A001387 The binary "look and say" sequence.

1, 11, 101, 111011, 11110101, 100110111011, 111001011011110101, 111100111010110100110111011, 100110011110111010110111001011011110101, 1110010110010011011110111010110111100111010110100110111011

Offset: 1

Author

Thomas L. York

Keywords

• nonn
• base

Comments

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)

Examples

			To get the 5th term, for example, note that 4th term has three (11 in binary!) 1's, one (1) 0 and two (10) 1's, giving 11 1 1 0 10 1.
		

Links

• John Cerkan, Table of n, a(n) for n = 1..17
• J. H. Conway, The weird and wonderful chemistry of audioactive decay, Eureka 46 (1986) 5-16, reprinted in: Open Problems in Communications and Computations, Springer, 1987, 173-188.
• Nathaniel Johnston, The Binary "Look-and-Say" Sequence
• Thomas Morrill, Look, Knave, arXiv:2004.06414 [math.CO], 2020.
• Torsten Sillke, The binary form of Conway's sequence

Crossrefs

Cf. A005150, A001391, A049190, A049194.

Programs

Mathematica

a[1] := 1; a[n_] := a[n] = FromDigits[Flatten[{IntegerDigits[Length[#],2], First[#]}& /@ Split[IntegerDigits[a[n-1]]]]]; Map[a, Range[20]] (* Peter J. C. Moses, Mar 24 2013 *)
Nest[Append[#, FromDigits@ Flatten@ Map[Reverse /@ IntegerDigits[Tally@ #, 2] &, Split@ IntegerDigits@ Last@ #]] &, {1}, 9] (* Michael De Vlieger, Dec 12 2017 *)

Python

from itertools import accumulate, groupby, repeat
def summarize(n, _): return int("".join(bin(len(list(g)))[2:]+k for k, g in groupby(str(n))))
def aupto(terms): return list(accumulate(repeat(1, terms), summarize))
print(aupto(11)) # Michael S. Branicky, Sep 18 2022

Extensions

New name from Andrey Zabolotskiy, Dec 13 2017