Given different primes a and p we know by Fermat's little theorem that
$$a^{p-1} \equiv 1 \pmod{p}. $$
In the case of a = 2, a sequence for
$$ 2^{p-i} \equiv j_i \pmod{p}$$
can be written as
$$ f(i)= \begin{cases} j_i = \frac{j_{i-1}}{2} ,&\text{if } j_{i-1}\equiv 0 \pmod{2}\\ j_i = \frac{j_{i-1}+p}{2}, & \text{otherwise} \end{cases} $$
for $$1 < i \leqslant n$$ if $p \mid 2^{n}-1$, or $$1 < i \leqslant (p-1)$$ if it doesn't, and if $i = 1, j_i = 1$ because of Fermat's.
Taking p = 11 as an example, $j_1$ is going to be 1, and since 1 isn't even, $j_2 = 6$ with $2^{9} \equiv 6 \pmod{11}$, $j_3 = 3$, $j_4 = 7$ and so on up until $j_{10} = 2$ for which $2^{1} \equiv 2 \pmod{11}$.
With a number like $p = 23$, since it divides $2^{11} -1$, it's only going to have 11 values of $j_i$.
Since $2^{(p-1)n - (i+1)} \equiv 2^{(p-i)} \pmod{p}$ and given the conditions above, any power of two is going to be congruent with at most $n$ or $p-1$ number of unique $j_i$ mod p.
Now my question is if there's any way to generalize this for whichever different primes a and p.
