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A192911
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Constant term in the reduction by (x^3 -> x + 1) of the polynomial F(n+1)*x^n, where F(n)=A000045 (Fibonacci sequence).
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3
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1, 0, 0, 3, 5, 16, 52, 147, 442, 1320, 3916, 11664, 34717, 103298, 307440, 914949, 2722885, 8103424, 24116008, 71769885, 213589298, 635647790, 1891705884, 5629770720, 16754357925, 49861446392, 148389084968, 441610143507
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OFFSET
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0,4
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COMMENTS
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Regarding polynomial reduction, see A192232 and A192744. In the case of the reduction at A192911, each term in the three resulting sequences is a product of a Fibonacci number and a tribonacci numbers
All three obey the same linear recurrence, shown below at Formula.
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LINKS
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FORMULA
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a(n) = a(n-1) + 4*a(n-2) + 5*a(n-3) + 2*a(n-4) - a(n-5) + a(n-6).
G.f.: (x+1)*(2*x^2+2*x-1)/(x^6-x^5+2*x^4+5*x^3+4*x^2+x-1). - Colin Barker, Aug 31 2012
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EXAMPLE
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The first six polynomials and reductions:
1 -> 1
x -> 2
2*x^2 -> 2*x^2
3*x^3 -> 3 + 3*x + 3*x^2
5*x^4 -> 5 + 10*x + 10*x^2
8*x^5 -> 16 + 24*x + 32*x^2
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MATHEMATICA
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q = x^3; s = x^2 + x + 1; z = 22;
p[0, x_] := 1; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2;
Table[Expand[p[n, x]], {n, 0, 7}]
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}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192911 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192912 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}] (* A192913 *)
LinearRecurrence[{1, 4, 5, 2, -1, 1}, {1, 0, 0, 3, 5, 16}, 28] (* Ray Chandler, Aug 02 2015 *)
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PROG
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(PARI) my(x='x+O('x^30)); Vec((x+1)*(2*x^2+2*x-1)/(x^6-x^5+2*x^4+5*x^3 +4*x^2+x-1)) \\ G. C. Greubel, Jan 12 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (x+1)*(2*x^2+2*x-1)/(x^6-x^5+2*x^4+5*x^3+4*x^2+x-1) )); // G. C. Greubel, Jan 12 2019
(Sage) ((x+1)*(2*x^2+2*x-1)/(x^6-x^5+2*x^4+5*x^3+4*x^2+x-1)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019
(GAP) a:=[1, 0, 0, 3, 5, 16];; for n in [7..30] do a[n]:=a[n-1]+4*a[n-2] + 5*a[n-3]+2*a[n-4]-a[n-5]+a[n-6]; od; a; # G. C. Greubel, Jan 12 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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