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 + 2 i + 2 j, p + i + j)

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

                                  Let b(m):=

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

                  n
             (1/4)  R(-m, n) R(p + m, n) R(1, m) R(1/2 + p, m)
             -------------------------------------------------,  .
                                        m
             R(1, n) R(1/2 + p, n) (1/4)  R(-m, m) R(p + 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

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

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

               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

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

which is the same operator annihilating,

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

                             the certificate being

       3      2           2       2                     2
4 (10 m  + 6 m  n + 18 p m  + 15 m  + 7 m n + 7 m + 10 p  m + 6 p m n + 19 p m

          2                              3      2
     + 5 p  + 5 p + 2 n + 3 p n + 1 + 2 p  + 2 p  n) (2 n - 1 + 2 p) n/(

    -m + n - 1),  QED .

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               The whole thing took, 8.438, seconds of CPU time

                                     8.516