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 - 6, steps, that start at, [4, 2, 0],  Then

                          A(n) = M(n - 2) - 2 M(n - 3)



                                     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

              4         3       3       2  2       2         2        3
3 (k - 1) (9 k  - 18 n k  + 18 k  + 15 n  k  + 75 k  - 45 n k  - 6 k n

           2                        3    4       2                /
     + 33 n  k - 87 n k + 26 k - 8 n  + n  + 25 n  - 10 n) k n!  /  ((k + 1) n
                                                                /

                                                2
    (n - 1) (n - 2) (n - 3) (n - 4) (n - 5) (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 - 2, k) + 2 F(n - 3, 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

      4       3       2                     3       2                    2
- k (n  - 10 n  + 29 n  + 4 n + 10 k - 4 k n  + 38 n  k - 146 n k + 143 k

            2      2  2       3        3      4       /
     - 6 n k  + 4 n  k  - 42 k  - 6 n k  + 9 k ) n!  /  (n (n - 1) (n - 2)
                                                    /

                                2
    (n - 3) (n - 4) (n - 5) (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, 0.326, second of CPU time