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 April 9th, 2010, 09:56 PM #1 Newbie   Joined: Nov 2009 Posts: 6 Thanks: 0 Index of a fibonacci number Given a fibonacci number, is there any efficient way to compute the its position in the sequence ? I found this on Wikipedia [img] http://upload.wikimedia.org/math/4/6/2/ ... ea259e.png [/img] But when F very large (more than 10000 digits) there wont be sufficient precision to get the answer correctly . Is there any way like matrix exponentiation or something that will get me the answer ? PS: if A= [ [0,1],[1,1] ] .. computing A^n (can be done in O(logn)) will give the nth fibonacci number Thank you
 April 10th, 2010, 04:21 AM #2 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: Index of a fibonacci number Hello abhijith, To find numbers of fibonacci, you can use this $f_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\cdot\sqrt{5}}$ see here for dutch page, quite easily explained, or here for english page in more detail. Will this do? Hoempa
 April 10th, 2010, 02:17 PM #3 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: Index of a fibonacci number I think you were trying to find the index of the Fibonacci number. Yes, there are a few ways of doing this. Perhaps the easiest way is, $If \ A \ is \ a \ Fibonacci \ number, \ then \ it's \ index \ can \ be \ found \ by \\ F_n = \left \lfloor log_\phi (A \sqrt{5} ) + 1/2 \right \rfloor$

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