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A192617 Coefficient of x in the reduction of the n-th Fibonacci polynomial by x^3->x^2+x+1.

%I A192617 #16 Feb 19 2025 02:42:04
%S A192617 0,1,0,3,2,10,16,43,92,213,486,1100,2522,5719,13068,29721,67772,
%T A192617 154334,351670,801137,1825184,4158219,9473244,21582392,49169220,
%U A192617 112018989,255203904,581412535,1324587918,3017709810,6875021540,15662845615
%N A192617 Coefficient of x in the reduction of the n-th Fibonacci polynomial by x^3->x^2+x+1.
%C A192617 For discussions of polynomial reduction, see A192232 and A192744.
%H A192617 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-1,-4,1,1).
%F A192617 a(n) = a(n-1)+4*a(n-2)-a(n-3)-4a(n-4)+a(n-5)+a(n-6).
%F A192617 G.f.: x^2*(x^2+x-1)/(x^6+x^5-4*x^4-x^3+4*x^2+x-1). [_Colin Barker_, Jul 27 2012]
%e A192617 The first five polynomials p(n,x) and their reductions are as follows:
%e A192617 F1(x)=1 -> 1
%e A192617 F2(x)=x -> x
%e A192617 F3(x)=x^2+1 -> x^2+1
%e A192617 F4(x)=x^3+2x -> x^2+3x+1
%e A192617 F5(x)=x^4+3x^2+1 -> 4x^2+2x+2, so that
%e A192617 A192616=(1,0,1,1,2,...), A192617=(0,1,0,3,2,...), A192651=(0,0,1,1,5,...)
%t A192617 (See A192616.)
%t A192617 LinearRecurrence[{1,4,-1,-4,1,1},{0,1,0,3,2,10},40] (* _Harvey P. Dale_, Feb 23 2021 *)
%Y A192617 Cf. A192744, A192232, A192616.
%K A192617 nonn,easy
%O A192617 1,4
%A A192617 _Clark Kimberling_, Jul 09 2011