cp's OEIS Frontend

It appears that a(n) gives the position of its own n-th 1 modulo 3 term, the n-th 2 modulo 3 term in A026602, and the n-th multiple of 3 in A026603. A026602 and A026603 appear to have analogous indexical properties. - Matthew Vandermast, Oct 06 2010

This follows directly from the generating morphism for A026600: a 1 in position k creates a 1 in position 3k-2, a 2 in position 3k-1, and a 3 in position 3k. Since each block of three terms in A026600 is a permutation of {1,2,3}, these created terms are the k-th terms of their respective index sequences. The proof for the other index sequences is similar. - Charlie Neder, Mar 10 2019

It appears that a(n) gives the position of its own n-th 1 modulo 3 term, the n-th 2 modulo 3 term in A026603, and the n-th multiple of 3 in A026601. A026601 and A026603 appear to have the same sort of indexical properties. - Matthew Vandermast, Oct 06 2010

It appears that a(n) gives the position of its own n-th 1 modulo 3 term, the n-th 2 modulo 3 term in A026601, and the n-th multiple of 3 in A026602. A026601 and A026602 appear to have the same sort of indexical properties. - Matthew Vandermast, Oct 06 2010

From Michel Dekking, Apr 16 2019: (Start)

(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

Old name was: Analog of A026600 using instead of 1: 0,1; instead of 2: 1,0; instead of 3: 0,1.

A nonperiodic sequence of 0 and 1, with one 0 and one 1 in every subsequence of three terms.

From Michel Dekking, Apr 17 2019: (Start):

(a(n)) is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of a morphism.

Let the morphism sigma be given by

1->123, 2->456, 3->345, 4->612, 5->561, 6->234,

and let the letter-to-letter map delta be given by

1->0, 2->1, 3->1, 4->0, 5->0, 6->1.

Then (a(n)) = delta(x), with x the fixed point of sigma starting with 1.

This representation can be obtained by doubling 1,2 and 3, and renaming the resulting six letters as 1,2,3,4,5,6.

(End)

This sequence essentially equals A026605, which is its standard form: a(n) = A026605(n)-1 for all n. - Michel Dekking, Apr 18 2019

Old name was: a(n) = n-th term of A026600 that is not a 3.

This sequence equals the standard form of A057215, i.e., a(n) = A057215(n)+ 1 for all n. This follows by noting that this sequence equals the [123->12, 231->21, 312->12]-transform of three-symbol Thue-Morse A026600. - Michel Dekking, Apr 18 2019

From Michel Dekking, Apr 16 2019: (Start)

{a(n)} is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of a morphism. This follows from a study of the return words of 1 in {a(n)}: the word 1 in {a(n)} has 7 return words. These are A:=1, B:=123, C:=12, D:=13, E:=12323, F:=1233, and G:=1223.

[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]

The sequence A026600 is a fixed point of the 3-symbol Thue-Morse morphism mu given by mu: 1->123, 2->231, 3->312.

This induces a morphism beta on the return words given by

beta: A->B, B->EDC, C->EA, D->FC, E->EDGDC, F->EDBC, G->EBDC.

Counting 2's in the return words yields the morphism gamma given by

gamma: A->0, B->1, C->1, D->0, E->2, F->1, G->2.

Let y = EDGDCFCEBDCf... be the unique fixed point of beta. Then clearly (a(n)) = gamma(y).

(End)

The frequencies of 0's, 1's and 2's in {a(n)} are 4/13, 5/13 and 4/13, despite the fact that the gamma above is different from the gamma in A026609. However, the languages of the words A026609 and {a(n)} are different. The word 20201 does appear in A026608, A026611, and A026612, but not in the other triple of sequences A026609, A026610 and A026613. - Michel Dekking, Apr 16 2019

From Michel Dekking, Apr 15 2019: (Start)

(a(n)) is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of a morphism. This follows from a study of the return words of 1 in (a(n)): the word 1 in (a(n)) has 7 return words. These are A:=1, B:=123, C:=12, D:=13, E:=12323, F:=1233, G:=1223.

The sequence A026600 is fixed point of the 3-symbol Thue-Morse morphism mu given by mu: 1->123, 2->231, 3->312.

This induces a morphism beta on the return words given by beta: A->B, B->EDC, C->EA, D->FC, E->EDGDC, F->EDBC, G->EBDC.

Counting 3's in the return words yields the morphism gamma given by gamma: A->0, B->1, C->0, D->1, E->2, F->2, G->1.

Let y = EDGDCFCEBDCFC... be the unique fixed point of beta. Then clearly (a(n)) = gamma(y).

(End)

The frequencies of 0's, 1's and 2's in (a(n)) are 4/13, 5/13 and 4/13. - Michel Dekking, Apr 15 2019

(a(n)) is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of a morphism. See A026608 and A026610. - Michel Dekking, Apr 16 2019

The frequencies of 0's, 1's and 2's in (a(n)) are 4/13, 5/13 and 4/13. See A026610. - Michel Dekking, Apr 16 2019