

A130256


Minimal index k of an odd Fibonacci number A001519 such that A001519(k) = Fib(2*k1) >= n (the 'upper' odd Fibonacci Inverse).


9



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
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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/(1x)*Sum_{k>=0} x^Fib(2*k1).
a(n) = ceiling(1/2*(1+log_phi(sqrt(5)*n1))) 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)*n1)/(log((1+sqrt(5))/2)))))), ", ")) \\ G. C. Greubel, Sep 12 2018
(MAGMA) [0, 0] cat [Ceiling((1/2)*(1 + Log(Sqrt(5)*n1)/(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: A084516 A084526 A081288 * A103586 A194847 A262070
Adjacent sequences: A130253 A130254 A130255 * A130257 A130258 A130259


KEYWORD

nonn


AUTHOR

Hieronymus Fischer, May 24 2007, Jul 02 2007


STATUS

approved



