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Re: Sum of squares of binomial coefficients
Posted:
Oct 12, 2012 6:13 PM


In message <121020120727017676%edgar@math.ohiostate.edu.invalid> Jérôme Collet <Jerome.Collet@laposte.net> wrote:
> So my question is > \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } > seems to be equivalent to > \sqrt{2\pi m} \binom{2m}{m}^2 > How can I prove it ?
A bit more attention; maybe not very useful. Suppose you have a 2melement set. Call a quadruple of subsets (A,B,U,V) such that 1) #U = #V = m 2) #A = #B 3) #(A\cap U) = #(B\cap V) "jolly". Then I believe that \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } is the number of jolly quadruples. Now \binom{2m}{m}^2 is the number of ways of choosing pairs of subsets (U,V) such that #U = #V = m. I am sceptical that all the rest of the jollity counts asymptotically for \sqrt{2\pi m}. I would have expected something a lot bigger, but I cannot give you any good reasons right now.
 Gavin Wraith (gavin@wra1th.plus.com) Home page: http://www.wra1th.plus.com/



