A036058 -id:A036058 - cp's OEIS frontend

A001155 Describe the previous term! (method A - initial term is 0).

0, 10, 1110, 3110, 132110, 1113122110, 311311222110, 13211321322110, 1113122113121113222110, 31131122211311123113322110, 132113213221133112132123222110, 11131221131211132221232112111312111213322110, 31131122211311123113321112131221123113111231121123222110

Offset: 1

N. J. A. Sloane

Method A = 'frequency' followed by 'digit'-indication.
a(n), A001140, A001141, A001143, A001145, A001151 and A001154 are all identical apart from the last digit of each term (the seed). This is because digits other than 1, 2 and 3 never arise elsewhere in the terms (other than at the end of each of them) of look-and-say sequences of this type (as is mentioned by Carmine Suriano in A006751). - Chayim Lowen, Jul 16 2015
a(n+1) - a(n) is divisible by 10^5 for n > 5. - Altug Alkan, Dec 04 2015

			The term after 3110 is obtained by saying "one 3, two 1's, one 0", which gives 132110.

S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 452-455.
I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 4.

T. D. Noe, Table of n, a(n) for n=1..20
J. H. Conway, The weird and wonderful chemistry of audioactive decay, in T. M. Cover and Gopinath, eds., Open Problems in Communication and Computation, Springer, NY 1987, pp. 173-188.
S. R. Finch, Conway's Constant [Broken link]
S. R. Finch, Conway's Constant [From the Wayback Machine]

Cf. A005150, A006751, A006715, A001140, A001141, A001143, A001145, A001151, A001154.

Cf. A036058.

A001155[1] := 0; A001155[n_] := A001155[n] = FromDigits[Flatten[{Length[#], First[#]}&/@Split[IntegerDigits[A001155[n-1]]]]]; Map[A001155,Range[15]] (* Peter J. C. Moses, Mar 21 2013 *)

A001155(n,a=0)={ while(n--, my(c=1); for(j=2,#a=Vec(Str(a)), if( a[j-1]==a[j], a[j-1]=""; c++, a[j-1]=Str(c,a[j-1]); c=1)); a[#a]=Str(c,a[#a]); a=concat(a)); a }  \\ M. F. Hasler, Jun 30 2011

from itertools import accumulate, groupby, repeat
def summarize(n, _): return int("".join(str(len(list(g)))+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

A036059 The summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

1, 1, 21, 1221, 3231, 233231, 533221, 15534221, 3514334231, 3534533241, 3544832231, 183544733221, 28172544634231, 2827162554535241, 2827265554337241, 2837267544338231, 3847264544637221, 3847362564636221, 2837662564536221, 2827863534537221, 3837564524538221

Offset: 0

Floor van Lamoen

After the 23rd term the sequence goes into a cycle of 16 terms.

			a(20) = 3837564524538221;
a(21) = 4837265534637221;
a(22+16*k) = 3837365544636221, k >= 0;
a(36) = a(20+16) = 3837265554834221 <> a(20);
a(37) = a(21+16) = 3837266544735221 <> a(21);
a(38) = a(22+16) = 3837365544636221 = a(22). - _Reinhard Zumkeller_, Aug 10 2014

Reinhard Zumkeller, Table of n, a(n) for n = 0..181, 10 periods.
Index to sequences related to say what you see

Cf. A036058, A036066.

import Data.List (sort, group)
a036059 n = a036059_list !! n
a036059_list = map (read . concatMap show) fss :: [Integer] where
   fss = [1] : [1] : zipWith h (tail fss) fss where
         h vs ws = concatMap (\us -> [length us, head us]) $
                   group $ reverse $ sort $ vs ++ ws
-- Reinhard Zumkeller, Aug 10 2014

a:= proc(n) option remember; `if`(n<2, 1, (p-> parse(cat(seq((c->
     `if`(c=0, [][], [c, 9-i][]))(coeff(p, x, 9-i)), i=0..9))))(
      add(x^i, i=map(x-> convert(x, base, 10)[], [a(n-1),a(n-2)]))))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 18 2022

a[0] = a[1] = 1; a[n_] := a[n] = FromDigits @ Flatten @ Reverse @ Select[ Transpose @ { DigitCount[a[n-1]] + DigitCount[a[n-2]], Append[ Range[9], 0]}, #[[1]] > 0&];
Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Dec 30 2017 *)

def aupton(nn):
  alst = [1, 1]
  for n in range(2, nn+1):
    prev2, anstr = sorted(str(alst[-2]) + str(alst[-1])), ""
    for d in sorted(set(prev2), reverse=True):
      anstr += str(prev2.count(d)) + d
    alst.append(int(anstr))
  return alst
print(aupton(20)) # Michael S. Branicky, Feb 02 2021

A244112 Reverse digit count of n in decimal representation.

Original entry on oeis.org

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1110, 21, 1211, 1311, 1411, 1511, 1611, 1711, 1811, 1911, 1210, 1211, 22, 1312, 1412, 1512, 1612, 1712, 1812, 1912, 1310, 1311, 1312, 23, 1413, 1513, 1613, 1713, 1813, 1913, 1410, 1411, 1412, 1413, 24, 1514, 1614, 1714

Offset: 0

Reinhard Zumkeller, Nov 11 2014

Frequencies of digits 0 through 9, occurring in n, are summarized in order of decreasing digits;

a(A010785(n)) = A047842(A010785(n)).

			101 contains two 1s and one 0, therefore a(101) = 2110;
102 contains one 2, one 1 and one 0, therefore a(102) = 121110.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A047842, A010785.

See A036058 for the orbit of 0 under this map.

import Data.List (sort, group); import Data.Function (on)
a244112 :: Integer -> Integer
a244112 n = read $ concat $
   zipWith ((++) `on` show) (map length xs) (map head xs)
   where xs = group $ reverse $ sort $ map (read . return) $ show n

f[n_] := Block[{s = Split@ IntegerDigits@ n}, FromDigits@ Reverse@ Riffle[Union@ Flatten@ s, Length@# & /@ s]]; Array[f, 48, 0] (* Robert G. Wilson v, Dec 01 2016 *)

A244112(n,c=1,S="")={for(i=2,#n=vecsort(digits(n),,4),n[i]==n[i-1]&&c++&&next;S=Str(S,c,n[i-1]);c=1);eval(Str(S,c,if(n,n[#n])))} \\ M. F. Hasler, Feb 25 2018

def A244112(n):
    return int(''.join([str(str(n).count(d))+d for d in '9876543210' if str(n).count(d) > 0])) # Chai Wah Wu, Dec 01 2016

A007890 Summarize the previous term! (in decreasing order).

Original entry on oeis.org

1, 11, 21, 1211, 1231, 131221, 132231, 232221, 134211, 14131231, 14231241, 24132231, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221, 14233221

Offset: 1

Mira Bernstein

			For example, the term after 131221 is obtained by saying "one 3, two 2's, three 1's", which gives 13-22-31, i.e. 132231.

Onno M. Cain, Sela T. Enin, Inventory Loops (i.e. Counting Sequences) have Pre-period 2 max S_1 + 60, arXiv:2004.00209 [math.NT], 2020.

Cf. A005150, A034003, A036058, A244112.

Nest[Append[#, FromDigits@ Flatten@ Map[Reverse, Tally@ ReverseSort@ IntegerDigits@ #[[-1]] ] ] &, {1}, 24] (* Michael De Vlieger, Jul 15 2020 *)

From Seiichi Manyama, Aug 18 2020: (Start)
a(1) = 1 and a(n) = A244112(a(n-1)) for n > 1.
a(n) = 14233221 for n >= 13. (End)

A036212 When the 'Summarize preceding term' sequence gets into a cycle.

Original entry on oeis.org

11, 13, 12, 13, 9, 11, 11, 11, 11, 11, 10, 12, 11, 12, 8, 10, 10, 10, 10, 10, 7, 11, 1, 10, 7, 7, 7, 7, 7, 7, 9, 12, 10, 11, 7, 9, 9, 9, 9, 9, 10, 8, 7, 7, 8, 9, 10, 10, 10, 10, 11, 10, 7, 9, 9, 8, 10, 11, 11, 11, 13, 10, 7, 9, 10, 10, 8, 11, 13, 13, 14, 10, 7, 9, 10, 11, 11, 8, 13, 14

Offset: 0

Floor van Lamoen

			a(0)=11 since the 'Summarize the previous term' sequence starting with 0 gets in a cycle from the 11th term.

Sean A. Irvine, Java program (github)

Cf. A007890, A036058.

A036225 Length of cycle the 'summarize the previous term' sequence gets into.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 2, 3, 3, 3, 2, 1, 1, 1, 2, 2, 1, 3, 2, 2, 2, 1, 1, 1, 2, 3, 3, 1, 2, 2, 2, 1, 1, 1, 2, 3, 2, 2, 1, 2, 2, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1

Offset: 0

Floor van Lamoen

For n <= 10^7, a(n) <= 3. - A.H.M. Smeets, Jun 24 2019

			a(0)=1 because the 'Summarize the previous term' starting with 0 becomes constant and a(40)=2 because starting with 40 it gets into a cycle of two.

A.H.M. Smeets, Table of n, a(n) for n = 0..20000

Cf. A007890, A036058, A036212.

A300191 Look and Say the digits of the last three terms, by increasing digit value. Start with a(1)=1, a(2)=2 and a(3)=3.

Original entry on oeis.org

1, 2, 3, 111213, 412223, 51423314, 7152433415, 515253543517, 613263548527, 5142634495263718, 515263549546372819, 516263648566373839, 515283648596374849, 414283649596376859, 3132837475106377869, 10413283746576677869, 20513283644596976859, 30514283545596976859

Offset: 1

Paolo P. Lava, Feb 28 2018

From the 1458th term the sequence goes into a cycle of 850 terms.

			1, 2, 3 => there are one '1', one '2' and one '3': 111213;
2, 3, 111213 => there are four '1', two '2' and two '3': 412223;
3, 111213, 412223 => there are five '1', four '2', three '3', one '4': 51423314.

Jinyuan Wang, Table of n, a(n) for n = 1..3200 (first 1937 terms from Paolo P. Lava)

Cf. A036058, A036059, A036066, A036067.

P:=proc(q) local a,b,c,d,k,n,y,x; y:=array(1..3); x:=array(0..9);
y[1]:=1; y[2]:=2; y[3]:=3;
print(1); print(2); print(3); a:=[]; b:=[1]; c:=[2];
for n from 1 to q do for k from 0 to 9 do x[k]:=0; od;
a:=b; b:=c; c:=convert(y[3],base,10);
for k from 1 to nops(a) do x[a[k]]:=x[a[k]]+1; od;
for k from 1 to nops(b) do x[b[k]]:=x[b[k]]+1; od;
for k from 1 to nops(c) do x[c[k]]:=x[c[k]]+1; od;
y[1]:=y[2]; y[2]:=y[3]; y[3]:=0;
for k from 0 to 9 do if x[k]>0 then if k=0 then d:=10*x[k];
else d:=10*x[k]+k; fi; y[3]:=y[3]*10^(ilog10(d)+1)+d; fi; od;
print(y[3]); od; end: P(10^2);

Nest[Append[#, FromDigits[Join @@ Map[Flatten@ Reverse@ # &, IntegerDigits@ Sort@ Tally[Join @@ IntegerDigits[#[[-3 ;; -1]]]]]]] &, {1, 2, 3}, 15] (* Michael De Vlieger, Mar 01 2018 *)

cp's OEIS Frontend

A001155 Describe the previous term! (method A - initial term is 0).

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A036059 The summarize Fibonacci sequence: summarize the previous two terms!.

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A244112 Reverse digit count of n in decimal representation.

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A007890 Summarize the previous term! (in decreasing order).

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A036212 When the 'Summarize preceding term' sequence gets into a cycle.

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A036225 Length of cycle the 'summarize the previous term' sequence gets into.

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A300191 Look and Say the digits of the last three terms, by increasing digit value. Start with a(1)=1, a(2)=2 and a(3)=3.

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Mathematica