cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A034003 A007890 expanded into single digits.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 3, 1, 2, 2, 1, 1, 3, 2, 2, 3, 1, 2, 3, 2, 2, 2, 1, 1, 3, 4, 2, 1, 1, 1, 4, 1, 3, 1, 2, 3, 1, 1, 4, 2, 3, 1, 2, 4, 1, 2, 4, 1, 3, 2, 2, 3, 1, 1, 4, 2, 3, 3, 2, 2, 1, 1, 4, 2, 3, 3, 2, 2, 1, 1, 4, 2, 3, 3, 2, 2, 1, 1, 4, 2, 3, 3, 2, 2, 1, 1, 4, 2, 3, 3, 2
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

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

Original entry on oeis.org

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

Views

Author

Floor van Lamoen

Keywords

Comments

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

Examples

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

Links

Crossrefs

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.

Programs

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

Formula

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

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

Views

Author

Floor van Lamoen

Keywords

Examples

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

Links

Crossrefs

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

Views

Author

Floor van Lamoen

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

Cf. A007890, A036058, A036212.

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

Views

Author

Abdul Gaffar Khan, Sep 28 2015

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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).)

Programs

  • Mathematica
    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}]
  • PARI
    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

Formula

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))).

A036110 A summarize Fibonacci sequence: summarize the previous two terms!.

Original entry on oeis.org

2, 2, 22, 32, 1332, 332211, 433231, 14533231, 1524632231, 162524534241, 263544336241, 363564435231, 463554733221, 17364544733221, 37263554634231, 37363554734231, 37364544933221, 1937263554933221, 3927263544835231, 391827264534836231, 293827363544836231
Offset: 0

Views

Author

Floor van Lamoen

Keywords

Comments

From the 26th term the sequence gets into a cycle of 46.

Crossrefs

Cf. A036059, A007890.

Programs

  • Python
    def aupton(nn):
  alst = [2, 2]
  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
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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