A026609 - OEIS

%I #13 Apr 15 2019 07:27:43

%S 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,

%T 1,1,0,2,0,2,1,1,1,0,1,2,1,1,1,0,2,0,2,1,0,2,0,2,1,1,1,0,1,2,1,1,1,0,

%U 2,0,2,1,1,0,2,0,2,0,2,1,1,1,0,2,0,2,0,2,1,1

%N a(n) = number of 3's between n-th 1 and (n+1)st 1 in A026600.

%C From _Michel Dekking_, Apr 15 2019: (Start)

%C (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.

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

%C 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.

%C 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.

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

%C (End)

%C 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

%e beta(B) = mu(123) = 123231312 = EDC.

%K nonn

%O 1,1

%A _Clark Kimberling_