A192978 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

0, 1, 4, 12, 29, 62, 122, 227, 406, 706, 1203, 2020, 3356, 5533, 9072, 14816, 24129, 39218, 63654, 103215, 167250, 270886, 438599, 709992, 1149144, 1859737, 3009532, 4869972, 7880261, 12751046, 20632178, 33384155, 54017326, 87402538

Offset: 0

Author

Clark Kimberling, Jul 13 2011

Keywords

• nonn

Comments

The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) + 1 + 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.

Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n)=1, and T(r,c) = T(r-1,c-1) + T(r-2,c-1). The sum of the terms in row(n) is a(n+1). - J. M. Bergot, Apr 14 2013

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.

Programs

GAP

List([0..40], n-> Lucas(1,-1, n+5)[2] -(n^2+5*n+11)); # G. C. Greubel, Jul 24 2019

Magma

[Lucas(n+5)-(n^2+5*n+11): n in [0..40]]; // G. C. Greubel, Jul 24 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^2 +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}] (* A027181 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192978 *)
(* Additional programs *)
CoefficientList[Series[x*(1+x^2)/((1-x-x^2)*(1-x)^3), {x, 0, 40}], x] (* Vincenzo Librandi, May 13 2014 *)
Table[LucasL[n+5] -(n^2+5*n+11), {n,0,40}] (* G. C. Greubel, Jul 24 2019 *)
LinearRecurrence[{4,-5,1,2,-1},{0,1,4,12,29},40] (* Harvey P. Dale, Dec 24 2023 *)

PARI

vector(40, n, n--; f=fibonacci; f(n+6)+f(n+4) -(n^2+5*n+11)) \\ G. C. Greubel, Jul 24 2019

Sage

[lucas_number2(n+5, 1,-1) -(n^2+5*n+11) 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)/((1-x-x^2)*(1-x)^3). - R. J. Mathar, May 12 2014

a(n) = Lucas(n+5) - n*(n+5) - 11. - Ehren Metcalfe, Jul 13 2019

From Stefano Spezia, Jul 13 2019: (Start)

a(n) = (1/2)*(-22 + (11 - 5*sqrt(5))*((1/2)*(1 - sqrt(5)))^n + 11*((1/2)* (1 + sqrt(5)))^n + 5*sqrt(5)*((1/2)*(1 + sqrt(5)))^n - 10*n - 2*n^2).

E.g.f.: (1/2)*(2 + sqrt(5))*((-47 + 21*sqrt(5))*exp(-(1/2)*(-1 + sqrt(5))*x) + (3 + sqrt(5))*exp((1/2)*(1 + sqrt(5))*x) - 2*(-2 + sqrt(5))*exp(x)*(11 + 6*x + x^2)).

(End)