

A130233


a(n) is the maximal k such that Fibonacci(k) <= n (the "lower" Fibonacci Inverse).


40



0, 2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
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OFFSET

0,2


COMMENTS

Inverse of the Fibonacci sequence (A000045), nearly, since a(Fibonacci(n)) = n except for n = 1 (see A130234 for another version). a(n) + 1 is equal to the partial sum of the Fibonacci indicator sequence (see A104162).


LINKS



FORMULA

a(n) = floor(log_phi((sqrt(5)*n + sqrt(5*n^2+4))/2)) where phi = (1+sqrt(5))/2 = A001622.
a(n) = floor(arcsinh(sqrt(5)*n/2) / log(phi)), with log(phi) = A002390.
G.f.: g(x) = 1/(1x) * Sum_{k>=1} x^Fibonacci(k).
a(n) = floor(log_phi(sqrt(5)*n+1)), n >= 0, where phi is the golden ratio.  Hieronymus Fischer, Jul 02 2007


EXAMPLE

a(10) = 6, since Fibonacci(6) = 8 <= 10 but Fibonacci(7) = 13 > 10.


MATHEMATICA

fibLLog[0] := 0; fibLLog[1] := 2; fibLLog[n_Integer] := fibLLog[n] = If[n < Fibonacci[fibLLog[n  1] + 1], fibLLog[n  1], fibLLog[n  1] + 1]; Table[fibLLog[n], {n, 0, 88}] (* Alonso del Arte, Sep 01 2013 *)


PROG



CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



