A248806 Number of 2's separating successive 1's in the Kolakoski sequence A000002.

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, 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, 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

Offset: 1

Author

Jean-Christophe Hervé, Oct 14 2014

Keywords

• nonn

Comments

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

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.

Links

• Jean-Christophe Hervé, Table of n, a(n) for n = 1..4995

Crossrefs

CF. A000002, A100429, A156256.