Notation:  R(a,k) denotes the rising factorial a(a+1)...(a+k-1)

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

 whose (i,j) entry is a(i,j):=, binomial(i + j, j) binomial(2 p - i - j, p - j)

                                 (0<=i,j<=m) .

                                  Let b(m):=

                                   m               2
                      (2 p)! (1/16)  R(-2 p - 1, m)
                    -----------------------------------,  .
                                                      2
                    R(-p + 1/2, m) R(-p - 1/2, m) (p!)

                       Then [det A(m)]/[det A(m-1)] = b(m)

                     and hence det [A(m)]=b(1)b(2)...b(m) .



                                Rigorous Proof:

                                 Let c(m,n) be

               R(-m, n) R(-2 p - 1 + m, n) R(1, m) R(-p, m)
               --------------------------------------------,  .
               R(1, n) R(-p, n) R(-m, m) R(-2 p - 1 + m, m)

                    Note that c(m,m)=1, c(m,n)=0, for n>m .

                                 I claim that

                (1) Sum(c(k,n)*a(m,n),n=0..m)=0 for 0<=k<m and

                 (2) Sum(c(m,n)*a(m,n),n=0..m)=b(m) for m>=0 .

                These two statements imply the theorem since (1)

           is the defining property for the cofactors of the m-th row

     up to normalization, and with the normalization such that  c(m,m)=1,

 which is tantamount to dividing the cofactors of the m-th row by the cofact\
    or

                    of the (m,m) entry, which is det(A(m-1)),

                   the left of (2) is det[A(m)]/det[A(m-1)].

    According to EKHAD, the left side of (1) is annihilated by the operator

                 -(m + 1) (m - p) + (m - k + 1) (m - 2 p + k) M

                                             2
                                            n  (m - p)
                     the certificate being, ----------,  .
                                              m + 1

               This implies that indeed the sum is zero for k<m .

     Also according to EKHAD, the sum on the left of (2) is annihilated by

                                            2
                               (m - 2 p - 1)
                    - --------------------------------- + M
                      4 (2 m - 2 p + 1) (2 m - 2 p - 1)

                                                          m               2
                                             (2 p)! (1/16)  R(-2 p - 1, m)
  which is the same operator annihilating, -----------------------------------
                                                                             2
                                           R(-p + 1/2, m) R(-p - 1/2, m) (p!)

                             the certificate being

         2                     2                                 2
     (5 m  - 12 m p - 2 m + 8 p  + 3 n m - n - 4 n p + 1 + 4 p) n
     -------------------------------------------------------------,  QED .
                              -m + n - 1

                      -----------------------------------

               The whole thing took, 14.763, seconds of CPU time

                                     14.858