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A143589
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Kolakoski fan based on A000034 with initial row 1.
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4
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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
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1,2
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COMMENTS
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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.)
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LINKS
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FORMULA
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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 n-1 tells the string-length (1 or 2) of the next string in row n, each term being either 1 or 2.
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EXAMPLE
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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
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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