
Sum of squares of binomial coefficients
Posted:
Oct 11, 2012 2:12 PM


ResentFrom: <bergv@illinois.edu> From: Jérôme Collet <Jerome.Collet@laposte.net> Subject: Sum of squares of binomial coefficients Date: October 11, 2012 10:37:25 AM MDT To: "scimathresearch@moderators.isc.org" <scimathresearch@moderators.isc.org>
I need to compute the sum : \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } I know, because I used Stirling formula, Taylorpolynomials, and ignored 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 compute a 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*mrs,mr))^2; Zeilberger(f, r, s);
To sum on diagonals (rs=2u or r+s=2t) load(zeilberger); f:(binomial(2*t,t+u)*binomial(2*m2*t,mtu))^2; Zeilberger(f, u, t); Zeilberger(f, t, u);
Thanks for your help.

