Version of July 16, 1996


       This version corrects a subtle bug discovered by Helmut Prodinger


           Previous versions benefited from comments by Paula Cohen,

                         Lyle Ramshaw, and Bob Sulanke.


                    This is EKHAD, One of the Maple packages

                             accompanying the book

                                     "A=B"

                   (published by A.K. Peters, Welesley, 1996)

              by Marko Petkovsek, Herb Wilf, and Doron Zeilberger.


                The most current version is available on WWW at:

                     http://www.math.temple.edu/~zeilberg .

        Information about the book, and how to order it, can be found in

               http://www.central.cis.upenn.edu/~wilf/AeqB.html .

             Please report all bugs to: zeilberg@math.temple.edu .

         All bugs or other comments used will be acknowledged in future

                                   versions.


            For general help, and a list of the available functions,

         type "ezra();". For specific help type "ezra(procedure_name)"


             Identity 4.2 of Geroge Andrews's papar Pfaff Method I


A PROOF OF A RECURRENCE

By Shalosh B. Ekhad, Temple University, ekhad@math.temple.edu

Theorem:Let   F(n,k)   be given by

(- 2 n - 2 + k)! (x + n + 1 + k)! (x + k)! (x - z - 1/2 + k)! (z + k)! (x - n)!

                              k   /
     (2 z + 1)! (2 x - 2 z)! 4   /  ((- 2 n - 2)! (x + n + 1)! x!
                                /

    (x - z - 1/2)! z! k! (x - n + 2 k)! (2 z + 1 + k)! (2 x - 2 z + k)!)


and let   SUM(n)   be the sum of   F(n,k)    with
respect to   k   .

SUM(n)    satisfies the following linear recurrence equation

(2 n + 3) (n + 2) (n + 1) (n - x + 2 z + 2) (n + x - 2 z + 1) SUM(n) - (n + 2)

            3      2          2                    2      3      2          2
(n - x) (4 n  + 4 n  x - 2 n x  + 10 n x z - 10 n z  - 2 x  + 8 x  z - 8 x z

       2                       2                2
 + 26 n  + 19 n x - 5 n z - 3 x  + 20 x z - 24 z  + 57 n + 23 x - 12 z + 42)

SUM(n + 1)

 + (n + z + 3) (n + x + 3) (n - x + 1) (n - x) (2 n + 2 x - 2 z + 5) SUM(n + 2)

                                      =0.


PROOF: We cleverly construct   G(n,k)   :=

                2                                                            2
- (3 + 5 n + 2 n ) (n + 2) (n + 6 n x + 9 n k + 15 x + 2 + 22 k + 3 k x + 2 x )

     k (x - n - 1 + 2 k) (x - n + 2 k) (2 z + 1 + k) (2 x - 2 z + k)

    (- 2 n - 2 + k)! (x + n + 1 + k)! (x + k)! (x - z - 1/2 + k)! (z + k)!

                                      k   /
    (x - n)! (2 z + 1)! (2 x - 2 z)! 4   /  ((- 2 n - 2 + k) (x + n + 2)
                                        /

    (- 2 n - 3 + k) (- 2 n - 4 + k) (- 2 n - 5 + k) (- 2 n - 2)! (x + n + 1)!

    x! (x - z - 1/2)! z! k! (x - n + 2 k)! (2 z + 1 + k)! (2 x - 2 z + k)!)

with the motive that

(2 n + 3) (n + 2) (n + 1) (n - x + 2 z + 2) (n + x - 2 z + 1) F(n, k) - (n + 2)

               3      2          2                    2      3      2
   (n - x) (4 n  + 4 n  x - 2 n x  + 10 n x z - 10 n z  - 2 x  + 8 x  z

          2       2                       2                2
   - 8 x z  + 26 n  + 19 n x - 5 n z - 3 x  + 20 x z - 24 z  + 57 n + 23 x

   - 12 z + 42) F(n + 1, k) +

  (n + z + 3) (n + x + 3) (n - x + 1) (n - x) (2 n + 2 x - 2 z + 5) F(n + 2, k)

=   G(n,k+1)-G(n,k)   (check!)

and the theorem follows upon summing with respect to   k   .QED.
                                   This took

                                    108.400

                              Seconds of CPU time