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A130259
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Maximal index k of an even Fibonacci number (A001906) such that A001906(k) = Fib(2k) <= n (the 'lower' even Fibonacci Inverse).
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10
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0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 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, 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
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OFFSET
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0,4
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COMMENTS
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Inverse of the even Fibonacci sequence (A001906), since a(A001906(n))=n (see A130260 for another version).
a(n)+1 is the number of even Fibonacci numbers (A001906) <=n.
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LINKS
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FORMULA
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a(n) = floor(arcsinh(sqrt(5)*n/2)/(2*log(phi))), where phi=(1+sqrt(5))/2.
G.f.: g(x) = 1/(1-x)*Sum_{k>=1} x^Fibonacci(2*k).
a(n) = floor(1/2*log_phi(sqrt(5)*n+1)) for n>=0.
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EXAMPLE
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MATHEMATICA
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Table[Floor[1/2*Log[GoldenRatio, (Sqrt[5]*n + 1)]], {n, 0, 100}] (* G. C. Greubel, Sep 12 2018 *)
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PROG
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(PARI) vector(100, n, n--; floor(log((sqrt(5)*n+1))/(2*log((1+sqrt(5))/2) ))) \\ G. C. Greubel, Sep 12 2018
(Magma) [Floor(Log((Sqrt(5)*n+1))/(2*Log((1+Sqrt(5))/2)))): n in [0..100]]; // G. C. Greubel, Sep 12 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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