A049190 -id:A049190 - cp's OEIS frontend

A001387 The binary "look and say" sequence.

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

Offset: 1

Thomas L. York

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)

			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.

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

Cf. A005150, A001391, A049190, A049194.

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 *)

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

New name from Andrey Zabolotskiy, Dec 13 2017

A056966 In binary: write what is described (putting a leading zero on numbers which have an odd number of binary digits).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 2, 2, 4, 5, 3, 3, 6, 7, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0

Offset: 0

Henry Bottomley, Jul 20 2000

			a(54)=2 because 54 = 110110 base 2, which can be read as one 1 followed by zero 1's followed by one 0, i.e., 10 base 2 = 2 base 10.

Cf. A014081, A049190, A056967.

def A056966(n):
    s = bin(n)[2:]
    s = '0'*(len(s)&1)+s
    return int('0'+''.join(s[i+1]*int(s[i])for i in range(0,len(s),2)),2) # Chai Wah Wu, Feb 12 2023

A259710 Describe previous term in binary, then convert to decimal (initial term is 0).

Original entry on oeis.org

0, 2, 14, 30, 38, 918, 31190, 1261014, 1925823958, 66980978212310, 88736005061423513046, 284201453518322998760440446422, 21227119308677323964491588521405117134126550, 1978879038810329481388510354539283798664092126944649356922961366

Offset: 1

Kade Robertson, Jul 03 2015

			E.g., the term after 14 can be obtained by changing 14 to binary 1110_2, then describing in binary as three (11_2) ones and one (1) zero, concatenate these to make 11110_2, then convert back to decimal to obtain 30.

Kade Robertson, Table of n, a(n) for n = 1..21

Cf. A049064, A049190 (initial term is 1).

a(n) = base-10(A049064(n)).

Offset corrected by Jianing Song, Mar 16 2019

A321226 Describe the binary representation of n in binary and convert back to decimal.

Original entry on oeis.org

2, 3, 14, 5, 28, 59, 22, 7, 30, 115, 238, 117, 44, 91, 30, 9, 56, 123, 462, 229, 476, 955, 470, 119, 46, 179, 366, 181, 60, 123, 38, 11, 58, 227, 494, 245, 924, 1851, 918, 231, 478, 1907, 3822, 1909, 940, 1883, 478, 233, 88, 187, 718, 357, 732, 1467, 726, 183

Offset: 0

Rémy Sigrist, Nov 10 2018

This sequence is a binary variant of the "Look and Say" sequence A045918.

There is only one fixed point: a(7) = 7.

			For n = 67:
- the binary representation of 67 is "1000011",
- we see, in binary: "1" "1", "100" "0", "10" "1",
- hence the binary representation of a(67) is "111000101",
- and a(67) = 453 in decimal.

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Cf. A020330, A045918, A049190.

a(n, b=2) = if (n==0, return (b)); my (d=digits(b*n, b), v=0, w=0); d[#d] = -1; for (i=1, #d-1, w++; if (d[i]!=d[i+1], v = b*(v*b^#digits(w, b) + w) + d[i]; w = 0)); v

a(2^n - 1) = 2*n + 1 for any n > 0.
a(4*n + 1) = 4*a(2*n) + 3 for any n > 0.
a(4*n + 2) = 4*a(2*n + 1) + 2 for any n >= 0.
a(A020330(2*n)) = A020330(a(2*n)) for any n > 0.
a(A049190(n)) = A049190(n+1) for any n > 0.

cp's OEIS Frontend

A001387 The binary "look and say" sequence.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Extensions

A056966 In binary: write what is described (putting a leading zero on numbers which have an odd number of binary digits).

Views

Author

Keywords

Examples

Crossrefs

Programs

Python

A259710 Describe previous term in binary, then convert to decimal (initial term is 0).

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A321226 Describe the binary representation of n in binary and convert back to decimal.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A049192 Start with 1. Read as base 10, convert to base 2, describe it in base 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions