

A130233


Maximal index k of a Fibonacci number such that Fib(k)<=n (the 'lower' Fibonacci Inverse).


35



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

Inverse of the Fibonacci sequence (A000045), nearly, since a(Fib(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

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000


FORMULA

a(n) = floor(log_phi((sqrt(5)*n + sqrt(5*n^2+4))/2)) = floor(arcsinh(sqrt(5)*n/2)/log(phi)) where phi = (1+sqrt(5))/2. Also true: a(n) = A130234(n+1)  1. G.f.: g(x) = 1/(1x)*sum{k>=1, x^Fib(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 Fib(6) = 8 <= 10 but Fib(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

(PARI) a(n)=log(sqrt(5)*n+1.5)\log((1+sqrt(5))/2) \\ Charles R Greathouse IV, Mar 21 2012


CROSSREFS

Partial sums: A130235. Other related sequences: A000045, A130234, A130237, A130239, A130255, A130259, A104162, A108852, A130255, A130259. Lucas inverse: A130241  A130248.
Sequence in context: A199769 A030601 A049839 * A131234 A204924 A172006
Adjacent sequences: A130230 A130231 A130232 * A130234 A130235 A130236


KEYWORD

nonn,easy


AUTHOR

Hieronymus Fischer, May 17 2007


STATUS

approved



