A001155 -id:A001155 - cp's OEIS frontend

A036058 Summarize digits of preceding number, by decreasing digit value. Start with a(0) = 0.

0, 10, 1110, 3110, 132110, 13123110, 23124110, 1413223110, 1423224110, 2413323110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110, 1433223110

Offset: 0

Floor van Lamoen

This kind of counting sequence is always eventually periodic with period 1, 2 or 3. - Herve Lehning (lehning(AT)noos.fr), Oct 01 2003

			The third term is 1110 because the second term contains one 1 and one 0.

Herve Lehning, Computer-aided or analytic proof?, College Mathematics Journal, Vol. 21, No. 3, 1990, pp. 228-239.
Index of eventually constant sequences.

Cf. A007890 (same as this, starting at 1), A001155 (same as this, but using method A047842: by increasing digit value), A005150 (as before, starting at 1), A036059 ("fibonacci" based on this), A036066.

a(n)=if(n>9,1433223110,[0,10,1110,3110,132110,13123110,23124110,1413223110, 1423224110,2413323110][n+1]) \\ Charles R Greathouse IV, Jul 24 2012

a(n,a=0)={for(k=1,n,a==(a=A244112(a))&&break);a} \\ M. F. Hasler, Feb 25 2018

a(n+1) = A244112(a(n)), a(0) = 0. - M. F. Hasler, Feb 25 2018

A209234 A two-digit Look-and-Say sequence starting with 10: each term summarizes the increasing two-digit substrings of the previous term.

Original entry on oeis.org

10, 110, 110111, 101110311, 101103210311131, 101203310311113121231132, 101203210411312213120121123431132133, 201103104210311512413220421122123331232133134141143, 101203204310411412313214115220421222223124431232333134341242143151

Offset: 0

Reinhard Zumkeller, Jan 13 2013

			a(0) = 10: 1x10 --> a(1)=110;
a(1) = 110: 1x10 and 1x11 --> a(2)=110111;
a(2) = 110111: 1x01, 1x10 and 3x11 -> a(3)=101110311;
a(3) = 101110311: 1x01, 1x03, 2x10, 3x11 and 1x31 -> a(4)=101103210311131.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Look and Say Sequence
Wikipedia, Look-and-say sequence
Reinhard Zumkeller, Haskell program for two-digit Look-and-Say sequences

Cf. A005151, A047842, A001155.

Cf. A209233 (start=11), A221368 (start=12), A221369 (start=13), A221372 (start=19), A221373 (start=99).

Haskell
```
-- See Link.
```

A022471 Length of n-th term of A022470.

Original entry on oeis.org

1, 2, 4, 4, 6, 10, 12, 14, 22, 26, 30, 44, 56, 70, 98, 130, 162, 216, 292, 358, 470, 628, 792, 1050, 1384, 1788, 2334, 3072, 3974, 5162, 6784, 8786, 11420, 14992, 19484, 25388, 33160, 43262, 56252, 73392, 95798, 124496, 162556, 212048, 275976, 360154

Offset: 1

Clark Kimberling

nonn

a(n) is the length of the n-th term of many sequences generated by methods A and B, including those shown here:
Method A, 1st term ... Method B, 1st term
A001155, 0
A006751, 2 ......... A022470, 2
A006715, 3 ......... A022499, 3
A001140, 4 ......... A022500, 4
A001141, 5 ......... A022501, 5
A001143, 6 ......... A022502, 6
A001145, 7 ......... A022503, 7
A001151, 8 ......... A022504, 8
A001154, 9 ......... A022505, 9
Clark Kimberling, Jun 14 2013

Peter J. C. Moses, Table of n, a(n) for n = 1..3000

Cf. A022470.

a[0] = 2; a[n_] := a[n] = FromDigits[Flatten[{First[#], Length[#]} & /@   Split[IntegerDigits[a[n - 1]]]]]; Map[Length[IntegerDigits[a[#]]] &, Range[0, 40]] (* Peter J. C. Moses, Jun 14 2013 *)
p = {9, -9, 6, -16, 5, 2, 0, -9, -1, -1, 20, 2, 6, -3, -15, -13, 15, 20, 15, -26, -28, 7, 6, 26, -27, -4, 9, -15, 3, 2, 8, 43, 9, -39, -24, -2, -24, 28, 9, 13, 13, -18, -12, -16, 14, 13, 16, 8, -36, 1, -6, -8, 15, 1, 14, 3, -6, -7, -3, 2, -2, 2, 2, 0, -1, -2, -1, 3, 3, -1, -1, -1}; q = {-6, 9, -9, 18, -16, 11, -14, 8, -1, 5, -7, -2, -8, 14, 5, 5, -19, -3, 6, 7, 6, -16, 7, -8, 22, -17, 12, -7, -5, -7, 8, -4, 7, 9, -13, 4, 6, -14, 14, -19, 7, 13, -2, 4, -18, 0, 1, 4, 12, -8, 5, 0, -8, -1, -7, 8, 5, 2, -3, -3, 0, 0, 0, 0, 2, 1, 0, -3, -1, 1, 1, 1, -1}; gf = Fold[x #1 + #2 &, 0, p]/Fold[x #1 + #2 &, 0, q]; CoefficientList[Series[gf, {x, 0, 100}], x] (* Peter J. C. Moses, Jun 16 2013 *)

A171773 This sequence is a relative of the audioactive sequences. We generate it by starting with a symbol * and describe sequentially: , 1, 111, 311, 13211*,...

Original entry on oeis.org

1, 111, 311, 13211, 111312211, 31131122211, 1321132132211, 111312211312111322211, 3113112221131112311332211, 13211321322113311213212322211, 1113122113121113222123211211131211121332211, 3113112221131112311332111213122112311311123112112322211

Offset: 1

Louis Hirsch Kauffman (kauffman(AT)uic.edu), Dec 18 2009

The interest of this is that if A_{n} is the n-th term of this sequence then A_{n} is a truncate of A_{n+3}. Thus the sequence gives rise to a triple A,B,C of infinite sequences of 1,2,3 such that B describes A, C describes B and A describes C.

This sequence serves as the initial portion of A001155, A001140, A001141, A001143, A001145, A001151, and A001154, as it is those sequences with the 'seed value' removed. - James E Davis, Apr 28 2016

			The term after 311 is one-three, two-one, one: i.e. 13211. - _James E Davis_, Apr 28 2016

Michael De Vlieger, Table of n, a(n) for n = 1..22

Cf. A005150, A001155, A001140, A001141, A001143, A001145, A001151, and A001154

NestList[FromDigits@ Append[Flatten@ Map[{Length@ #, First@ #} &, Split@ IntegerDigits@ #], 1] &, 1, 10] (* Michael De Vlieger, Apr 28 2016 *)

Each term can be found by doing a look-and-say on the previous term and appending a 1. - James E Davis, Apr 28 2016

More terms from James E Davis, Apr 28 2016

A109973 First repeating NA iterates. The NA (Noun-Adjective) function of a finite sequence s of nonnegative integers is the finite sequence 0a1b2c...mz, where a = #0's in s, b = #1's in s, ..., z = #m's in s, m = greatest term in s.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 2, 0, 1, 1, 3, 2, 1, 3, 3, 0, 1, 1, 3, 2, 1, 3, 3, 0, 1, 1, 3, 2, 2, 3, 3, 4, 1, 0, 1, 1, 4, 2, 2, 3, 2, 4, 2, 5, 1, 0, 1, 1, 5, 2, 2, 3, 2, 4, 1, 5, 2, 6, 1, 0, 1, 1, 5, 2, 4, 3, 1, 4, 1, 5, 1, 6, 2, 7, 1, 0, 1, 1, 6, 2, 4, 3, 1, 4, 1, 5, 1, 6, 1, 7, 2, 8, 1, 0, 1, 1, 7, 2, 4, 3, 1, 4, 1, 5

Offset: 1

Clark Kimberling, Jul 06 2005

nonn

This is a concatenation of finite segments.  The first segment is 01122332, obtained by writing the NA iterates of 0 until repetition occurs: 0, 01, 0111, 0113, 01122031, 02132231, 01122332, after which 01122332 repeats.  It helps to speak your way through: write 0 and say 0 one time - that's 01; then say 0 one time and 1 one time - that's 0111; then say 0 one time and 1 3 times, and so on, until reaching the repeating segment 01122332.  This segment is a fixed point of the NA function.
The second segment is obtained by writing the NA iterates of 1 until repetition occurs: 1, 0011, 0212, 011122, 011322, 01122231, 01132331, 01132133, after which 01132133 repeats, so that the second segment is 01132133.
Third segment (from initial 2): 01132133.
Fourth segment (from initial 3): 0113223341.
Fifth segment (from initial 4): 011422324251.  Here, for the first time, the repetition does not occur immediately after the first occurrence.  Indeed, iteration never reaches a fixed point of the NA function.  Instead, the iterates oscillate between 011422324251 and 011324314251.
These observations prompt questions: (1) what initial segments generate fixed points?  (2) do segments eventually occur periodically, regardless of the choice of initial segment?
- Clark Kimberling, May 08 2011

Cf. A001155, A055168, A191654 (repeating AN iterates).

(* Program computes the NA segment starting with 0 *)
nounAdjective[s_] := Flatten@Transpose@({#1, (Count[s, #1] &) /@ #1} &)[Range[0, Max[s]]];
NestList[nounAdjective[#1] &, nounAdjective[{0}], 7]
(* Next program: the NA segment starting with 1 *)
nounAdjective[s_] := Flatten@Transpose@({#1, (Count[s, #1] &) /@ #1} &)[Range[0, Max[s]]];
NestList[nounAdjective[#1] &, nounAdjective[{1}], 7]
(* ...and so on.  By Peter J. C. Moses, Jun 03 2011 *)

A355316 Stuttering Look and Say sequence with seed 0.

Original entry on oeis.org

0, 10, 1110, 333110, 333322110, 4444322222110, 444441355555222110, 5555541113555555333222110, 5555551433311366666653333333222110, 6666665111433332211366666661577777773333222110, 66666661533311444443222221137777777611157777777744443333222110

Offset: 1

Jonathan Comes, Jun 28 2022

If we let L(n) denote the number of digits in the n-th term, then the limit of L(n+1)/L(n) is an algebraic integer of degree 415. This limit is a stuttering analog of Conway's constant (see A014715).

			E.g., to obtain the term after 1110, we look at 1110 and see "three 1's and one 0". We then say what we saw by stuttering the counts as many times as the count prescribes: we stutter the "three" 3 times and the "one" 1 time (no stutter); so we say "three three three 1's and one 0" to get 333110.

Michael S. Branicky, Table of n, a(n) for n = 1..18
Jonathan Comes, Stuttering look and say sequences and a challenger to Conway's most complicated algebraic number from the silliest source, arXiv:2206.11991 [math.HO], 2022.

Stuttering variant of A001155.

Cf. A014715.

first(n) = my(c, d=[0], x, res=vector(n)); for(i=2, n, c=1; x=""; for(j=1, #d, if(j<#d && d[j]==d[j+1], c++, x=concat(x, concat(vector(c+1, k, Str(if(k==c+1, d[j], c))))); c=1)); res[i]=eval(x); d=digits(res[i])); res \\ Iain Fox, Jun 30 2022

from itertools import accumulate, groupby, repeat
def summarize(n, _): return int("".join(str(c:=len(list(g)))*c+k for k, g in groupby(str(n))))
def aupton(terms): return list(accumulate(repeat(0, terms), summarize))
print(aupton(11)) # Michael S. Branicky, Jun 28 2022

A363054 Look and say sequence: describe the previous term (method A, starting with 20).

Original entry on oeis.org

20, 1210, 11121110, 31123110, 132112132110, 11131221121113122110, 311311222112311311222110, 1321132132211213211321322110, 11131221131211132221121113122113121113222110, 3113112221131112311332211231131122211311123113322110

Offset: 1

Julia Zimmerman, May 15 2023

			The term after 1210 is given by saying "I see one 1, one 2, one 1, and one 0", and then writing down the digits as 11-12-11-10, yielding 11121110.

Michael De Vlieger, Table of n, a(n) for n = 1..21
Julia Witte Zimmerman, Denis Hudon, Kathryn Cramer, Alejandro J. Ruiz, Calla Beauregard, Ashley Fehr, Mikaela Irene Fudolig, Bradford Demarest, Yoshi Meke Bird, Milo Z. Trujillo, Christopher M. Danforth, and Peter Sheridan Dodds, Tokens, the oft-overlooked appetizer: Large language models, the distributional hypothesis, and meaning, arXiv:2412.10924 [cs.CL], 2024. See pp. 21, 28.

Other look-and-say sequences: A005150, A001155, A006751, A006715, A001140, A001141, A001143, A001145, A001151, A007651, A001387, A098595, A045918.

NestList[FromDigits@ Flatten@ Map[Reverse@ Tally[#][[1]] &, Split@ IntegerDigits[#] ] &, 20, 12] (* Michael De Vlieger, Jul 05 2023 *)

from itertools import count, groupby, islice
def LS(n): return int(''.join(str(len(list(g)))+k for k, g in groupby(str(n))))
def agen(an=20): yield an; yield from (an:=LS(an) for n in count(1))
print(list(islice(agen(), 10))) # Michael S. Branicky, May 15 2023

A260387 Numbers n = d_0d_1...d_n (n < 10) such that d_i is the number of digits equal to i in n (base b), where b is less than 10.

Original entry on oeis.org

12, 13, 320, 3201, 72200, 89000, 132110, 345000, 643000, 2320200, 3121300, 10103111, 11300130, 42430000, 51340000, 64030000, 72300000, 86300000, 125102000, 130213000, 211220001, 220101111, 323111000, 431130000, 614110000, 667000000, 2153100000, 2521002000, 3021211100

Offset: 1

Pieter Post, Jul 24 2015

The only terms having the same number of digits as the base are 13, 10103111, 211220001 and 220101111. For example, 13 is 1101_2,  which has 1 zero and 3 ones.
The least term with 10 digits that describes itself is 2153100000.
2153100000 is 104233022322_7, so it has 2 zeros, 1 one, 5 twos, 3 threes, 1 four, 0 fives, 0 sixes, 0 sevens, 0 eights and 0 nines in base 7.

			12 = 110_3, which has 1 zero and 2 ones.
13 = 1101_2, which has 1 zero and 3 ones.
320 = 11000_4, which has 3 zeros, 2 ones and 0 twos.
3201 = 100301_5, which has 3 zeros, 2 ones, 0 twos and 1 three.
72200 = 10200001002_3
89000 = 10101101110101000_2
132110 = 13211420_5
345000 = 122112020210_3
643000 1012200000211_3
42430000 = 2201312320300_4
51340000 = 3003312023200_4
64030000 = 3310100110300_4
72300000 = 122002100000_5
86300000 = 20000101111100022_3
431130000 = 110440340120_6
614110000 = 2224203010000_5
667000000 = 1201111002002222201_3
2153100000 = 104233022322_7

Cf. A001155, A001387, A005150, A046043, A138480.

a(10)-a(13), a(19)-a(23), a(28)-a(29) added by Giovanni Resta, Jul 26 2015

A336752 The 'Look and Say' sequence of the concatenation of the prime numbers A033308.

Original entry on oeis.org

1, 2, 1, 3, 1, 5, 1, 7, 3, 1, 1, 3, 1, 1, 1, 7, 1, 1, 1, 9, 1, 2, 1, 3, 1, 2, 1, 9, 1, 3, 1, 1, 1, 3, 1, 7, 1, 4, 1, 1, 1, 4, 1, 3, 1, 4, 1, 7, 1, 5, 1, 3, 1, 5, 1, 9, 1, 6, 1, 1, 1, 6, 2, 7, 1, 1, 1, 7, 1, 3, 1, 7, 1, 9, 1, 8, 1, 3, 1, 8, 2, 9, 1, 7, 1, 1, 1, 0, 2, 1, 1, 0, 1, 3, 1, 1, 1, 0, 1, 7

Offset: 1

Scott R. Shannon, Aug 15 2020

Concatenate all the decimal prime numbers, see A033308, then describe the resulting infinite string using the 'Look and Say' method of A005150.

			The concatenation of the primes starts "23571113171923293137...".
a(1) = 1, a(2) = 2 as there is one '2' at the start of the string.
a(9) = 3, a(10) = 1 as the primes '11' and '13' from the substring '1113'. which starts with three 1's.

J. H. Conway and Brady Haran, Look-and-Say Numbers (2014), Numberphile video.
Wikipedia, Look-and-say sequence.

Cf. A033308, A005150, A001155, A056815, A000040.

cp's OEIS Frontend

A036058 Summarize digits of preceding number, by decreasing digit value. Start with a(0) = 0.

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A209234 A two-digit Look-and-Say sequence starting with 10: each term summarizes the increasing two-digit substrings of the previous term.

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Haskell

A022471 Length of n-th term of A022470.

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Mathematica

A171773 This sequence is a relative of the audioactive sequences. We generate it by starting with a symbol * and describe sequentially: *, 1*, 111*, 311*, 13211*,...

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A109973 First repeating NA iterates. The NA (Noun-Adjective) function of a finite sequence s of nonnegative integers is the finite sequence 0a1b2c...mz, where a = #0's in s, b = #1's in s, ..., z = #m's in s, m = greatest term in s.

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Mathematica

A355316 Stuttering Look and Say sequence with seed 0.

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PARI

Python

A363054 Look and say sequence: describe the previous term (method A, starting with 20).

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Mathematica

Python

A260387 Numbers n = d_0d_1...d_n (n < 10) such that d_i is the number of digits equal to i in n (base b), where b is less than 10.

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A171773 This sequence is a relative of the audioactive sequences. We generate it by starting with a symbol * and describe sequentially: , 1, 111, 311, 13211*,...