|
I conjecture that the ratio r(n) of the number of "1"s to the number of "0"s in a(n) converges to 5/3 (or some nearby limit). - Joseph L. Pe, Jan 31 2003
The ratio r(n) of the number of "1"s to the number of "0"s in a(n) actually converges to ((101 - 10*sqrt(93))*a^2 + (139 - 13*sqrt(93))*a - 76)/108, where a = (116 + 12*sqrt(93))^(1/3). This ratio has decimal expansion 1.6657272222676... - Nathaniel Johnston, Nov 07 2010 [Corrected by Kevin J. Gomez, Dec 12 2017]
Reading terms as binary numbers and converting to decimal gives A049190. - Andrey Zabolotskiy, Dec 12 2017
From Jianing Song, Oct 05 2022: (Start)
"000" or "11111" never appear in any a(n). Proof:
When "000" appears for the first time in a(n),
- if it reads as "..00 0's", then a(n-1) must contain at least 4 consecutive 0's, which is impossible;
- if it reads as "...000... 0's" or "...000... 1's", then a(n-1) must contain at least 8 consecutive 0's or at least 8 consecutive 1's.
In conclusion, a(n-1) must contain at least 8 consecutive 1's.
When "11111" appears for the first time in a(n),
- if it reads as "...1111 1's", then a(n-1) must contain at least 15 consecutive 1's, which is impossible;
- if it reads as "...111 1's, 1... 0's", then a(n-1) must contain at least 7 consecutive 1's, which is impossible;
- if it reads as "...11 1's, 11... 0's", then a(n-1) must contain at least 3 consecutive 0's;
- if it reads as "...1 1's, 111... 0's", then a(n-1) must contain at least 7 consecutive 0's;
- if it reads as "... 1's, 1111... 0's", then a(n-1) must contain at least 15 consecutive 0's;
- if it reads as "...11111... 0's" or "...11111... 1's", then a(n-1) must contain at least 31 consecutive 0's or at least 31 consecutive 1's.
In conclusion, a(n-1) must contain at least 3 consecutive 0's. Combining these two results, one can easily show that "000" or "11111" cannot appear. (End)
| 