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

                                                       1
                    whose (i,j) entry is a(i,j):=, ---------
                                                   i + j + p

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

                                  Let b(m):=

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

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

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

                   the certificate being, (n - 1 + 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

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

which is the same operator annihilating,

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

                             the certificate being

    3      2      2        2                                        2
(5 m  + 6 m  + 3 m  n + 9 m  p + 3 m p n + 8 m p + 3 m n + 2 m + 5 p  m + n

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

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               The whole thing took, 13.663, seconds of CPU time