Date: Oct 11, 2012 2:12 PM
Author: Jérôme Collet
Subject: Sum of squares of binomial coefficients

Resent-From: <bergv@illinois.edu>From: Jérôme Collet <Jerome.Collet@laposte.net>Subject: Sum of squares of binomial coefficientsDate: October 11, 2012 10:37:25 AM MDTTo: "sci-math-research@moderators.isc.org"<sci-math-research@moderators.isc.org>I need to compute the sum :\sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r})^2  }I know, because I used Stirling formula, Taylor-polynomials, andignored some problems on the borders, that this sum should be close to\sqrt{2\pi m}.The convergence is very fast, error is less than .5% if m>7.Nevertheless, I do not know how to prove it correctly.I am sure that two features of the problem are difficult : we have 2 indexes, and the summand is squared.I read "A=B", from Zeilgerger, Petkovsek and Wilf, and tried to use the Maxima tools developed to implement these ideas.The result was very complicated, and I think useless. The Maxima input is at the end of my post.I think the cause is I try to compute a sum on 2 indexes. If I computea  partial sum (on a column or a row, or even on a diagonal), the recurrence is necessarily more complicated.Do you think using multivariate versions of these methods (using Maple or Mathematica) would help ?The Maxima input :To sum on lines:load(zeilberger);f:(binomial(r+s,r)*binomial(2*m-r-s,m-r))^2;Zeilberger(f, r, s);To sum on diagonals (r-s=2u or r+s=2t)load(zeilberger);f:(binomial(2*t,t+u)*binomial(2*m-2*t,m-t-u))^2;Zeilberger(f, u, t);Zeilberger(f, t, u);Thanks for your help.