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 2-3 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

                           (1/2 k (k + 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       n 2  4    n  2
  a2 (q  q - 1) (1 + q  q) ((q )  q  + q  q  + 2) SUM(n)                    n
- ------------------------------------------------------ + a2 (- 2 - 2 q - q  q
          n 2  2   n  2         n 2  2        n
        (q )  q  (q  q  - 1) ((q )  q  + 2 + q  q)

           n 2  3    n  3       n 2  4     n 4  6     n 2  5     n 3  6
     - 3 (q )  q  - q  q  - 3 (q )  q  + (q )  q  - (q )  q  + (q )  q

         n 4  7     n 3  5     n 2  2     n 3  4       n 5  9       n 5  8
     + (q )  q  + (q )  q  - (q )  q  + (q )  q  + 2 (q )  q  + 2 (q )  q

          7   n 5     n 6  9     n 6  10     n 7  11
     + 2 q  (q )  + (q )  q  + (q )  q   + (q )  q  ) SUM(n + 1)

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

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

                                      =0.


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

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

        k   n 4  7      k   n 3  6    k   n 4  6      k   n 3  4    k  n  3
     + q  (q )  q  + 2 q  (q )  q  + q  (q )  q  + 2 q  (q )  q  - q  q  q

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

           k 2  3  n     k 2   n 2  3     k 2  2  n       k 2
     + 2 (q )  q  q  - (q )  (q )  q  - (q )  q  q  - 2 (q )  q) a2

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

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

with the motive that

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

         n         n 2  3    n  3       n 2  4     n 4  6     n 2  5     n 3  6
      - q  q - 3 (q )  q  - q  q  - 3 (q )  q  + (q )  q  - (q )  q  + (q )  q

          n 4  7     n 3  5     n 2  2     n 3  4       n 5  9       n 5  8
      + (q )  q  + (q )  q  - (q )  q  + (q )  q  + 2 (q )  q  + 2 (q )  q

           7   n 5     n 6  9     n 6  10     n 7  11
      + 2 q  (q )  + (q )  q  + (q )  q   + (q )  q  ) b(n + 1, k)

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

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

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

and the theorem follows upon summing with respect to   k   .

    
                                   This took

                                     8.650

                              Seconds of CPU time


Then, L and B in 2-5 can be found after multiplying the last equation throughout by

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