Hi mathgeeks,

Given an extremely large fibonacci number X, and it's position in the fibonacci sequence N, is there any way to determine the N-1th fibo num, WITHOUT doing the bruteforce dance of counting from 1 upwards?

I've tried attacking the problem by dividing the number by two, and determining the difference between the N-1th, and N-2th, but that would require calculating Fib(n-3)/2, which makes the problem circular.

Also would be interested, if this can be proven not to be possible, thanks


closed as off topic by Todd Trimble, Mark Sapir, José Figueroa-O'Farrill, Andrés E. Caicedo, Qiaochu Yuan Dec 26 '10 at 16:49

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    $\begingroup$ Just divide by the golden mean and round to the nearest integer. This gives the correct (n-1)st Fibonacci provided that F_n is greater than 1. $\endgroup$ – Todd Trimble Dec 26 '10 at 15:34
  • $\begingroup$ @Todd, that's a very similar approach to the one suggested below by J.C. Ottem. $\endgroup$ – Mariano Suárez-Álvarez Dec 26 '10 at 15:41
  • $\begingroup$ Yes, I see that now. Anyway, I vote to close this question as being not anywhere near research level. $\endgroup$ – Todd Trimble Dec 26 '10 at 15:42

For large $n$, $\frac{f_{n+1}}{f_n} \approx \phi = 1.618..$. Hence you can get $f_n$ from $f_{n+1}$ by rounding $\frac{f_{n+1}}{\phi}$ to the nearest integer.


The n-th Fibonacci number can be calculated in log n steps, see, e.g., here


Yes, mathgeeks can compute a large fibonacci number without doing any kind of dance.


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    $\begingroup$ But here the idea is that you know one of the numbers and want the previous one. Using Binet's formula in the obvious way forgets that piece of information. $\endgroup$ – Mariano Suárez-Álvarez Dec 26 '10 at 15:25

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