6

Let $T$ be the set of nonnegative integers, I need to prove that $$\limsup_{n\to\infty}|\{(p,q)\in T\times T,p!q!=n\}|=6$$

It's really easy to show that

$$\limsup_{n\to\infty}|\{(p,q)\in T\times T,p!q!=n\}|\ge6$$ since for $n>2$ $$(n!)!=0!(n!)!=1!(n!)!=n!(n!-1)!=(n!-1)!n!=(n!)!1!=(n!)!0!$$

So I'm looking for a way to prove the apposite inequality.

  • maybe this link can help :http://math.stackexchange.com/questions/991728/equal-products-of-consecutive-integers – Elaqqad Feb 05 '15 at 14:00
  • A generalization for your equation $n!(n!-1)!=(n!)!$ may be : $$a!(\frac{a!}{b!}-1)!=\frac{a!}{b!}!b!$$ for $a\geq b$ – Elaqqad Feb 08 '15 at 23:40
  • The problem here is related to solving the equation $p!q!=r!s!$, and this equation can be simplified in all cases to one of the form $\frac{q!}{r!}=\frac{s!}{p!}$ which rises the problem of EQUAL PRODUCTS OF CONSECUTIVE INTEGER and this problem was conjectured to have only finite solution by Erdos and was confirmed in this paper http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.25.727 – Elaqqad Feb 09 '15 at 00:29

1 Answers1

7

This equality was proved by Daniel M. Kane in this paper.

Elaqqad
  • 13,725
  • Thank you very much and also Daniel. Do you know him? –  Feb 15 '15 at 15:16
  • No, I don't know him, I was working in your problem and one of my friends told me that he have seen this problem somewhere else and so i made a research and I found it, good luck! – Elaqqad Feb 15 '15 at 15:24