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

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

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



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



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

                  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:=

                                1 + n + k + m
                              - ------------- + M
                                    1 + m

                                 or equivalenty

                            -1 - n - k - m + M + 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

                   The sequence of successive commutators is

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

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

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

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

                                      QED .

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