A192921 Constant term in the reduction by (x^2->x+1) of the polynomial p(n,x) defined below at Comments.
1, 2, 2, 7, 16, 44, 113, 298, 778, 2039, 5336, 13972, 36577, 95762, 250706, 656359, 1718368, 4498748, 11777873, 30834874, 80726746, 211345367, 553309352, 1448582692, 3792438721, 9928733474, 25993761698, 68052551623, 178163893168, 466439127884, 1221153490481
Offset: 0
Examples
The coefficients in the polynomials p(n,x) are Fibonacci numbers. The first seven and their reductions: ... 1 -> 1 1 + x^2 -> 2 + x x + x^2 + x^3 -> 2 + 4*x 2*x^2 + x^3 + 2*x^4 -> 7 + 10*x 3*x^3 + 2*x^4 + 3*x^5 -> 16 + 27*x 5*x^4 + 3*x^5 + 5*x^6 -> 44 + 70*x 8*x^5 + 5*x^6 + 8*x^7 -> 113 + 184*x, so that A192921=(1,2,2,7,16,44,113,...).
Links
- Alois P. Heinz, 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
List([0..30], n -> Fibonacci(n-2)^2 +Fibonacci(n)*Fibonacci(n+1)); # G. C. Greubel, Feb 06 2019
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Magma
[Fibonacci(n-2)^2 + Fibonacci(n)*Fibonacci(n+1): n in [0..30]]; // G. C. Greubel, Feb 06 2019
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Maple
a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,2,2>>)[1,1]: seq(a(n), n=0..30); # Alois P. Heinz, Sep 28 2016
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Mathematica
q = x^2; s = x + 1; z = 28; p[0, x_] := 1; p[1, x_] := x^2 + 1; 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}] (* A192921 *) u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192879 *) LinearRecurrence[{2,2,-1}, {1,2,2}, 30] (* G. C. Greubel, Feb 06 2019 *)
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PARI
a(n) = round((2^(-n)*(-3*(-2)^n-(-4+sqrt(5))*(3+sqrt(5))^n+(3-sqrt(5))^n*(4+sqrt(5))))/5) \\ Colin Barker, Oct 01 2016
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PARI
Vec(-(2*x-1)*(1+2*x)/((1+x)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Oct 01 2016
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PARI
{a(n) = fibonacci(n-2)^2 +fibonacci(n)*fibonacci(n+1)}; \\ G. C. Greubel, Feb 06 2019
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Sage
[fibonacci(n-2)^2 +fibonacci(n)*fibonacci(n+1) for n in range(30)] # G. C. Greubel, Feb 06 2019
Formula
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-2*x)*(1+2*x) / ( (1+x)*(1-3*x+x^2) ). - R. J. Mathar, May 08 2014
a(n) = F(n-4)*F(n) + F(n-1)*F(n+2), where F(-4)=-3, F(-3)=2, F(-2)=-1, F(-1)=1. - Bruno Berselli, Nov 03 2015
a(n) = (2^(-n)*(-3*(-2)^n-(-4+sqrt(5))*(3+sqrt(5))^n+(3-sqrt(5))^n*(4+sqrt(5))))/5. - Colin Barker, Oct 01 2016
Comments