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):=, (i + j + p)!

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

                                  Let b(m):=

                          p! R(1, m) R(p + 1, 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(1, m) R(p + 1, m)
                       ----------------------------,  .
                       R(1, n) R(p + 1, n) R(-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 + 1) + (m - k + 1) M

                     the certificate being, n (n + 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 + p + 1) + M

        which is the same operator annihilating, p! R(1, m) R(p + 1, m)

                             the certificate being

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

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               The whole thing took, 4.704, seconds of CPU time