# An interesting sequence,Look and say sequnce

Why in this sequence $$1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221$$

333 never appears? It is a Look and say sequence.

• Half an answer: This is a tricky sequence and it works this way, You begin with $1$ and then the next number is $11$ to indicate there is one times $1$. Now it's two times $1$ so the next number is $21$, then we have one time $2$ and one time $1$ so you get $1211$ and so on. – Yanko Jun 23 '18 at 10:18
• Hint: if 333 appears at some point of the sequence, we must have read three 3s in a row in the previous sequence... – Wojowu Jun 23 '18 at 10:18
• This would be the reason, but Why we must have read that? – user122424 Jun 23 '18 at 10:32
• this is sequence $A005150$ at OEIS – Claude Leibovici Jun 23 '18 at 10:58
• The answers given so far don't really do it for me. Here's an argument that satisfies me. Let $f$ be the given sequence. Assume $333$ is a substring of an element of this sequence, then there's one such minimal element, say $f_n$. Since $333$ is a substring of $f_n$, then three $3$s come up as a substring of $f_{n-1}$ (note that $n-1$ is necessarily a natural number), but this means that $f_n$ was not the first element of $f$ to contain $333$, a contradiction. – Git Gud Jun 23 '18 at 11:46

If a sequence $abc$ appears in some term, then either the previous term had $a$ consecutive occurrences of $b$, or $b$ consecutive occurrences of $c$. In particular, if $aaa$ occurs, the previous term has $a$ consectutive $a$s. So for $333$ to occur, the previous term has three consecutive $3$s, so a $333$ within.
Say $3332$ occurs in a term. Then the previous term must contain $333222$, and the term prior to that would have to contain $33322222$ making the next term $3352$ rather than $333222$.
$333222$ is not achievable, and so neither is $3332$.
• "Making the next term $3352$", this is wrong. You can only conclude that it would contain $3352$, but that doesn't contradict anything by itself. – Git Gud Jun 23 '18 at 11:36