%I #12 Sep 08 2022 08:45:58
%S 0,1,1,4,11,27,58,115,215,386,673,1149,1932,3213,5301,8696,14207,
%T 23143,37622,61071,99035,160486,259941,420889,681336,1102777,1784713,
%U 2888140,4673555,7562451,12236818,19800139,32037887,51839018,83877961
%N Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.
%C The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) - n + n^2, with p(0,x)=1. For an introduction to reductions of polynomials by substitutions such as x^2->x+1, see A192232 and A192744.
%H G. C. Greubel, <a href="/A192965/b192965.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,1,2,-1).
%F a(n) = 4*a(n-1) - 5*a(n-2) + a(n-3) + 2*a(n-4) - a(n-5).
%F G.f.: x*(1 -3*x +5*x^2 -x^3)/((1-x-x^2)*(1-x)^3). - _R. J. Mathar_, May 11 2014
%F a(n) = Fibonacci(n+4) + 3*Fibonacci(n+2) - (n^2 + 3*n + 6). - _G. C. Greubel_, Jul 11 2019
%t (* First program *)
%t q = x^2; s = x + 1; z = 40;
%t p[0, x]:= 1;
%t p[n_, x_]:= x*p[n-1, x] + n(n-1);
%t Table[Expand[p[n, x]], {n, 0, 7}]
%t reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
%t t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
%t u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192964 *)
%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192965 *)
%t (* Second program *)
%t With[{F=Fibonacci}, Table[F[n+4]+3*F[n+2] -(n^2+3*n+6), {n,0,40}]] (* _G. C. Greubel_, Jul 11 2019 *)
%o (PARI) vector(40, n, n--; f=fibonacci; f(n+4)+3*f(n+2)-(n^2+3*n+6)) \\ _G. C. Greubel_, Jul 11 2019
%o (Magma) F:=Fibonacci; [F(n+4) +3*F(n+2) -(n^2+3*n+6): n in [0..40]]; // _G. C. Greubel_, Jul 11 2019
%o (Sage) f=fibonacci; [f(n+4) +3*f(n+2) -(n^2+3*n+6) for n in (0..40)] # _G. C. Greubel_, Jul 11 2019
%o (GAP) F:=Fibonacci;; List([0..40], n-> F(n+4) +3*F(n+2) -(n^2+3*n+6)); # _G. C. Greubel_, Jul 11 2019
%Y Cf. A000045, A192232, A192744, A192951, A192964.
%K nonn
%O 0,4
%A _Clark Kimberling_, Jul 13 2011