A052995 Expansion of 2*x*(1 - x)/(1 - 3*x + x^2).
0, 2, 4, 10, 26, 68, 178, 466, 1220, 3194, 8362, 21892, 57314, 150050, 392836, 1028458, 2692538, 7049156, 18454930, 48315634, 126491972, 331160282, 866988874, 2269806340, 5942430146, 15557484098, 40730022148, 106632582346, 279167724890, 730870592324
Offset: 0
References
- A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 30.
- B. Muslu, Sayılar ve Bağlantılar 2, Luna, 2021, pages 60-61.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Younseok Choo, Some Results on the Infinite Sums of Reciprocal Generalized Fibonacci Numbers, International Journal of Mathematical Analysis, 12(12) (2018), 621-629.
- Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1072.
- Youngwoo Kwon, Binomial transforms of the modified k-Fibonacci-like sequence, arXiv:1804.08119 [math.NT], 2018.
- Index entries for linear recurrences with constant coefficients, signature (3,-1).
Crossrefs
Programs
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Maple
spec := [S, S=Prod(Sequence(Union(Prod(Sequence(Z),Z),Z)),Union(Z,Z))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
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Mathematica
LinearRecurrence[{3, -1}, {0, 2, 4}, 30] (* or *) Nest[Append[#, 3 #[[-1]] - #[[-2]]] &, {0, 2, 4}, 27] (* or *) CoefficientList[Series[-2 x (-1 + x)/(1 - 3 x + x^2), {x, 0, 29}], x] (* Michael De Vlieger, Jul 18 2018 *)
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PARI
concat(0, Vec(2*x*(1-x)/(1-3*x+x^2) + O(x^50))) \\ Colin Barker, Mar 30 2016
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PARI
a(n) = fibonacci(max(0,2*n-1))<<1; \\ Kevin Ryde, Mar 25 2021
Formula
G.f.: -2*x*(-1 + x)/(1 - 3*x + x^2).
a(0) = 0, a(1) = 2, a(2) = 4; for n > 0, a(n) - 3*a(n+1) + a(n+2) = 0.
a(n) = A069403(n-1)+1.
a(n) = Sum(2/5*(-1 + 4*_alpha)*_alpha^(-1-n), _alpha = RootOf(_Z^2 - 3*_Z + 1)).
a(n) = 2*Fibonacci(2*n-1) = 2*A001519(n) for n > 0. - Vladeta Jovovic, Mar 19 2003
a(n+2) = F(n)^2 + F(n+3)^2 = 2*F(n+1)^2 + 2*F(n+2)^2, where F = A000045. - N. J. A. Sloane, Feb 20 2005
a(n) = 1/2*(F(2*n+8) mod F(2*n+2)) for n > 2. - Gary Detlefs, Nov 22 2010
a(n) = F(n-3)*F(n-1) + F(n)*F(n+2) for n > 0, F(-2) = -1, F(-1) = 1. - Bruno Berselli, Nov 03 2015
a(n) = (2^(-n)*((3 - sqrt(5))^n*(1 + sqrt(5)) + (-1 + sqrt(5))*(3 + sqrt(5))^n))/sqrt(5) for n > 0. - Colin Barker, Mar 30 2016
a(n) = Fibonacci(2*n-2) + Lucas(2*n-2) for n > 0. - Bruno Berselli, Oct 13 2017
a(n) = Lucas(2*n) - Fibonacci(2*n) for n > 0. - Diego Rattaggi, Mar 08 2023
Extensions
More terms from James Sellers, Jun 05 2000
Comments