A006999 Partitioning integers to avoid arithmetic progressions of length 3.

0, 1, 2, 4, 7, 11, 17, 26, 40, 61, 92, 139, 209, 314, 472, 709, 1064, 1597, 2396, 3595, 5393, 8090, 12136, 18205, 27308, 40963, 61445, 92168, 138253, 207380, 311071, 466607, 699911, 1049867, 1574801, 2362202, 3543304, 5314957, 7972436

Offset: 0

Author

N. J. A. Sloane, D. R. Hofstadter, and James Propp, Jul 15 1977

Keywords

• nonn
• easy

Comments

a(n) = A006997(3^n-1).

It appears that, aside from the first term, this is the (L)-sieve transform of A016789 ={2,5,8,11,...,3n+2....}. This has been verified up to a(30)=311071. See A152009 for the definition of the (L)-sieve transform. - John W. Layman, Nov 20 2008

a(n) is also the largest-index square reachable in n jumps if we start at square 0 of the Infinite Sidewalk. - Jose Villegas, Mar 27 2023

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• B. Chen, R. Chen, J. Guo, S. Lee et al, On Base 3/2 and its sequences, arXiv:1808.04304 [math.NT], 2018.
• Joseph Gerver, James Propp and Jamie Simpson, Greedily partitioning the natural numbers into sets free of arithmetic progressions Proc. Amer. Math. Soc. 102 (1988), no. 3, 765-772.
• D. R. Hofstadter, Eta-Lore [Cached copy, with permission]
• D. R. Hofstadter, Pi-Mu Sequences [Cached copy, with permission]
• D. R. Hofstadter and N. J. A. Sloane, Correspondence, 1977 and 1991
• James Propp and N. J. A. Sloane, Email, March 1994
• Jane Street, Traversing the Infinite Sidewalk (2023).

Crossrefs

Cf. A061419, A061418, A005428 (first differences), A083286.

Cf. A006997, A016789, A152009.

Cf. A003312.

Programs

Haskell

a006999 n = a006999_list !! n
a006999_list = 0 : map ((`div` 2) . (+ 2) . (* 3)) a006999_list
-- Reinhard Zumkeller, Oct 26 2011

Mathematica

a[0] = 0; a[n_] := a[n] = Floor[(3 a[n-1] + 2)/2];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Aug 01 2018 *)

PARI

a(n)=if(n<1,0,floor((3*a(n-1)+2)/2))

Formula

a(n) = A061419(n) - 1.

a(n) = A061418(n) - 2.

a(n) = floor((3a(n-1)+2)/2).

a(n) = -1 + floor(c*(3/2)^n) where c=1.0815136... - Benoit Cloitre, Jan 10 2002; this constant c is 2/3*K(3) (see A083286). - Ralf Stephan, May 29 2003

a(n+1) = (3*a(n))/2+1 if a(n) is even. a(n+1) = (3*a(n)+1)/2 if a(n) is odd. - Miquel Cerda, Jun 15 2019

Extensions

More terms from James Sellers, Feb 06 2000