A192962 Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

1, 2, 7, 15, 30, 55, 97, 166, 279, 463, 762, 1247, 2033, 3306, 5367, 8703, 14102, 22839, 36977, 59854, 96871, 156767, 253682, 410495, 664225, 1074770, 1739047, 2813871, 4552974, 7366903, 11919937, 19286902, 31206903, 50493871, 81700842

Offset: 1

Author

Clark Kimberling, Jul 13 2011

Keywords

• nonn

Comments

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.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Crossrefs

Cf. A000045, A192232, A192744, A192951, A192963.

Programs

GAP

F:=Fibonacci;; List([1..40], n-> 3*F(n+1) +4*F(n) -2*(n+2)); # G. C. Greubel, Jul 12 2019

Magma

F:=Fibonacci; [3*F(n+1) +4*F(n) -2*(n+2): n in [1..40]]; // G. C. Greubel, Jul 12 2019

Mathematica

(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + n(n+1);
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192962 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192963 *)
(* Additional programs *)
CoefficientList[Series[(1-x+3x^2-x^3)/((1-x-x^2)(1-x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
With[{F=Fibonacci}, Table[3*F[n+1]+4*F[n] -2*(n+2), {n,1,40}]] (* G. C. Greubel, Jul 12 2019 *)

PARI

vector(40, n, f=fibonacci; 3*f(n+1)+4*f(n)-2*(n+2)) \\ G. C. Greubel, Jul 12 2019

Sage

f=fibonacci; [3*f(n+1) +4*f(n) -2*(n+2) for n in (1..40)] # G. C. Greubel, Jul 12 2019

Formula

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).

From R. J. Mathar, May 09 2014: (Start)

G.f.: x*(1 -x +3*x^2 -x^3)/((1-x-x^2)*(1-x)^2).

a(n) -2*a(n-1) + a(n-2) = A022120(n-4). (End)

a(n) = 3*Fibonacci(n+1) + 4*Fibonacci(n) - 2*(n+2). - G. C. Greubel, Jul 12 2019