We have an $n$ natural number. Prove, that $2(3n)!$ is divisible with $n!(n+1)!(n+2)!.$
I tried to make it look simplier, but can't prove that this is true.
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Is there a generalization to this problem for $n$ and $k$ is :$$ \frac{k!(n(k+1))!}{n!(n+1)!\cdots(n+k)!}$$ an integer? – Elaqqad Mar 04 '15 at 13:38
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1@Elaqqad: Note that it is not an integer at least for $(k,n)=(3,1)$. – mathlove Mar 04 '15 at 15:21
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@mathlove, it's true if we replace the $k!$ by $(k+1)!$ it's true but it's not the minimal factor – Elaqqad Mar 04 '15 at 16:22
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@Elaqqad: For $k=3$, you can replace $k!$ by $12$ which is the minimal factor for $k=3$. See this OEIS. – mathlove Mar 04 '15 at 16:26
