|
| |
|
|
A287674
|
|
{0->1, 1->001}-transform of the infinite Fibonacci word A003849.
|
|
4
|
|
|
|
1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0
(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. In fact, one can take the alphabet {1,2,3,4} with the morphism
mu: 1->12341, 2->1, 3->2, 4->34,
and the letter-to-letter map g defined by
g: 1->1, 2->0, 3->0, 4->1.
Then (a(n)) = g(x), where x = 123411234... is the unique fixed point of the morphism mu.
In my paper "Morphic words, Beatty sequences and integer images of the Fibonacci language" the transform map 0->1, 1->001 is called a decoration. 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 1+3 = 4 symbols. In general there are several choices for mu. Here we have chosen mu such that it is the SAME morphism as used to generate A286749 as a morphic sequence.
This will make it possible to prove the remarkable conjecture by R. J. Mathar from June 14, 2017. Note that A286749 is a letter-to-letter projection of the SAME sequence x = 123411234... but by a different letter-to-letter map f:
f: 1->1, 2->1, 3->0, 4->0.
How does one prove that A286749(n+1) = a(n), for n > 0?
The crucial observation is that x is a concatenation of the two words 12341 and 1234. This follows directly from mu(12341) = 12341123412341, and mu(1234) = 123411234. Next one observes that
f(12341) = 11001 = 1g(12341)1^{-1},
f(1234) = 1100 = 1g(1234)1^{-1}.
This property implies then that A286749(n+1) = a(n).
(End)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
As a word, A003849 = 0100101001001010010100100..., and replacing each 0 by 1 and each 1 by 001 gives 100111001100111001110011001...
|
|
|
MATHEMATICA
|
s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "001"}]
st = ToCharacterCode[w1] - 48 (* this sequence *)
Flatten[Position[st, 0]] (* A287675 *)
Flatten[Position[st, 1]] (* A287676 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
