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A026610
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a(n) = number of 1's between n-th 2 and (n+1)st 2 in A026600.
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3
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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, 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, 1, 1, 0, 1, 2, 1, 1, 1, 0, 2, 0, 2, 0, 2, 1, 1, 1, 0, 2, 0, 2, 1
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OFFSET
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1,2
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COMMENTS
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(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.
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.
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
1j->12,23,3j, 2j ->23,31,1j , 3j->31,12,2j for j=1,2,3.
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
A:=12,21, B:=12,23,32,23,31,13,31, C:=12,23,32,23,31,11,
D:=12,23,31, E:=12,23,33,31, F:=12,22,23,31,13,31,
G:=12,23,32,23,31, and H:=12,23,31,13,31.
[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]
The morphism mu_2 induces a morphism beta on the return words given by
A->C, B->BFAEA, C->BFAD, D->BA, E-> BDA, F->GHAEA, G->BFA, H->BAEA.
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
delta: A->1, B->02, C->02, D->1, E->1, F->02, G->01, H->2.
Let y = BFAEAGHAEACBD... be the unique fixed point of beta. Then clearly (a(n)) = delta(y).
(End)
The frequencies of 0's, 1's and 2's in (a(n)) are 4/13, 5/13 and 4/13.
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
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REFERENCES
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J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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