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A192781 Coefficient of x in the reduction of the n-th Fibonacci polynomial by x^3->x^2+1.

%I A192781 #12 Feb 18 2025 11:54:49
%S A192781 0,1,0,2,1,4,6,12,25,46,96,183,368,720,1424,2809,5536,10930,21545,
%T A192781 42516,83846,165404,326257,643550,1269440,2503983,4939232,9742752,
%U A192781 19217952,37908017,74774848,147495906,290940561,573890084,1132017286,2232942124
%N A192781 Coefficient of x in the reduction of the n-th Fibonacci polynomial by x^3->x^2+1.
%C A192781 For discussions of polynomial reduction, see A192232 and A192744.
%H A192781 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-1,-3,1,1).
%F A192781 a(n) = a(n-1)+3*a(n-2)-a(n-3)-3*a(n-4)+a(n-5)+a(n-6).
%F A192781 G.f.: x^2*(x^2+x-1)/(x^6+x^5-3*x^4-x^3+3*x^2+x-1). [_Colin Barker_, Nov 23 2012]
%e A192781 The first five polynomials p(n,x) and their reductions:
%e A192781 F1(x)=1 -> 1
%e A192781 F2(x)=x -> x
%e A192781 F3(x)=x^2+1 -> x^2+1
%e A192781 F4(x)=x^3+2x -> x^2+2x+1
%e A192781 F5(x)=x^4+3x^2+1 -> 4x^2+1x+2, so that
%e A192781 A192777=(1,0,1,1,2,...), A192778=(0,1,0,2,1,...), A192779=(0,0,1,1,4,...)
%t A192781 q = x^3; s = x^2 + 1; z = 40;
%t A192781 p[n_, x_] := Fibonacci[n, x];
%t A192781 Table[Expand[p[n, x]], {n, 1, 7}]
%t A192781 reduce[{p1_, q_, s_, x_}] :=
%t A192781 FixedPoint[(s PolynomialQuotient @@ #1 +
%t A192781        PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
%t A192781 t = Table[reduce[{p[n, x], q, s, x}], {n, 1, z}];
%t A192781 u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
%t A192781   (* A192780 *)
%t A192781 u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
%t A192781   (* A192781 *)
%t A192781 u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}]
%t A192781   (* A192782 *)
%Y A192781 Cf. A192744, A192232, A192616, A192780, A192782.
%K A192781 nonn,easy
%O A192781 1,4
%A A192781 _Clark Kimberling_, Jul 09 2011