Erdős Problem #401 — Large Factorial Divisors

About 46 years old · traced to

Does there exist a function f:N→Rf:\mathbb{N}\to\mathbb{R} with f(r)→∞f(r)\to\infty as r→∞r\to\infty such that, for every r∈Nr\in\mathbb{N} with r≥1r\geq1, there are infinitely many n∈Nn\in\mathbb{N} for which there exist a1,a2∈Na_1,a_2\in\mathbb{N} with a1>0a_1>0, a2>0a_2>0,

a1+a2>n+f(r)log⁡n,a_1+a_2>n+f(r)\log n,

and

a1!a2!∣n!(∏i=0r−1pi)n,a_1!a_2!\mid n!\left(\prod_{i=0}^{r-1}p_i\right)^n,

where pip_i denotes the iith prime?

References

Progress summary

Refreshed
Claimed solved

A formal proof now confirms that the requested function exists, so the problem is solved, although a stronger version requiring the condition for every sufficiently large number is false.

Problem 401401 asks whether there is an unbounded function f(r)f(r) for which infinitely many nn satisfy the stated factorial divisibility and size conditions.

January 2026 formal resolution

A formal result proves that, for infinitely many triples (a,b,n)(a,b,n), one has a!b!∣n!(a+b−n)!a!b!\mid n!(a+b-n)! with a+b−na+b-n comparable to log⁡n\log n; the paper explains that these examples imply the required unbounded f(r)f(r). The proof uses pp-adic inequalities, Kummer’s theorem, and carry-counting, and was formalized in Lean. The stronger interpretation requiring the assertion for all sufficiently large nn is false.

Current status (as of April 2026): Problem 401401 is resolved by a formal Lean proof; only the stronger all-nn variant remains false rather than open.

  • AristotleHarmonicsolved2026-01-01evidence
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