A156256 - OEIS

%I #20 May 27 2021 09:52:54

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

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

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

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

%C After deleting 0's in this sequence it remains the bisection of Kolakoski sequence A000002(2n+1) n>=1 given by A100428.

%C This is because A100428 gives the lengths of runs of 1's in Kolakoski sequence. - _Jean-Christophe Hervé_, Oct 14 2014

%C The Kolakovski sequence can be obtained back (except the initial 1) by the following substitution rules: insert 2 between two successive nonzero values and 0 -> 22, 1 -> 1, 2 -> 11. - _Jean-Christophe Hervé_, Oct 14 2014

%F a(n) = A078649(n+1)-A078649(n)-2.

%e The Kolakoski sequence begins with 122112122122, thus this one begins 0, 2, 1, 0, 1, 0. - _Jean-Christophe Hervé_, Oct 14 2014

%Y Cf. A000002, A078649, A100428, A248806.

%K nonn

%O 1,2

%A _Benoit Cloitre_, Feb 07 2009

%E Better name from _Jean-Christophe Hervé_, Oct 15 2014