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A192904 Constant term in the reduction by (x^2->x+1) of the polynomial p(n,x) defined below at Comments. 7
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

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,...)

LINKS

Table of n, a(n) for n=0..26.

Index entries for linear recurrences with constant coefficients, signature (3,0,1,1).

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]

EXAMPLE

The first six polynomials and reductions:

1 -> 1

x -> x

x+x^3 -> 1+3x

x^2+x^3+x^5 -> 5+8x

x^2+2x^4+x^5+x^7 -> 16+25x

x^3+2x^4+3x^6+x^7+x^9 -> 49+79x, so that

A192904=(1,0,1,5,16,49,...) and

A192905=(0,1,3,8,25,79,...)

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 *)

CROSSREFS

Cf. A192232, A192744, A192905, A192872.

Sequence in context: A244410 A052909 A037536 * A082001 A084356 A007806

Adjacent sequences:  A192901 A192902 A192903 * A192905 A192906 A192907

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Jul 12 2011

STATUS

approved

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Last modified June 23 21:29 EDT 2017. Contains 288675 sequences.