A005151 -id:A005151 - cp's OEIS frontend

A118628 "Say what you see".

3, 13, 1113, 3113, 2123, 112213, 312213, 212223, 114213, 31121314, 41122314, 31221324, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314, 21322314

Offset: 1

Parthasarathy Nambi, May 09 2006

nonn

			3 = "one three" --> 13
13 = "one one, one three" --> 1113
1113 = "three ones, one three" --> 3113
3113 = "two ones, two threes" --> 2123

Madras College, Descriptive Numbers

Cf. A005151.

Cf. A047842, A023989.

import Data.List (group, sort, transpose)
a118628 n = a118628_list !! (n-1)
a118628_list = 3 : f [3] :: [Integer] where
   f xs = (read $ concatMap show ys) : f (ys) where
          ys = concat $ transpose [map length zss, map head zss]
          zss = group $ sort xs
-- Reinhard Zumkeller, Jan 26 2014

a(n) = 21322314 for n > 12; a(n) = A005151(n) for n > 6. - Reinhard Zumkeller, Jan 26 2014

a(n) = A047842(a(n-1)). - Pontus von Brömssen, Jun 04 2023

A037220 Summarize the previous term!.

Original entry on oeis.org

0, 10, 1011, 1031, 102113, 10311213, 10411223, 1031221314, 1041222314, 1031321324, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314, 1031223314

Offset: 0

N. J. A. Sloane

			a(0) is given as 0;
a(1) is one zero -> 10;
a(2) is one zero and one one -> 1011;
a(3) is one zero and three ones -> 1031;
a(7) and onward is 1031223314.

Cf. A005151.

NestList[ FromDigits@ Flatten@ Map[IntegerDigits@ Reverse@# &, Sort@ Tally@ Flatten@ IntegerDigits@#] &, 0, 20] (* Robert G. Wilson v, Nov 28 2019, adapted from code by Michael De Vlieger in A177359 *)

A098153 Summarize the previous term in binary (in increasing order).

Original entry on oeis.org

1, 11, 101, 10101, 100111, 1001001, 1000111, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001, 1101001

Offset: 1

Rick L. Shepherd, Aug 29 2004

Similar to A005151 but uses base 2: Let a(1)=1. Describing a(1) as "one 1" again gives a(2)=11 (same digit string as A005151 and similar sequences), but describing a(2) as "two 1's" gives a(3)=101 when the frequency of digit occurrence is written in binary and followed by the digit counted.

			Summarizing a(8) = 1101001 in increasing digit order, there are "three 0's, four 1's", so concatenating 11 0 100 1 gives a(9) = 1101001 (=a(10)=a(11)=...).

Cf. A098154 (ternary), A098155 (base 4), A005151 (decimal and digit strings for all other bases b >= 5).

a(n) = 1101001 for all n >= 8 (see example).

A098154 Summarize the previous term in ternary (in increasing order).

Original entry on oeis.org

1, 11, 21, 1112, 10112, 1010112, 2011112, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122, 1011122

Offset: 1

Rick L. Shepherd, Aug 29 2004

Similar to A005151 but uses base 3: Let a(1)=1. Describing a(1) as "one 1" again gives a(2)=11 (same digit string as A005151 and similar sequences). Likewise, a(3) and a(4) have same digit strings as all but the binary sequence, but describing a(4) as "three 1's, one 2" gives a(5)=10112 when the frequency of digit occurrence is written in ternary and followed by the digit counted.

			Summarizing a(8) = 1011122 in increasing digit order, there are "one 0, four 1's, two 2s", so concatenating 1 0 11 1 2 2 gives a(9) = 1011122 (=a(10)=a(11)=...).

Cf. A098153 (binary), A098155 (base 4), A005151 (decimal and digit strings for all other bases b >= 5).

a(n) = 1011122 for all n >= 8 (see example).

A098155 Summarize the previous term in base 4 (in increasing order).

Original entry on oeis.org

1, 11, 21, 1112, 3112, 211213, 312213, 212223, 1110213, 101011213, 201111213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213, 101112213

Offset: 1

Rick L. Shepherd, Aug 29 2004

Let a(1)=1. Describing a(1) as "one 1" again gives a(2)=11 (same digit string as A005151 and similar sequences). Likewise, a(3) through a(8) have the same digit strings as the corresponding terms of A005151, but describing a(8) as "one 1, four 2s, one 3" gives a(9)=1110213 when the frequency of digit occurrence is written in base 4 and followed by the digit counted.

			Summarizing a(12) = 101112213 in increasing digit order, there are "one 0, five 1's, two 2s, one 3", so concatenating 1 0 11 1 2 2 1 3 gives a(13) = 101112213 (=a(14)=a(15)=...).

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. A098153 (binary), A098154 (ternary), A005151 (decimal and digit strings for all other bases b >= 5).

Nest[Append[#, FromDigits[Flatten@ Map[IntegerDigits[#, 4] & /@ Reverse@ # &, Tally@ Sort@ IntegerDigits@ #[[-1]] ] ]] &, {1}, 24] (* Michael De Vlieger, Jul 15 2020 *)

a(n) = 101112213 for all n >= 12 (see example).

A138485 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 1.

Original entry on oeis.org

1, 11, 21, 1112, 1231, 211312, 223113, 232122, 421113, 13311214, 14411223, 13223124, 14322123, 23322114, 14213223, 23322114, 14213223, 23322114, 14213223, 23322114, 14213223, 23322114, 14213223, 23322114, 14213223

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {23322114,14213223}

			To get the term after 211312, we say: two 2's, three 1's, one 3's, so 223113.

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484, A138486-A138493.

A138486 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 2.

Original entry on oeis.org

2, 12, 1211, 3112, 122113, 133122, 222123, 134211, 31121413, 23411412, 22312413, 23211432, 32231421, 21321423, 23321421, 21321423, 23321421, 21321423, 23321421, 21321423, 23321421, 21321423, 23321421, 21321423

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {21321423,23321421}

			To get the term after 122113, we say: one 3's, three 1's, two 2's, so 133122

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484, A138485, A138487-A138493.

Duplicate term 21321423 removed by Georg Fischer, Sep 18 2023

A138487 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 3.

Original entry on oeis.org

3, 13, 1311, 3113, 2321, 112213, 133122, 222123, 134211, 31121413, 23411412, 22312413, 23211432, 32231421, 21321423, 23321421, 21321423, 23321421, 21321423, 23321421, 21321423, 23321421, 21321423, 23321421, 21321423

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {21321423,23321421}

			To get the term after 112213, we say: one 3's, three 1's, two 2's, so 133122

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138486, A138488-A138493.

A138488 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 4.

Original entry on oeis.org

4, 14, 1411, 3114, 142113, 13311214, 14411223, 13223124, 14322123, 23322114, 14213223, 23322114

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {23322114,14213223}

			To get the term after 142113, we say: one 3's, three 1's, one 2's, one 4's, so 13311214

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138487, A138489-A138493.

A138489 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 5.

Original entry on oeis.org

5, 15, 1511, 3115, 152113, 13311215, 15411223, 1322311415, 1541142322, 3213243115, 1531331422, 2214313315, 1531331422, 2214313315, 1531331422, 2214313315, 1531331422, 2214313315, 1531331422, 2214313315, 1531331422

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {1531331422,2214313315}

			To get the term after 152113, we say: one 3's, three 1's, one 2's, one 5's, so 13311215

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138488, A138490-A138493.

cp's OEIS Frontend

A118628 "Say what you see".

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A037220 Summarize the previous term!.

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A098153 Summarize the previous term in binary (in increasing order).

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A098154 Summarize the previous term in ternary (in increasing order).

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A098155 Summarize the previous term in base 4 (in increasing order).

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A138485 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 1.

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A138486 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 2.

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A138487 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 3.

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A138488 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 4.

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A138489 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 5.

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