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Kolakoski Sequence
MontyCarloHall
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sequence 1 2 2 1 1 2 1 2 2 1
run lens 1 2-- 2-- 1 1 2-- 1
Read out the bottom sequence of run lengths, and be amazed that it's the same as the first 7 digits as the top sequence. Extend the bottom to continue to recapitulate the top sequence, and add terms to the top sequence accordingly to reflect the run lengths in the bottom sequence. Repeat infinitely.
AnotherGoodName
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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?
MontyCarloHall
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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--
What are the values of x and y?>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.
AnotherGoodName
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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).