[seqfan] Re: when does n+1 fail to divide binomial(n^2,n+1) ?
William Keith
wjk26 at drexel.edu
Sat May 8 22:16:37 CEST 2010
> %N A177234 a(n) = binomial(n^2, n)/(n+1) if n+1 divides binomial(n^2, n), otherwise a(n) = -1.
> Can someone find further terms (beyond n=1) where
> the division fails?
>
> Neil
It will always be an integer for n>1.
binomial(n^2, n) * 1/(n+1)
= (n^2)(n^2-1)(n^2-2)!/((n^2-n)!n(n-1)(n-2)!) * 1/(n+1)
= n (n^2-2)!/((n^2-n)!(n-2)!) = n * binomial(n^2-2,n-2).
William J. Keith
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