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

Author

James Propp

Keywords

• nonn
• easy
• nice

Comments

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

Examples

			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).
		

References

• 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).

Links

• 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.

Crossrefs

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).

Programs

Maple

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

Mathematica

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 *)

PARI

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

PARI

{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

Python

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

Formula

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