A192919 Constant term in the reduction by (x^2 -> x+1) of the polynomial F(n+4)*x^n, where F=A000045 (Fibonacci sequence).
3, 0, 8, 13, 42, 102, 275, 712, 1872, 4893, 12818, 33550, 87843, 229968, 602072, 1576237, 4126650, 10803702, 28284467, 74049688, 193864608, 507544125, 1328767778, 3478759198, 9107509827, 23843770272, 62423801000, 163427632717, 427859097162, 1120149658758
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
Programs
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GAP
F:=Fibonacci;; List([0..30], n-> F(n-1)*F(n+4)); # G. C. Greubel, Jul 28 2019
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Magma
F:=Fibonacci; [F(n-1)*F(n+4): n in [0..30]]; // G. C. Greubel, Jul 28 2019
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Maple
with(combinat):seq(fibonacci(n-1)*fibonacci(n+4), n=0..27);
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Mathematica
(* First program *) q = x^2; s = x + 1; z = 28; p[0, x_]:= 3; p[1, x_]:= 5 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}] (* A192919 *) u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192920 *) (* Second program *) With[{F=Fibonacci}, Table[F[n-1]*F[n+4], {n,0,30}]] (* G. C. Greubel, Jul 28 2019 *) -
PARI
a(n) = round((2^(-n)*(11*(-2)^n-(3-sqrt(5))^n*(-2+sqrt(5))+(2+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Oct 01 2016
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PARI
Vec((3+2*x^2-6*x)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Oct 01 2016
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PARI
vector(30, n, n--; f=fibonacci; f(n-1)*f(n+4)) \\ G. C. Greubel, Jul 28 2019
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Sage
f=fibonacci; [f(n-1)*f(n+4) for n in (0..30)] # G. C. Greubel, Jul 28 2019
Formula
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = Fibonacci(n-1)*Fibonacci(n+4). - Gary Detlefs, Oct 19 2011
G.f.: (3 -6*x +2*x^2)/((1+x)*(1-3*x+x^2)). - R. J. Mathar, May 08 2014
a(n) + a(n+1) = A001906(n+1). - R. J. Mathar, May 08 2014
a(n) = (2^(-n)*(11*(-2)^n-(3-sqrt(5))^n*(-2+sqrt(5))+(2+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Oct 01 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=2} 1/a(n) = (1/5) * A290565 - 17/150.
Sum_{n>=2} (-1)^n/a(n) = 1/phi - 83/150, where phi is the golden ratio (A001622). (End)
Comments