For a finite ergodic Markov chain, there exists $ m\in\mathbb{N}$ such that $P_{ij}^{m}>0$ for any 2 (not necessarily different) states $i$ and $j$, where $P_{ij}$ is the transition probability between state $i$ and $j$.
My question is: what is the minimum $m$ such that $P_{ij}^m>0$ for ALL aperiodic and irreducible Markov chains that have $k$ states?
Is it just $k$? Because if we consider the graph representation of a Markov chain, the shortest path from $i$ to $j$ must contain no circle so the length of the path is at most $k-1$ if $i \neq j$ and $k$ if $i=j$, and it is easy to construct a directed graph such that the minimum number of steps needed for an arbitrary vertex $i$ to reach $j$ is $k$ (say a big clockwise circle and a self loop).
I am asking this because I just watched an online video where the professor stated without proof that the property $P_{ij}^{m}>0$ holds for an ergodic Markov chain for all $m>(k-1)^2$.
EDIT:
The shortest path from $i$ to $j$ of length $l$ only guarantee that $P^{nl}_{ij}\neq0$ for $n\in\mathbb{N^{+}}$, there must be another path of length coprime to $l$ to make $P^{k}_{ij}>0,\ \forall k\ge m$.
For any two positive coprime integers $p$ and $q$, $p<q$, there exists a minimum number $r$ such that $\forall k\ge r, \exists i, j\in\mathbb{N}$, $k=ip+jq$. In other words, $r$ is the minimum number such that all integers no less than $r$ can be expressed as a non-negative integral combination of $p$ and $q$. It is easy to show that $r=(q-1)(p-1)$. So if the shortest path from $i$ to $i$ is $k$ and there is a loop along the path (excluding $i$) of length $k-1$, then $P_{ii}^n>0\quad \forall n=k+ak+b(k-1), a, b\in\mathbb{N}$, so the minimum number $m$ such that all number $n$ no less than $m$ satisfies $P^n_{ij}>0$ is $k+(k-1)(k-2)=k^2-2k+2.$
The graph structure of the ergodic Markov chain that reaches this bound can be as follows: vertices $1, 2, \dots, k$ form a circle, i.e. a directed edge from $i$ to $i+1$ and from $k$ to $1$. There is also a directed edge from $k$ to $2$.