``
                Let, F([m[1], m[2]]),  be the number of ways of

                                of walking from

                                     [0, 0]

                        to the point , m = [m[1], m[2]]

                    in the , 2, -dimensional cubic lattice

                       using the following allowed steps:

                            {[0, 2], [2, 0], [1, 1]}



           F(m[1], m[2]), satisfies the following linear recurrences

                                    equation

                          with polynomial coefficients

                     in the, m[1], direction, it satisfies

    (m[2] + m[1] + 4) (m[2] + m[1] + 2) F(m[1], m[2])
3/4 -------------------------------------------------
                  (m[1] + 3) (m[1] + 4)

           (m[2] + 7 m[1] + 18) (m[2] + m[1] + 4) F(2 + m[1], m[2])
     - 1/4 --------------------------------------------------------
                            (m[1] + 3) (m[1] + 4)

     + F(m[1] + 4, m[2]) = 0

                     in the, m[2], direction, it satisfies

    (m[2] + m[1] + 4) (m[2] + m[1] + 2) F(m[1], m[2])
3/4 -------------------------------------------------
                  (3 + m[2]) (4 + m[2])

           (m[2] + m[1] + 4) (7 m[2] + m[1] + 18) F(m[1], 2 + m[2])
     - 1/4 --------------------------------------------------------
                            (3 + m[2]) (4 + m[2])

     + F(m[1], 4 + m[2]) = 0

        Let F(n) be the number of ways of walking from , [0, 0], to the

            point , [n, n], in the , 2, -dimensional cubic lattice

                       using the following allowed steps:

                            {[0, 2], [2, 0], [1, 1]}



            F(n) satisfies the following linear recurrence equation

                          with polynomial coefficients

                3 (1 + n) F(n)   (2 n + 3) F(1 + n)
              - -------------- - ------------------ + F(n + 2) = 0
                    n + 2              n + 2



                       subject to the initial conditions

                          F(0) = 1, F(1) = 1, F(2) = 3

            This implies, thanks to Birkhoff-Trijinski, that F(n) is

                        asymptotically a constant times

                              n /     3       1        135  \
                  0.48860251 3  |1 - ---- + ------ + -------|
                                |    16 n        2         3|
                                \           512 n    8192 n /
                  -------------------------------------------
                                      1/2
                                     n

                           which is roughly equal to

                   n /     0.1875000000   0.001953125000   0.01647949219\
      0.48860251 3.  |1. - ------------ + -------------- + -------------|
                     |          n                2               3      |
                     \                          n               n       /
      -------------------------------------------------------------------
                                      1/2
                                     n



             For the record, the first 31 terms of the sequence are

[1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227,

    1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729,

    3241135527, 9513228123, 27948336381, 82176836301, 241813226151,

    712070156203, 2098240353907, 6186675630819, 18252025766941]

               The whole thing took, 4.708, seconds of CPU time