Theorem: Let F(m,n) be the coefficient of x^m*y^n in the Maclaurin series of

                                       1
                               -----------------
                               1 - x - y + 2 x y

                F(m,n) satisfies the following pure recurrences

        2 (1 + m) F(m, n)   (-4 + n - 3 m) F(1 + m, n)
        ----------------- + -------------------------- + F(2 + m, n) = 0
              2 + m                   2 + m



        2 (1 + n) F(m, n)   (-4 + m - 3 n) F(m, 1 + n)
        ----------------- + -------------------------- + F(m, 2 + n) = 0
              2 + n                   2 + n

                  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):=f(m+1,n), Nf(m,n):=f(m,n+1) )

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

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

                                 or equivalenty

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

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

                                M N - N - M + 2

                   The sequence of successive commutators is

                         [(M - 1) M (M N - N - M + 2)]

As you can see, the last entry, (M - 1) M (M N - N - M + 2), is a multiple of ,

    M N - N - M + 2

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

                                      QED .

----------------------------------------------------------------------------\
    ----

               The whole thing took , 0.908, seconds of CPU time