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Problem 1161

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claims/: The 1 claim page of Problem 1161, one per claimant's result; the problem's standing derives from them.


Statement. Let fk(n)f_k(n) count the number of elements of SnS_n of order kk. For which values of kk will fk(n)f_k(n) be maximal?

Formulation. The question is read for large nn, as the paper the site credits and the site's label read it. Beker takes up the question of Erdős and Turán (1968, p. 414), restated by Acan, Burnette, Eberhard, Schmutz and Thomas, as one about the most probable order of a random permutation, and calls his answer, which holds for all sufficiently large nn, essentially complete. Read for every nn, the question is settled only beyond a threshold the argument does not make explicit, and some small nn have other maximizers (Remark 1.3). The page's standing targets the large-nn reading.

Status. Solved.

Source. erdosproblems.com/1161, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #1161, https://www.erdosproblems.com/1161.

References.

  • [Be25d] A. Beker, The most probable order of a random permutation. arXiv:2510.11698 (2025).

Formalization. None recorded.

Current assessment

The standing judges the site's formulation of 2026-09-04, read for large nn as the Formulation states. Beker [Be25d] answers it for all sufficiently large nn: the largest of the counts fk(n)f_k(n) is (1+o(1)) (n−1)!(1+o(1))\,(n-1)!, and it is attained at exactly one order, the least k≥1k\geq1 divisible by every integer from 11 to n−kn-k; the paper's own Remark 1.3 says that small nn behave differently and that the bound its argument could give is most probably not small enough to check the remaining cases by a naive method. The accepted claim page Beker 2025 carries the acceptance evidence, which is the site's curator crediting the preprint as the solution; no refereed publication is recorded. Search scope: the site's problem page, discussion thread and proof-claims page, and the arXiv record of the preprint, accessed 2026-10-07; they record no other claim. Nothing on this page is independently reviewed.

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