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a(n+1) -3*a(n) + a(n-1) = (2/3)*(1+w^(n+1)+w^(2*n+2)); a(1) = 0, a(2) = 1; where w is the cubic root of unity.
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%I #14 Jul 31 2015 11:53:13

%S 0,1,5,14,37,99,260,681,1785,4674,12237,32039,83880,219601,574925,

%T 1505174,3940597,10316619,27009260,70711161,185124225,484661514,

%U 1268860317,3321919439,8696898000,22768774561,59609425685,156059502494

%N a(n+1) -3*a(n) + a(n-1) = (2/3)*(1+w^(n+1)+w^(2*n+2)); a(1) = 0, a(2) = 1; where w is the cubic root of unity.

%C w = exp(2Pi*I/3) = (-1-Sqrt(-3))/2.

%C The sequence (2/3)*(1+w^(n+1)+w^(2*n+2)) is "Period 3: repeat [0,2,0]."

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3, -1, 1, -3, 1).

%F G.f.: x^2*(1+x)*(1+x-x^2)/((1-x)*(1-3*x+x^2)*(1+x+x^2)). [Colin Barker, Jan 14 2012]

%F a(1)=0, a(2)=1, a(3)=5, a(4)=14, a(5)=37, a(n)=3*a(n-1)- a(n-2)+ a(n-3)-3*a(n-4)+a(n-5). - _Harvey P. Dale_, Aug 19 2012

%t a[1] = 0; a[2] = 1; w = Exp[2Pi*I/3]; a[n_] := (2/3)(1 + w^n + w^(2n)) + 3a[n - 1] - a[n - 2]; Table[ Simplify[ a[n]], {n, 1, 28}]

%t LinearRecurrence[{3,-1,1,-3,1},{0,1,5,14,37},30] (* _Harvey P. Dale_, Aug 19 2012 *)

%Y Cf. A071618.

%K nonn

%O 1,3

%A _Robert G. Wilson v_, Jun 24 2002