_____________
sequence 1 2 2 1 1 2 1 2 2 1
run lens 1 2-- 2-- 1 1 2-- 1
1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 4, 4,
Goes to
1, 3, 3, 3, 2... Etc.
I could extend this trivially too since the bottom sequence trails the sequence we write up top. If i wanted another '2' down the bottom whatever number i choose up top i just write twice right?
So there's nothing about this particular sequence? I can just create any such sequences trivially; Whenever you start a new count, choose a random number and repeat for a many times as needed for the trailing sequence to match the top sequence.
It seems that this particular variant is uninteresting in the broader picture right? I could write another similar one
2, 2, 1, 1, 2, 1
2, 2, 1, 1, ... etc.
I don't get the specialness here?
Kolakoski's sequence is conjectured to be "uniformly recurrent", which means that every block of contiguous symbols appearing once in the sequence appears infinitely often and with bounded gaps. This is clearly not true for any sequence constructed like this. And the fact that we don't know how to prove uniform recurrence for Kolakoski's, which is arguably the simplest sequence defined by this method, is remarkable. There are other conjectures about frequencies of symbols, etc.
No, because there's no deterministic way to infinitely extend that sequence. In your first example:
1 3 3 3 2 2 2 1 1 1 4 4 x x y y
1 3---- 3---- 3---- 2-- 2-- 2--
>Whenever you start a new count, choose a random number and repeat for a many times as needed for the trailing sequence to match the top sequence.
You answered your own question. The Kolakoski sequence is special because it does not just pick a random number: the sequence is deterministically encoded by the run lengths, and vice versa.
It's not like i had any difficulty coming up with that sequence i wrote to that point.
1 3 3 3 2 2 2 1 1 1 4 4 5 5 6 6 8 1 2... as an example of how trivial it is to continue to this.
I get that if you limit yourself to '1' or '2' you force the choice of next number but even then there's two possibilities of this sequence (start on 1 vs start on 2).
2, 2, 1, 1, 2, 1, 2...
2, 2, 1, 1,
Again no randomness. Just a variant of the above and trivial to continue this.
[0] https://en.wikipedia.org/wiki/William_Kolakoski#Personal_sig... [1] https://en.wikipedia.org/wiki/Look-and-say_sequence [2] https://esolangs.org/wiki/Kolakoski_sequence
although the sequence is not completely random (there are around n^alpha length-n blocks of contiguous symbols, for some alpha in [2,3) ), what people have done with other sequences is to take two simple PRNG (e.g. from a linear recurrence mod p) and alternate them according to the symbol you read from the sequence.
and the sequence does repeat. it is conjectured that any finite block of contiguous symbols occurs infinitely many times and moreover the gap between consecutive occurrences is uniformly bounded.
> More generally, the sequence is cube-free, i.e., has no substring of the form w w w with w some nonempty finite string.
from wiki