A353651 -id:A353651 - cp's OEIS frontend

A003605 Unique monotonic sequence of nonnegative integers satisfying a(a(n)) = 3n.

0, 2, 3, 6, 7, 8, 9, 12, 15, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120

Offset: 0

James Propp

Another definition: a(0) = 0, a(1) = 2; for n > 1, a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is a multiple of 3". - Benoit Cloitre, Feb 14 2003
Yet another definition: a(0) = 0, a(1)=2; for n > 1, a(n) is the smallest integer > a(n-1) satisfying "if n is in the sequence, a(n)==0 (mod 3)" ("only if" omitted).
This sequence is the case m = 2 of the following family: a(1, m) = m, a(n, m) is the smallest integer > a(n-1, m) satisfying "if n is in the sequence, a(n, m) == 0 (mod (2m-1))". The general formula is: for any k >= 0, for j = -m*(2m-1)^k, ..., -1, 0, 1, ..., m*(2m-1)^k, a((m-1)*(2*m-1)^k+j) = (2*m-1)^(k+1)+m*j+(m-1)*abs(j).
Numbers whose base-3 representation starts with 2 or ends with 0. - Franklin T. Adams-Watters, Jan 17 2006
This sequence was the subject of the 5th problem of the 27th British Mathematical Olympiad in 1992 (see link British Mathematical Olympiad, reference Gardiner's book and second example for the answer to the BMO question). - Bernard Schott, Dec 25 2020

			9 is in the sequence and the smallest multiple of 3 greater than a(9-1)=a(8)=15 is 18. Hence a(9)=18.
a(1992) = a(2*3^6+534) = 3^7+3*534 = 3789 (answer to B.M.O. problem).

A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, pages 5 and 113-114 (1992).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Yifan Xie, Table of n, a(n) for n = 0..10000 (first 1000 terms from Vincenzo Librandi)
J.-P. Allouche, N. Rampersad and J. Shallit, On integer sequences whose first iterates are linear, Aequationes Math. 69 (2005), 114-127.
J.-P. Allouche and J. Shallit, The Ring of k-regular Sequences, II
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
British Mathematical Olympiad 1992, Problem 5
Benoit Cloitre, N. J. A. Sloane and Matthew J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2. (math.NT/0305308)
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47.
J. Shallit, Number theory and formal languages, in D. A. Hejhal, J. Friedman, M. C. Gutzwiller and A. M. Odlyzko, eds., Emerging Applications of Number Theory, IMA Volumes in Mathematics and Its Applications, V. 109, Springer-Verlag, 1999, pp. 547-570.
Index entries for sequences of the a(a(n)) = 2n family
Index to sequences related to Olympiads.

Cf. A079000, A007378, A080588, A079351.

Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3: A080637 (k=2), this sequence (k=3), A353651 (k=4), A353652 (k=5), A353653 (k=6).

filter:= n ->  (n mod 3 = 0) or (n >= 2*3^floor(log[3](n))):
select(filter, [$0..1000]); # Robert Israel, Oct 15 2014

a[n_] := a[n] = Which[ Mod[n, 3] == 0, 3 a[n/3], Mod[n, 3] == 1, 2*a[(n-1)/3] + a[(n-1)/3 + 1], True, a[(n-2)/3] + 2*a[(n-2)/3 + 1]]; a[0]=0; a[1]=2; a[2]=3; Table[a[n], {n, 0, 67}] (* Jean-François Alcover, Jul 18 2012, after Michael Somos *)

a(n)=if(n<3,n+(n>0),(3-(n%3))*a(n\3)+(n%3)*a(n\3+1))

{A(n)=local(d,w,l3=log(3),l2=log(2),l3n);
           l3n = log(n)/l3;
           w   = floor(l3n);         \\ highest exponent w such that 3^w <= n
           d   = frac(l3n)*l3/l2+1;  \\ first digit in base-3 repr. of n
              if ( d<2 , d=1 , d=2 );\\   make d an integer either 1 or 2
           if(d==1, n = n + 3^w , n = (n - 3^w)*3);
           return(n);}
\\ Gottfried Helms, Jan 11 2012

from sympy import integer_log
def A003605(n): return max(n+(m:=3**integer_log(n,3)[0]),3*(n-m)) if n else 0  # Chai Wah Wu, Feb 03 2025

For any k>=0, a(3^k - j) = 2*3^k - 3j, 0 <= j <= 3^(k-1); a(3^k + j) = 2*3^k + j, 0 <= j <= 3^k.
From Michael Somos, May 03 2000: (Start)
a(3*n) = 3*a(n), a(3*n+1) = 2*a(n) + a(n+1), a(3*n+2) = a(n) + 2a(n+1), n > 0.
a(n+1) - 2*a(n) + a(n-1) = {2 if n=3^k, -2 if n=2*3^k, otherwise 0}, n > 1. (End)
a(n) = n + A006166(n). - Vladeta Jovovic, Mar 01 2003
a(n) = abs(2*3^floor(log_3(n)) - n) + 2n - 3^floor(log_3(n)) for n>=1. - Theodore Lamort de Gail, Sep 12 2017
For any k >= 0, a(2*3^k + j) = 3^(k+1) + 3*j, 0 <= j <= 3^k. - Bernard Schott, Dec 25 2020

A080637 a(n) is the smallest positive integer which is consistent with the sequence being monotonically increasing and satisfying a(1)=2, a(a(n)) = 2n+1 for n > 1.

Original entry on oeis.org

2, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 37, 39, 41, 43, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 96, 97, 98, 99, 100, 101, 102

Offset: 1

N. J. A. Sloane and Benoit Cloitre, Feb 28 2003

Sequence is the unique monotonic sequence satisfying a(a(n)) = 2n+1.

Except for the first term, numbers (greater than 2) whose binary representation starts with 11 or ends with 1. - Yifan Xie, May 26 2022

			From _Yifan Xie_, May 02 2022: (Start)
a(8) = 12 because 2*2^2 <= 8 < 3*2^2, hence a(8) = 8 + 2^2 = 12;
a(13) = 19 because 3*2^2 <= 13 < 4*2^2, hence a(13) = 2*(13 - 2^2) + 1 = 19. (End)

Yifan Xie, Table of n, a(n) for n = 1..10000
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Index entries for sequences of the a(a(n)) = 2n family

Except for first term, same as A079905. Cf. A079000.
A007378, A079905, A080637, A080653 are all essentially the same sequence.
Equals A007378(n+1)-1. First differences give A079882.
Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3: this sequence (k=2), A003605 (k=3), A353651 (k=4), A353652 (k=5), A353653 (k=6).

t := []; for k from 0 to 6 do for j from -2^k to 2^k-1 do t := [op(t), 4*2^k - 1 + 3*j/2 + abs(j)/2]; od: od: t;

b[n_] := b[n] = If[n<4, n+1, If[OddQ[n], b[(n-1)/2+1]+b[(n-1)/2], 2b[n/2]]];
a[n_] := b[n+1]-1;
a /@ Range[70] (* Jean-François Alcover, Oct 31 2019 *)

a(3*2^k - 1 + j) = 4*2^k - 1 + 3*j/2 + |j|/2 for k >= 0, -2^k <= j < 2^k.
a(2n+1) = 2*a(n) + 1, a(2n) = a(n) + a(n-1) + 1.
From Yifan Xie, May 02 2022: (Start)
For n in the range 2*2^i <= n < 3*2^i, for i >= 0:
  a(n) = n + 2^i.
  a(n) = 1 + a(n-1).
Otherwise, for n in the range 3*2^i <= n < 4*2^i, for i >= 0:
  a(n) = 2*(n - 2^i) + 1.
  a(n) = 2 + a(n-1). (End)

A353652 Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 5.

Original entry on oeis.org

2, 3, 8, 9, 10, 11, 12, 13, 18, 23, 28, 33, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 68, 73, 78, 83, 88, 93, 98, 103, 108, 113, 118, 123, 128, 133, 138, 143, 148, 153, 158, 163, 168, 173, 178, 183, 188

Offset: 1

Yifan Xie, Jul 15 2022

Numbers m such that the base-5 representation of (2*m-1) starts with 3 or 4 or ends with 0.

First differences give a run of 5^i 1's followed by a run of 5^i 5's, for i >= 0.

			a(7) = 12 because (5^1 + 1)/2 <= 7 < (3*5^1 + 1)/2, hence a(7) = 7 + 5^1 = 12;
a(11) = 28 because (3*5^1 + 1)/2 <= 11 < (5*5^1 + 1)/2, hence a(11) = 5*(11 - 5^1) - 2 = 28.

For other values of k: A080637 (k=2), A003605 (k=3), A353651 (k=4), this sequence (k=5), A353653 (k=6).

a[n_] := Module[{n2 = 2n, p}, p = 5^Floor@Log[5, n2]; If[n2 < 3p, n+p, 5(n-p)-2]];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Sep 22 2023, after Kevin Ryde *)

a(n) = my(n2=n<<1, p=5^logint(n2, 5)); if(n2 < 3*p, n+p, 5*(n-p)-2); \\ Kevin Ryde, Apr 18 2022
(C++)
/* program used to generate the b-file */
#include
using namespace std;
int main(){
    int cnt1=1, flag=0, cnt2=1, a=2;
    for(int n=1; n<=10000; n++) {
        cout<

For n in the range (5^i + 1)/2 <= n < (3*5^i + 1)/2, for i >= 0:
  a(n) = n + 5^i.
  a(n+1) = 1 + a(n).
Otherwise, for n in the range (3*5^i + 1)/2 < n <= (5*5^i + 1)/2, for i >= 0:
  a(n) = 5*(n - 5^i) - 2.
  a(n+1) = 5 + a(n).

A353653 Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 6.

Original entry on oeis.org

2, 3, 9, 10, 11, 12, 13, 14, 15, 21, 27, 33, 39, 45, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 93, 99, 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165

Offset: 1

Yifan Xie, Jul 15 2022

Numbers m such that the base-6 representation of (5*m-3) starts with 11 or 12 or 13 or 14 or 15 or ends with 0.

First differences give a run of 6^i 1's followed by a run of 6^i 6's, for i >= 0.

			a(5) = 11 because (2*6^1 + 3)/3 <= 5 < (7*6^1 + 3)/5, hence a(5) = 5 + 6^1 = 11;
a(10) = 21 because (7*6^1 + 3)/5 <= 10 < (12*6^1 + 3)/5, hence a(10) = 6*(10 - 6^1) - 3 = 21.

For other values of k: A080637 (k=2), A003605 (k=3), A353651 (k=4), A353652 (k=5), this sequence (k=6).

okQ[m_] := With[{id = IntegerDigits[5 m - 3, 6] }, MatchQ[id[[1 ;; 2]], {1, 1}|{1, 2}|{1, 3}|{1, 4}|{1, 5}] || id[[-1]] == 0];
Join[{2}, Select[Range[3, 1000], okQ]] (* Jean-François Alcover, Sep 22 2023 *)

For n in the range (2*6^i + 3)/5 <= n < (7*6^i + 3)/5, for i >= 0:
  a(n) = n + 6^i.
  a(n+1) = 1 + a(n).
Otherwise, for n in the range (7*6^i + 3)/5 <= n < (12*6^i + 3)/5, for i >= 0:
  a(n) = 6*(n - 6^i) - 3.
  a(n+1) = 6 + a(n).

cp's OEIS Frontend

A003605 Unique monotonic sequence of nonnegative integers satisfying a(a(n)) = 3n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Python

Formula

A080637 a(n) is the smallest positive integer which is consistent with the sequence being monotonically increasing and satisfying a(1)=2, a(a(n)) = 2n+1 for n > 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A353652 Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 5.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A353653 Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 6.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula