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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/(1-x)*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: A030601 A049839 * A131234 A204924 A172006 A172005

Adjacent sequences:  A130230 A130231 A130232 * A130234 A130235 A130236

KEYWORD

nonn,easy

AUTHOR

Hieronymus Fischer, May 17 2007

STATUS

approved

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Last modified December 18 18:54 EST 2018. Contains 318243 sequences. (Running on oeis4.)