Kourovka Notebook Problem 21.89

Definitions and mathematical statement

For a nonnegative integer $n$, an integer partition of $n$ is a finite nonincreasing list $\lambda_1\ge\cdots\ge\lambda_r\ge1$ with $\sum_i\lambda_i=n$; for $n=0$ allow the empty list. Let $p(n)$ be the number of these lists. For positive $n$, the factorial is $n!=1\cdot2\cdots n$.

The problem asks whether \[ \forall n\in\mathbb Z,\quad n\ge40\ \Longrightarrow\ p(n)\nmid n!. \] Divisibility here is integer divisibility: $p(n)\mid n!$ would mean that $n!=a\,p(n)$ for some positive integer $a$. Thus one counterexample at any integer $n\ge40$ would refute the assertion, while checking only an initial finite interval does not settle it.

In group-theoretic terms, let $S_n$ be the group of all permutations of $n$ points. It has order $n!$. Every permutation decomposes into disjoint cycles, and the multiset of their lengths is a partition of $n$. Two permutations are conjugate exactly when these length multisets coincide, so $k(S_n)=p(n)$. The question is consequently whether the number of conjugacy classes of $S_n$ ever divides its order beyond degree $39$. This equivalence fixes both the counting convention for partitions and the strict lower threshold on $n$.

Short English statement

For every integer $n\ge40$, does the partition number $p(n)$ fail to divide $n!$?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2598 (KOU-21.89), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, version 47 (30 September 2026), new problems, Problem 21.89. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.