A192904 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) defined below at Comments.

1, 0, 1, 5, 16, 49, 153, 480, 1505, 4717, 14784, 46337, 145233, 455200, 1426721, 4471733, 14015632, 43928817, 137684905, 431542080, 1352570689, 4239325789, 13287204352, 41645725825, 130529073953, 409113752000, 1282274186177

Offset: 0

Author

Clark Kimberling, Jul 12 2011

Keywords

• nonn
• easy

Comments

The titular polynomial is defined by p(n,x) = (x^2)*p(n-1,x) + x*p(n-2,x), with p(0,x) = 1, p(1,x) = x. The resulting sequence typifies a general class which we shall describe here. Suppose that u,v,a,b,c,d,e,f are numbers used to define these polynomials:

...

q(x) = x^2

s(x) = u*x + v

p(0,x) = a, p(1,x) = b*x + c

p(n,x) = d*(x^2)*p(n-1,x) + e*x*p(n-2,x) + f.

...

We shall assume that u is not 0 and that {d,e} is not {0}. The reduction of p(n,x) by the repeated substitution q(x) -> s(x), as defined and described at A192232 and A192744, has the form h(n) + k(n)*x. The numerical sequences h and k are linear recurrence sequences, formally of order 5. The Mathematica program below, with first line deleted, shows initial terms and recurrence coefficients, which imply these properties:

(1) the recurrence coefficients depend only on u,v,d,e; the parameters a,b,c,f affect only the initial terms.

(2) if e=0 or v=0, the order of recurrence is <= 3;

(3) if e=0 and v=0, the recurrence coefficients are 1+d*u^2 and -d*u^2 (cf. similar results at A192872).

...

Examples:

u v a b c d e f... seq h.....seq k

1 1 1 1 1 1 0 0... A001906..A001519

1 1 1 1 0 0 1 0... A103609..A193609

1 1 1 1 0 1 1 0... A192904..A192905

1 1 1 1 1 1 0 0... A001519..A001906

1 1 1 1 1 1 1 0... A192907..A192907

1 1 1 1 1 1 0 1... A192908..A069403

1 1 1 1 1 1 1 1... A192909..A192910

The terms of these sequences involve Fibonacci numbers, F(n)=A000045(n); e.g.,

A001906: even-indexed F(n)

A001519: odd-indexed F(n)

A103609: (1,1,1,1,2,2,3,3,5,5,8,8,...)

Examples

			The first six polynomials and reductions:
1 -> 1
x -> x
x + x^3 -> 1 + 3*x
x^2 + x^3 + x^5 -> 5 + 8*x
x^2 + 2*x^4 + x^5 + x^7 -> 16 + 25*x
x^3 + 2*x^4 + 3*x^6 + x^7 + x^9 -> 49 + 79*x, so that
A192904 = (1,0,1,5,16,49,...) and
A192905 = (0,1,3,8,25,79,...)
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,0,1,1).

Crossrefs

Cf. A192232, A192744, A192905, A192872.

Programs

GAP

a:=[1,0,1,5];; for n in [5..40] do a[n]:=3*a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 10 2019

Magma

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4) )); // G. C. Greubel, Jan 10 2019

Mathematica

(* To obtain general results, delete the next line. *)
u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 0;
q = x^2; s = u*x + v; z = 24;
p[0, x_] := a; p[1, x_] := b*x + c;
p[n_, x_] :=  d*(x^2)*p[n - 1, x] + e*x*p[n - 2, x] + f;
Table[Expand[p[n, x]], {n, 0, 8}]
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}];
u0 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192904 *)
u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192905 *)
Simplify[FindLinearRecurrence[u0]] (* recurrence for 0-sequence *)
Simplify[FindLinearRecurrence[u1]] (* recurrence for 1-sequence *)
LinearRecurrence[{3,0,1,1}, {1,0,1,5}, 40] (* G. C. Greubel, Jan 10 2019 *)

PARI

my(x='x+O('x^40)); Vec((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)) \\ G. C. Greubel, Jan 10 2019

Sage

((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 10 2019

Formula

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

G.f.: (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4). - Colin Barker, Aug 31 2012