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 A130241 Maximal index k of a Lucas number such that Lucas(k) <= n (the 'lower' Lucas (A000032) Inverse). 24
 1, 1, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 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, 9, 9, 9, 9 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Inverse of the Lucas sequence (A000032), nearly, since a(Lucas(n))=n for n>=1 (see A130242 and A130247 for other versions). For n>=2, a(n)+1 is equal to the partial sum of the Lucas indicator sequence (see A102460). Identical to A130247 except for n=2. LINKS G. C. Greubel, Table of n, a(n) for n = 1..5000 FORMULA a(n) = floor(log_phi((n+sqrt(n^2+4))/2)) = floor(arcsinh((n+1)/2)/log(phi)) where phi=(1+sqrt(5))/2. a(n) = A130242(n+1) - 1 for n>=2. a(n) = A130247(n) except for n=2. G.f.: g(x) = 1/(1-x) * Sum{k>=1, x^Lucas(k)}. a(n) = floor(log_phi(n+1/2)) for n>=2, where phi is the golden ratio. EXAMPLE a(10)=4, since Lucas(4)=7<=10 but Lucas(5)=11>10. MATHEMATICA Join[{1}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *) PROG (PARI) for(n=1, 50, print1(floor(log((2*n+1)/2)/log((1+sqrt(5))/2)), ", ")) \\ G. C. Greubel, Sep 09 2018 (MAGMA) [Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [2..50]]; // G. C. Greubel, Sep 09 2018 CROSSREFS For partial sums see A130243. Other related sequences: A000032, A130242, A130245, A130247, A130249, A130255, A130259. Indicator sequence A102460. Fibonacci inverse see A130233 - A130240, A104162. Sequence in context: A324728 A327008 A316846 * A130247 A209869 A087839 Adjacent sequences:  A130238 A130239 A130240 * A130242 A130243 A130244 KEYWORD nonn AUTHOR Hieronymus Fischer, May 19 2007, Jul 02 2007 STATUS approved

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Last modified October 22 18:08 EDT 2019. Contains 328319 sequences. (Running on oeis4.)