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

%I A079729 #38 Aug 06 2025 11:55:12
%S A079729 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,
%T A079729 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,
%U A079729 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
%N A079729 Kolakoski-(1,2,3) sequence: a(n) is the length of the n-th run.
%C A079729 Old name was: Kolakoski variation using (1,2,3) starting with 1,2.
%C A079729 Partial sum sequence is expected to be asymptotic to 2*n.
%C A079729 From _Michel Dekking_, Jan 31 2018: (Start)
%C A079729 (a(n)) is the unique fixed point of the 3-block substitution beta given by
%C A079729     111 -> 123,            112 -> 1233,
%C A079729     122 -> 12233,          123 -> 122333,
%C A079729     222 -> 112233,         223 -> 1122333,
%C A079729     231 -> 112223,         233 -> 11222333,
%C A079729     311 -> 11123,          312 -> 111233,
%C A079729     331 -> 1112223,        333 -> 111222333.
%C A079729 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).
%C A079729   Note: In general, a 3-block substitution beta maps a word w(1)...w(3n) to the word
%C A079729   beta(w(1)w(2)w(3))...beta(w(3n-2)w(3n-1)w(3n)).
%C A079729   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)
%C A079729 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
%H A079729 Ivan Neretin, <a href="/A079729/b079729.txt">Table of n, a(n) for n = 1..10000</a>
%H A079729 Kerry Mitchell, <a href="http://www.archive.bridgesmathart.org/2025/bridges2025-239.html">Creating Unicursal Curves Using Tiles</a>, Bridges Conf. Proc. (2025) 239-246.
%H A079729 Ulrich Reitebuch, Henriette-Sophie Lipschütz, and Konrad Polthier, <a href="https://archive.bridgesmathart.org/2023/bridges2023-481.html">Visualizing the Kolakoski Sequence</a>, Bridges Conf. Proc.; Math., Art, Music, Architecture, Culture (2023) 481-484.
%F A079729 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
%e A079729 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.
%t A079729 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 *)
%o A079729 (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
%Y A079729 Cf. A000002.
%K A079729 nonn
%O A079729 1,2
%A A079729 _Benoit Cloitre_, Feb 17 2003
%E A079729 More terms from _Philippe Deléham_, Sep 24 2006
%E A079729 Name changed and text edited by _Michel Dekking_, Jan 31 2018