Theorem: Let M(n) be the number of walks of length n

            from  [0,0,0], using unit positive steps and staying in

          the region x>=y>=z>=0. These are the famous Motzkin numbers,

                    as first proved by Amitai Regev in 1981.

                      Let A(n) be the number of such walks

             with , n - 10, steps, that start at, [6, 4, 0],  Then

 A(n) = M(n - 4) - 4 M(n - 5) + 3 M(n - 6) + 3 M(n - 7) - 2 M(n - 8) - M(n - 9)



                                     Proof:

             Thanks to Regev's seminal result (that can be reproved

                    with our approach), the summand for M(n),

                            let's call it F(n,k), is

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

                             The summand for A(n),

                            let's call it G(n,k), is

                                                  3       5      6         2
(k - 1) (k - 2) (k - 3) (-1560 n + 3240 k - 2425 n  - 95 n  + 5 n  + 3330 n

                          2              2         3            2  2
     - 12744 n k + 13195 n  k - 23893 n k  - 5112 n  k + 13188 n  k

                3         3  2         2  3           4       5          4  2
     - 15080 n k  - 2550 n  k  + 4210 n  k  - 3497 n k  - 50 n  k + 215 n  k

            3  3        2  4          5          2        4        4          6
     - 508 n  k  + 697 n  k  - 526 n k  + 12054 k  + 745 n  + 775 n  k + 171 k

             5         4          3         /
     + 1161 k  + 6495 k  + 14319 k ) k n!  /  ((k + 1) n (n - 1) (n - 2)
                                          /

                                                                2
    (n - 3) (n - 4) (n - 5) (n - 6) (n - 7) (n - 8) (n - 9) (k!)  (n - 2 k)!)

                    To prove the Theorem, we must show that

                      the sum (w.r.t. k), whose summand is

G(n, k) - F(n - 4, k) + 4 F(n - 5, k) - 3 F(n - 6, k) - 3 F(n - 7, k)

     + 2 F(n - 8, k) + F(n - 9, k)

                              is identically zero

        But this is a telescoping sum, the summand being H(n,k+1)-H(n,k)

                                where H(n,k) is

                                3        5       6       6          5  2
(k - 1) (-720 n + 2160 k + 856 n  + 580 n  - 84 n  - 56 n  k + 300 n  k

            4  3         3  4        2  5          6        2
     - 797 n  k  + 1114 n  k  - 792 n  k  + 270 n k  + 744 n  - 384 n k

              2              2          3            2  2             3
     - 13126 n  k + 43744 n k  + 17556 n  k - 70941 n  k  + 115386 n k

              3  2          2  3            4         5           4  2
     + 35804 n  k  - 77479 n  k  + 78524 n k  + 1140 n  k - 5763 n  k

              3  3          2  4           5         2         4         4
     + 13986 n  k  - 16836 n  k  + 8760 n k  - 4212 k  - 1380 n  - 7626 n  k

          7       7         6          5          4          3         /
     + 4 n  - 57 k  - 1011 k  - 29763 k  - 66561 k  - 39660 k ) k n!  /  (n
                                                                     /

    (n - 1) (n - 2) (n - 3) (n - 4) (n - 5) (n - 6) (n - 7) (n - 8) (n - 9)

        2
    (k!)  (n - 2 k)!)

        Check! (or believe me). This concludes the proof of our theorem

                       that is an analog of Amitai Regev's

          original conjecture (whose starting point was [2,1,0]). QED.

                      this took, 0.615, second of CPU time