Erdős Problem #373

declared status 'open'. Formalized: yes. Show that the equation$$n! = a_1!a_2!\cdots a_k!,$$with $n-1>a_1\geq a_2\geq \cdots \geq a_k\geq 2$, has only finitely many solutions. Current best: More generally, Hickerson (as reported in [Er76d]) conjectured that the only non-trivial solutions to the equation in the problem statement are$$9!=2!3!3!7!,$$$$10!=6!7!,$$$$10!=3!5!7!,$$and$$16!=14!5!2!.$$Luca [Lu07b] has shown that there are only finitely many solutions, conditional on the ABC conjecture, and proved unconditionally that the number of $n\leq x$ which admit a non-trivial solution is$$\leq \exp \bigg(f(x)\frac{\log (x)}{\log\log (x)}\bigg)$$for any function $f(x)$ which tends to infinity. In the case when $k=2$, Erd\H{o}s [Er93] proved that if $n!=a_1!a_2!$ with $n-1>a_1\geq a_2$ then$$a_1\geq n-5\log\log n,$$and says it 'would be nice' to prove $a_1\geq n-o(\log\log n)$. Bhat and Ramachandra [BhRa10] replace the $5$ with $(1+o(1))\frac{1}{\log 2}$, and also prove that the same bound holds for arbitrary $k\geq 2$. Numerical investigations on solutions to $n!=a_1!a_2!$ have been carried out by Caldwell [Ca94] and Habsieger [Ha], and it is known that there are no solutions aside from $10!=6!7!$ for $n\leq 10^{3000}$. Prize: no. OEIS: A003135. Tags: factorials, number theory.

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