TITLE: Periodic sequences on finite alphabet
QUESTION [5 upvotes]: Let $\Sigma=\{A,B,C\}$ be an alphabet, and let $\Sigma^{\mathbb{N}}$ be the set of infinite sequences on $\Sigma$ (ie $ABCBCCCBABC...$). By outside conditions, I have several subsequences that are disallowed, namely $AA, BB, CC, ABAB, ACAC, BABA, BCBC, CACA, CBCB, ACB, BAC, CBA$, and I wish to show that all $\sigma\in\Sigma^{\mathbb{N}}$ are eventually periodic. I'd like to know if there are any results on this subject or any suggestions of ways to proceed on a proof.
Previous attempts: My original thought was that I could create a counterexample by making blocks of length 3 to represent 0 and 1 and then to create some sort of irrational decimal with these (say $\pi$ in binary, for example). I went through the allowable combinations and found that this was impossible. Doing this again for larger blocks seems possible, but I'd like to find some general results on this problem.
(Apologies for the tags, I can't quite figure out what good ones are for this question.)

REPLY [2 votes]: It is seen that the restrictions are cyclically-invariant, so it's better to specify the sequence of symbols $\{A,B,C\}$ using some "delta coding": say, $\{-, =, +\}$ where $+$ means "change to the symbol to the right" (cyclically) ($A \to B$, $B \to C$, $C \to A$), $=$ means "repeat same symbol", $-$ means "symbol to the left". This (together with the first symbol) fully specifies any sequence. For example, instead of $ABBCBAC$ we would write $A+=+--$
Actually the first three disallowed subsequences (AA, BB, CC) implicitly disallow the $=$ symbol (repetitions are forbidden), so we can restrict ourselves to  $\{- +\}$, or $\{0,1\}$.
The rest of restrictions disallow the following subsequences: 
$ ABAB, BCBC, CACA \equiv 101$  
$ ACAC, BABA, CBCB \equiv 010$ 
$ CBA, BAC, ACB \equiv 00$
So we are left with the much simpler problem of building a sequence of $\{0,1\}$ with the subsequences $00, 010, 101$ disallowed. It's easy to see that the only allowed infinite sequences are:
$11111 \cdots$  and $011111 \cdots$ 
So that, after the first two symbols, the sequence will repeat cyclically "to the right" ( ....ABCABCABC...)
Specifically the only possible infinite sequences are
ABCABCABC....   (A111111....)

BCABCABC....    (B111111....)

CABCABC....     (C111111....)

ACABCABC....    (A0111111....)

BABCABCABC....  (B0111111....)

CBCABCABC....   (C0111111....)