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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 $n$?

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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 $p\nmid 3!5!\cdots (2n+1)!$.

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  • $\begingroup$ 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

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