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A130255
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Maximal index k of an odd Fibonacci number (A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).
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12
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1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
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OFFSET
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1,2
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COMMENTS
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Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=0 (see A130256 for another version). a(n)+1 is the number of odd Fibonacci numbers (A001519) <= n (for n >= 1).
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LINKS
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FORMULA
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a(n) = floor((1 + arcsinh(sqrt(5)*n/2)/log(phi))/2).
a(n) = floor((1 + arccosh(sqrt(5)*n/2)/log(phi))/2).
a(n) = floor((1 + log_phi(sqrt(5)*n))/2) for n >= 1, where phi = (1 + sqrt(5))/2.
G.f.: g(x) = 1/(1-x)*Sum_{k>=1} x^Fibonacci(2k-1).
a(n) = floor((1/2)*(1 + log_phi(sqrt(5)*n + 1))) for n >= 1.
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EXAMPLE
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MATHEMATICA
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Table[Floor[(1 +ArcSinh[Sqrt[5]*n/2]/Log[GoldenRatio])/2], {n, 1, 100}] (* G. C. Greubel, Sep 09 2018 *)
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PROG
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(PARI) phi=(1+sqrt(5))/2; vector(100, n, floor((1 +asinh(sqrt(5)*n/2)/log(phi))/2)) \\ G. C. Greubel, Sep 09 2018
(Magma) phi:=(1+Sqrt(5))/2; [Floor((1 +Argsinh(Sqrt(5)*n/2)/Log(phi))/2): n in [1..100]]; // G. C. Greubel, Sep 09 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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