Thanks for sharing! So what about the 1-to-2 ratio? The long running average looks like:

lra = Accumulate[N[ko]]/Range[Length[ko]];
ListPlot[lra[[;; ;; 10000]], DataRange -> {0, Length[ko]},Joined -> True]

This result is for n=40. Pretty close to 1.5 indeed…

Thanks for pointing out this excinting sequence! I could not resist playing around!

Using Mathematica I am always amazed how much one can do with how little code: Using the code above

kolakoskiSeq[iter_Integer] := Prepend[Nest[Flatten[Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, iter], 1]

then with

kHead[l_] := Length[#] @@ # & /@ SplitBy[l, Head]
koH = Length[#] @@ # & /@ Split[kolakoskiSeq[9]];
koTree = Nest[kHead, koH, 14];
Manipulate[TreeForm["Kolakoski" @@ Level[koTree, {level}], ImageSize -> 900, PlotStyle -> Blue], {level, 16, 3, -1}]

one gets:

@Todd Rowland: I like very much your approach of coding it differently! This seems to work:

kolakoskiStep[seq_List] := Flatten@MapIndexed[ConstantArray[Mod[First[#2] + First[seq], 2] + 1, #1] &, seq]
kSeq = Nest[kolakoskiStep, {1, 2}, 9]

With this it becomes obvious that there are only two possible sequences of this type, one starting with the "seed" {1,2}, the other one with {2}.