A307243 An irregular fractal binary sequence embedding three copies of itself. See comments for precise definition.

1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0

Offset

1

Comments

The sequence has the following property: if a(n) = 0, underline a(n+1); if a(n) = 1, underline a(n+2). The terms underlined twice are the original sequence (copy #1), the terms underlined once are also (copy #2) and the non-underlined also (copy #3).

From Lars Blomberg: (Start)

I think this idea can be generalized:

Select some k >= 1, for example 2.

Initialize S = (k)(k-1)...0, for example S=210, and evaluate the underlining: the underlined amount is shown after '.', S=210.3

Initialize the "fetch pointers", f0=k, f1...fk=0 (using 0-based indexing);

Let p be the pointer to the next term to be evaluated, start with p=k=2;

The term indexed by p is shown in [], and the values of f0...fk are shown in ();

So for 210.[3]-(200), the underlined count is 3 - we should take the value from f2 which is 2, and calculate its underlining:

2102.[0]01-(201). Next is 0 so we take from f0:

21020.[1]1-(301) and so on.

Note that before «.» are the terms, and after «.» are the underlined, so a single array will suffice to store everything.

For k=2, I get:

2 1 0 2 2 2 2 1 0 1 2 2 2 2 2 2 1 0 0 1 1 2 2 0 2 2 1 2 1 2 0 2 2 2 1 0 2 2 2 0 2 1 2 2 2 1 2 2 1 2 2 0 1 0 0 2 0 2 1 1 1 2 1 1 2 2 2 0 2 2 2 2 2 1 0 2 0 2 2 2 2 0 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 1 2 0 1 2 2 0 0 1 2 0 1 0 1 2 0 2 0 2 1 2 2 1 1 2 1 0 2 2 2 2 1 1 0 2 2 0 2 2 2 2 2 0 1 2 2 1 2 2 2 2 2 1 0 2 2 2 2 0 1 2 2 1 2 2 2 2

And for k=3:

3 2 1 0 3 3 3 3 3 2 1 0 2 3 3 3 2 3 3 3 1 3 2 3 0 3 2 2 3 1 1 0 2 3 3 2 2 3 0 3 3 3 3 3 2 3 3 3 3 3 1 3 3 2 1 3 1 3 0 3 2 3 3 3 0 2 2 2 3 1 1 0 0 2 3 2 3 2 1 3 3 1 1 0 3 0 3 2 2 3 2 2 3 3 3 2 0 3 3 2 3 3 2 0 3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 1 3 1 3 2 1 3 3 3 2 3 1 3 3 0 3 2 3 3 2 3 3 3 3 3 0 0 3 2 3 3 2 1 2 3 3 2 3 3 2 2 1 3 1 1

(End)

Links

Lars Blomberg, Table of n, a(n) for n = 1..1000

Example

The sequence starts with 1,0,1,1,1,0,0,1,1,1,1,0,1,1,0,1,...
Instead of underlining them, we will put () around the terms we want to emphasize.
a(1) = 1 produces parentheses around a(3):
1,0,(1),1,1,0,0,1,1,1,1,0,1,1,0,1,...
a(2) = 0 produces parentheses around a(3) again:
1,0,((1)),1,1,0,0,1,1,1,1,0,1,1,0,1,...
a(3) = 1 produces parentheses around a(5):
1,0,((1)),1,(1),0,0,1,1,1,1,0,1,1,0,1,...
a(4) = 1 produces parentheses around a(6):
1,0,((1)),1,(1),(0),0,1,1,1,1,0,1,1,0,1,...
a(5) = 1 produces parentheses around a(7):
1,0,((1)),1,(1),(0),(0),1,1,1,1,0,1,1,0,1,...
a(6) = 0 produces parentheses around a(7) again:
1,0,((1)),1,(1),(0),((0)),1,1,1,1,0,1,1,0,1,...
a(7) = 0 produces parentheses around a(8):
1,0,((1)),1,(1),(0),((0)),(1),1,1,1,0,1,1,0,1,...
a(8) = 1 produces parentheses around a(10):
1,0,((1)),1,(1),(0),((0)),(1),1,(1),1,0,1,1,0,1,...
a(9) = 1 produces parentheses around a(11):
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),0,1,1,0,1,...
a(10) = 1 produces parentheses around a(12):
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),(0),1,1,0,1,...
a(11) = 1 produces parentheses around a(13):
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),(0),(1),1,0,1,...
a(12) = 0 produces parentheses around a(13) again:
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),(0),((1)),1,0,1,...
a(13) = 1 produces parentheses around a(15):
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),(0),((1)),(1),(0),1,...
a(14) = 1 produces parentheses around a(16):
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),(0),((1)),(1),(0),(1),...
a(15) = 0 produces parentheses around a(16) again:
1,0,((1)),1,(1),(0),((0)),(1),1,(1),(1),(0),((1)),(1),(0),((1)),...
Etc.
We see in this small example that the doubly parenthesized terms of the last line slowly reconstruct the starting sequence:
((1)), ((0)), ((1)), ((1)), ...
The same holds for the singly parenthesized terms:
(1), (0), (1), (1), (1), (0), (1), (0), ...
And again by the non-parenthesized terms:
1, 0, 1, 1, ...

Crossrefs

Cf. A307183 (no distinction between underlined once or twice: two copies only of the starting sequence).

Sequence in context: A099076 A282339 A175479 * A120530 A078616 A267800

Adjacent sequences: A307240 A307241 A307242 * A307244 A307245 A307246

Keyword

base,nonn,nice

Author

Eric Angelini and Lars Blomberg, Mar 30 2019

Status

approved