4 ms·

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

Can't you trivially force this to happen for any sequence? 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 8mo ago
>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 8mo ago
  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).
  - MontyCarloHall 8mo ago
    You are still choosing random numbers here. The Kolakoski sequence has zero randomness whatsoever.
    - AnotherGoodName 8mo ago
      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.
      - MontyCarloHall 8mo ago
        This is just the Kolakoski sequence starting from the second term.
        AnotherGoodName 8mo ago
        Ahh that makes sense. Thank you!
  - flufluflufluffy 8mo ago
    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)
robotpepi 8mo ago
> So there's nothing about this particular sequence? 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.

MontyCarloHall 8mo ago

>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 8mo ago
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).
- MontyCarloHall 8mo ago
  You are still choosing random numbers here. The Kolakoski sequence has zero randomness whatsoever.
  - AnotherGoodName 8mo ago
    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.
    - MontyCarloHall 8mo ago
      This is just the Kolakoski sequence starting from the second term.
      - AnotherGoodName 8mo ago
        Ahh that makes sense. Thank you!
- flufluflufluffy 8mo ago
  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)

robotpepi 8mo ago

> So there's nothing about this particular sequence? 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.