|
|
A285952
|
|
{0->1, 1->10}-transform of the Thue-Morse word A010060.
|
|
6
|
|
|
1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1
|
|
COMMENTS
|
The morphism {0->1, 1->10} has the infinite Fibonacci word A005614 as fixed point.
Since the Thue-Morse word is a concatenation of the 2-blocks 01 and 10, this word is a concatenation of 3-blocks 110 and 101. In particular the frequency of 0 is 1/3. This also follows from the fact that this sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism. The morphism on {1,2,3} can be chosen as 1->123, 2->23, 3->1, and the letter-to-letter map as 1->1, 2->1, 3->0. - Michel Dekking, Sep 20 2020
Actually this sequence is an automatic sequence. It follows directly from the observations above that (a(n)) is the decoration by the map 0->110, 1->101 of the Thue-Morse word A010060. - Michel Dekking, Jun 18 2022
|
|
LINKS
|
|
|
EXAMPLE
|
As a word, A010060 = 0110100110010110100101100..., and replacing each 0 by 1 and each 1 by 10 gives 110101101110101110110101...
|
|
MATHEMATICA
|
s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Thue-Morse, 010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "10"}] (* A285952, word *)
st = ToCharacterCode[w1] - 48 (* A285952, sequence *)
Flatten[Position[st, 0]] (* A285953 *)
Flatten[Position[st, 1]] (* A285954 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|