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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; fibLLog := 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

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Last modified October 21 04:40 EDT 2019. Contains 328291 sequences. (Running on oeis4.)