Number of zeros in Fibonacci sequences mod $p$

Question

We know that Fibonacci sequences are periodic in mod $m$. For example, for $p\equiv \pm 1 \pmod 5$ and $p\equiv \pm 2 \pmod 5$ the periods for Fibonacci sequences modulo $p$ divide $p-1$ and $2p+2$ respectively. I want to find the number of zeros in the Fibonacci sequences modulo $p$ which have maximal period for these special cases. For example, if we take $p=11$ then the period is $10$ and the sequence is $\{0,1,1,2,3,5,8,2,10,1\}$ and there is only one zero. For $p=19$ the period is $18$ and the sequence is $\{0,1,1,2,3,5,8,13,2,15,17,13,11,5,16,2,18,1\}$ and there is only one zero. But, for $p=41$ the period is $40$ and the sequence is $\{0,1,1,2,3,5,8,13,21,34,14,7,21,28,8,36,3,39,1,40,0,40,40,39,38,36,33,28,20,7,27,34,20,13,33,5,38,2,40,1\}$ there are two zeros. My question is that: "Can we give an implicit formula, which count the number of zeros in a sequence for given maximal periods?"

Answer 1

New edit:I forgot the $\text{ord}_n(F_{k+1}) $ part actually tells you how many 0's there are before the end of the period. Thus here is new, edited answer:

Find $k$ where $k$ is the smallest positive integer such that $n\mid F_k$ ($n$ being the index of the highest Fibonacci Number ($F_n$). Now, the amout of 0's is $\lfloor {p \over k} \rfloor + 1$, where $\lfloor x \rfloor$ is the floor rounding function. If you start the sequence from $F_0$ otherwise if it is $F_1$ it would just be $\lfloor {n \over k} \rfloor$ obviously.

However, this doesn't tell us much does it? It just says that the distance between $0$'s in $F_n \mod p$ is always the same.

There may be a few special cases, but one of the most interesting cases is when $p = F_k$. That means that when you are modding to a Fibonacci number, the amount of $0$'s in the OTHER Fibonacci sequence is just $\lfloor {n \over k} \rfloor$. (Note that $k$ has a different context in this setting.)

Thus there is no easy way to calculate them without finding that initial $0$, however we know that since $\pi (p) < p^2$ The maximum "tries" to find it is going to be $p^2$.

Finally, in the entire context of my answer, $p$ does not necessarily prime.