How to prove that the following sequence will never contains number greater than 3 You may now the following sequence:
1
11
21
1211
111221
312211
13112221

Explanation of the sequence (I've put an hint just for the ones who want to search a bit ;) )

Where, in each iteration you count the number of occurrence of a digit then put the count and the digit in the new child. (First you've 1, then you've got one 1 (11), then you've got two 1(21)...)

I've the feeling that the numbers of the sequence will never contains a digit >3 but my mathematics knowledge are way to poor to prove it. So I've made a little program that compute the terms for me to found out if there is a number greater than 3 but up to the 42th term it doesn't find any.
As the length of the string is growing very fast, I can't compute a lot more terms and, even if I was able to do so, this would not be a proof.
So here's the question: How can I prove my feeling?

Each number contains pairs of type $xy$ where x is the number of corresponding consecutive y digit in the previous number. Consider the first number in sequence containing digit 4. Then the previous number must have 4 equal consecutive digits, let's say $dddd$.Now, $dddd$ cannot represents two pair of d repetition of d, therefore we must have something like k repetitions of d and d repetitions of d which also is not possible.
