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\topmatter
{\noindent\smcp the electronic journal of combinatorics 5 (1998), \#R10\hfill} 
\vskip .6in
\title A binomial coefficient identity associated to a conjecture
of Beukers
\endtitle
\rightheadtext{A binomial coefficient identity}
\author Scott Ahlgren, Shalosh B. Ekhad, Ken Ono and 
       Doron Zeilberger\endauthor
\address Department of Mathematics, Penn State University,
University Park, Pennsylvania 16802 \endaddress
\email ahlgren\@math.psu.edu \endemail
 
 
\address Department of Mathematics, Temple University,
Philadelphia, Pennsylvania 19122\endaddress
\email ekhad\@math.temple.edu; http://www.math.temple.edu/
 \~{}ekhad \endemail
 
\address Department of Mathematics, Penn State University,
University Park, Pennsylvania 16802 \endaddress
\email ono\@math.psu.edu; http://www.math.psu.edu/ono/  \endemail
 
\address Department of Mathematics, Temple University,
Philadelphia, Pennsylvania 19122
\endaddress
\email zeilberg\@math.temple.edu; http://www.math.temple.edu/\~{} zeilberg
\endemail
\thanks 
The  third author is supported by NSF grant DMS-9508976 and
NSA grant MSPR-Y012. The last author is supported in part by
the NSF.
\endthanks
%  Math Subject Classifications 
%\subjclass Primary 11F67 ; Secondary  11F37 \endsubjclass
\abstract Using the WZ method, a binomial coefficient identity is proved. This identity is noteworthy since its truth is known to imply a conjecture of Beukers.
\endabstract
 
\endtopmatter

 \document
\vskip -10pt
\centerline{Received: January 28, 1998; Accepted: February 1, 1998}

\vskip 10pt

If $n$ is a positive integer, then let
  $\dsize{A(n):=\sum_{k=0}^{n}{n\choose k}^2{n+k\choose k}^2}$,
and define integers $a(n)$ by
  $$\dsize{\sum_{n=1}^{\infty}a(n)q^n:=q\prod_{n=1}^{\infty}
   (1-q^{2n})^4(1-q^{4n})^4=q-4q^3-2q^5+24q^7- \cdots.}$$ 
Beukers conjectured that
if $p$ is an odd prime, then
$$
  A\left ( \frac{p-1}{2}\right ) \equiv a(p) \pmod{p^2}. \tag{1}
$$
 
 
In [A-O] it is shown that (1) is implied by the truth
of the following identity.
\proclaim{Theorem} If $n$ is a positive integer, then
$$
  \sum_{k=1}^{n}k{n\choose k}^2{n+k \choose k}^2
  \left \{ \frac{1}{2k}+\sum_{i=1}^{n+k}\frac{1}{i}+
 \sum_{i=1}^{n-k}\frac{1}{i}-2\sum_{i=1}^{k}\frac{1}{i}\right \}=0.
$$
\endproclaim
 
\medskip
 
\noindent
{\bf Remark.} 
This identity is easily verified using the WZ method, in a generalized form
[Z] that applies when the summand is a hypergeometric term times a WZ
potential function.
It holds for all positive $n$,
since it holds for $n$=1,2,3 (check!), and since the
sequence defined by the sum satisfies a certain 
(homog.) third order linear recurrence
equation. To find the recurrence, and its proof,
download the Maple package EKHAD and the Maple program zeilWZP
from {\eighttt http://www.math.temple.edu/\~{} zeilberg }.
Calling the quantity inside the braces $c(n,k)$, compute
the WZ pair $(F,G)$, where $F=c(n,k+1)-c(n,k)$ and
$G=c(n+1,k)-c(n,k)$. 
Go into Maple, and type 
{\eighttt read zeilWZP;
zeilWZP(k*(n+k)!**2/k!**4/(n-k)!**2,F,G,k,n,N):
}
 
\Refs
\widestnumber\key{DS-PPP}
 
\ref \key A-O \by S. Ahlgren and K. Ono
\paper A Gaussian hypergeometric series evaluation and
Ap\'ery number congruences
\vol (in preparation) 
\endref
 
\ref \key B \by F. Beukers
\paper Another congruence for Ap\'ery numbers
\jour J. Number Th. \vol  25 \yr 1987 \pages 201-210
\endref
 
\ref \key Z \by D. Zeilberger
\paper Closed Form (pun intended!)
\jour Contemporary Mathematics \vol  143 \yr 1993 \pages 579-607
\endref
 
 
 
\endRefs
\enddocument