A128052 a(n) = (F(2*n-1) + F(2*n+1))*(5/6 - cos(2*Pi*n/3)/3), where F(n) = Fibonacci(n).

1, 3, 7, 9, 47, 123, 161, 843, 2207, 2889, 15127, 39603, 51841, 271443, 710647, 930249, 4870847, 12752043, 16692641, 87403803, 228826127, 299537289, 1568397607, 4106118243, 5374978561, 28143753123, 73681302247, 96450076809, 505019158607, 1322157322203

Offset: 0

Author

Paul Barry, Feb 13 2007

Keywords

• easy
• nonn

Comments

The a(n+1) are the numerators of A178381(4*n+3)/A178381(4*n+2). For the denominators see A179133(n). - Johannes W. Meijer, Jul 01 2010

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,18,0,0,-1).

Crossrefs

Cf. A128053.

Cf. A179134. Trisection: A023039.

Programs

Magma

I:=[1,3,7,9,47,123]; [n le 6 select I[n] else 18*Self(n-3)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, Jul 17 2019

Maple

with(combinat): nmax:=25; for n from 0 to nmax do a(n):= (fibonacci(2*n-1)+fibonacci(2*n+1))*(5/6-cos(2*Pi*n/3)/3) od: seq(a(n),n=0..nmax); # Johannes W. Meijer, Jul 01 2010

Mathematica

LinearRecurrence[{0, 0, 18, 0, 0, -1}, {1, 3, 7, 9, 47, 123}, 40] (* Vincenzo Librandi, Jul 17 2019 *)

Formula

Lim_{n->infinity} A128052(n+1)/A179133(n) = 1 + cos(Pi/5). - Johannes W. Meijer, Jul 01 2010

a(n) = Lucas(2*n)*(Fibonacci(n) mod 2 + 1)/2, Lucas(n)=A000032, Fibonacci(n)=A000045. - Gary Detlefs, Jan 19 2001

From Colin Barker, Jun 27 2013: (Start)

a(n) = 18*a(n-3) - a(n-6).

G.f: -(3*x^5 + 7*x^4 + 9*x^3 - 7*x^2 - 3*x - 1) / ((x^2 - 3*x + 1)*(x^4 + 3*x^3 + 8*x^2 + 3*x + 1)). (End)

With L(n) the Lucas number A000032, a(n) = L(2*n)/2 or L(2*n) according as n is, or is not, divisible by 3. - David Callan, Jul 17 2019