A192982 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

0, 1, 1, 3, 8, 20, 44, 89, 169, 307, 540, 928, 1568, 2617, 4329, 7115, 11640, 18980, 30876, 50145, 81345, 131851, 213596, 345888, 559968, 906385, 1466929, 2373939, 3841544, 6216212, 10058540, 16275593, 26335033, 42611587, 68947644, 111560320

Offset: 0

Author

Clark Kimberling, Jul 14 2011

Keywords

• nonn
• easy

Comments

The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) + (n-1)^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. A000032, A000045, A192232, A192744, A192951, A192981.

Programs

GAP

F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2) -(n^2+2*n+5)); # G. C. Greubel, Jul 25 2019

Magma

F:=Fibonacci; [F(n+4)+2*F(n+2) -(n^2+2*n+5): n in [0..40]]; // G. C. Greubel, Jul 25 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-1)^2;
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}] (* A192981 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192982 *)
(* Additional programs *)
CoefficientList[Series[x*(1-3*x+4*x^2)/((1-x-x^2)*(1-x)^3), {x, 0, 40}], x] (* Vincenzo Librandi, May 13 2014 *)
Table[LucasL[n+3]+Fibonacci[n+2]-(n^2+2*n+5), {n,0,40}] (* G. C. Greubel, Jul 25 2019 *)

PARI

vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2) -(n^2+2*n+5)) \\ G. C. Greubel, Jul 25 2019

Sage

f=fibonacci; [f(n+4)+2*f(n+2) -(n^2+2*n+5) for n in (0..40)] # G. C. Greubel, Jul 25 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*(1-3*x+4*x^2)/((1-x-x^2)*(1-x)^3). - R. J. Mathar, May 12 2014

a(n) = Lucas(n+3) + Fibonacci(n+2) - (n^2 + 2*n + 5). - G. C. Greubel, Jul 25 2019