A156256 Number of 1's separating successive 2's in the Kolakoski sequence A000002.

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, 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, 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

Offset: 1

Author

Benoit Cloitre, Feb 07 2009

Keywords

• nonn

Comments

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

This is because A100428 gives the lengths of runs of 1's in Kolakoski sequence. - Jean-Christophe Hervé, Oct 14 2014

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

Examples

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

Crossrefs

Cf. A000002, A078649, A100428, A248806.

Formula

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

Extensions

Better name from Jean-Christophe Hervé, Oct 15 2014