%I #35 Jan 01 2024 11:49:37
%S 1,1,11,71,481,3241,21851,147311,993121,6695281,45137291,304300151,
%T 2051487361,13830424921,93239986331,628592042591,4237752187201,
%U 28569473336161,192605600952971,1298480972398631,8753913839156641
%N a(0) = 1, a(1) = 1, a(n) = 6*a(n-1) + 5*a(n-2) for n > 1.
%C Hankel transform is [1, 10, 0, 0, 0, 0, 0, 0, 0, 0, ...]. - _Philippe Deléham_, Dec 04 2008
%H G. C. Greubel, <a href="/A123362/b123362.txt">Table of n, a(n) for n = 0..1000</a>
%H Lucyna Trojnar-Spelina, Iwona Włoch, <a href="https://doi.org/10.1007/s40995-019-00757-7">On Generalized Pell and Pell-Lucas Numbers</a>, Iranian Journal of Science and Technology, Transactions A: Science (2019), 1-7.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,5).
%F a(n) = Sum_{k = 0..n} 5^(n - k)*A122542(n, k).
%F G.f.: (1 - 5*x)/(1 - 6*x - 5*x^2).
%t LinearRecurrence[{6, 5}, {1, 1}, 50] (* _G. C. Greubel_, Oct 12 2017 *)
%o (PARI) x='x+O('x^50); Vec((1-5*x)/(1 - 6*x - 5*x^2)) \\ _G. C. Greubel_, Oct 12 2017
%K nonn,easy
%O 0,3
%A _Philippe Deléham_, Oct 12 2006