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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 (list; graph; refs; listen; history; text; internal format)
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
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/(1-x)*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: A105513 A004233 A363832 * A068549 A132173 A023968
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, May 25 2007, Jul 02 2007
STATUS
approved

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Last modified April 19 21:09 EDT 2024. Contains 371798 sequences. (Running on oeis4.)