A130255 Maximal index k of an odd Fibonacci number (A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).

1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 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, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 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

Offset

1,2

Comments

Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=0 (see A130256 for another version). a(n)+1 is the number of odd Fibonacci numbers (A001519) <= n (for n >= 1).

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

a(n) = floor((1 + arcsinh(sqrt(5)*n/2)/log(phi))/2).

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

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

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

a(n) = floor((1/2)*(1 + log_phi(sqrt(5)*n + 1))) for n >= 1.

Example

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

Mathematica

Table[Floor[(1 +ArcSinh[Sqrt[5]*n/2]/Log[GoldenRatio])/2], {n, 1, 100}] (* G. C. Greubel, Sep 09 2018 *)

Prog

(PARI) phi=(1+sqrt(5))/2; vector(100, n, floor((1 +asinh(sqrt(5)*n/2)/log(phi))/2)) \\ G. C. Greubel, Sep 09 2018
(Magma) phi:=(1+Sqrt(5))/2; [Floor((1 +Argsinh(Sqrt(5)*n/2)/Log(phi))/2): n in [1..100]]; // G. C. Greubel, Sep 09 2018

Crossrefs

Cf. partial sums A130257. Other related sequences: A000045, A130233, A130237, A130239, A130256, A130259, A104160. Lucas inverse: A130241 - A130248.

Sequence in context: A330558 A237657 A244317 * A082527 A294235 A186188

Adjacent sequences: A130252 A130253 A130254 * A130256 A130257 A130258

Keyword

nonn

Author

Hieronymus Fischer, May 24 2007, Jul 02 2007

Status

approved