%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(n1,x)+x*p(n2,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(n1,x)+e*x*p(n2,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: evenindexed F(n)
%C A001519: oddindexed 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(n1)+a(n3)+a(n4).
%F G.f.: (1x)*(12*xx^2)/(13*xx^3x^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 0sequence *)
%t Simplify[FindLinearRecurrence[u1]] (* recurrence for 1sequence *)
%Y Cf. A192232, A192744, A192905, A192872.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Jul 12 2011
