G_2\lambda^2/2$, we have $F/\lambda\ge\log(1/\lambda)+1+a-G_2\lambda_t/2$. \item \emph{The quantity $z$.} From $F'=\log\frac{1+\lambda}{\lambda}+g'$ and $|g'-a|\le G_2\lambda$ we get $z=e^{-F'}\in\bigl[\lambda Ee^{-G_2\lambda}/(1+\lambda),\ \lambda\bar z\bigr]$. \item \emph{The $X$ term.} Using $\log(1-y)\ge -y/(1-y)$ in $\log X=\log(1-M)+\log(1-z)/(1-M)$ gives the stated lower bound for $\log X/\lambda$. \item \emph{A lower bound for $u$.} Using $\log(1-y)\le-y$ instead gives $u:=-\log X\ge M+z\ge\lambda k_-=:u_-$, and also $u_-\le u_{\max}$. \item \emph{The $Y$ term, Erd\H{o}s--Szekeres case.} Here $\log Y=\log(1-e^{-u})\ge\log u_-+\log(1-u_-/2)$. \item \emph{The $Y$ term, Lemma~\ref{lem:15} case.} The point $x=e^{-u}$ lies in the far branch $x>b$, where $Y=A(t)$ and $u=\varphi(t):=f(t)-tf'(t)=\log(1+t)+g_p(t)-tg_p'(t)$. The function $\varphi$ is increasing and satisfies $\varphi(t)\le t+B_2t^2/2$. Put $t_{\rm low}=u_-/(1+B_2u_-/2)\le u_-$. Then $\varphi(t_{\rm low})\le u_-\le u$, so $t\ge t_{\rm low}$. Since $f'$ is decreasing, $\log Y=-f'(t)\ge-f'(t_{\rm low})\ge-t_{\rm low}-\log(1/t_{\rm low})-b-B_2t_{\rm low}$. Combining this with $\log(1/t_{\rm low})\le-\log\lambda-\log k_-+B_2u_-/2$ gives $\log Y\ge\log\lambda+\log k_- +T_Y$. \item \emph{The $M$ term.} $\tfrac12\log M\ge\tfrac12\log c_-+\tfrac12\log\lambda$. \end{itemize} Adding the bounds, the terms $\log(1/\lambda)+\tfrac12\log\lambda+\tfrac12\log\lambda$ cancel. \end{proof} For $\lambda_t=10^{-7}$, both programs evaluate this bound in interval arithmetic for every step. On GNNW's own steps the bound gives $0.0209565$ and $0.145956$, just below the limits $0.0209569\ldots$ and $0.1459569\ldots$ reported in~\cite[Appendix~A]{GNNW}. \subsection{Results} Table~\ref{tab:results} lists the outcome. Every step is verified on all of $(0,1]$ by both programs. \begin{table}[h] \centering\small \begin{tabular}{lcccc} \toprule step & $e^{F(1)}\le$ & min.\ cert.\ $\Phi/\lambda$ on $[10^{-6},1]$ & tail bound on $(0,10^{-7}]$ & Arb cells\\ \midrule GNNW step 0 & 3.8697319 & $2.5\cdot10^{-3}$ & $2.10\cdot10^{-2}$ & 20\,756\\ GNNW step 1 & 3.7992028 & $2.9\cdot10^{-4}$ & $1.46\cdot10^{-1}$ & 26\,745\\ Remark 17 ($F_{AI}$) & 3.7823288 & $3.3\cdot10^{-7}$ & $6.20\cdot10^{-5}$ & 832\,915\\ link 1 & 3.8518222 & $4.7\cdot10^{-5}$ & $5.83\cdot10^{-4}$ & 411\,386\\ link 2 & 3.7837987 & $2.6\cdot10^{-6}$ & $6.22\cdot10^{-4}$ & 404\,848\\ link 3 & 3.7726166 & $1.8\cdot10^{-6}$ & $8.30\cdot10^{-4}$ & 270\,513\\ link 4 & 3.7703296 & $1.0\cdot10^{-4}$ & $1.18\cdot10^{-3}$ & 168\,875\\ link 5 & 3.7698513 & $1.1\cdot10^{-4}$ & $1.09\cdot10^{-3}$ & 168\,514\\ link 6 & 3.7697479 & $1.1\cdot10^{-4}$ & $1.09\cdot10^{-3}$ & 168\,760\\ link 7 & 3.7697261 & $1.1\cdot10^{-4}$ & $1.10\cdot10^{-3}$ & 168\,845\\ link 8 & \textbf{3.7697215} & $1.1\cdot10^{-4}$ & $1.10\cdot10^{-3}$ & 168\,849\\ \bottomrule \end{tabular} \caption{Verified steps. The third column is the mean-value certifier's worst certified lower bound, which is conservative. The tail column is Lemma~\ref{lem:tail} with $\lambda_t=10^{-7}$. The gap $[10^{-7},10^{-6}]$ is covered by separate cell runs.} \label{tab:results} \end{table} \begin{proof}[Proof of Theorems~\ref{thm:main} and~\ref{thm:ai}] For $n=1,\dots,8$, link $n$ satisfies the hypotheses of Theorem~\ref{thm:14} in the form of Proposition~\ref{prop:reg}. Here $F_n$ is $C^1$, and $F_n'$ is Lipschitz on $[\eta,1]$ because $g_n''$ is bounded. The required $(X,Y)\in\cR$ comes from Observation~10(1) for $n=1$. For $n\ge2$ it comes from Lemma~\ref{lem:15}, applied to $F_{n-1}$ using Lemma~\ref{lem:15pw} and the conclusion for link $n-1$. Theorem~\ref{thm:14} for link~8 gives $R(k,\ell)\le e^{F_8(\ell/k)k+o(k)}$, and $\ell=k$ gives Theorem~\ref{thm:main}. Theorem~\ref{thm:ai} follows in the same way from GNNW's steps 0 and 1 followed by the $F_{AI}$ step. \end{proof} \section{The limit of the method}\label{sec:limit} Iterating the optimal step of \S\ref{sec:ode}, starting from $Y=1-X$, produces the following values of $e^{F(1)}$: $3.85163$, $3.78361$, $3.77224$, $3.76958$, $3.76910$, $3.76899$, $3.76897$, $3.76897$. The explicit links of Table~\ref{tab:results} trail these values by about $7.5\cdot10^{-4}$, because of the safety shifts. As $\lambda\to0$ the optimal step has a closed form. If $g'(0)=a$ for the new exponent and $b$ for the previous one, then the best $M\sim c\lambda$ has $1/c+1/(c+e^{-a})=1$, and \[ \lim_{\lambda\to0}\Phi/\lambda=1+a-\tfrac12(c+e^{-a})+\tfrac12\log c+\tfrac12\bigl(\log(c+e^{-a})-b\bigr). \] For GNNW's $F_{AI}$ ($a=-0.3864$, $b=-0.25$) this limit equals $6.31\cdot10^{-5}$, so that function already sits close to the edge of what one step allows near $0$. These numbers are not proved. They come from a discretised ODE, and the iteration's start value near $\lambda=0$ is taken from a proved bound rather than optimised. We nevertheless expect the true limit of repeated applications of Theorem~\ref{thm:14} and Lemma~\ref{lem:15} to be close to $3.769$. Going substantially further presumably needs a change to the framework itself, for instance a better analogue of Lemma~\ref{lem:15} or a stronger inductive statement. \section{Limitations} \begin{itemize} \item The argument relies on the correctness of~\cite[Theorem~14 and Lemma~15]{GNNW}. We have checked how their hypotheses are used (\S\ref{sec:gnnw}), but we have not re-verified the whole of their proof. \item The two verification programs are independent of each other. Neither has been formally verified, and human review of the code is welcome. \item The improvement over~\cite{GNNW} is a refinement of their method, which they anticipated could be pushed further. \end{itemize} \section*{Acknowledgements and use of AI} This work was carried out by the author with extensive use of the AI assistant Claude (Anthropic), through Claude Code. The assistant wrote the software, ran the computations and drafted this manuscript under the author's direction. The author is responsible for the content. GNNW's Remark~17 likewise reports an AI-generated candidate, and Theorem~\ref{thm:ai} confirms it. \appendix \section{Reproducibility}\label{app:repro} The code, data and Lean files are deposited with this preprint. To reproduce, install the package with \texttt{pip install -e .[dev,audit]} and run \begin{center}\ttfamily python scripts/certify\_all.py\\ python scripts/certify\_tails.py\\ python scripts/audit\_all.py \end{center} The tests run with \texttt{pytest}. Software: Python~3.12, mpmath~1.3.0, python-flint~0.9.0, SciPy~1.18.0, NumPy~2.4.6, Lean~4.35.0 with Mathlib~v4.35.0. SHA-256 checksums of the certified data files (LF line endings) are listed in \texttt{data/SHA256SUMS}. \begin{thebibliography}{99} \bibitem{Arb} F.~Johansson, \emph{Arb: efficient arbitrary-precision midpoint-radius interval arithmetic}, IEEE Trans.\ Comput.\ \textbf{66} (2017), 1281--1292. \bibitem{BBCGHMST} P.~Balister, B.~Bollob\'as, M.~Campos, S.~Griffiths, E.~Hurley, R.~Morris, J.~Sahasrabudhe, M.~Tiba, \emph{Upper bounds for multicolour Ramsey numbers}, J.\ Amer.\ Math.\ Soc.\ \textbf{39} (2026), 765--780; arXiv:2410.17197. \bibitem{CGMS} M.~Campos, S.~Griffiths, R.~Morris, J.~Sahasrabudhe, \emph{An exponential improvement for diagonal Ramsey}, Ann.\ of Math.\ \textbf{203} (2026), no.~3; arXiv:2303.09521. \bibitem{EP77} T.~F.~Bloom, \emph{Erd\H{o}s Problem \#77}, \url{https://www.erdosproblems.com/77}. \bibitem{Erd47} P.~Erd\H{o}s, \emph{Some remarks on the theory of graphs}, Bull.\ Amer.\ Math.\ Soc.\ \textbf{53} (1947), 292--294. \bibitem{ES35} P.~Erd\H{o}s, G.~Szekeres, \emph{A combinatorial problem in geometry}, Compositio Math.\ \textbf{2} (1935), 463--470. \bibitem{GNNW} P.~Gupta, N.~Ndiaye, S.~Norin, L.~Wei, \emph{Optimizing the CGMS upper bound on Ramsey numbers}, arXiv:2407.19026v2. \end{thebibliography} \end{document}