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a(n) = 4*a(n-1)+3*a(n-2) for n>1, a(0)=2, a(1)=4.
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%I #24 Jun 17 2023 07:25:58

%S 2,4,22,100,466,2164,10054,46708,216994,1008100,4683382,21757828,

%T 101081458,469599316,2181641638,10135364500,47086382914,218751625156,

%U 1016265649366,4721317472932,21934066839826,101900219778100

%N a(n) = 4*a(n-1)+3*a(n-2) for n>1, a(0)=2, a(1)=4.

%C This is the Lucas sequence V(4,-3). [_Bruno Berselli_, Jan 09 2013]

%H Vincenzo Librandi, <a href="/A080042/b080042.txt">Table of n, a(n) for n = 0..300</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence#Specific_names">Lucas sequence: Specific names</a>.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4, 3).

%F G.f.: (2-4*x)/(1-4*x-3*x^2).

%F a(n) = (2+sqrt(7))^n+(2-sqrt(7))^n.

%F G.f.: G(0)/x -2/x, where G(k)= 1 + 1/(1 - x*(7*k-4)/(x*(7*k+3) - 2/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 03 2013

%F a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 28*x^2))/2 )^n for n >= 1. - _Peter Bala_, Jun 23 2015

%t CoefficientList[Series[(2 - 4 t)/(1 - 4 t - 3 t^2), {t, 0, 25}], t]

%o (Sage) [lucas_number2(n,4,-3) for n in range(0, 22)] # _Zerinvary Lajos_, May 14 2009

%Y Cf. A015530: Lucas sequence U(4,-3).

%K nonn,easy

%O 0,1

%A Mario Catalani (mario.catalani(AT)unito.it), Jan 21 2003