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 indeedÂ…
