OFFSET
1,2
COMMENTS
After a(24)=19, there are no more terms because a(24)-24 = -5 is not positive and a(24)+24 = 43 is equal to a(21).
If we only require that a(n+1) be either a(n)-n or a(n)+n, is there a sequence that contains every positive integer exactly once? I.e., can we take a walk on the positive integers starting at 1 and always moving (either left or right) a distance n on the n-th step so that we hit every positive integer exactly once?
A356080 is targeted to be such a sequence, but starting from 0. Its definition incorporates a limited look-ahead condition that is clearly a necessary condition for the sequence not to encounter a dead end (i.e., be finite) and is conjectured to be a sufficient condition. - Peter Munn, Feb 09 2023
EXAMPLE
a(9)=13, so a(10) is either 13-9=4 or 13+9=22. But 4 is not available since it equals a(3), so a(10)=22.
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Leroy Quet, Dec 15 2002
EXTENSIONS
Edited by Dean Hickerson, Dec 18 2002
STATUS
approved