Erdős Problem #166
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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
The requested lower bound is $R(4,k)\gg k^3/(\log k)^{O(1)}$. We establish the known stronger specified estimate \[ R(4,k)\gg\frac{k^3}{(\log k)^4}. \] Here $R(4,k)$ is the least order forcing a $K_4$ or an independent $k$-set. Mattheus and Verstraete proved this bound. We reconstruct the sampling and alteration step, including its quantifier for every large $k$, from their finite-geometric graph and independent-set estimate.
Restatement from Math-5471-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-5471
Erdős Problem 166: the off-diagonal Ramsey number $R(4,k)$
1) Statement and scope
The requested lower bound is $R(4,k)\gg k^3/(\log k)^{O(1)}$. We establish the known stronger specified estimate \[ R(4,k)\gg\frac{k^3}{(\log k)^4}. \] Here $R(4,k)$ is the least order forcing a $K_4$ or an independent $k$-set. Mattheus and Verstraete proved this bound. We reconstruct the sampling and alteration step, including its quantifier for every large $k$, from their finite-geometric graph and independent-set estimate.2) The external construction input
The construction and counting argument in Section 4.2 of the cited paper have the following consequence, with absolute constants $A,B>0$. For every sufficiently large prime power $q$ there is a $K_4$-free graph $G_q$ on at least $q^4/2$ vertices such that, for \[ s=s(q)=\lceil Aq(\log q)^2\rceil, \] the number of independent $s$-subsets is at most $(Bq/(\log q)^2)^s$. The paper obtains order $q^2(q^2-q+1)$ and permits an absolute $A$ of order $2^{30}$; the formulation above absorbs harmless rounding into the constants. We use the graph construction and this counting bound as external theorems. They are the substantial input, not conclusions of sampling.3) Sampling and alteration
Include each vertex independently with probability $p=(\log q)^2/(2Bq)$, which is at most one for large $q$. Let $X$ count sampled vertices and $Y$ count independent $s$-sets wholly inside the sample. Then \[ \mathbb E X\geq \frac{q^3(\log q)^2}{4B},\qquad \mathbb E Y\leq \left(\frac{Bq}{(\log q)^2}\right)^sp^s=2^{-s}. \] Some sample therefore has $X-Y$ at least the difference of these expectations. Enumerate its original independent $s$-sets and, whenever such a set still survives, delete one of its vertices. At most $Y$ vertices are deleted. Any independent $s$-set in the final graph would have appeared in the original list, so none survives.The final graph is still $K_4$-free, has independence number less than $s$, and has at least $c q^3(\log q)^2$ vertices for an absolute $c>0$. Thus \[ R(4,s(q))>c q^3(\log q)^2. \tag{1} \]
4) Passing from prime powers to every parameter
Restrict $q$ to powers of $2$. For a sufficiently large integer $k$, choose the largest such $q$ with $s(q)\leq k$. Then $k<s(2q)\leq Cq(\log q)^2$ for an absolute $C$. In particular $q\geq k/(C(\log k)^2)$, after increasing the lower threshold so $q\leq k$. This implies $\log q\geq\frac12\log k$ for all sufficiently large $k$. By monotonicity and $1$, \[ R(4,k)\geq R(4,s(q)) \gg q^3(\log q)^2 \gg\frac{k^3}{(\log k)^4}. \] This step is needed: a construction only along a sparse sequence would not by itself establish the stated bound for every $k$.5) Outcome and sources
Outcome: SOLVED using the external finite-geometric construction and its independent-set count. The alteration argument and interpolation are complete. No independent reconstruction of the finite geometry, container estimates, or original breakthrough is claimed.- S. Mattheus and J. Verstraete, The asymptotics of $r(4,t)$, Section 4.2,
https://arxiv.org/abs/2306.04007. - T. F. Bloom, Problem 166,
https://www.erdosproblems.com/166, checked 23 September 2026.