%I #5 Mar 30 2012 18:39:12
%S 1,8,1,4,1,4,6,4,1,6,1,4,8,1,4,1,8,1,4,1,8,4,1,8,1,4,8,1,4,1,8,1,8,1,
%T 4,1,4,6,4,1,6,1,4,1,4,8,1,4,6,4,1,8,4,1,8,1,4,1,8,1,4,1
%N Iterate the process: "sum of terms of n-th run" on the Kolakoski sequence until all runs have length 1.
%C Is it enough to apply the process three times only to obtain all runs of length 1? Then all terms should have values among 1,4,6 or 8.
%e Kolakoski sequence begins 1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2, When process is applied once: 1,2+2,1+1,2,1,2+2,1,2+2, --> 1,4,2,2,1,4,1,4,...Twice: 1,4,2+2,1,4...->1,4,4,1,4 ... Three times -> 1,8,1,4...
%Y Cf. A000002.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Feb 15 2003