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I have a few questions about Andy Loo's proof (get it here):

  • why, for example, if $2n<p\le3n$, then $p$ does not divide $\binom{4n}{3n}$? Same situation for $\frac{4n}{3}<p\le\frac{3n}{2}$...
    • ${s \brace r}$.. what is it for?

Thank you in advance...

virnoy
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1 Answers1

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If $2n<p\le 3n$ then $(3n)!=(3n)(3n-1)\cdots(p+1)p(p-1)\cdots(2)(1)$ is divisible by $p$ but not $p^2$ since $2p>4n>3n.$

If $2n<p\le 3n$ then $(4n)!=(4n)(4n-1)\cdots(p+1)p(p-1)\cdots(2)(1)$ is divisible by $p$ but not $p^2$ since $2p>4n.$

If $2n<p\le 3n$ then $n!$ is not divisible by $p$ since $n<p/2<p.$

Hence ${4n}\choose{3n}$ is $p$ times something not divisible by $p$ divided by $p$ and two somethings not divisible by $p$, resulting in a number not divisible by $p$: $$ \frac{ps_1}{ps_2s_3}=\frac{s_1}{s_2s_3}. $$

Charles
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  • Unfortunately, here ${s \brace r}$ doesn't represent a Stirling number of second kind. – virnoy Jul 14 '15 at 19:50
  • @virnoy: It looks like the author defines the notation on p. 4. I won't copy it here. – Charles Jul 14 '15 at 19:53
  • Yes, I've seen it, but I've no idea what is has to the approximation of $P_2$. It seems for me like slightly modified binomial coefficient? – virnoy Jul 14 '15 at 19:55