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Problem 231

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claims/: The 2 claim pages of Problem 231, one per claimant's result; the problem's standing derives from them.


Statement. Let SS be a string of length 2k−12^k-1 formed from an alphabet of kk characters. Must SS contain an abelian square: two consecutive blocks xx and yy such that yy is a permutation of xx?

Statement (corrected). Let SS be a string of length 2k2^k formed from an alphabet of kk characters. Must SS contain an abelian square: two consecutive blocks xx and yy such that yy is a permutation of xx?

Notes. The site's wording fails for k=1,…,4k=1,\ldots,4: the strings 11, 121121, 12131211213121 and 121312141213121121312141213121 have length 2k−12^k-1 and no abelian square (the strings 121121 and 12131211213121 are noted in the formal-conjectures file, and the string of length 1515 is Alexeev's, below). At length 2k2^k the question holds for k≤3k\le3, as Erdős reports in [Er61] (below) and as the formal-conjectures variant erdos_231.variants.two_pow_small proves for k=2,3k=2,3 by kernel computation. It fails at k=4k=4, by the site's string 12131214121321241213121412132124, which has no abelian square.

The change replaces "2k−12^k-1" by "2k2^k"; nothing else changes. The evidence is Erdős's own words about instances of the question. [Er61], Part II, item 2 (Some unsolved problems, 1961), calls two consecutive blocks "identical" when each symbol occurs equally often in both, and continues: "I conjectured that in a sequence of length 2k−12^k-1 formed from kk symbols there must be two “identical” blocks. This is true for k≤3k \leq 3, but for k=4k=4 de BRUIJN and I disproved it". Both reports are true at length 2k2^k and false at length 2k−12^k-1: at that length the conjecture already fails for k≤3k\le3, and k=4k=4 is not its first failure. [Er57], item 28, printed p. 298 (Some unsolved problems, 1957), defines N(k)N(k) as the least NN such that every sequence of length NN over {1,…,k}\{1,\ldots,k\} contains two adjacent blocks, each a rearrangement of the other, and reports that his "earliest conjecture, that N(k)=2k−1N(k) = 2^k - 1, has been disproved by Bruijn and myself"; the strings above give N(k)≥2kN(k)\ge2^k for k≤3k\le3, so this print carries the same slip. The misprint is therefore already in both of the poser's texts, and the site's wording repeats it; his words about the instances hold only for length 2k2^k. The site's commentary suggests that Erdős may have meant 2k2^k; that suggestion is not the evidence.

Results about the site's wording are credited here and count for nothing. Boris Alexeev gave the ruler sequence 121312141213121121312141213121 on the site's discussion thread on 2026-02-15, and the formal-conjectures statement file notes the strings 121121 and 12131211213121. A Lean file in Alexeev's lean-proofs collection, whose header names de Bruijn and Erdős as informal authors and AxiomProver as formal author, published by Axiom Math, proves the negation of the site's wording, not_erdos_231, (pinned file; copy at lean-proofs) from an abelian-square-free string of length 1515 over four characters; it is linked from the de Bruijn–Erdős page and settles no instance of the corrected Statement.

Status. The site shows DISPROVED (LEAN), crediting the negative answer for all k≥4k\geq4 to Keränen's infinite abelian-square-free word on four letters [Ke92], a label that describes the corrected Statement. The corrected Statement is disproved: the accepted claim is infinite abelian-square-free words on four letters, and the finite disproof at k=4k=4 that Erdős attributes to de Bruijn and himself, published without a construction, is a pending claim, reported finite disproof at four letters. The site's wording, with length 2k−12^k-1, already fails for k≤3k\le3, as the Notes record. The Lean artifacts behind the label's qualifier are described under Formalization.

Source. erdosproblems.com/231, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #231, https://www.erdosproblems.com/231.

References.

  • [Er57] Erdős, P., Some unsolved problems. Michigan Math. J. 4 (1957), 291-300; item 28. Library home: erdos_1957_unsolved_problems.
  • [Er61] Erdős, P., Some unsolved problems. Magyar Tud. Akad. Mat. Kutató Int. Közl. 6 (1961), 221-254; section II, item 2. Library home: erdos_1961_unsolved_problems.
  • [FiPu23] Fici, Gabriele and Puzynina, Svetlana, Abelian combinatorics on words: a survey. Comput. Sci. Rev. 47 (2023), Paper No. 100532, 21 pp. Library home: fici_2023_abelian_combinatorics_words_survey.
  • [Ke92] Keränen, Veikko, Abelian squares are avoidable on 44 letters. Automata, languages and programming (Vienna, 1992), Lecture Notes in Comput. Sci. 623, Springer (1992), 41-52.

Formalization. Statement in formal-conjectures, added 2026-09-19 and marked solved there. Its main theorem erdos_231 states the site's wording, for k≥2k\ge2, and links as its formal disproof the length-1515 counterexample in Boris Alexeev's lean-proofs collection described in the Notes; it gives no evidence for the corrected Statement. The same file's variant erdos_231.variants.two_pow states the corrected Statement, for k≥2k\ge2, and proves its negation from the string 12131214121321241213121412132124, and erdos_231.variants.two_pow_small proves the cases k=2,3k=2,3 by kernel computation. The community database lists the site's label with its Lean qualifier as of its last update of 2026-05-14, after Lorenzo Luccioli's formalization of Keränen's theorem with Aristotle, linked from Keränen's claim page. None of these developments was built or audited here.

Current assessment

The corrected Statement asks whether every string of length 2k2^k over kk characters contains an abelian square, two consecutive blocks that are permutations of each other. It holds for k≤3k\le3 and fails for every k≥4k\ge4, so the answer is no. The problem's standing, solved with claim disproved, derives from Keränen's infinite abelian-square-free word on four letters, credited by the site's curator and recorded as a theorem in the refereed survey of Fici and Puzynina: its prefixes give abelian-square-free strings of every length on four letters, so for every k≥4k\geq4 a string of length 2k2^k over kk characters can avoid abelian squares. At k=4k=4 the site's string 12131214121321241213121412132124 already answers the corrected Statement in the negative. Keränen's theorem settles more: no length forces an abelian square on four letters, so N(4)N(4) is infinite in the notation of [Er57], which answers Erdős's 1957 remark that this was not known, and the infinite form of the question, Problem 192 on the site, is answered.

Erdős's original conjecture [Er57], [Er61] was the corrected Statement for every kk, which he reported disproved at k=4k=4 with de Bruijn but never published with a construction; that report is the pending de Bruijn–Erdős page. The site's wording, with length 2k−12^k-1, already fails for k≤3k\le3, as the Notes record.

Search scope: the site's page and discussion thread (six comments, no proof claims), the community database (teorth/erdosproblems), the formal-conjectures statement file, the lean-proofs collection, the KE92ErdosProblems repository and Crossref. No other claim on the problem was found. A set of partial Lean files toward Keränen's theorem posted on the thread on 2026-02-14 is not a claim and is not linked.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.