A079729 Kolakoski-(1,2,3) sequence: a(n) is the length of the n-th run.

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

Offset: 1

Author

Benoit Cloitre, Feb 17 2003

Keywords

• nonn

Comments

Old name was: Kolakoski variation using (1,2,3) starting with 1,2.

Partial sum sequence is expected to be asymptotic to 2*n.

From Michel Dekking, Jan 31 2018: (Start)

(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

Examples

			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.
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000
• Kerry Mitchell, Creating Unicursal Curves Using Tiles, Bridges Conf. Proc. (2025) 239-246.
• Ulrich Reitebuch, Henriette-Sophie Lipschütz, and Konrad Polthier, Visualizing the Kolakoski Sequence, Bridges Conf. Proc.; Math., Art, Music, Architecture, Culture (2023) 481-484.

Crossrefs

Cf. A000002.

Programs

Mathematica

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 *)

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

Formula

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

Extensions

More terms from Philippe Deléham, Sep 24 2006

Name changed and text edited by Michel Dekking, Jan 31 2018