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A192781
Coefficient of x in the reduction of the n-th Fibonacci polynomial by x^3->x^2+1.
3
0, 1, 0, 2, 1, 4, 6, 12, 25, 46, 96, 183, 368, 720, 1424, 2809, 5536, 10930, 21545, 42516, 83846, 165404, 326257, 643550, 1269440, 2503983, 4939232, 9742752, 19217952, 37908017, 74774848, 147495906, 290940561, 573890084, 1132017286, 2232942124
OFFSET
1,4
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^2*(x^2+x-1)/(x^6+x^5-3*x^4-x^3+3*x^2+x-1). [Colin Barker, Nov 23 2012]
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 09 2011
STATUS
approved