Theorem: Let F(m,n,k) be the number of lattice paths from (0,0,0) to (m,n,k)

        in the positive quadrant of the three-dimensional cubic lattice

                        where the Fundamental Steps are

                       {[0, 0, 1], [0, 1, 0], [1, 0, 0]}

Let M, N and K be the shift operators in the m, n and k directions, respecti\
    vely

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

             Then 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

We have to prove that F(m,n) 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 combinatorics, that F(m,n,k) is annihilated by

                            K N M - N M - K M - K N

                   The sequence of successive commutators is

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

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

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

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

                                      QED .

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