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a(n) = 3*a(n-1) + a(n-2), with a(0)=4, a(1)=11.
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%I #21 Sep 08 2022 08:46:18

%S 4,11,37,122,403,1331,4396,14519,47953,158378,523087,1727639,5706004,

%T 18845651,62242957,205574522,678966523,2242474091,7406388796,

%U 24461640479,80791310233,266835571178,881298023767,2910729642479,9613486951204,31751190496091,104867058439477,346352365814522

%N a(n) = 3*a(n-1) + a(n-2), with a(0)=4, a(1)=11.

%H Vincenzo Librandi and Indranil Ghosh, <a href="/A282702/b282702.txt">Table of n, a(n) for n = 0..1922</a>

%H Sergio Falcon, <a href="http://dx.doi.org/10.1016/j.chaos.2016.03.038">The k-Fibonacci difference sequences</a>, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,1).

%F G.f.: (4-x) / (1-3*x-x^2). - _Vincenzo Librandi_, Feb 26 2017

%F a(n) = (2^(-n)*((3-sqrt(13))^n*(-5+2*sqrt(13)) + (3+sqrt(13))^n*(5+2*sqrt(13)))) / sqrt(13). - _Colin Barker_, Feb 26 2017

%t LinearRecurrence[{3, 1}, {4, 11}, 28] (* _Indranil Ghosh_, Feb 26 2017 *)

%t RecurrenceTable[{a[0]==4, a[1]==11, a[n]==3 a[n-1] + a[n-2]}, a, {n, 40}] (* or *) CoefficientList[Series[(4 - x)/(1 - 3 x - x^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, Feb 26 2017 *)

%o (Magma) I:=[4,11]; [n le 2 select I[n] else 3*Self(n-1)+Self(n-2): n in [1..40]]; // _Vincenzo Librandi_, Feb 26 2017

%o (PARI) Vec((4-x) / (1-3*x-x^2) + O(x^30)) \\ _Colin Barker_, Feb 26 2017

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_, Feb 25 2017