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

%I #25 Jan 01 2024 23:50:42

%S 1,4,14,41,122,364,1093,3280,9842,29525,88574,265720,797161,2391484,

%T 7174454,21523361,64570082,193710244,581130733,1743392200,5230176602,

%U 15690529805,47071589414,141214768240,423644304721

%N a(n) = 3*a(n-1) - a(n-3) + 3*a(n-4).

%H Paolo Xausa, <a href="/A132357/b132357.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-1,3).

%F O.g.f.: -(1+x+2*x^2)/((3*x-1)*(x+1)*(x^2-x+1)) = -(3/2)/(3*x-1)+(1/3)*(x-2)/(x^2-x+1)+(1/ 6)/(x+1). - _R. J. Mathar_, Nov 28 2007

%F a(n) = (1/2)*3^(n+1) + (1/6)*(-1)^n - (2/3)*cos(Pi*n/3). Or, a(n) = (1/2)*3^(n+1) + (1/2)*[ -1; -1; 1; 1; 1; -1]. - _Richard Choulet_, Jan 02 2008

%F a(n+1) - 3a(n) = A132367(n+1). - _Paul Curtz_, Dec 02 2007

%F 6*a(n) = (-1)^n +3^(n+2) -2*A057079(n+1). - _R. J. Mathar_, Oct 03 2021

%t LinearRecurrence[{3,0,-1,3},{1,4,14,41},50] (* _Paolo Xausa_, Dec 05 2023 *)

%o (PARI) a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 3,-1,0,3]^n*[1;4;14;41])[1,1] \\ _Charles R Greathouse IV_, Oct 08 2016

%Y First differences of A132353.

%Y Cf. A129339.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Nov 24 2007