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%I A026610 #21 Oct 03 2019 09:00:15
%S A026610 0,2,0,2,1,1,1,0,1,2,1,1,1,0,2,0,2,1,1,0,2,0,2,0,2,1,0,2,1,1,1,0,2,0,
%T A026610 2,1,1,0,2,0,2,0,2,1,1,0,2,0,2,1,1,1,0,2,1,0,2,0,2,0,2,1,1,0,2,0,2,1,
%U A026610 1,1,0,1,2,1,1,1,0,2,0,2,0,2,1,1,1,0,2,0,2,1
%N A026610 a(n) = number of 1's between n-th 2 and (n+1)st 2 in A026600.
%C A026610 From _Michel Dekking_, Apr 16 2019: (Start)
%C A026610 (a(n)) is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of a morphism. A proof of this is more involved than the proof for the case of the closely related sequence A026609. The reason is that 2 is not the first letter of A026600.
%C A026610 There are several ways to tackle this. We first remark that it suffices to prove that (a(n)) is the image of a fixed point of a morphism by a morphism delta (instead of a letter-to-letter projection), see Corollary 7.7.5 in the book by Allouche and Shallit.
%C A026610 The sequence A026600 is fixed point of the 3-symbol Thue-Morse morphism mu given by mu: 1->123, 2->231, 3->312. Since the first 2 in (a(n)) is at position 2, we consider the 2-block 3-symbol Thue-Morse morphism mu_2 defined on the set of all nine 2-blocks ij by
%C A026610 1j->12,23,3j, 2j ->23,31,1j , 3j->31,12,2j for j=1,2,3.
%C A026610 We then consider the unique fixed point x = 12,23,32,23,31,13,... of mu_2. The return words of the 'letter' 12 in x are
%C A026610 A:=12,21, B:=12,23,32,23,31,13,31, C:=12,23,32,23,31,11,
%C A026610 D:=12,23,31, E:=12,23,33,31, F:=12,22,23,31,13,31,
%C A026610 G:=12,23,32,23,31, and H:=12,23,31,13,31.
%C A026610 [See Justin & Vuillon (2000) for definition of return word. - _N. J. A. Sloane_, Sep 23 2019]
%C A026610 The morphism mu_2 induces a morphism beta on the return words given by
%C A026610 A->C, B->BFAEA, C->BFAD, D->BA, E-> BDA, F->GHAEA, G->BFA, H->BAEA.
%C A026610 Counting 3's between 2's in the j's of the ij's in the return words followed by 12 yields the morphism delta given by
%C A026610 delta: A->1, B->02, C->02, D->1, E->1, F->02, G->01, H->2.
%C A026610 Let y = BFAEAGHAEACBD... be the unique fixed point of beta. Then clearly (a(n)) = delta(y).
%C A026610 (End)
%C A026610 The frequencies of 0's, 1's and 2's in (a(n)) are 4/13, 5/13 and 4/13.
%C A026610 This follows from an eigenvector computation, but can also be deduced from the frequency result for the sequence A026609: since the 3-symbol Thue Morse morphism generating sequence A026600 is symmetric under the permutation 1->2->3->1, the two sequences A026609 and A026610 generate the same language, and in particular all subwords have the same frequencies. - _Michel Dekking_, Apr 16 2019
%D A026610 J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
%H A026610 Michael De Vlieger, Table of n, a(n) for n = 1..19683
%H A026610 Jacques Justin and Laurent Vuillon, Return words in Sturmian and episturmian words, RAIRO-Theoretical Informatics and Applications 34.5 (2000): 343-356.
%e A026610 delta(B)=02, since there is no 11, 21, or 31 between B(1)=12 and B(3)=32, and there is twice a 31 among B(4)=23,...,B(7)=31.
%t A026610 Rest@ Map[Count[#, 1] &, DeleteCases[SplitBy[#, # == 2 &], _?(# == {2} &)]] &@ Nest[Flatten[# /. {1 -> {1, 2, 3}, 2 -> {2, 3, 1}, 3 -> {3, 1, 2}}] &, {1}, 6] (* _Michael De Vlieger_, Apr 16 2019, after _Robert G. Wilson v_ at A026600 *)
%K A026610 nonn
%O A026610 1,2
%A A026610 _Clark Kimberling_
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