>Can't you trivially force this to happen for any sequence?

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 7 hours ago | parent [-]

Pick any number that's not 4 and repeat it twice. For the next 2 after that pick any number that's not that previous number and repeat it twice. So on.

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).

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flufluflufluffy 6 hours ago | parent | next [-]

I guess you could say the Kolakoski sequence is special in being the “simplest” version of such a sequence (ignoring the finite trivial case {1} xD)

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MontyCarloHall 7 hours ago | parent | prev [-]

You are still choosing random numbers here. The Kolakoski sequence has zero randomness whatsoever.

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AnotherGoodName 7 hours ago | parent [-]

Do we care similarly about the version of this that starts 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.

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MontyCarloHall 7 hours ago | parent [-]

This is just the Kolakoski sequence starting from the second term.

	▲	AnotherGoodName 5 hours ago \| parent [-]
		Ahh that makes sense. Thank you!