Wiki
Wiki

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

Updated


Claim. The conjectured asymptotic of Problem 1131,

min⁡I=2−(1+o(1))1n,\min I=2-(1+o(1))\frac{1}{n},

is false. Liam Price posted the claim in the site's discussion thread on 26 April 2026, writing that GPT-5.5 Pro claims a disproof of the asymptotic the problem displays, with a link to a write-up on an online editor. The read link exposes only the editor's shell and no document to a reader without an account, so the corpus records the target of the claim from the thread post: the displayed asymptotic is the only asymptotic the problem states, and a disproof of it says that n(2−min⁡I)n(2-\min I) does not tend to 11. Price is the claimant, with the system named as the post names it.

Submission note. Posted to the site's forum by Liam Price on 26 April 2026:

GPT-5.5 Pro claims a disproof of the claimed asymptotic here.

Covers. The second question of Problem 1131, whether min⁡I=2−(1+o(1))/n\min I=2-(1+o(1))/n: the answer claimed is no. The first question, the minimal value of II, is not determined. The bounds 2−O((log⁡n)2/n)≤min⁡I≤2−2/(2n−1)2-O((\log n)^2/n)\le\min I\le 2-2/(2n-1) of Erdős, Szabados, Varma and Vértesi, which the site's commentary records, are consistent with either answer; Brutman and Toledano's numerical study (Comput. Math. Appl. 34 (1997), no. 12, 37--47) had pointed to a limit of n(2−min⁡I)n(2-\min I) near 1.091.09, as the problem page records.

Depends on. No page of this wiki.

Acceptance. None recorded. Nat Sothanaphan replied in the thread on 27 April 2026 that a standard check found no issues, and noted that the Brutman--Toledano paper had earlier given numerical evidence against the asymptotic; Sothanaphan is not the site's curator, so the reply is recorded here and not listed as reviewed evidence. The site labels the problem OPEN (page last edited 2026-01-23) and no journal publication, referee report or Lean proof of the result is recorded, so no evidence is listed and the claim is pending.