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A192780
Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+1. See Comments.
3
1, 0, 1, 1, 2, 5, 8, 19, 34, 71, 137, 272, 537, 1056, 2089, 4112, 8121, 16009, 31586, 62301, 122888, 242411, 478146, 943183, 1860433, 3669792, 7238769, 14278720, 28165265, 55556896, 109587889, 216165713, 426394178, 841076725, 1659052040
OFFSET
1,5
COMMENTS
For discussions of polynomial reduction, see A192232 and A192744.
FORMULA
a(n)=a(n-1)+3*a(n-2)-a(n-3)-3*a(n-4)+a(n-5)+a(n-6).
G.f.: -x*(x-1)*(1+x)*(x^2+x-1) / ( -1+x+3*x^2-x^3-3*x^4+x^5+x^6 ). - R. J. Mathar, May 06 2014
EXAMPLE
The first five polynomials p(n,x) and their reductions:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+2x+1
F5(x)=x^4+3x^2+1 -> 4x^2+1x+2, so that
A192777=(1,0,1,1,2,...), A192778=(0,1,0,2,1,...), A192779=(0,0,1,1,4,...)
MATHEMATICA
q = x^3; s = x^2 + 1; z = 40;
p[n_, x_] := Fibonacci[n, x];
Table[Expand[p[n, x]], {n, 1, 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, 1, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
(* A192780 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
(* A192781 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}]
(* A192782 *)
LinearRecurrence[{1, 3, -1, -3, 1, 1}, {1, 0, 1, 1, 2, 5}, 40] (* Harvey P. Dale, Nov 07 2021 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jul 09 2011
STATUS
approved