cp's OEIS Frontend

Except the first term, equals first difference of A249942.

The only possible values are 3, 5, 6, 9 and 10: a(n) cannot take the value 2, because it would imply the word 21212 which does not appear in the OK sequence; a(n) = 3 with the word 212-212, a(n) = 5 with 212-11-212, a(n) = 6 with 212-211-212 or 212-112-212, a(n) = 9 with 212-211-211-212 or 212-112-112-212, a(n) = 10 with 212-2112112-212. And from this it is easily seen that a(n) and a(n+1) cannot be equal unless a(n) = a(n+1) = 6 with the word 21-21122-1-22112-12.

The single 1's that are considered in this sequence are the 1's between two 2's in the OK sequence A000002 (the first term of A000002 which is indeed a single 1 but not between two 2's is thus not considered here). Each such single 1 is generated by a preceding 1 in the OK sequence that could be single or double, but each single 1 has at least a double 1 in its ancestors since the first 1 of the OK sequence has no descendance except itself. This sequence gives the number of generations between the n-th single 1 in A000002 and its nearest double 1 ancestor. A249942 gives the position of the single 1's in A000002.

The single 1's of the OK sequence are associated with iterated words which develop themselve around each single 1 in two branches; for a description of the iterated words, see comments in A249507 which gives their lengths.

The length of the iterated word around a single 1 is equal to A249507(2*a(n)+1) or to A249507(2*a(n)+2).

Conjecture: this sequence takes all integers k >= 1 as values, so there is no bound to the length of the iterated words that appear in the Kolakoski sequence; the limiting frequency of k is 2/3^k.

It is not known if this sequence is defined for any n > 0; showing that it is would solve some of the problems posed by Clark Kimberling about the Kolakoski sequence A000002 (see the link), at least the 2nd and 3rd problems.