OFFSET
0,1
COMMENTS
a(n) = 2k+1, where apparently k = 8m, m odd for n > 3.
More generally, a(k) is congruent to a(n) modulo 2^(n-1) for any k > n. - Charlie Neder, Mar 09 2019
LINKS
Hao Pan, Congruences for q-Lucas Numbers, Electron. J. Combin., 20, Issue 2 (2013), P29.
FORMULA
Recurrence: a(n) = a(n-1) + 2^(n-2)*a(n-2), starting 2, 1.
PROG
(Sage) # sage -i ore_algebra
from ore_algebra import *
R.<x> = QQ['x']; A.<Qx> = OreAlgebra(R, 'Qx', q=2)
print((Qx^2 - Qx - x).to_list([2, 1], 10))
CROSSREFS
KEYWORD
nonn
AUTHOR
Ralf Stephan, Apr 24 2014
STATUS
approved