A192970 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

0, 1, 3, 9, 21, 44, 85, 156, 276, 476, 806, 1347, 2230, 3667, 6001, 9787, 15923, 25862, 41955, 68006, 110170, 178406, 288828, 467509, 756636, 1224469, 1981455, 3206301, 5188161, 8394896, 13583521, 21978912, 35562960, 57542432, 93105986

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) + n(n+3)/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, A192970.

Programs

GAP

F:=Fibonacci;; List([0..40], n-> 2*F(n+4)+F(n+2)-(n^2+7*n+14)/2); # G. C. Greubel, Jul 24 2019

Magma

[Fibonacci(n+4)+Lucas(n+3)-(n^2+7*n+14)/2: n in [0..40]]; // Vincenzo Librandi, Jul 13 2019

Mathematica

(* First progream *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + n*(n+3)/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}] (* A192969 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192970 *)
(* Additional programs *)
CoefficientList[Series[x*(1-x+2*x^2-x^3)/((1-x-x^2)*(1-x)^3), {x,0,40}], x] (* Vincenzo Librandi, Jul 13 2019 *)
Table[LucasL[n+3]+Fibonacci[n+4]-(n^2+7*n+14)/2, {n,0,40}] (* G. C. Greubel, Jul 24 2019 *)

PARI

vector(40, n, n--; f=fibonacci; 2*f(n+4)+f(n+2)-(n^2+7*n+14)/2) \\ G. C. Greubel, Jul 24 2019

Sage

f=fibonacci; [2*f(n+4)+f(n+2)-(n^2+7*n+14)/2 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-x^3)/((1-x-x^2)*(1-x)^3). - R. J. Mathar, May 11 2014

a(n) = Fibonacci(n+4) + Lucas(n+3) - (n^2 + 7*n + 14)/2. - Ehren Metcalfe, Jul 13 2019