A130256 Minimal index k of an odd Fibonacci number A001519 such that A001519(k) = Fibonacci(2*k-1) >= n (the 'upper' odd Fibonacci Inverse).

0, 0, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset

0,3

Comments

Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=1 (see A130255 for another version).

a(n+1) is the number of odd Fibonacci numbers (A001519) <= n (for n >= 0).

Links

G. C. Greubel, Table of n, a(n) for n = 0..10000

Formula

a(n) = ceiling((1+arccosh(sqrt(5)*n/2)/log(phi))/2), where phi=(1+sqrt(5))/2.

G.f.: (x/(1-x))*Sum_{k>=0} x^Fibonacci(2*k-1).

a(n) = ceiling((1/2)*(1+log_phi(sqrt(5)*n-1))) for n >= 2, where phi=(1+sqrt(5))/2.

Example

a(10)=4 because A001519(4) = 13 >= 10, but A001519(3) = 5 < 10.

Mathematica

Join[{0, 0}, Table[Ceiling[1/2*(1 + Log[GoldenRatio, (Sqrt[5]*n - 1)])], {n, 2, 100}]] (* G. C. Greubel, Sep 12 2018 *)

Prog

(PARI) for(n=0, 100, print1(if(n==0, 0, if(n==1, 0, ceil((1/2)*(1 + log(sqrt(5)*n-1)/(log((1+sqrt(5))/2)))))), ", ")) \\ G. C. Greubel, Sep 12 2018
(Magma) [0, 0] cat [Ceiling((1/2)*(1 + Log(Sqrt(5)*n-1)/(Log((1+Sqrt(5))/2)))): n in [2..100]]; // G. C. Greubel, Sep 12 2018

Crossrefs

Cf. partial sums A130258.

Other related sequences: A000045, A001906, A130234, A130237, A130239, A130255, A130260.

Lucas inverse: A130241 - A130248.

Sequence in context: A084526 A356593 A081288 * A335741 A103586 A194847

Adjacent sequences: A130253 A130254 A130255 * A130257 A130258 A130259

Keyword

nonn

Author

Hieronymus Fischer, May 24 2007, Jul 02 2007

Status

approved