|
| |
|
|
A285960
|
|
{01->1}-transform of the Thue-Morse word A010060.
|
|
4
|
|
|
|
1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1
|
|
|
COMMENTS
|
This sequence is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism mu. Here is a proof.
The Thue-Morse sequence x:=A010060 is a concatenation of the four return words A=011, B=010, C=0110, and D=01, of the word 01 in x. (These are the words occurring in x starting with 01, and having no other occurrences of 01 in them). By applying the Thue-Morse morphism 0->01, 1->10 to the return words one induces the derived morphism
tau: A -> AB, B -> CD, C -> ABD, D -> C.
It is clear that to obtain (a(n)) from x one has to apply the morphism delta to x written as ABCDABD... given by
A-> 11, B-> 10, C-> 110, D-> 1.
In the paper "Morphic words, Beatty sequences and integer images of the Fibonacci language" delta is called a decoration map. It is well-known that decorated fixed points are morphic sequences, and the 'natural' algorithm to achieve this yields a morphism on an alphabet of 2+2+3+1 = 8 symbols. In this particular case one can reduce the number of symbols to 4, say {a,b,c,d}, and obtain the morphism mu
mu: a->ab, b->cd, c->abd, d->c.
Let y be the fixed point of mu starting with the letter a. Then (a(n)) = lambda(y), where the letter-to-letter map lambda is defined by
lambda: a->1, b->1, c->1, d->0.
Proof of the bisection formula below: the derived morphisms for the return words of 01, respectively 011 in A285960 and A286046 both are the same morphism tau. Moreover, the decoration morphism A-> 11, B-> 10, C-> 110, D-> 1 to obtain (a(n)) corresponds exactly to the letters in the decoration morphism A-> 1010, B-> 1001, C-> 101001, D->10 occurring at the odd positions. This implies a(n) = A286046(2n-1), n = 1,2,...
(End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
EXAMPLE
|
As a word, A010060 = 0110100110010110100101100..., and replacing each 01 by 1 gives 1110110111101110111011011101110...
|
|
|
MATHEMATICA
|
s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 8] (* Thue-Morse, A010060 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"01" -> "1"}]
st = ToCharacterCode[w1] - 48 (* A285960 *)
Flatten[Position[st, 0]] (* A285961 *)
Flatten[Position[st, 1]] (* A285962 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
