Look and say... forever?

Sorry about the sloppy writing, there was just too much to write about. I hope it serves as a general idea.

Facts:

1- there can only be runs of length 3 or less of any digit, because of the nature of the dynamical rule. To understand, if we have $2222$ as part of the element $a_n$ of the sequence, it means that the corresponding part in $a_{n-1}$ has been $2222$ , which should be translate into $42$ rather than $2222$ .

As a consequence digits of the elements of the sequence can only be $\{1,2,3\}$ , given the seed is $1$ .

2- If we index the digits of a number as $a_0a_1 \dots a_t$ , then a run of length $3$ can only start at an even position.

3- runs of length $1,2,3$ in $a_n$ would cause the following number digits in $a_{n+1}$ : $2,2,2$ .

4- No element $a_n$ would start with a run of length $2$ or $3$ of the digits $2$ or $3$ . Because, if there is such an element, going backwards in the sequence, we would always have elements that would start with $2$ or $3$ , but obviously the seed is $1$ (starts with 1).

Now we put facts together. If $a_n$ consists of only runs of length 1, the obviously $a_n <a_{n+1}$ .

If $a_n$ consists of both runs of length 1 and 2 and no run of length 3, then the number of digits of $a_{n+1}$ is more than the number of digits $a_n$ , hence $a_n < a_{n+1}$ .

If $a_n$ consists only of runs of length 2,the length is maintained, but it should start with $11$ which leads to $a_{n+1}$ starting with $21$ , so $a_{n} < a_{n+1}$ .

If $a_{n}$ contains any run of length 3, then it would be followed by a run of length $1$ , which automatically compensate the reduction of the number of digits in $a_{n+1}$ that is caused by the run of length $3$ . So, the number of digits of $a_{n+1}$ would not be less than the number of digits of $a_n$ . Now, in case $a_n$ and $a_{n+1}$ have the same number of digits, if $a_n$ starts with $11$ , then $a_n < a_{n+1}$ , for $a_{n+1}$ start with $21$ . Note that if it starts with a run of $1$ , then the number of digits of $a_{n+1}$ would be definitely more than the number digits of $a_n$ .

It's stating the no. Of particular digit followed by the digit itself stated in the previous no. Like 1 and then one 1 = 11 and then two 1=21.... So it's ever expanding

This look and say sequence is always a increasing one as for you need to say how many numbers are there in the previous term to get the next one.And so if you have 2 digit in term then you will get at least 2 or 4 digits in the next term.If you get 2 digit then those 2 digits will never be same in the next term and will form out as a 4 digital term.And so the answer is TRUE