A248806 - OEIS

%I #14 Nov 22 2017 01:15:04

%S 2,0,1,2,2,0,1,0,2,1,0,1,2,0,1,0,1,2,2,0,1,2,1,0,1,0,2,2,0,1,2,2,0,1,

%T 0,1,2,1,0,2,2,0,1,2,2,0,1,0,2,1,0,1,2,2,0,1,0,1,2,0,1,0,2,1,0,1,0,2,

%U 2,0,1,2,1,0,2,2,1,0,1,0,2,2,0,1,2,1,0,1,0,2,1,0,1,2,0,1,0,1,2,1

%N Number of 2's separating successive 1's in the Kolakoski sequence A000002.

%C Without the zeros, this sequence is equal to the bisection of the Kolakoski sequence A100429 = lengths of runs of 2's in OK sequence.

%C The Oldenburger-Kolakovski sequence can be obtained back (except the initial 1) by the following substitution rules: insert 1 between two successive nonzero values and 0 -> 11, 1 -> 2, 2 -> 22.

%H Jean-Christophe Hervé, <a href="/A248806/b248806.txt">Table of n, a(n) for n = 1..4995</a>

%Y CF. A000002, A100429, A156256.

%K nonn

%O 1,1

%A _Jean-Christophe Hervé_, Oct 14 2014