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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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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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  • $\begingroup$ @Mark - Don't you mean $\frac{\log \sqrt{5} x}{\log G}$ ? $\endgroup$ Oct 11, 2010 at 9:59
  • $\begingroup$ @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. $\endgroup$
    – user6976
    Oct 11, 2010 at 10:21

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