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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The research report Erdős Problem #554: odd-cycle Ramsey ratios versus triangles, dated 1 August 2026 and revised 2 August 2026, a GitHub gist in nine revisions from 2026-08-01T20:41Z to 2026-08-02T19:48Z (the last revision is linked and is the one described), was linked from the discussion thread of Problem 554 on 1 August 2026 by the forum user mysticflounder, who is the claimant. Its main statement, Section 4: "If the Chapter 9 lower bound and the ACJMR25 upper bound are correct, then #554 holds for every fixed n ≥ 4." The two inputs are the lower bound Rk(K3)≥(ck1/3/log⁡k)kR_k(K_3)\ge(ck^{1/3}/\log k)^k of Chapter 9, Theorem 1.1 of OpenAI's 2026 report, and the upper bound Rk(C2n+1)≤(4n−2)kkk/n+1R_k(C_{2n+1})\le(4n-2)^kk^{k/n}+1 of Axenovich, Cames van Batenburg, Janzer, Michel and Rundström (Theorem 1.1, J. Combin. Theory Ser. B 179 (2026), 293--298). The derivation is the two-term comparison

Rk(C2n+1)Rk(K3)≤((4n−2) k1n−13log⁡kc)k+(log⁡kc k1/3)k,\frac{R_k(C_{2n+1})}{R_k(K_3)} \le\Bigl(\frac{(4n-2)\,k^{\frac1n-\frac13}\log k}{c}\Bigr)^k +\Bigl(\frac{\log k}{c\,k^{1/3}}\Bigr)^k,

whose first base tends to 00 exactly when 13−1n>0\frac13-\frac1n>0, that is for n≥4n\ge4, and whose second term tends to 00 for every nn (the report keeps the +1+1 of the upper bound as this second term). For n=2n=2 and n=3n=3 the report says the same comparison decides nothing, with a k1/6k^{1/6} exponent gap for C5C_5 and a logarithmic gap for C7C_7, names what sharper bounds would close them, finds no disproof, and records the problem as open. The report also notes, crediting a reply in the thread by a coauthor of the upper-bound paper, that before that paper the best unconditional upper bounds had exponent kk/2k^{k/2} for every nn, so the comparison reached no nn at all. The problem page's own comparison reproduces this derivation.

Covers. The problem's statement for every fixed n≥4n\ge4: the limit of Rk(C2n+1)/Rk(K3)R_k(C_{2n+1})/R_k(K_3) as k→∞k\to\infty is 00. Not covered: n=2n=2 and n=3n=3, which the report leaves open.

Depends on. OpenAI 2026, the accepted claim page of Problem 183 recording the lower bound, which the report takes as its denominator input; the upper bound is the refereed theorem linked above.

Hypotheses. The report states its result conditionally on the correctness of its two inputs, and classifies the OpenAI input as proved tentatively on a source-level inspection of the release's Lean files without a build or axiom audit. Both inputs are accepted on this wiki: the upper bound is refereed, and the lower bound is reviewed on Problem 183's claim page by the site's curator and through Rob Morris's exposition hosted there, from an unrefereed report. The comparison itself is elementary.

AI systems. The thread post of 1 August 2026 attributes the reduction to Claude and says GPT 5.6 audited it adversarially; the report's own method line says the Lean bundle was inspected at source level with no build, axiom audit, comparator run or adversarial audit. Both statements are recorded as the sources give them.

Standing. Claimed. The reply of 2 August 2026 by a coauthor of the upper-bound paper agrees that the two papers together dispose of every n≥4n\ge4, up to n=2n=2 or 33; the site's label for the problem is OPEN, its commentary (last edited 8 February 2026) does not mention the report, and its proof-claim tab is empty, so the thread carries no acceptance. The report has no refereed or reviewed version, and nothing on this page is independently reviewed; the problem page records the same deduction as its own and unreviewed.