the Gessel-Xin identities

      (5.47) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

          whose (i,j) entry is a(i,j):=, binomial(3 i + 3 j + 2, i + j)

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                              9 (6 m + 5) (3 m + 5) (3 m + 4) (6 m + 7)
      recurrence operator , - ----------------------------------------- + M
                                                                 2
                                  4 (4 m + 7) (4 m + 3) (5 + 4 m)

                   with the initial condition b(1)= , 3,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (m - n) (2 n + 3 + 2 m) (2 m + 1 - 2 n) (n + 1 + m)
the operator, -----------------------------------------------------
                     81 (3 n + 5) (3 n + 7) (n + 2) (n + 1)

                           2                        2          2       3
       2 (-35 + 81 m + 54 m  - 133 n + 54 m n + 36 m  n - 126 n  - 36 n ) N
     + --------------------------------------------------------------------
                          27 (n + 2) (3 n + 7) (3 n + 5)

        2
     + N ,  .

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .

      (5.48) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

                                         binomial(3 i + 3 j + 1, i + j)
          whose (i,j) entry is a(i,j):=, ------------------------------
                                                 3 i + 3 j + 1

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                              9 (6 m + 5) (3 m + 2) (3 m + 4) (6 m + 1)
      recurrence operator , - ----------------------------------------- + M
                                                                 2
                                  4 (5 + 4 m) (4 m + 1) (4 m + 3)

                   with the initial condition b(1)= , 2,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (2 n + 3) (n + 1 + m) (2 m - 1 - 2 n) (2 m + 2 n + 1) (m - n)
the operator, ---------------------------------------------------------------
                     81 (3 n + 5) (2 n + 1) (3 n + 4) (n + 2) (n + 1)

                           2                        2          2         2
     + 4 (-30 + 12 m + 24 m  - 134 n + 36 m n + 72 m  n - 211 n  + 18 m n

           2  2        3       4
     + 36 m  n  - 144 n  - 36 n ) N/(27 (n + 2) (3 n + 4) (2 n + 1) (3 n + 5))

        2
     + N ,  .

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .

      (5.49) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

                                       binomial(3 i + 3 j + 4, j + 1 + i)
        whose (i,j) entry is a(i,j):=, ----------------------------------
                                                 3 i + 3 j + 4

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                              9 (6 m + 5) (3 m + 5) (3 m + 4) (6 m + 7)
      recurrence operator , - ----------------------------------------- + M
                                                                 2
                                  4 (4 m + 7) (4 m + 3) (5 + 4 m)

                   with the initial condition b(1)= , 3,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (2 m - 1 - 2 n) (n + 2 + m) (m - n) (2 n + 3 + 2 m)
the operator, -----------------------------------------------------
                     81 (3 n + 5) (3 n + 7) (n + 2) (n + 1)

                         2                          2
       2 (2 n + 3) (-18 n  - 54 n - 35 + 27 m + 18 m ) N    2
     + ------------------------------------------------- + N ,  .
                27 (n + 2) (3 n + 7) (3 n + 5)

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .

      (5.50) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

                                       binomial(3 i + 3 j + 2, j + 1 + i)
        whose (i,j) entry is a(i,j):=, ----------------------------------
                                                 3 i + 3 j + 2

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                              9 (6 m + 5) (3 m + 5) (3 m + 4) (6 m + 7)
      recurrence operator , - ----------------------------------------- + M
                                                                 2
                                  4 (4 m + 7) (4 m + 3) (5 + 4 m)

                   with the initial condition b(1)= , 3,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (2 n + 3) (2 n + 3 + 2 m) (m - n) (n + 1 + m) (2 m + 1 - 2 n)
the operator, ---------------------------------------------------------------
                     81 (3 n + 5) (2 n + 1) (3 n + 4) (n + 2) (n + 1)

                           2                        2          2         2
     + 4 (-15 + 45 m + 30 m  - 98 n + 108 m n + 72 m  n - 193 n  + 54 m n

           2  2        3       4
     + 36 m  n  - 144 n  - 36 n ) N/(27 (n + 2) (3 n + 4) (2 n + 1) (3 n + 5))

        2
     + N ,  .

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .

      (5.51) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

                                       binomial(3 i + 3 j + 5, j + 2 + i)
        whose (i,j) entry is a(i,j):=, ----------------------------------
                                                 3 i + 3 j + 5

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                             9 (3 m + 5) (3 m + 7) (6 m + 7) (6 m + 11)
     recurrence operator , - ------------------------------------------ + M
                                                                 2
                                  4 (4 m + 9) (5 + 4 m) (4 m + 7)

                  with the initial condition b(1)= , 11/2,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (n + 2 + m) (m - n) (2 n + 5 + 2 m) (2 m + 1 - 2 n)
the operator, -----------------------------------------------------
                     81 (3 n + 8) (3 n + 4) (n + 2) (n + 1)

                         2                          2
       2 (2 n + 3) (-18 n  - 54 n - 14 + 45 m + 18 m ) N    2
     + ------------------------------------------------- + N ,  .
                27 (n + 2) (3 n + 4) (3 n + 8)

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .

      (5.52) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

                                      2 binomial(3 i + 3 j + 1, j + 1 + i)
       whose (i,j) entry is a(i,j):=, ------------------------------------
                                                 3 i + 3 j + 1

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                             9 (3 m + 5) (3 m + 7) (6 m + 7) (6 m + 11)
     recurrence operator , - ------------------------------------------ + M
                                                                 2
                                  4 (4 m + 9) (5 + 4 m) (4 m + 7)

                  with the initial condition b(1)= , 11/2,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (2 n + 3) (m - n) (n + 1 + m) (2 m + 3 - 2 n) (2 n + 5 + 2 m)
the operator, ---------------------------------------------------------------
                     81 (3 n + 5) (2 n + 1) (3 n + 4) (n + 2) (n + 1)

                          2                        2          2         2
     + 4 (21 + 60 m + 24 m  + 10 n + 180 m n + 72 m  n - 139 n  + 90 m n

           2  2        3       4
     + 36 m  n  - 144 n  - 36 n ) N/(27 (n + 2) (3 n + 4) (2 n + 1) (3 n + 5))

        2
     + N ,  .

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .

      (5.53) in Theorem 31 of Krathenthaler's Adv. Det. Cal.: A complement

                 Theorem: Let A(m) be the (m+1) by (m+1) matrix

                                      2 binomial(3 i + 3 j + 4, j + 2 + i)
       whose (i,j) entry is a(i,j):=, ------------------------------------
                                                 3 i + 3 j + 4

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

           Let  b(m) (m=0,1,2, ...) be the sequene annihilated by the

                             9 (6 m + 11) (3 m + 8) (3 m + 7) (6 m + 13)
     recurrence operator , - ------------------------------------------- + M
                                                                  2
                                  4 (4 m + 11) (4 m + 7) (4 m + 9)

                  with the initial condition b(1)= , 26/3,  .

                      Then [det A(m)]/[det A(m-1)] = b(m) .



                             Semi-Rigorous Proof:

                 Let c(m,n) be the unique vector of length m+1

                             (c(m,0), ..., c(m,m))

           satisfying c(m,m)=1, c(m,n)=0, for n>m, and annihilated by

              4 (m - n) (2 m + 3 - 2 n) (n + 2 + m) (2 n + 7 + 2 m)
the operator, -----------------------------------------------------
                     81 (3 n + 5) (3 n + 7) (n + 2) (n + 1)

                         2                          2
       2 (2 n + 3) (-18 n  - 54 n + 19 + 63 m + 18 m ) N    2
     + ------------------------------------------------- + N ,  .
                27 (n + 2) (3 n + 7) (3 n + 5)

                                 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) .

                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(n,n)=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)].

               But both c(m,n) and b(m) are holonomic functions,

     hence (1) and (2) are routinely provable within the Holonomic Paradigm.

           Since we know for sure that a proof exists (if it is true),

         and we know how to find it, we don't feel that it is necessary

 to actually bother the computer with the time- and memory- consuming chore \
    of finding it.

                     Instead,  we will be content to verify

                           (1) for 0<=k < m <=, 40, ,

                          and (2) for 0<=m <=, 40,  .

                        For this range these are indeed

                               true,  QED(sr) .