

A130259


Maximal index k of an even Fibonacci number (A001906) such that A001906(k) = Fib(2k) <= n (the 'lower' even Fibonacci Inverse).


10



0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 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, 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
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OFFSET

0,4


COMMENTS

Inverse of the even Fibonacci sequence (A001906), since a(A001906(n))=n (see A130260 for another version).
a(n)+1 is the number of even Fibonacci numbers (A001906) <=n.


LINKS

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


FORMULA

a(n) = floor(arcsinh(sqrt(5)*n/2)/(2*log(phi))), where phi=(1+sqrt(5))/2.
a(n) = A130260(n+1)  1.
G.f.: g(x) = 1/(1x)*Sum_{k>=1} x^Fibonacci(2*k).
a(n) = floor(1/2*log_phi(sqrt(5)*n+1)) for n>=0.


EXAMPLE

a(10)=3 because A001906(3)=8<=10, but A001906(4)=21>10.


MATHEMATICA

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


PROG

(PARI) vector(100, n, n; floor(log((sqrt(5)*n+1))/(2*log((1+sqrt(5))/2) ))) \\ G. C. Greubel, Sep 12 2018
(MAGMA) [Floor(Log((Sqrt(5)*n+1))/(2*Log((1+Sqrt(5))/2)))): n in [0..100]]; // G. C. Greubel, Sep 12 2018


CROSSREFS

Cf. partial sums A130261. Other related sequences: A000045, A001519, A130233, A130237, A130239, A130255, A130260, A104160. Lucas inverse: A130241  A130248.
Sequence in context: A077430 A105513 A004233 * A068549 A132173 A023968
Adjacent sequences: A130256 A130257 A130258 * A130260 A130261 A130262


KEYWORD

nonn


AUTHOR

Hieronymus Fischer, May 25 2007, Jul 02 2007


STATUS

approved



