OFFSET
1,1
COMMENTS
In calculating terms of this sequence, use the convention that a(n)=0 for n<=-702.
Sequences like this are more naturally considered with the first nonzero term in position 1. But this sequence would then begin with 702 terms consisting entirely of alternating 4 and 702.
This sequence has exactly 11969 terms, since a(11969)=0 and computing a(11970) would refer to itself.
From Chai Wah Wu, Jul 26 2020: (Start)
a(n) = 2*a(n-4) - a(n-8) for 8 < n <= 702.
a(n) = a(n-4) + a(n-8) - a(n-12) for 713 < n <= 1410.
a(n) = a(n-4) + a(n-12) - a(n-16) for 1423 < n <= 2106.
(End)
LINKS
Nathan Fox, Table of n, a(n) for n = 1..11969
MAPLE
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Nathan Fox, Mar 19 2017
STATUS
approved