|
| |
|
|
A284793
|
|
Difference sequence of A284775.
|
|
9
|
|
|
|
1, -1, 0, 1, 0, -1, 1, -1, 1, -1, 0, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, 0, -1, 1, -1, 0, 1, 0, -1, 1, -1, 1, -1, 0, 1, 0, -1, 0, 1, 0, -1, 1, -1, 1, -1, 0, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, 0, -1, 1, -1, 1, -1, 0, 1, 0, -1, 0, 1, 0, -1, 1, -1, 0, 1, 0, -1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1
|
|
|
COMMENTS
|
This is a morphic sequence, i.e., the letter-to-letter image of a fixed point of a morphism. The way to find this morphism is to compute the 2-block substitution induced by the defining morphism
sigma: 0->01, 1->0011.
If we code the 2-blocks by A:=00, B:=01, C:=10, D:=11, the 2-block morphism tau is given by
A -> BC, B->BC, C->ABDC, D->ABDC.
The fixed point of tau is x = BCABDCBC.... The letter-to letter map is
A->0, B->1, C->-1, D->0.
With a little more work can show that (a(n)) is even an automatic sequence, using that A284775 is automatic.
The generating morphism for A284775 is the uniform length morphism mu given by
a -> abc, b -> deb,c -> aba, d -> bcd, e -> ebc.
The 2-block substitution of this morphism is a morphism on an alphabet of 6 letters, corresponding to the words ab, bc, cd, de, ba and eb occurring in the fixed point of mu. One can merge the 2 letters corresponding to bc and ba, and then obtains modulo a coding the morphism mu again.
So (a(n)) is the fixed point of mu starting with the letter a, followed by the letter-to-letter map
a->1, b->-1, c->0, d->1, e->0.
(End)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
A284775 = (0,1,0,0,1,1,0,1,0,1,0,0,1,1,0,0,...), with differences (1,-1,0,1,0,-1,1,-1,1,...).
|
|
|
MATHEMATICA
|
s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 0, 1, 1}}] &, {0}, 7]; (* A284775 *)
Differences[s]
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,sign
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
