A192389 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

0, 1, 3, 9, 22, 48, 96, 181, 327, 573, 982, 1656, 2760, 4561, 7491, 12249, 19966, 32472, 52728, 85525, 138615, 224541, 363598, 588624, 952752, 1541953, 2495331, 4037961, 6534022, 10572768, 17107632, 27681301, 44789895, 72472221, 117263206

Offset: 0

Author

Clark Kimberling, Jul 13 2011

Keywords

• nonn
• easy

Comments

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.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-5,1,2,-1).

Crossrefs

Cf. A000045, A192232, A192744, A192951.

Programs

GAP

List([0..40], n-> 3*Fibonacci(n+2)-n*(n+4)-9); # G. C. Greubel, Jul 24 2019

Magma

[3*Fibonacci(n+2)-n*(n+4)-9: n in [0..40]]; // G. C. Greubel, Jul 24 2019

Mathematica

(* First program *)
p[0, x]:= 1; p[n_, x_]:= x*p[n-1, x] +n^2 +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}] (* A192953 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192389 *)
(* Additional programs *)
CoefficientList[Series[x*(1-x+2*x^2)/((1-x)^3*(1-x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, May 13 2014 *)
Table[3*Fibonacci[n+4] -n*(n+4)-9, {n,0,40}] (* G. C. Greubel, Jul 24 2019 *)

PARI

Vec(x*(1-x+2*x^2)/((1-x)^3*(1-x-x^2)) + O(x^40)) \\ Colin Barker, May 12 2014

PARI

vector(40, n, n--; 3*fibonacci(n+2)-n*(n+4)-9) \\ G. C. Greubel, Jul 24 2019

Sage

[3*fibonacci(n+2)-n*(n+4)-9 for n in (0..40)] # G. C. Greubel, Jul 24 2019

Formula

a(n) = 4*a(n-1) - 5*a(n-2) + a(n-3) + 2*a(n-4) - a(n-5).

G.f.: x*(1-x+2*x^2)/((1-x)^3*(1-x-x^2)). - Colin Barker, May 12 2014

a(n) = 3*Fibonacci(n+4) - n*(n+4) - 9. - Ehren Metcalfe, Jul 13 2019