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Current version
Let $C>0$ be a constant. Are there infinitely many integers $a,b,n$ with $a+b> n+C\log n$ such that the denominator of\[\frac{n!}{a!b!}\]contains only primes $\ll_C 1$?
Erdős
[Er68c] proved that if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$. The proof is easy, and can be done with powers of $2$ alone: Legendre's formula implies that if $2^k$ is the highest power of $2$ dividing $n!$ then $k=n+O(\log n)$, and hence if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$.
This problem is asking if $a!b!\mid n!$ 'ignoring what happens on small primes' still implies $a+b\leq n+O(\log n)$.
See also
[728], and
[401] for a later problem in a similar spirit.
This has been proved in the affirmative by Barreto and Leeham, using ChatGPT and Aristotle, with a modification of the argument used for
[728].
2025-10-20 00:00:00
Let $C>0$ be a constant. Are there infinitely many integers $a,b,n$ with $a+b> n+C\log n$ such that the denominator of\[\frac{n!}{a!b!}\]contains only primes $\ll_C 1$?
Erdős
[Er68c] proved that if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$. The proof is easy, and can be done with powers of $2$ alone: Legendre's formula implies that if $2^k$ is the highest power of $2$ dividing $n!$ then $k=n+O(\log n)$, and hence if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$.
This problem is asking if $a!b!\mid n!$ 'ignoring what happens on small primes' still implies $a+b\leq n+O(\log n)$.
See also
[728].
