Theorem: Let h(r):=

                                  r! (m - r)!

                      Let H(n,r) be the Hankel determinant

                          det (h(r+i+j)), 0<=i,j<=n-1

                      Then we have the explicit evaluation

                                    H(n,r)=

               /    n      /
               | --------' |
             n |'  |  |    |
(r! (m - r)!)  |   |  |    |
               |   |  |    |
               |   |  |    |
               \ n[1] = 1  \

       n[1]
     --------'
    '  |  |      (n[2] - 1) (-m - 3 + n[2]) (n[2] - 1 + r) (n[2] - 3 - m + r)
       |  |    ----------------------------------------------------------------
       |  |                                                                   2
       |  |    (2 n[2] - 3 - m + r) (2 n[2] - 5 - m + r) (-4 + 2 n[2] + r - m)
     n[2] = 2

    \\
    ||
    ||
    ||
    ||
    ||
    //



             Proof: Let's call the expression on the r.h.s. G(n,r).

                      Obviously  H(n,r)=G(n,r), for n=0,1

                           It is readily checked that

                                                                   2
                      G(n - 1, r) G(n - 1, r + 2) - G(n - 1, r + 2)
            G(n, r) = ----------------------------------------------
                                     G(n - 2, r + 2)

                           Since, by Dodgson's rule:

                                                                   2
                      H(n - 1, r) H(n - 1, r + 2) - H(n - 1, r + 2)
            H(n, r) = ----------------------------------------------
                                     H(n - 2, r + 2)

                It follows by induction on n that, for all n>=0

                               H(n, r) = G(n, r)

                                      QED

                      It took, 39.141, seconds of CPU time

                        This does MacMahon's determinant

            An Explicit Evaluation Of a Certain Toeplitz Determinant

         By S. B. Ekhad (c/o D. Zeilberger, zeilberg@math.rutgers.edu)



                              Theorem: Let h(r):=

                                       1
                                    --------
                                    (m + r)!

                     Let H(n,r) be the Toeplitz determinant

                          det (h(r+i-j)), 0<=i,j<=n-1

                      Then we have the explicit evaluation

                                    H(n,r)=

                         /    n      /   n[1]                    \\
                         | --------' | --------'                 ||
             /   1    \n |'  |  |    |'  |  |        n[2] - 1    ||
             |--------|  |   |  |    |   |  |    ----------------||
             \(m + r)!/  |   |  |    |   |  |    m - 1 + r + n[2]||
                         |   |  |    |   |  |                    ||
                         \ n[1] = 1  \ n[2] = 2                  //



             Proof: Let's call the expression on the r.h.s. G(n,r).

                      Obviously  H(n,r)=G(n,r), for n=0,1

                           It is readily checked that

                                 2
                      G(n - 1, r)  - G(n - 1, 1 + r) G(n - 1, r - 1)
            G(n, r) = ----------------------------------------------
                                       G(n - 2, r)

                           Since, by Dodgson's rule:

                                 2
                      H(n - 1, r)  - H(n - 1, 1 + r) H(n - 1, r - 1)
            H(n, r) = ----------------------------------------------
                                       H(n - 2, r)

                It follows by induction on n that, for all n>=0

                               H(n, r) = G(n, r)

                                      QED

                       It took, .699, seconds of CPU time

                       This does the Hilbert  determinant

             An Explicit Evaluation Of a Certain Hankel Determinant

         By S. B. Ekhad (c/o D. Zeilberger, zeilberg@math.rutgers.edu)



                              Theorem: Let h(r):=

                                       1
                                     -----
                                     1 + r

                      Let H(n,r) be the Hankel determinant

                          det (h(r+i+j)), 0<=i,j<=n-1

                      Then we have the explicit evaluation

                                    H(n,r)=

         /    n
         | --------'
/  1  \n |'  |  |
|-----|  |   |  |
\1 + r/  |   |  |
         |   |  |
         \ n[1] = 1

    /   n[1]                                                        \\
    | --------'                       2               2             ||
    |'  |  |                (n[2] - 1)  (n[2] - 1 + r)              ||
    |   |  |    ----------------------------------------------------||
    |   |  |                                                       2||
    |   |  |    (2 n[2] - 1 + r) (2 n[2] - 3 + r) (-2 + 2 n[2] + r) ||
    \ n[2] = 2                                                      //



             Proof: Let's call the expression on the r.h.s. G(n,r).

                      Obviously  H(n,r)=G(n,r), for n=0,1

                           It is readily checked that

                                                                   2
                      G(n - 1, r) G(n - 1, 2 + r) - G(n - 1, 2 + r)
            G(n, r) = ----------------------------------------------
                                     G(n - 2, 2 + r)

                           Since, by Dodgson's rule:

                                                                   2
                      H(n - 1, r) H(n - 1, 2 + r) - H(n - 1, 2 + r)
            H(n, r) = ----------------------------------------------
                                     H(n - 2, 2 + r)

                It follows by induction on n that, for all n>=0

                               H(n, r) = G(n, r)

                                      QED

                       It took, .420, seconds of CPU time