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4 times the Lucas number A000032(n).
4

%I #36 Jan 03 2024 08:48:15

%S 8,4,12,16,28,44,72,116,188,304,492,796,1288,2084,3372,5456,8828,

%T 14284,23112,37396,60508,97904,158412,256316,414728,671044,1085772,

%U 1756816,2842588,4599404,7441992

%N 4 times the Lucas number A000032(n).

%C This is a second kind "autosequence" whose first kind companion is A022087. - _Jean-François Alcover_, Aug 20 2022

%H Vincenzo Librandi, <a href="/A156279/b156279.txt">Table of n, a(n) for n = 0..1000</a>

%H OEIS Wiki, <a href="https://oeis.org/wiki/Autosequence">Autosequence</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).

%F a(n) = 4*A000032(n).

%F a(n) = a(n-1) + a(n-2).

%F a(n) = A014217(n+3) - A014217(n-3), with A014217(-5) = -11, A014217(-4) = 6, A014217(-3) = -4, A014217(-2) = 2, A014217(-1) = -1 extended as proposed in A153263.

%F G.f. 4*(-2 + x) / (-1 + x + x^2). - _R. J. Mathar_, Mar 11 2011

%F a(n) = Lucas(n+3) - Lucas(n-3), where Lucas(i) for i = 0..2 gives -4, 3, -1. - _Bruno Berselli_, Jul 27 2017

%t Table[4*LucasL[n], {n,0,30}] (* _G. C. Greubel_, Dec 21 2017 *)

%o (PARI) a(n)=4*(fibonacci(n-1)+fibonacci(n+1)) \\ _Charles R Greathouse IV_, Oct 16 2015

%o (Magma) [4*Lucas(n): n in [0..30]]; // _G. C. Greubel_, Dec 21 2017

%Y Cf. A000032, A000045, A014217.

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Feb 07 2009