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. Let A(n) be the number of such walks with n-3 steps, that start at (2,1,0). Then A(n)=M(n-1)-M(n-3) . Proof: 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 2 2 2 (n - 2 n - 2 n k - 1 + k + 4 k) k n! ------------------------------------------ 2 (k + 1) n (n - 1) (n - 2) (k!) (n - 2 k)! To prove Amitai Regev's conjecture we must show that Sum(F(n-1,k)-F(n-3,k),k=0..n)=Sum(G(n,k),k=0..n). Hence the summand of the difference, F(n-1,k)-F(n-3,k)-G(n,k), is 2 2 2 k (n - n + 1 - 4 n k + 2 k + 3 k ) n! ------------------------------------------ 2 n (n - 1) (n - 2) (k + 1) (k!) (n - 2 k)! But this is a telescoping sum, the summand being H(n,k+1)-H(n,k) where H(n,k) is 2 (k - 1) k n! ---------------------------------- 2 n (n - 1) (n - 2) (k!) (n - 2 k)! Check! (or believe me). This concludes the proof of Amitai Regev's conjecture. QED. this took, 0.094, second of CPU time