Let $L$ be a regular language. Let's say we sort $L$ by length and then lexicographically; then let $L_p \subset L$ be every $p$th word in $L$ according to this sort. Is $L_p$ regular as well?


For example, let's use $L = a^*b^* = \{\epsilon, a, b, aa, ab, bb, aaa, aab, abb, ...\}$. Then $L_2 = \{\epsilon, b, ab, aaa, abb, aaaa, aabb, bbbb, ...\}$, which is a regular language: $$L_2 = (aaaa)^*((\epsilon + b) + a(b + bb)+aa(bb+bbb)+aaa(bbb+\epsilon))(bbbb)^*$$

Put another way, $L_2 = \{a^mb^n \mid n = m \lor n = m+1 \mod 4\}$.


1 Answer 1


The answer is yes: Theorem 4 in "Numeration systems on a regular language" (Leconte and Rigo) shows that any arithmetic progression in a regular language is regular.

  • 2
    $\begingroup$ For future readers, the text is available on arXiv. $\endgroup$
    – Nathaniel
    Sep 4, 2022 at 7:31

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