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(2 p + i + j, p - i + j)

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

                                  Let b(m):=

                                             3 p           3 p
        (2 p)! R(1, m) R(1 + 2 p, m) R(1/2 + ---, m) R(1 + ---, m)
                                              2             2
       -------------------------------------------------------------,  .
                                                                   2
       R(1 + 3 p, m) R(1 + p/2, m) R(1/2 + p/2, m) R(1 + p, 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

                                           3 p
R(-m, n) R(1 + p, n) R(1 + 3 p, n) R(3/2 + ---, n) R(1, m) R(1 + 2 p, m)
                                            2

            3 p                         / /
    R(1/2 + ---, m) R(m + 2 + 3 p, m)  /  |R(1, n) R(1 + 2 p, n)
             2                        /   \

            3 p
    R(1/2 + ---, n) R(m + 2 + 3 p, n) R(-m, m) R(1 + p, m) R(1 + 3 p, m)
             2

            3 p    \
    R(3/2 + ---, m)|,  .
             2     /

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

                            (p - m + n) (k + n + 1 + 3 p) (n + 2 p) n
     the certificate being, -----------------------------------------,  .
                                          2 n + 1 + 3 p

               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

            (m + 1) (m + 2 p + 1) (2 m + 3 p + 1) (2 m + 3 p + 2)
          - ----------------------------------------------------- + M
            (2 m + p + 2) (m + p + 1) (m + 3 p + 1) (2 m + p + 1)

which is the same operator annihilating,

                                          3 p           3 p
     (2 p)! R(1, m) R(1 + 2 p, m) R(1/2 + ---, m) R(1 + ---, m)
                                           2             2
    -------------------------------------------------------------
                                                                2
    R(1 + 3 p, m) R(1 + p/2, m) R(1/2 + p/2, m) R(1 + p, m) (p!)

                             the certificate being

      2                     2
2 (4 m  + 6 m + 12 m p + 9 p  + 9 p + 2) (m + p + 1) (p - m + n) (n + 2 p) n/(

    (-m + n - 1) (2 n + 1 + 3 p)),  QED .

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

               The whole thing took, 83.676, seconds of CPU time

                                     83.769