Is $(aaaaa)^n aa (aaaaa)^n$ a regular language?

Question

Is the language $\lbrace (aaaaa)^n aa (aaaaa)^n \mid n \in \mathbb{N} \rbrace$ regular? It looks like I need infinitely many states so it is not regular.

Answer 1

The language consists of those strings that contain $5 n + 2 + 5 n = 10 n + 2$ symbols $a$ for some $n \in \mathbb{N}$, i.e., $$\lbrace(aaaaa)^n aa (aaaaa)^n \mid n \in \mathbb{N}\rbrace = \lbrace (aaaaaaaaaa)^n aa \mid n \in \mathbb{N} \rbrace.$$ It is regular because it is accepted by the automaton which has twelve states $s_0, \ldots, s_{11}$ with a starting state $s_0$, accepting state $s_2$, and transitions $s_i \stackrel{a}{\longrightarrow} s_{i+1}$ for each $i = 0, \ldots, 10$, and one more transition $s_{11} \stackrel{a}{\longrightarrow} s_0$. In essence, this automaton counts the number of letters modulo 10 and accepts if the answer is 2.

Answer 2

We can prove a more general theorem. The language $\{ a^n : n \in A \}$ is regular if and only if the set $A$ is eventually periodic, i.e. there exist $m$ and $k$ such that for all $n \geq m$, $n \in A$ iff $n+k \in A$.

Indeed, a language of this form is clearly regular. For the other direction, consider a DFA accepting the language. It has outdegree $1$ and so must consist of a path leading into a cycle. Given that representation, we see that it can be represented using an eventually periodic $A$.