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Erdős Problem #1161

SOLVEDdatabase updated 2026-01-26

Resolved in some way other than a straight proof or disproof — usual for open-ended questions, or for multi-part problems whose parts went different ways.

What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).

`f k n` is \(f_k(n)\), the number of elements of \(S_n\) of order exactly \(k\). Here \(S_n\) is `Equiv.Perm (Fin n)`. Since every element of a finite group has positive order, `f 0 n = 0`.

`maxF n` is \(\max_{k \geq 1} f_k(n)\). Only the orders that occur in \(S_n\) are considered, because `f k n = 0` for every other `k`.

The answer: the least \(k\geq 1\) with \([1,\ldots,n-k]\mid k\) (as a set with one element).

Let \(f_k(n)\) count the number of elements of \(S_n\) of order \(k\). For which values of \(k\) will \(f_k(n)\) be maximal? [Va99, 5.72] Beker [Be25d, Theorem 1.2] proved that, for all large \(n\), \(f_k(n)\) is maximal if and only if \(k\geq 1\) is minimal such that \([1,\ldots,n-k]\mid k\), where \([\cdot]\) is the least common multiple. The last clause on erdosproblems.com/1161 reads "\(f_k(n)=(n-1)!\)" instead of "\(f_k(n)\) is maximal"; we follow [Be25d]. For \(k > n\) we read \(n-k\) as \(0\), which does not change the least such \(k\), because \(k = n\) qualifies.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 2 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/1161.
Prizenone
Topicsgroup theory
Statement in LeanFormal Conjectures
Solution in Leanno
OEISA057731
Original pageerdosproblems.com/1161

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