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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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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Let $G:=(1+\sqrt{5})/2;g:=(1\sqrt{5})/2$, then the $n$th Fibonacci number is $\frac{1}{\sqrt{5}}(G^ng^n)$. 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). 

