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The problem:

               Given a Fibonacci number,find its index.

I am aware of the standard solution 'generate-hash-find'. I am just curious if there is some inverse system of matrix exponentiation or some other mathematical method that gives the solution.

J. M. ain't a mathematician
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Quixotic
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    $\lceil 1+\log_{\phi}(x)\rceil$ should be a good starting point. – J. M. ain't a mathematician Dec 10 '10 at 12:03
  • To expand a bit on J.M.'s comment, the idea is to (1) find an $n$ such that $F_n \leq x \leq F_{n+k}$ for some small $k$ - i.e. you want your first guess to be good - (2) use the matrix multiplication method to compute $F_n$ and $F_{n+1}$ and (3) if you haven't hit $x$ yet, then just keep computing $F_{n+2},F_{n+3},\ldots, F_{n+k}$ until you hit $x$ – kahen Dec 10 '10 at 12:16
  • In any event... – J. M. ain't a mathematician Dec 10 '10 at 12:23
  • @J.M. As for the downvote on this question: I downvoted it, because I expect the asker to do some rudimentary research before posting a question. A question about Fibonacci numbers that can be answered within 2 minutes by looking up the term "Fibonacci number" on Wikipedia is, in my mind, a bad question. I have flagged your comment by the way. – Alex B. Dec 10 '10 at 13:31
  • http://math.stackexchange.com/questions/9999/checking-if-a-number-is-a-fibonacci-or-not – Qiaochu Yuan Dec 10 '10 at 17:03

2 Answers2

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From wikipedia: "Similarly, if we already know that the number F is a Fibonacci number, we can determine its index within the sequence by

$$n = \bigg\lfloor \log_\varphi \left(F\cdot\sqrt{5} + \frac{1}{2} \right)\bigg\rfloor.$$

Alex B.
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  • Do you have a link? There are two different indices that both have the value of 1, so there can't be any formula that doesn't deal with that special case. – Acccumulation Aug 07 '19 at 18:05
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Using this other my answer
Using integers only, I would use Binary Search. Certainly you can compute $F_n$ only with integers, the simplest way is matrix exponentiation. Using Binary Search you can find numbers ``near'' your number $x$ and you will find $x = F_n$ (and $n$). I suppose this method is generic for anything monotone you can compute fast. To initiliaze the binary search, just keep doubling $ F_{2n} $

Binary search allows you to search for a number x in a sorted "array" F[] (in the programming sense). Use this method to search for your number. When you need F[n] just compute $F_n$. This will work because the sequence is strictly increasing except for the initial 1,1.

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