1,3

That is, numbers n such that there is a choice of signs s_1, s_2, ..., s_n (each +1 or -1) so that (i) 0 <= Sum_{i = 1..j } i*s_i <= n for all 1 <= j <= n-1 and (ii) Sum_{i = 1..n } i*s_i = n. (This forces s_1 = s_2 = s_n = +1.)

It has been shown by Dick Hess and Benji Fisher that a number n >= 20 is in the sequence iff n == 0 or 1 mod 4. (For a proof see the Applegate link.) It is easy to see that n == 0 or 1 mod 4 is a necessary condition.

Further comments from David Applegate and njas, Jul 14 2008: (Start) An obvious greedy algorithm (working backwards) does the following: For j = n, n-1, ..., 1, let target_j = n - Sum_{i = j+1..n} i * s_i, and set s_j = +1 if target_j >= j, and s_j = -1 otherwise. This works unless we hit one of five exceptions, in which we must set s_j = -1 instead of +1.

The five exceptions are when (j, target_j) is (5,5), (6,9), (7,14), (8,8), or (9,13). The algorithm also works for the more general case when the target total target_n is different from n, with the additional exception of (8,20). (End)

Ivan Moscovich, "MATH - Isn't It Beautiful !", 2009 (to appear)

D. Applegate, Notes on A141000

Ivan Moscovich, Grasshop Puzzle-Game

4 is a member because we can take s_1 = s_2 = s_4 = +1, s_3 = -1. Note in particular that 1+2-3+4 = 4. (See illustration.)

Cf. A000980, A063865.

Adjacent sequences: A140997 A140998 A140999 this_sequence A141001 A141002 A141003

Sequence in context: A033662 A136640 A068949 this_sequence A140485 A010397 A020684

nonn,nice

Ivan Moscovich (i.moscovich2(AT)chello.nl), Jul 07 2008