A005151 -id:A005151 - cp's OEIS frontend

A138490 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 6.

6, 16, 1611, 3116, 162113, 13311216, 16411223, 1322311416, 1641142322, 3213243116, 1631331422, 2214313316, 1631331422, 2214313316, 1631331422, 2214313316, 1631331422, 2214313316, 1631331422, 2214313316, 1631331422

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {1631331422,2214313316}

			To get the term after 162113, we say: one 3's, three 1's, one 2's, one 6's, so 13311216

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138489, A138491-A138493.

A138491 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 7.

Original entry on oeis.org

7, 17, 1711, 3117, 172113, 13311217, 17411223, 1322311417, 1741142322, 3213243117, 1731331422, 2214313317, 1731331422, 2214313317, 1731331422, 2214313317, 1731331422, 2214313317, 1731331422, 2214313317, 1731331422

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {1731331422,2214313317}

			To get the term after 172113, we say: one 3's, three 1's, one 2's, one 7's, so 13311217

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138490, A138492, A138493.

A138492 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 8.

Original entry on oeis.org

8, 18, 1811, 3118, 182113, 13311218, 18411223, 1322311418, 1841142322, 3213243118, 1831331422, 2214313318, 1831331422, 2214313318, 1831331422, 2214313318, 1831331422, 2214313318, 1831331422, 2214313318, 1831331422

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 20 2008

After a while sequence has period 2 -> {1831331422,2214313318}

			To get the term after 182113, we say: one 3's, three 1's, one 2's, one 8's, so 13311218

Cf. A063850, A022482, A005150, A005151, A006751, A006715, A006711, A022506-A022513, A138484-A138491, A138493.

A320890 a(1) = 11. For all subsequent terms a(n), take a(n-1) and substitute for the k-th digit the binary number of times that digit has appeared in a(n-1), reading left to right from the 1st to k-th digit.

Original entry on oeis.org

11, 110, 1101, 110111, 110111100101, 11011110010111010111111001000, 1101111001011101011111100100010011010101101111011001101111011111000010001111100010010100110101011

Offset: 1

Thomas Anton, Oct 23 2018

Each term is an initial segment of all of its successors.
There are always more 1's than 0's in a term.
The proportion of 0's or 1's in the n-th term approaches 1/2 as n approaches infinity.
Starting with any binary integer apart from 0 or 1 and applying the same process to yield a sequence s(n), we have that, for a sufficiently large x, a(n) is always an initial segment of s(n+x). The constancy and uniqueness of the limiting behavior of initial segments in base 2 is unique among all bases, unless the tally system is considered as a degenerate case.

			a(1) = 11
The first 1 is replaced with 1, and the second 1 is replaced with 10 (two), so a(2) = 110 (1|10)
The first 1 is replaced with 1, the second 1 with 10, and the first 0 with 1, so a(3) = 1101 (1|10|1)
The first 1 is replaced with 1, the second 1 with 10, the first 0 with 1, and the third 1 with 11 (three), so a(4) = 110111 (1|10|1|11)
The first 1 is replaced with 1, the second 1 with 10, the first 0 with 1, the third 1 with 11, the fourth 1 with 100, and the fifth 1 with 101, so a(5) = 110111100101 (1|10|1|11|100|101)
The first 1 is replaced with 1, the second 1 with 10, the first 0 with 1, the third 1 with 11, the fourth 1 with 100, the fifth 1 with 101, the sixth 1 with 110, the second 0 with 10, the third 0 with 11, the seventh 1 with 111, the fourth 0 with 100, and the eighth 1 with 1000, so a(6) = 11011110010111010111111001000 (1|10|1|11|100|101|110|10|11|111|100|1000)

Chai Wah Wu, Table of n, a(n) for n = 1..8

Cf. A005150, A007651, A001387, A005151.

FromDigits /@ Nest[Append[#, Flatten[IntegerDigits[#, 2] & /@ Table[Count[#, Last@ #] &@ #[[1 ;; k]], {k, Length@ #}]] &[#[[-1]] ] ] &, {{1, 1}}, 6] (* Michael De Vlieger, Oct 23 2018 *)

eva(n) = subst(Pol(n), x, 10)
replace(v) = my(w=[], zeros=0, ones=0); for(k=1, #v, if(v[k]==0, zeros++; w=concat(w, binary(zeros))); if(v[k]==1, ones++; w=concat(w, binary(ones)))); w
terms(n) = my(v=[1, 1], i=0); while(i < n, print1(eva(v), ", "); i++; v=replace(v))
/* Print initial 7 terms as follows: */
terms(7) \\ Felix Fröhlich, Oct 23 2018

A320890_list = [11]
while len(A320890_list)<10:
    a0,a1,s = 0,0,''
    for d in str(A320890_list[-1]):
        if d == '0':
            a0 += 1
            s += bin(a0)[2:]
        else:
            a1 += 1
            s += bin(a1)[2:]
    A320890_list.append(int(s)) # Chai Wah Wu, Nov 30 2018

A103618 Number of steps required for n to reach a digit count invariant or cycle loop under the 'Look and Say' function A045918.

Original entry on oeis.org

10, 12, 11, 12, 8, 10, 10, 10, 10, 10, 9, 11, 10, 11, 7, 9, 9, 9, 9, 9, 6, 10, 0, 9, 6, 6, 6, 6, 6, 6, 8, 11, 9, 10, 6, 8, 8, 8, 8, 8, 9, 7, 6, 6, 7, 8, 9, 9, 9, 9, 10, 9, 6, 8, 8, 7, 9, 10, 10, 10, 12, 9, 6, 8, 9, 9, 7, 10, 12, 12, 13, 9, 6, 8, 9, 10, 10, 7, 12, 13, 14, 9, 6, 8, 9, 10, 12, 12, 7, 13

Offset: 1

Lekraj Beedassy, Mar 25 2005

			a(10)=9 because we have the 9-step chain 10 -> 1011 -> 1031 -> 102113 -> 103112113 -> 10411223 -> 1031221314 -> 1041222314 -> 1031321324 -> 1031223314, the latter being autobiographical, hence invariant. Also, a(40)=9 since we have the digit count iteration 40 -> 1014 -> 102114 -> 10311214 -> 1041121314 -> 1051121324 -> 104122131415 -> 105122132415 -> 104132131425 -> 104122232415 <-> 103142132415, ending in a 2-cycle loop.

Madras Math., Descriptive Numbers: Table of results

Cf. A005151, A047841, A104785, A104786, A104787, A308781 (duplicate?).

Corrected and extended by Sean A. Irvine, Feb 08 2010

A121808 Say what you see in previous term, reporting each digit encountered followed by frequency.

Original entry on oeis.org

1, 11, 12, 1121, 1321, 122131, 132231, 122232, 112431, 13213141, 14213241, 13223142, 12233241, 12233241, 12233241, 12233241, 12233241, 12233241, 12233241, 12233241, 12233241, 12233241

Offset: 1

Ross Presser (rpresser(AT)gmail.com), Sep 09 2006

			To get the term after 13213141, we say: four 1's, one 2, two 3's, one 4; therefore 14213241 (digit named before frequency).

Another variant of A005150, A005151, A063850, etc.

RunLengthEncode[ x_List ] := (Through[ {First, Length}[ #1 ] ] &) /@ Split@ Sort@x; LookAndSay[ n_, d_:1 ] := NestList[ Flatten[ Reverse /@ RunLengthEncode[ # ] ] &, {d}, n - 1 ]; F[ n_ ] := LookAndSay[ n, 1 ][ [ n ] ]; Table[ FromDigits[ F[ n ] ], {n, 1, 15} ] (* Robert G. Wilson v Sep 11 2006 *)

Constant at 12233241 after 13 terms.

A228530 Summarize the previous two terms!

Original entry on oeis.org

1, 11, 31, 3113, 3133, 3153, 215315, 31123335, 41225335, 3132631435, 313263243516, 413283242536, 31527334253618, 3152733435261728, 4152832445263728, 3172634445263738, 2162636435363738, 2142934425663728, 216273442566272819, 319233542556372829

Offset: 1

Edison Y. He, Sep 14 2013

a(k) is found by counting the frequency of the digits in terms a(k-1) and a(k-2). Digits with zero frequency are not counted.

At n=54 the sequence enters a cycle of 46 terms so that for n>=100 we have a(k) = a(k-46). - Lars Blomberg, Jan 04 2014

			For n=5, a(5) is found by counting the frequency of the digits in the last two terms; there are three 1s and three 3s, so you get "three one three three", or 3133.

Lars Blomberg, Table of n, a(n) for n = 1..145 containing the beginning and two full cycles.
Original reddit post where this sequence was found.

Like A005151, but uses the previous two terms instead of just the previous term.

Corrected a(8)-a(15), added a(16)-a(20) by Lars Blomberg, Jan 04 2014

A262721 Modified Look and Say sequence: compute sum of digits of previous term, square it, and apply the "Say What You See" process.

Original entry on oeis.org

1, 11, 14, 1215, 1811, 111211, 1419, 2215, 1120, 1116, 1811, 111211, 1419, 2215, 1120, 1116, 1811, 111211, 1419, 2215, 1120, 1116, 1811, 111211, 1419, 2215, 1120, 1116, 1811, 111211, 1419, 2215, 1120, 1116, 1811, 111211, 1419, 2215, 1120, 1116

Offset: 0

Abdul Gaffar Khan, Sep 28 2015

1. Generated with the help of sequence generated as follows:
c(0)=b, c(n)=k-th power of sum of digits of c(n-1).
Example: c(0)=1, c(n)=1 for all k hence convergent.
Example: c(0)=2, k=2, c(1)=4, c(2)=16, c(3)=49 as (1+6)^2=49.
Example: c(0)=3, k=2, c(n)=81 for all n, hence convergent.
In fact, for c(0)=3 and any k, a sequence generated by using this method converges.
(Methods G, V1 and V2 are explain in Link attached, namely "Generalization of A262721")
2. Every sequence generated by c(0)=b and any k, by using G, V1 or V2 has at least two convergent subsequences or in other words sequence generated by method G, V1 or V2 never converges for any b and k.
(2.1) For a(0)=1, k=2, and method G has 6 convergent subsequences with initial terms 1, 11, 14 and 1215 and converging to 1811, 111211, 1419, 2215, 1120, or 1116.
(2.2) For c(0)=1, k=2, and method V1 has 2 convergent subsequences with initial terms 1, 11, 14, 1215, 1118, 2112 and converging to 1316 and 2112.
(2.3) For c(0)=1, k=2, and method V2 has 2 convergent subsequences with initial terms 1, 11, 14, 15125, 1811, 1221 and converging to 1613 and 1221.
3. For any b and k, the least 'g-th' term of the sequence generated by methods G, V1 or V2 reaches a point at which one of the convergent subsequence of generated sequence,converges. That is, c(g) will be the converging point of one of the subsequence of generated sequence, but there is no subsequence which converges to the term c(m),m=0,...,g-1,with initial term read as c(0).
(3.1) For a(0)=1, k=2, method G we have g=4 with converging point 1811 by refering (2.1).
(3.2) For c(0)=1, k=2, method V1 we have g=5, with converging point 2112.
(3.3) For c(0)=1, k=2, method V2 we have g=5, with converging point 1221.
(3.4) For any b and k, the value of g for methods V1 and V2 is the same.
4. For method G, V1 or V2 with c(0)=b and k chosen randomly, the following holds:
(I) g<=k*23, for b=1 and for k<=100
(II) g<=k*23*b for k<=100
(III) g<=k*(23^(b+1)) for large values of k.
5. In the manner of A083671, sequence become periodic from 5th row with period of 6.

			a(0) = 1 has 1 digit, and the sum of digits is 1, and the square of the sum of digits is 1. So a(1) = 11, that is, one times 1.
a(1) = 11 has 2 digits, and the sum of digits is 1+1=2 and the square of the sum of digits is 4. So a(2) = 14, that is, one times 4.
Since a(2)=14, we compute 1+4=5, 5^2 = 25, where we see one 2 and one 5, so a(3)=1215.

Abdul Gaffar Khan, Generalization of A262721

Cf. A005150 (Look and Say).
Cf. A118881 (square of sum of digits of n).
Cf. A005151 (Summarize the previous term! (in increasing order)).
Cf. A007890 (Summarize the previous term! (in decreasing order)).
Cf. A045918 (Describe n. Also called the "Say What You See" or "Look and Say" sequence LS(n).)

A262721[0] := 1;
A262721[n_] :=
A262721[n] =
  FromDigits[
   Flatten[{Length[#], First[#]} & /@
     Split[IntegerDigits[
       Total[IntegerDigits[A262721[n - 1]]]^2]]]]; Table[
A262721[n], {n, 0, 100}]

say(n) = {d = digits(n); da = d[1]; na = 1; s = ""; for (k=2, #d, if (d[k] == da, na++, s = concat(s, Str(na, da)); na = 1; da = d[k]);); s = concat(s, Str(na, da)); eval(s);}
lista(nn) = {print1(a=1, ", "); for (k=2, nn, a = say(sumdigits(a)^2); print1(a, ", "););} \\ Michel Marcus, Sep 29 2015

1. a(0) = 1, a(n) = 'frequency' of digits in the square of the sum of digits of a(n-1) followed by 'digit'-indication.
2. a(0) = 1, a(n) = A005150(A118881(a(n-1))). Here first deal with the type of operations of A118881 on a(n-1)-th term and then deal with the operation of A005150 on obtained value from A118881(a(n-1)) in last step, instead of following a(n-1) term of A118881 and A118881(a(n-1)) as a member of sequences A118881 and A005150 respectively.
a(0) = 1, a(n) = A045918(A118881(a(n-1))).

A037192 Summarize the previous term!, starting with 8899.

Original entry on oeis.org

8899, 2829, 221819, 21221819, 31321819, 3112231819, 4122231819, 313213141819, 511233141819, 51122314151819, 61221314251819, 5132131415161819, 7112231425161819, 613213141516171819, 811223141526171819

Offset: 0

Dave Ellis (djellis(AT)hotmail.com)

Cf. A005151, A023989, A037220.

More terms from Erich Friedman.

A067818 Start with 1. To get a(n+1), describe a(n) in terms of the number of 0's, 1's, 2's,...,9's it has. Do not ignore leading 0's in the computation, but ignore them when listing the terms.

Original entry on oeis.org

1, 110203040506070809, 100211213141516171819, 201012213141516171819, 20913213141516171819, 10812223141516171829, 10714213141516172819, 10812213241516271819, 10714213141516172819, 10812213241516271819, 10714213141516172819, 10812213241516271819

Offset: 1

Joseph L. Pe, Feb 08 2002

If leading 0's are not included in the computation, the sequence becomes constant after the third term.

Including leading 0's, the sequence oscillates between 10714213141516172819 and 10812213241516271819 from a(7) onward. - Sean A. Irvine, Jan 08 2024

			1 has 0 0's, 1 1's, 0 2's, 0 3's, 0 4's, 0 5's, 0 6's, 0 7's, 0 8's, 0 9's, so the term following 1 is 00110203040506070809. Ignore the two leading zeros when listing this term, but include them in the computation of the third term. The second term has 10 0's, 2 1's, 1 2's, 1 3's, 1 4's, 1 5's, 1 6's, 1 7's, 1 8's, 1 9's, so the third term is 100211213141516171819.

Pickover, C. "Wonders of Numbers", Oxford Univ. Press, 2001.

C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review

Cf. A005150, A005151.

More terms from Sean A. Irvine, Jan 08 2024

cp's OEIS Frontend

A138490 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 6.

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A138491 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 7.

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A138492 Say what you see in previous term, from the right, reporting total number for each digit encountered. Initial term is 8.

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A320890 a(1) = 11. For all subsequent terms a(n), take a(n-1) and substitute for the k-th digit the binary number of times that digit has appeared in a(n-1), reading left to right from the 1st to k-th digit.

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A103618 Number of steps required for n to reach a digit count invariant or cycle loop under the 'Look and Say' function A045918.

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A121808 Say what you see in previous term, reporting each digit encountered followed by frequency.

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A228530 Summarize the previous two terms!

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A262721 Modified Look and Say sequence: compute sum of digits of previous term, square it, and apply the "Say What You See" process.

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A037192 Summarize the previous term!, starting with 8899.

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A067818 Start with 1. To get a(n+1), describe a(n) in terms of the number of 0's, 1's, 2's,...,9's it has. Do not ignore leading 0's in the computation, but ignore them when listing the terms.

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