After reading the question and various good answers on the post

Find $n$, where its factorial is a product of factorials

I wonder if $3! \cdot 5! \cdot 7! \cdots (2n+1)!$ would evaluate to a factorial of some expression of $n$.

$n=1$, ans $=3!$
$n=2$, ans $=6!$
$n=3$, ans $=10!$ (the one sought in the related post)

What about higher values of $n$?


As you've found, $n=1,2,3$ give solutions. In fact, the only solutions.

If $n=4$, then $3!5!7!9!=k!> 11!$, so $11\mid k!$ but $11\nmid 3!5!7!9!$.

If $n=5$, then $3!5!7!9!11!=k!> 13!$, so $13\mid k!$ but $13\nmid 3!5!7!9!11!$.

If $n\ge 6$, then $3!5!\cdots (2n+1)!=k!>(4n+2)!$. By Bertrand's Postulate exists a prime $2n+1<p<4n+2$. But then $p\mid k!$ and $p\nmid 3!5!\cdots (2n+1)!$.

  • $\begingroup$ Thank you. So, there's no general expression the product would evaluate to - although it will take me a while to understand the workings :-) $\endgroup$ – William Ha Oct 3 '15 at 4:09
  • $\begingroup$ Thanks for editing for me. $\endgroup$ – William Ha Oct 3 '15 at 4:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.