Fix $n \in \Bbb N, n> 1$. Now choose a two digit base-$n$ number $ab $ say. There's $n^2$ choices for this.
Consider the number $0.c_1 c_2 c_3 \ldots$ where the $c_i$ are defined recursively:
$c_1:= a$, $c_2:=b $ and $c_i:= c_{i-1}+c_{i-2} \pmod n $
First note that this number is rational. There can be at most $n^2$ different two digit numbers found in the expansion, paired as $c_i c_{i+1}$. Therefore by the digit $c_{n^2+1}$ all pairs are exhausted so it must recur at worst by this point. In fact because the way the digits are related it has a period which starts with $c_1$.
Now here's the question. Suppose we have two rational numbers $p, q$ defined in the above way with $p=\overline{p_1 p_2\ldots p_k}$ and $q= \overline{q_1 q_2\ldots q_k} $ for some $k$ in $\Bbb N$. Then we say $p\sim q$, $p$ is related to $q$ if $ p_1 p_2\ldots p_k $ is a cyclic permutation of $q_1 q_2\ldots q_k$.
How many distinct rationals are there for a given $n$ with respect to the equivalence relation $\sim$? Suppose $\rho$ is a function which gives the number of solutions for a given $n$, what is $\rho (n)$?
Eg. $n=2$
There are four possible numbers $0$, $0.\overline {011}$, $0.\overline{101}$, $0.\overline {110} $. The last three however are equivalent. Thus $\rho (2)=2$.
It's easy to show by hand that $\rho (3)=2$, $\rho (4)=4$, $\rho (5)=3$, $\rho (6)=4$, $\rho (7)=4$, $\rho (8)=8$.
Wondering if there's a way to predict this.