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%I #19 Sep 08 2022 08:45:58
%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 G. C. Greubel, <a href="/A192904/b192904.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">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 + 3*x
%e x^2 + x^3 + x^5 -> 5 + 8*x
%e x^2 + 2*x^4 + x^5 + x^7 -> 16 + 25*x
%e x^3 + 2*x^4 + 3*x^6 + x^7 + x^9 -> 49 + 79*x, 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_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + 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}] (* A192904 *)
%t u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192905 *)
%t Simplify[FindLinearRecurrence[u0]] (* recurrence for 0-sequence *)
%t Simplify[FindLinearRecurrence[u1]] (* recurrence for 1-sequence *)
%t LinearRecurrence[{3,0,1,1}, {1,0,1,5}, 40] (* _G. C. Greubel_, Jan 10 2019 *)
%o (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
%o (Magma) m:=40; R<x>:=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
%o (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
%o (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
%Y Cf. A192232, A192744, A192905, A192872.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Jul 12 2011
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