A192917 Coefficient of x in the reduction by (x^2 -> x+1) of the polynomial C(n)*x^n, where C=A022095.
0, 5, 6, 22, 51, 140, 360, 949, 2478, 6494, 16995, 44500, 116496, 304997, 798486, 2090470, 5472915, 14328284, 37511928, 98207509, 257110590, 673124270, 1762262211, 4613662372, 12078724896, 31622512325, 82788812070, 216743923894, 567442959603, 1485584954924
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(2*n+1) +2*F(n)^2 -(-1)^n); # G. C. Greubel, Jul 29 2019
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Magma
F:=Fibonacci; [F(2*n+1) +2*F(n)^2 -(-1)^n: n in [0..30]]; // G. C. Greubel, Jul 29 2019
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Mathematica
(* First program *) q = x^2; s = x + 1; z = 28; p[0, x_]:= 1; 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}] (* A192914 *) u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* see A192878 *) (* Second program *) With[{F=Fibonacci}, Table[F[2*n+1] +2*F[n]^2 -(-1)^n, {n,0,30}]] (* G. C. Greubel, Jul 28 2019 *) -
PARI
a(n) = round((2^(-1-n)*(-9*(-1)^n*2^(1+n)-(3-sqrt(5))^n*(-9+sqrt(5))+(3+sqrt(5))^n*(9+sqrt(5))))/5) \\ Colin Barker, Oct 01 2016
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PARI
concat(0, Vec((-x*(-5+4*x))/((1+x)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Oct 01 2016
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PARI
vector(30, n, n--; f=fibonacci; f(2*n+1) +2*f(n)^2 -(-1)^n) \\ G. C. Greubel, Jul 29 2019
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Sage
f=fibonacci; [f(2*n+1) +2*f(n)^2 -(-1)^n for n in (0..30)] # G. C. Greubel, Jul 29 2019
Formula
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(5 -4*x)/((1+x)*(1-3*x+x^2)). - R. J. Mathar, May 08 2014
a(n) = -3*a(n-1) +a(n-2) = 9*(-1)^(n+1). - R. J. Mathar, May 08 2014
a(n) = (2^(-1-n)*(-9*(-1)^n*2^(1+n)-(3-sqrt(5))^n*(-9+sqrt(5))+(3+sqrt(5))^n*(9+sqrt(5))))/5. - Colin Barker, Oct 01 2016
a(n) = Fibonacci(2*n+1) + 2*Fibonacci(n)^2 - (-1)^n. - G. C. Greubel, Jul 29 2019
Comments