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Re: Sum of squares of binomial coefficients
Posted:
Oct 18, 2012 4:28 AM


I stated, by trial and errors that: \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } = \binom{4m+1}{2m} It is true for m from 1 to 250, with infinite precision calculations, so I think it is true for all m.
But it is not a proof. I will try do do it using manually Sister Celine's method, but if anyone has a combinatorial proof, I am interested.
Le 13/10/2012 00:13, Gavin Wraith a écrit : > 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. >



