Theorem: Let F(m,n,k) be the coefficient of x^m*y^n*z^k in the Maclaurin ser\
    ies of

                                       1
                            -----------------------
                            1 - x - y - z + 4 x y z

               F(m,n,k) satisfies the following pure recurrences

2 (k + 1 + m - n) (k - 1 - m - n) F(m, n, k)
-------------------------------------------- -
          (m + 2) (-m - 1 + n + k)

                 2                  2                            2
    (-4 + 3 n + n  + 3 k - 2 k n + k  - 7 m + 2 m n + 2 m k - 3 m )

    F(m + 1, n, k)/((m + 2) (-m - 1 + n + k)) + F(m + 2, n, k) = 0



2 (k + 1 + n - m) (k - 1 - m - n) F(m, n, k)
-------------------------------------------- -
          (n + 2) (-n - 1 + m + k)

                 2                  2                            2
    (-4 + 3 m + m  + 3 k - 2 m k + k  - 7 n + 2 m n + 2 k n - 3 n )

    F(m, n + 1, k)/((n + 2) (-n - 1 + m + k)) + F(m, n + 2, k) = 0



2 (k + 1 + m - n) (m - 1 - k - n) F(m, n, k)
-------------------------------------------- -
          (k + 2) (-k - 1 + n + m)

                 2                  2                            2
    (-4 + 3 n + n  + 3 m - 2 m n + m  - 7 k + 2 k n + 2 m k - 3 k )

    F(m, n, k + 1)/((k + 2) (-k - 1 + n + m)) + F(m, n, k + 2) = 0

                  Proof: We will only do the first recurrence

   Let M and N be the shift operators in the m and n directions, respectively

     (M f(m,n,k):=f(m+1,n,k), Nf(m,n):=f(m,n+1,k), Kf(m,n,k):=f(m,n,k+1) )

We have to prove that F(m,n,k) is annihilated by the operator, let's call it\
     Q:=

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

                    2                  2                            2
       (-4 + 3 n + n  + 3 k - 2 k n + k  - 7 m + 2 m n + 2 m k - 3 m ) M    2
     - ----------------------------------------------------------------- + M
                           (m + 2) (-m - 1 + n + k)

                                 or equivalenty

   2              2                2                    2
2 k  - 4 k n + 2 n  - 2 - 4 m - 2 m  + 4 M - 3 M n - M n  - 3 M k + 2 M k n

          2                                  2        2      2        2
     - M k  + 7 m M - 2 M m n - 2 M m k + 3 m  M - 2 M  + 2 M  n + 2 M  k

            2    2        2        2  2
     - 3 m M  + M  m n + M  m k - m  M

        We know, by the obvious algebra, that F(m,n,k) is annihilated by

                          M N K - N K - M K - M N + 4

                   The sequence of successive commutators is

                                                                        2
[-4 K (k - 1 - m - n) M + (4 k - 4 n + 4 + 4 m) M N + 4 K (-1 + k - m) M

                                               2                         3
     + (-4 K - 2 K k - 2 K n - 4 - 4 m + 4 n) M  N - K (-m - 1 + n + k) M

                                           3
     + (-n + K k + m - k + 2 K + 1 + K n) M  N,

                  2
    2 K (-2 + M) M  N (M N K - N K - M K - M N + 4)]

                                              2
As you can see, the last entry, 2 K (-2 + M) M  N (M N K - N K - M K - M N + 4),

    is a multiple of , M N K - N K - M K - M N + 4

The proof follows by backwards induction, after checking the boundary condit\
    ions

                                      QED .

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               The whole thing took , 19.505, seconds of CPU time