[seqfan] Irregular triple fractal

[Éric Angelini]

Hello SeqFans,
I cannot resist posting this private e-mail
from Lars Blomberg -- as it deals with
a beautiful (!) sequence (which will
be submitted tomorrow to the OEIS)
-- sequence with a large amount of
unsolved questions (regularity, runs 
of 0s and 1s, possible loop, etc.)
Explanation after the e-mail.

[Lars]:
> I agree, the sequence is interesting and surprisingly simple to generate (no backtracking required!)
 
> I find the first 1000 terms to be

10110111010111011011110101101011110110111010011111011101011101100111111011110100111101011001111111
01011110100011111011011100111111110100111110110001011111011101111001011111111010001011111011100011
01111110111101011110011011111111101000011010111110111100011101011111101111101001011110011101111111
11101100001110110101111101111100011110110101111110101011110110011011111001011101011111111110111000
01010110111011011111101111110001111101110100111111101001101111101110011101011111001001010110110111
11111111011110000110100111011110101101111111011111110001010111101111011001111111101100111011111101
11100101110100101111100110011011011101011011111111111101010101100001011011001111011111011010110111
11111011111111000100110111110111110111001010111111101011001111011111110111110011011110110011010111
11001010001011010110111101101110111111111111101001101101110000100101001110011111010111110111011011
10111111111011111111100010001110111111011111101111001001101011111110110111001111101010111111011111
10010110111110111001
 
> For e=(3,4,5,6,7,8,9) the 10^e terms contain (70,696,6928,69249,693289,6941990,69507187,694529637) ones.
The fraction of ones seem to be rather constant, but the are subtle differences from any regular trend.
 
>Furthermore, if runs of 0’s and 1’s are counted it turns out that from 10^3 terms and upward there is never more than 4 in a run of zeros and from 10^5 terms and upwards never more than 17 in a run of ones.
 
>I cannot imagine why this should be so, and one might suspect the algorithm, but as I said above, it is not at all complicated and I can’t find anything wrong with it.

[Explanation]:
Start reading the sequence:
---if a(n) = 0, underline a(n+1);
---if a(n) = 1, underline a(n+2).
The underlined terms will reproduce 
the starting sequence.
The not underlined terms too.

(A term underlined twice is considered 
to be just "underlined").

Many thanks to Lars -- a true friend
and skilled person!

Best,
É.