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a(n) = Lucas(3*n) - 1.
4

%I #25 Sep 08 2022 08:45:15

%S 1,3,17,75,321,1363,5777,24475,103681,439203,1860497,7881195,33385281,

%T 141422323,599074577,2537720635,10749957121,45537549123,192900153617,

%U 817138163595,3461452808001,14662949395603,62113250390417,263115950957275,1114577054219521

%N a(n) = Lucas(3*n) - 1.

%C Main diagonal of triangle A100232.

%H Colin Barker, <a href="/A100233/b100233.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-3,-1).

%F a(n) = A014448(n) - 1.

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

%F G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-4*x-x^2)).

%F a(n) = [x^n] ( 1 + 2*x + sqrt(1 + 2*x + 5*x^2) )^n. Cf. A016064. - _Peter Bala_, Jun 23 2015

%F From _Colin Barker_, Jun 02 2016: (Start)

%F a(n) = -1+(2-sqrt(5))^n+(2+sqrt(5))^n.

%F a(n) = 5*a(n-1)-3*a(n-2)-a(n-3) for n>2.

%F G.f.: (1-2*x+5*x^2) / ((1-x)*(1-4*x-x^2)).

%F (End)

%t Table[LucasL[3*n] - 1, {n,0,50}] (* or *) LinearRecurrence[{5,-3,-1}, {1,3,17}, 30] (* _G. C. Greubel_, Dec 21 2017 *)

%o (PARI) a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-4*x-x^2)+x*O(x^n)),n))

%o (PARI) Vec((1-2*x+5*x^2)/((1-x)*(1-4*x-x^2)) + O(x^40)) \\ _Colin Barker_, Jun 02 2016

%o (Magma) I:=[1, 3, 17]; [n le 3 select I[n] else 5*Self(n-1) -3*Self(n-2) -Self(n-3): n in [1..30]]; // _G. C. Greubel_, Dec 21 2017

%Y Cf. A000032, A100230, A100232, A016064.

%K nonn,easy

%O 0,2

%A _Paul D. Hanna_, Nov 29 2004

%E New definition from _Ralf Stephan_, Dec 01 2004