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Look on wikipedia for more information on the Kolakoski sequence if you're unfamiliar with it. The Kolakoski sequence is supposed to be a fractal because you can get the same sequence by taking the length of each "run" in the sequence. I was in sage math, messing around with collapsing the Kolakoski sequence to a single digit by taking the length of the runs in the sequences. ex.
$[1,2,2,1,1,2,1,2,2,1]$
$[1,2,2,1,1,2,1]$
$[1,2,2,1,1]$
$[1,2,2]$
$[1,2]$
$[1,1]$
$[2]$

Now, sometimes this works out perfectly normal, but not always. It's quite frequent that by doing this you'll actually end up with sequences containing runs of lengths of 3 or greater. Here's an example of a length which does not collapse properly.
$[1,2,2,1,1,2,1,2,2]$
$[1,2,2,1,1,2]$
$[1,2,2,1]$
$[1,2,1]$
$[1,1,1]$
$[3?]$

So, I wrote a program in python to find which length sequences would actually collapse properly. The first ones go as follows: $1,2,3,5,7,10,$ $11,15,17,23,25,$ $34,37,50,55,75,82$. All of this brings up lots of questions for me. First of all, does this mean that the Kolaski sequence isn't really a fractal since it doesn't really collapse properly as has been shown? Is there some algorithm that can calculate the sequence which consists of all the lengths that do properly collapse?

Cadyn Bombaci
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    The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards. – Somos Jul 13 '18 at 17:19
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    Nice first question. Welcome to Math Stack Exchange! – qwr Jul 13 '18 at 17:31
  • Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski). – Joffan Jul 13 '18 at 18:04
  • Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s? – Caleb Stanford Jan 11 '19 at 20:02

1 Answers1

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Let $$S(p)=K_1+K_2+...K_p$$

the length will properly collapse if

$$n=S(S(S...(S(2)))))).$$

for example, if $n=S(S(2))$, it will collapse after three encoding operations.

hamam_Abdallah
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