This is qEKHAD, one of the Maple packages

                    accompanying the forthcoming book "A=B"

                   (soon to be published by A.K.Peters, Ltd.)

             by Marko Petkovsek, Herb Wilf, and Doron Zeilberger.

             Also try out the other packages: EKHAD and multiEKHAD



            The latest versions are always available by anon. ftp to

                              ftp.math.temple.edu

                    in the directory: /pub/zeilberg/programs

            or, on the Web, at http://www.math.temple.edu/~zeilberg

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

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


                        Warning: q is a global variable

                 qfac(k) means (1-q)*(1-q^2)*...*(1-q^k), and

                       NOT (1)(1+q)*...*(1+q+...+q^(k-1))



                             Version of  April 1996

        We wish to thank Tewodros Amdeberhan for suggesting improvements



                   The CF in 1-3 satisfies a 3-term recurrence

A PROOF OF THE   NAME   RECURRENCE

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

I will give a short proof of the following result(   REF   ).

Theorem:Let   b(n,k)   be

                           k  (1/2 k (k + 1))
                       (-1)  q                qfac(n + k)
                       ----------------------------------
                                     2
                              qfac(k)  qfac(n - k)


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

SUM(n)    satisfies the following linear recurrence equation

    n          n  2
  (q  q - 1) (q  q  + 2) a2 SUM(n)          6   n 4    n  3             n 2  4
  -------------------------------- + a2 (2 q  (q )  + q  q  + 2 q - 4 (q )  q
           n 2  2   n
         (q )  q  (q  q + 2)

            n  2        5   n 3       n 2  3    n       n 5  8      7   n 4
       - 4 q  q  + 2 + q  (q )  - 4 (q )  q  + q  q + (q )  q  + 2 q  (q ) )

                                                        n  2
                   /    n 2  3   n           a2 (- 1 + q  q ) SUM(n + 2)
      SUM(n + 1)  /  ((q )  q  (q  q + 2)) + ---------------------------
                 /                                      3   n 2
                                                       q  (q )

                                      =0.


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

    n  2        k  5   n 3       n 2  4  k       n 2  3  k      k  n  2    k
(- q  q  - 2 + q  q  (q )  + 2 (q )  q  q  + 2 (q )  q  q  + 3 q  q  q  - q  q

        k     k 2  2  n       k 2               k  n        k  (1/2 k (k + 1))
     - q  - (q )  q  q  - 2 (q )  q) a2 (- 1 + q  q  q) (-1)  q

                  /      k    n      n               2
    qfac(n + k)  /  ((- q  + q  q) (q  q + 2) qfac(k)  qfac(n - k))
                /

with the motive that

   n          n  2
 (q  q - 1) (q  q  + 2) a2 b(n, k)          6   n 4    n  3             n 2  4
 --------------------------------- + a2 (2 q  (q )  + q  q  + 2 q - 4 (q )  q
          n 2  2   n
        (q )  q  (q  q + 2)

           n  2        5   n 3       n 2  3    n       n 5  8      7   n 4
      - 4 q  q  + 2 + q  (q )  - 4 (q )  q  + q  q + (q )  q  + 2 q  (q ) )

                                                        n  2
                   /    n 2  3   n           a2 (- 1 + q  q ) b(n + 2, k)
     b(n + 1, k)  /  ((q )  q  (q  q + 2)) + ----------------------------
                 /                                      3   n 2
                                                       q  (q )

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

and the theorem follows upon summing with respect to   k   .

                                   This took

                                     4.066

                              Seconds of CPU time


Then, L and B in 1-5 can be found after multiplying the last equation throughout
by
        
                   n 2  3   n
                 (q )  q  (q  q + 2) /a2.