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A079729
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Kolakoski-(1,2,3) sequence: a(n) is the length of the n-th run.
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2
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1, 2, 2, 3, 3, 1, 1, 1, 2, 2, 2, 3, 1, 2, 3, 3, 1, 1, 2, 2, 3, 3, 3, 1, 2, 2, 3, 3, 3, 1, 1, 1, 2, 3, 1, 1, 2, 2, 3, 3, 3, 1, 1, 1, 2, 2, 2, 3, 1, 1, 2, 2, 3, 3, 3, 1, 1, 1, 2, 2, 2, 3, 1, 2, 3, 3, 1, 1, 1, 2, 3, 1, 1, 2, 2, 3, 3, 3, 1, 1, 1, 2, 2, 2, 3, 1, 2, 3, 3, 1, 1, 2, 2, 3, 3, 3, 1, 2, 3, 3, 1, 1, 2, 2, 2
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OFFSET
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1,2
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COMMENTS
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Old name was: Kolakoski variation using (1,2,3) starting with 1,2.
Partial sum sequence is expected to be asymptotic to 2*n.
(a(n)) is the unique fixed point of the 3-block substitution beta given by
111 -> 123, 112 -> 1233,
122 -> 12233, 123 -> 122333,
222 -> 112233, 223 -> 1122333,
231 -> 112223, 233 -> 11222333,
311 -> 11123, 312 -> 111233,
331 -> 1112223, 333 -> 111222333.
Here BL3 := {111, 112, 122, 123, 222, 223, 231, 233, 311, 312, 331, 333} is the set of all words of length 3 occurring at a position 1 mod 3 in (a(n)). This can be seen by splitting the words beta(B) into words of length 3, and looking at the possible extensions of those words beta(B) that have a length which is not a multiple of 3. For example, beta(122) = 12233 can only be extended to 122331 or to 122333, and both words 331 and 333 are in BL3. Interestingly, BL3 is invariant for the permutation 1->3, 2->1, 3->2 (and its square).
Note: In general, a 3-block substitution beta maps a word w(1)...w(3n) to the word
beta(w(1)w(2)w(3))...beta(w(3n-2)w(3n-1)w(3n)).
If the length of a word w is 3n+r, with r=1 or r=2, then the last letter, respectively last 2 letters are ignored.
(End)
Conjecture: the frequencies of 1's, 2's and 3's in (a(n)) exist and are all equal to 1/3. This conjecture implies the conjecture of Benoit Cloitre on the partial sum sequence. - Michel Dekking, Jan 31 2018
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LINKS
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Ulrich Reitebuch, Henriette-Sophie Lipschütz, and Konrad Polthier, Visualizing the Kolakoski Sequence, Bridges Conf. Proc.; Math., Art, Music, Architecture, Culture (2023) 481-484.
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FORMULA
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Iterate beta: 122 -> 12233 ~ 122331 -> 122331112223 -> 12233111222312331122333, etc. Here a(6)=1 has been added to 12233 in step two to continue the iteration. - Michel Dekking, Jan 31 2018
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EXAMPLE
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Sequence begins: 1,2,2,3,3,1,1,1,2,2,2,3,1,2,3,3,1,1,... read it as: (1),(2,2),(3,3),(1,1,1),(2,2,2),(3),(1),(2),(3,3),(1,1),... then count the terms in parentheses to get: 1,2,2,3,3,1,1,1,2,2,... which is the same sequence.
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MATHEMATICA
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seed = {1, 2, 3}; w = {}; i = 1; Do[w = Join[w, Array[seed[[Mod[i - 1, Length[seed]] + 1]] &, If[i > Length[w], seed, w][[i]]]]; i++, {n, 53}]; w (* Ivan Neretin, Apr 02 2015 *)
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PROG
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(PARI) a=[1, 2, 2]; for(n=3, 100, for(i=1, a[n], a=concat(a, 1+((n-1)%3)))); a; \\ Benoit Cloitre, Feb 13 2009
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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