Here, I am talking about sequences where the factor of multiplication is $k^m$ and the first term is also $k^m$ like:
$$k^m, k^{2m}, k^{3m}, \cdots \pmod{m}$$
Is it possible that none of the elements after $k^m$ are congruent to $k^m \pmod{m}$? I don't think that this is possible for a prime $m$ since by Fermat's Little Theorem $k^m \equiv k$ and $k^{m^2} \equiv k^m \equiv k$. I am not so sure for other $m$. In all the cases I've tried till now, there always exists a power congruent to the first term. Also, could you please give examples of sequences if they exist.
Thanks.
Write $m=ab$ where $a$ is the greatest divisors of $m$ coprime to $k$. For any $p\mid b$, if $p^r\mid m$ then certainly also $p^r\mid k^m$ (becasue $p\mid k$ and $m\ge r$). Hence $k^{im}\equiv 0\pmod b$ for all $i$. Whereas $\gcd(k,a)=1$ implies that $k^m$ is invertible $\bmod a$, hence the sequence is periodic. So for suitable $i>1$, we have $k^{im}\equiv k^m\pmod a$. As also $k^{im}\equiv k^m\pmod b$ and $\gcd(a,b)=1$, we indeed have $k^{im}\equiv k^m\pmod m$.