A192969 Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

1, 2, 6, 12, 23, 41, 71, 120, 200, 330, 541, 883, 1437, 2334, 3786, 6136, 9939, 16093, 26051, 42164, 68236, 110422, 178681, 289127, 467833, 756986, 1224846, 1981860, 3206735, 5188625, 8395391, 13584048, 21979472, 35563554, 57543061, 93106651

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) + 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

• Daniel Suteu, Table of n, a(n) for n = 0..999
• Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Crossrefs

Cf. A000045, A192232, A192744, A192951, A192970.

Programs

GAP

F:=Fibonacci;; List([0..40], n-> 2*F(n+2)+3*F(n+1)-n-4); # G. C. Greubel, Jul 11 2019

Magma

F:=Fibonacci; [2*F(n+2)+3*F(n+1)-n-4: n in [0..40]]; // G. C. Greubel, Jul 11 2019

Maple

F:= gfun:-rectoproc({a(0) = 1, a(1) = 2, a(n) = 1 + n + a(n-1) + a(n-2)},a(n),remember):
map(F, [$0..100]); # Robert Israel, Jan 18 2016

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 *)
(* Second program *)
Table[2*Fibonacci[n+2]+3*Fibonacci[n+1]-n-4, {n,0,40}] (* G. C. Greubel, Jul 11 2019 *)

PARI

vector(40, n, n--; f=fibonacci; 2*f(n+2)+3*f(n+1)-n-4) \\ G. C. Greubel, Jul 11 2019

Sage

f=fibonacci; [2*f(n+2)+3*f(n+1)-n-4 for n in (0..40)] # G. C. Greubel, Jul 11 2019

Sidef

func a((0)) { 1 }
func a((1)) { 2 }
func a(n) is cached { 1 + n + a(n-1) + a(n-2) }
100.times { |i| say a(i-1) }
# Daniel Suteu, Jan 12 2016

Formula

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

G.f.: (1 - x + 2*x^2 - x^3)/((1-x-x^2)*(1-x)^2). - R. J. Mathar, May 11 2014

a(0) = 1; a(1) = 2; a(n) = 1 + n + a(n-1) + a(n-2). - Daniel Suteu, Jan 12 2016

a(n) = 2*Fibonacci(n+2) + 3*Fibonacci(n+1) - n - 4. - G. C. Greubel, Jul 11 2019