

A143589


Kolakoski fan based on A000034 with initial row 1.


4



1, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1
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OFFSET

1,2


COMMENTS

Conjecture (following Benoit Cloitre's conjecture at A111090): if L(n) is the number (assumed finite) of terms in row n of K, then L(n)*(2/3)^n approaches a constant. (L= A143590.)


LINKS

Table of n, a(n) for n=1..105.


FORMULA

Introduced here is an array K called the "Kolakoski fan based on a sequence s with initial row w": suppose that s=(s(1),s(2),...) is a sequence of 1's and 2's and that w=(w(1),w(2),...) is a finite or infinite sequence of 1's and 2's. Assume that s(1)=w(1) and that if w(1)=1 then s contains at least one 2. Row 1 of the array K is w. Subsequent rows are defined inductively: the first term of row n is s(n) and the remaining terms are defined by Kolakoski substitution; viz., each number in row n1 tells the stringlength (1 or 2) of the next string in row n, each term being either 1 or 2.


EXAMPLE

s=(1,2,1,2,1,2,1,2,...) and w=1, so the first 7 rows are
1
2
1 1
2 1
1 1 2
2 1 2 2
1 1 2 1 1 2 2


CROSSREFS

Cf. A000002, A143477, A143490.
Sequence in context: A097305 A120675 A072699 * A003651 A073203 A073204
Adjacent sequences: A143586 A143587 A143588 * A143590 A143591 A143592


KEYWORD

nonn,tabf


AUTHOR

Clark Kimberling, Aug 25 2008


STATUS

approved



