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 coefficients
Date: October 11, 2012 10:37:25 AM MDT
To: "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, 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*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.