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For a large number x>0, how many Fibonacci numbers are there in the interval [1,x]? I have saw the corresponding results in certain places but I have forgotten now. Can anyone help me? Thanks!

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If you know that $F_n=a\omega^n+b\omega^{-n}$ with $a,b$ constant and $\omega=(1+\sqrt5)/2$, you can conclude. –  Denis Serre Oct 11 '10 at 9:14
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Very closely related problem - mathoverflow.net/questions/39124/fibonacci-sequence-inversion –  Nurdin Takenov Oct 11 '10 at 10:02
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I think this question should be closed, because the first hit on Google for "Fibonacci number" gives you the answer: en.wikipedia.org/wiki/… –  S. Carnahan Oct 11 '10 at 11:22
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closed as too localized by Gjergji Zaimi, Yemon Choi, Wadim Zudilin, Felipe Voloch, S. Carnahan Oct 11 '10 at 11:22

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1 Answer

Let $G:=(1+\sqrt{5})/2;g:=(1-\sqrt{5})/2$, then the $n$-th Fibonacci number is $\frac{1}{\sqrt{5}}(G^n-g^n)$. Note that $|g|<1$. Hence the number of Fibonacci numbers $\le x$ is $\frac{\log \sqrt{5}x}{\log G}$ plus or minus 1 (and it is easy to see when you need to add or subtract 1).

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@Mark - Don't you mean $\frac{\log \sqrt{5} x}{\log G}$ ? –  Moshe Schwartz Oct 11 '10 at 9:59
    
@Moshe: Yes, of course. When I first typed the answer, I denoted $G$ by $g$ and $g$ by $G$, but then I decided that $g$ cannot be bigger than $G$, so I switched the notation, but not everywhere. –  Mark Sapir Oct 11 '10 at 10:21
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