A265316 First row of A262057.

0, 2, 7, 21, 23, 64, 69, 71, 193, 207, 209, 214, 579, 581, 622, 627, 629, 643, 1737, 1739, 1744, 1866, 1868, 1882, 1887, 1889, 1930, 5211, 5213, 5218, 5232, 5234, 5599, 5604, 5606, 5647, 5661, 5663, 5668, 5790, 5792, 15634, 15639, 15641, 15655, 15696, 15698

Offset: 1

Author

Max Barrentine, Dec 06 2015

Keywords

• nonn

Comments

From Robert Israel, Feb 03 2016: (Start)

a(n) is the first member of the n-th sequence in the greedy partition of the nonnegative integers into sequences that contain no 3-term arithmetic progression.

As a special case (proved by Roth in 1953) of Szemerédi's theorem, sequences with no 3-term arithmetic progressions must have density 0. In particular, the nonnegative integers can't be partitioned into finitely many such sequences. Therefore this sequence is infinite.

a(n+1) >= a(n) + 2. There seem to be many cases where this is an equality. (End)

It can be deduced from the main result of Gerver, Propp, Simpson (below) that a(3n+1) = 3a(2n+1), a(3n+2) = 2 + 3a(2n+1), and a(3n) = 1 + 3a(2n). This implies infinitely many cases where a(n+1) = a(n) + 2. - C. Kenneth Fan, Dec 09 2018

Indices of records in A006997. - Rémy Sigrist, Jan 06 2024

Links

• Robert Israel, Table of n, a(n) for n = 1..140
• Matvey Borodin, Hannah Han, Kaylee Ji, Tanya Khovanova, Alexander Peng, David Sun, Isabel Tu, Jason Yang, William Yang, Kevin Zhang, and Kevin Zhao, Variants of Base 3 over 2, arXiv:1901.09818 [math.NT], 2019.
• J. Gerver, J. Propp, and J. Simpson, Greedily partitioning the natural numbers into sets free of arithmetic progressions, Proc. of the Amer. Math. Soc. 102 (1988), no. 3, pp. 765-772.
• Tanya Khovanova and Kevin Wu, Base 3/2 and Greedily Partitioned Sequences, arXiv:2007.09705 [math.NT], 2020.
• K. F. Roth, On certain sets of integers, Journal of the London Mathematical Society s1-28 (1953), 104-109.
• Wikipedia, Szemerédi's theorem.

Crossrefs

Cf. A006997, A262057.

Programs

Maple

M:= 100: # to get a(1) to a(M)
for i from 1 to M do B[i]:= {}: F[i]:= {}: od:
for x from 0 do
  for i from 1 to M do
     if not member(x,F[i]) then
       F[i]:= F[i] union map(y -> 2*x-y, B[i]);
     B[i]:= B[i] union {x};
     if not assigned(A[i]) then A[i]:= x fi;
     break
    fi
  od;
  if i = M+1 then break fi;
od:
seq(A[i],i=1..M); # Robert Israel, Feb 03 2016

Formula

A006997(a(n)) = n - 1. - Rémy Sigrist, Jan 06 2024