A001387 The binary "look and say" sequence.

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

Offset

1,2

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)

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A005150, A001391, A049190, A049194.

Sequence in context: A156668 A103992 A185949 * A358198 A247863 A180280

Adjacent sequences: A001384 A001385 A001386 * A001388 A001389 A001390

Keyword

nonn,base

Author

Thomas L. York

Extensions

New name from Andrey Zabolotskiy, Dec 13 2017

Status

approved