Let a(n) be the maximum number of steps required to reach 0 0 0 0 with starting values between 0 and n-1, includng at least one 0 and one n-1. Rather suprisingly, this sequence is not monotonic. It starts: 0,4,4,6,7,6,7,7,7,8,7,9,8,10,8,9,9,8,9,10,9,9,10,9,10,10,10,9,10,9,10,11, 10, 10,10,10,10,12,11,10,10,10,11,10,13,11,10,10,10,11,12,11,11,11,12,11,11, 11,11,11,11,12,11,12,13,11,11,12,12,11,11,12,11,11,12,13,12,11,12,11 This sequence is not in the OEIS, although its running maximums are A065677. The record-setting foursomes up to 600 are: 0 0 0 0 1 0 0 0 3 1 0 0 4 2 1 0 9 4 1 0 11 5 2 0 13 6 2 0 31 14 5 0 37 17 6 0 44 20 7 0 105 48 17 0 125 57 20 0 149 68 24 0 355 162 57 0 423 193 68 0 504 230 81 0 These are all unique, up to the obvious symmetries: rotation and reversal, plus complement (subtract each term from n-1). The maximums in this sequence are A045794, essentially the same as A065678. The second and third values are in A034803 and A034804. See those sequences for formulas; these are related to the tribonacci numbers. Franklin T. Adams-Watters -----Original Message----- From: Cordwell, William R <wrcordw@sandia.gov> Suppose that one has four nonnegative integers in a row, A B C D. Create a subsequent row by abs(A-B) abs(B-C) abs(C-D) abs(D-A). Repeating this process will eventually produce 0 0 0 0, the end state. The object is to create a "long" chain of rows ending in 0 0 0 0. I can "backup" chains to create arbitrarily long chains, but the process seems a bit unwieldy. Is there any formulaic way of specifying an initial row that will produce a chain of length n? Bill C.