

A284749


{001>2}transform of the infinite Fibonacci word A003849.


7



0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 0, 1, 2, 2, 0, 1, 2, 2, 0, 1, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

This word is the fixed point of the morphism 0>2, 1>01, 2>2201 with the finite string 0, 1, 2, 0, 1 appended to the beginning. This morphism comes from taking the 'proper' version of the Fibonacci morphism 0>01, 1>1, given by 0>001, 1>01 (A189661 but with the rightmost 0 moved to the left of each image word), then replacing 001 with 2 and noting that the new symbol 2 should map to 00100101 = 2201 in order to be consistent.
The finite string appended to the beginning comes from the process of finding a proper version of the Fibonacci morphism using a return word encoding and taking conjugates which causes a shift of the respective fixed points.
(End)
This sequence is the unique fixed point of the morphism 0>01, 1>2, 2>0122. See the paragraph following Lemma 23 in the paper by Allouche and me.  Michel Dekking, Oct 05 2018


LINKS



EXAMPLE

As a word, A003849 = 01001010010010100..., and replacing each 001 by 2 gives 01201220120...


MATHEMATICA

s = Nest[Flatten[# /. {0 > {0, 1}, 1 > {0}}] &, {0}, 13] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"001" > "2"}]
st = ToCharacterCode[w1]  48 (* A284749 *)
Flatten[Position[st, 0]] (* A214971 *)
Flatten[Position[st, 1]] (* A284624 *)
Flatten[Position[st, 2]] (* A284625 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



