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q-Lucas numbers with q=2.
0

%I #27 Jan 04 2025 07:37:52

%S 2,1,3,5,17,57,329,2153,23209,298793,6240297,159222313,6549286441,

%T 332636583465,27158513845801,2752117405591081,447717208255194665,

%U 90629100354663736873,29432224060567101302313,11908369665747052420720169,7727389313799049256214259241,6251142704628989668810750223913

%N q-Lucas numbers with q=2.

%C a(n) = 2k+1, where apparently k = 8m, m odd for n > 3.

%C More generally, a(k) is congruent to a(n) modulo 2^(n-1) for any k > n. - _Charlie Neder_, Mar 09 2019

%H Hao Pan, <a href="https://doi.org/10.37236/2557">Congruences for q-Lucas Numbers</a>, Electron. J. Combin., 20, Issue 2 (2013), P29.

%F Recurrence: a(n) = a(n-1) + 2^(n-2)*a(n-2), starting 2, 1.

%o (Sage) # sage -i ore_algebra

%o from ore_algebra import *

%o R.<x> = QQ['x']; A.<Qx> = OreAlgebra(R, 'Qx', q=2)

%o print((Qx^2 - Qx - x).to_list([2,1], 10))

%Y Cf. A015459, A241497.

%K nonn

%O 0,1

%A _Ralf Stephan_, Apr 24 2014