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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

%I

%S 1,0,1,5,16,49,153,480,1505,4717,14784,46337,145233,455200,1426721,

%T 4471733,14015632,43928817,137684905,431542080,1352570689,4239325789,

%U 13287204352,41645725825,130529073953,409113752000,1282274186177

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

%C 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:

%C ...

%C q(x)=x^2

%C s(x)=u*x+v

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

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

%C ...

%C 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:

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

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

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

%C ...

%C Examples:

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

%C 1 1 1 1 1 1 0 0... A001906..A001519

%C 1 1 1 1 0 0 1 0... A103609..A193609

%C 1 1 1 1 0 1 1 0... A192904..A192905

%C 1 1 1 1 1 1 0 0... A001519..A001906

%C 1 1 1 1 1 1 1 0... A192907..A192907

%C 1 1 1 1 1 1 0 1... A192908..A069403

%C 1 1 1 1 1 1 1 1... A192909..A192910

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

%C A001906: even-indexed F(n)

%C A001519: odd-indexed F(n)

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

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,1,1).

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

%F G.f.: (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4). [_Colin Barker_, Aug 31 2012]

%e The first six polynomials and reductions:

%e 1 -> 1

%e x -> x

%e x+x^3 -> 1+3x

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

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

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

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

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

%t (* To obtain general results, delete the next line. *)

%t u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 0;

%t q = x^2; s = u*x + v; z = 24;

%t p[0, x_] := a; p[1, x_] := b*x + c;

%t p[n_, x_] := d*(x^2)*p[n - 1, x] + e*x*p[n - 2, x] + f;

%t Table[Expand[p[n, x]], {n, 0, 8}]

%t reduce[{p1_, q_, s_, x_}] :=

%t FixedPoint[(s PolynomialQuotient @@ #1 +

%t PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]

%t t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];

%t u0 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]

%t (* A192904 *)

%t u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]

%t (* A192905 *)

%t Simplify[FindLinearRecurrence[u0]] (* recurrence for 0-sequence *)

%t Simplify[FindLinearRecurrence[u1]] (* recurrence for 1-sequence *)

%Y Cf. A192232, A192744, A192905, A192872.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Jul 12 2011

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