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.