Timeline for DFA that rejects $a^{23}$ but accepts $\{a^i|i\geq 24\}$

8 events

when toggle format	what		by	license	comment
Nov 22, 2018 at 16:55	vote	accept	Mark Regev
Nov 22, 2018 at 16:54	vote	accept	Mark Regev
Nov 22, 2018 at 16:54
Nov 22, 2018 at 15:11	comment	added	JohEker		@Solomonoff'sSecret Must all states really accept once you enter the cycle? Is it not enough to prove using LTL that `GF(Final)` consider `L = {a^2n \| n>0}`
Nov 22, 2018 at 15:06	comment	added	John L.		Slightly more generally, instead of accepting all $a^m$ for $m>n$, it is enough that the DFA accepts $a^{k}, a^{k+1}, \cdots, a^{k+n-1}$ for some $k\ge0$.
Nov 22, 2018 at 1:33	comment	added	Hendrik Jan		If you want, you can add the remark to your own answer.
Nov 22, 2018 at 1:19	comment	added	Reinstate Monica		@HendrikJan BTW, this would be better posted as an answer. Even though it doesn't technically answer the question, it suggests the question should have said NFA, not DFA, and it answers that question.
Nov 22, 2018 at 0:27	comment	added	Hendrik Jan		Finally I found a possibility to do it non-deterministically. If we have loops of length 5 and 6 with only the start state accepting we get strings of the length 0,5,6,10,11,12,15,16,17,18, [20,21,22,23,24,..] Last missing length is 19, so if we add a path to the loop with 4 letters we get a solution. The loops can be combined into 6 states so total of 10 states. This was trial and error.
Nov 21, 2018 at 23:40	history	answered	Reinstate Monica	CC BY-SA 4.0