A combinatorial formula for finding the $N$-th object of a Fibonacci sequence?

Asked Feb 22 '18 at 09:12

Active Feb 22 '18 at 09:20

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How can I determine the general term of a Fibonacci Sequence? Is it possible for any one to calculate $F_{2013}$ and large numbers like this? Is there a general formula for the nth term of the Fibonacci Sequence? But since a Fibonacci sequence can be interpreted as representing the all possible combinations of ordering $n$ things without moving things more than two places. If so there must be some combinatorial solution? Something by using factorial or something? $( N/k)$?

edited Feb 22 '18 at 09:20

Robert Z

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asked Feb 22 '18 at 09:12

bilanush

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https://en.wikipedia.org/wiki/Fibonacci_number – Alexander Burstein Feb 22 '18 at 09:14
I don't see it. – bilanush Feb 22 '18 at 09:15
I think there's a Mathologer video where they even derive the formula. I'm not completely sure ... But you can search on YouTube :) – Matti P. Feb 22 '18 at 09:17
@bengy, look under "closed form expressions". – Mees de Vries Feb 22 '18 at 09:17
But I don't want a recursive one. Nor do I want just a formula. Show me the combinatorial one. Wouldn't it be shorter for you guys to just post it here. Thx – bilanush Feb 22 '18 at 09:21
@bengy Scroll down to "Closed-form expression" and "Computation by rounding". To see how it's derived, try this book, for example: https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf – Alexander Burstein Feb 22 '18 at 09:21
For an explicit formula see: https://proofwiki.org/wiki/Euler-Binet_Formula
BTW $F_{2013}$ is a "big" number: wolframalpha.com/input/?i=fibonacci(2013)
– Robert Z Feb 22 '18 at 09:27
With a bit of research, you can easily figure it out. Or, if you are familiar with the Lucas Numbers, even better – Mr Pie Feb 22 '18 at 10:41

A combinatorial formula for finding the $N$-th object of a Fibonacci sequence?

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