login
A248806
Number of 2's separating successive 1's in the Kolakoski sequence A000002.
2
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
1,1
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
KEYWORD
nonn
AUTHOR
STATUS
approved