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I would like to know about the Fibonacci numbers $F_n = 1,1,2,3,5,8, \dots$ in $\mathbb{Z}/p^k\mathbb{Z}$.

$$ \mathbb{P}[p^k \text{ divides } F_n ] = \frac{\#\{1 \leq n\leq N: F_n \equiv 0 \mod p^k \} }{N} $$

For various primes, $p$ and $N \to \infty$ does this have a limit in explicit form?

This amounts to just looking at the Fibonacci sequence $F_{n+1} = F_n + F_{n-1}$ over all prime powers $p^k$ at once. I am asking how often is it $0$ ?

Possibly related:

In light of all this, perhaps I should be asking for the order of $\frac{1 + \sqrt{5}}{2}$ in $\mathbb{F}_p$ or $\mathbb{F}_{p^2}$. The smallest number $k$ such that $(\tfrac{1 + \sqrt{5}}{2})^k = 1 \text{ mod }p$

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cactus314
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It is relatively easy to show that the Fibonacci sequence must be eventually periodic modulo any integer. From there, it suffices to take the repeating string of integers and count the zeros.

There seems to be a way to find the period modulo $p^k$, but I do not know of any way to go about counting the zeros except directly.

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Ben Grossmann
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    see http://webspace.ship.edu/msrenault/fibonacci/fib.htm – Will Jagy May 11 '14 at 19:29
  • @WillJagy So my question is related to the order of the Golden ration $\phi = \frac{1 + \sqrt{5}}{2}$ modulo $p^k$. We know $ (\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic but not the order of any particular element... – cactus314 May 11 '14 at 19:43