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 + 1)!

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

                                  Let b(m):=

                                   /-1\m
                                   |--|
                                   \16/
                            -------------------,  .
                            R(1/2, m) R(3/2, 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(1 + m, n) R(1, m)
                       ----------------------------,  .
                       R(1, n) R(-m, m) R(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

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

                         the certificate being, 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

                                     1
                           --------------------- + M
                           4 (2 m + 3) (2 m + 1)

                                                          /-1\m
                                                          |--|
                                                          \16/
          which is the same operator annihilating, -------------------
                                                   R(1/2, m) R(3/2, m)

                             the certificate being

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

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

               The whole thing took, 2.434, seconds of CPU time