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A192956 Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

%I A192956 #17 Sep 08 2022 08:45:58
%S A192956 1,0,4,9,20,38,69,120,204,341,564,926,1513,2464,4004,6497,10532,17062,
%T A192956 27629,44728,72396,117165,189604,306814,496465,803328,1299844,2103225,
%U A192956 3403124,5506406,8909589,14416056,23325708,37741829,61067604,98809502
%N A192956 Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.
%C A192956 The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) +- 1 + 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 A192956 G. C. Greubel, <a href="/A192956/b192956.txt">Table of n, a(n) for n = 0..1000</a>
%H A192956 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-1,1).
%F A192956 a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).
%F A192956 From _R. J. Mathar_, May 09 2014: (Start)
%F A192956 G.f.: (1 -3*x +6*x^2 -2*x^3)/((1-x-x^2)*(1-x)^2).
%F A192956 a(n) -2*a(n+1) +a(n+2) = A022096(n-3). (End)
%F A192956 a(n) = Fibonacci(n+3) + 4*Fibonacci(n+1) - (2*n+5). - _G. C. Greubel_, Jul 12 2019
%t A192956 (* First program *)
%t A192956 q = x^2; s = x + 1; z = 40;
%t A192956 p[0, x]:= 1;
%t A192956 p[n_, x_]:= x*p[n-1, x] + n^2 - 1;
%t A192956 Table[Expand[p[n, x]], {n, 0, 7}]
%t A192956 reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
%t A192956 t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
%t A192956 u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192956 *)
%t A192956 u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192957 *)
%t A192956 (* Second program *)
%t A192956 With[{F=Fibonacci}, Table[F[n+3]+4*F[n+1]-(2*n+5), {n,0,40}]] (* _G. C. Greubel_, Jul 12 2019 *)
%o A192956 (PARI) vector(40, n, n--; f=fibonacci; f(n+3)+4*f(n+1)-(2*n+5)) \\ _G. C. Greubel_, Jul 12 2019
%o A192956 (Magma) F:=Fibonacci; [F(n+3)+4*F(n+1)-(2*n+5): n in [0..40]]; // _G. C. Greubel_, Jul 12 2019
%o A192956 (Sage) f=fibonacci; [f(n+3)+4*f(n+1)-(2*n+5) for n in (0..40)] # _G. C. Greubel_, Jul 12 2019
%o A192956 (GAP) F:=Fibonacci;; List([0..40], n-> F(n+3)+4*F(n+1)-(2*n+5)); # _G. C. Greubel_, Jul 12 2019
%Y A192956 Cf. A000045, A192232, A192744, A192951, A192957.
%K A192956 nonn
%O A192956 0,3
%A A192956 _Clark Kimberling_, Jul 13 2011