A102905 a(n) = A113655(Fibonacci(n+1)).

3, 3, 2, 1, 5, 8, 15, 19, 36, 57, 89, 142, 233, 377, 612, 985, 1599, 2586, 4181, 6763, 10946, 17711, 28659, 46366, 75027, 121395, 196418, 317809, 514229, 832040, 1346271, 2178307, 3524580, 5702889, 9227465, 14930350, 24157817, 39088169

Offset

0,1

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,0,0,0,1,-1,-1).

Formula

a(n) = f(Fibonacci(n+1)), where f(n) = n-2 if (n mod 3) = 0, f(n) = n+2 if (n mod 3) = 1, otherwise f(n) = n.

a(n) = A113655(Fibonacci(n+1)).

G.f.: (3-4*x^2-4*x^3+2*x^4+2*x^5+2*x^6-4*x^7-x^8+2*x^9) / ((1-x)*(1+x)*(1+x^2)*(1-x-x^2)*(1+x^4)). - Colin Barker, Dec 11 2012

a(n) = (1 + 3*(-1)^n)/4 + Fibonacci(n+1) + (3/2)*(-1)^floor(n/2) * (n mod 2) + A014017(n) + A014017(n-1) - A014017(n-2). - G. C. Greubel, Dec 09 2022

Mathematica

f[n_]:= If[Mod[n, 3]==0, n-2, If[Mod[n, 3]==1, n+2, n]]; (* f=A113655 *)
Table[f[Fibonacci[n+1]], {n, 0, 50}]

Prog

(Magma)
A113655:= func< n | 6*Floor((n+2)/3) -(n+2) >;
A102905:= func< n | A113655(Fibonacci(n+1)) >;
[A102905(n): n in [0..50]]; // G. C. Greubel, Dec 09 2022
(SageMath)
def A113655(n): return 6*((n+2)//3) -(n+2)
def A102905(n): return A113655(fibonacci(n+1))
[A102905(n) for n in range(51)] # G. C. Greubel, Dec 09 2022

Crossrefs

Cf. A000045, A014017, A113655.

Sequence in context: A377787 A140259 A175644 * A020862 A131589 A338113

Adjacent sequences: A102902 A102903 A102904 * A102906 A102907 A102908

Keyword

nonn

Author

Roger L. Bagula, Mar 16 2005

Extensions

Edited by G. C. Greubel, Dec 09 2022

Status

approved