= Solution
After a jump from one to $k$, the first return to one occurs exactly $k$ steps later, with <probability> $\alpha_k$. Later return times are sums of such positive-support values. Thus their greatest common divisor is
$$
d=\gcd\{k:\alpha_k>0\}.
$$
In an irreducible chain this is the common <period of a state in a Markov chain>[period] of every state. Hence \b[the chain is <aperiodic> exactly when $d=1$]. A self-loop at one is sufficient but not necessary; support at two and three also has greatest common divisor one.
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