A396815 a(n) is the maximum size of a subset of the cyclic group Z/nZ that can be partitioned into two sum-free sets.

1, 2, 3, 4, 4, 4, 6, 6, 8, 8, 9, 8, 9, 12, 12, 12, 12, 12, 16, 14, 16, 16, 18, 20, 17, 18, 21, 20, 24, 20, 24, 24, 24, 28, 27, 24, 25, 26, 32

Offset

2,2

Comments

A subset S of an abelian group is sum-free if there are no x, y, z in S (not necessarily distinct) with x + y = z. Equivalently, a(n) is the largest m such that some m-element subset of Z/nZ can be 2-colored so that neither color class contains a solution of x + y = z (mod n).

a(n) is not monotonic: a(13) = 8 < a(12) = 9.

For odd n, a(n) = 2*A211316(n), twice the maximum size of a single sum-free subset (see Formula). Each color class is sum-free, so a(n) <= 2*A211316(n) for every n. Conversely, let S be a maximum sum-free subset of Z/nZ and 2S = {2x : x in S}. For odd n the doubling map x -> 2x is a bijection of Z/nZ, so |2S| = |S|; 2S is sum-free, since 2x + 2y = 2z implies x + y = z after multiplying by the inverse of 2; and S and 2S are disjoint, since z = 2x with x, z in S would be the forbidden solution x + x = z. Hence S U 2S attains the bound. For even n the identity always fails: 2*A211316(n) = n, but a(n) <= n - 1 because no sum-free set contains 0.

More generally, for a finite abelian group G let S(G,2) be the maximum size of a subset that can be partitioned into two sum-free sets. For 2 <= n <= 48, S(G,2) is the same for every abelian group of order n, so it depends only on the order |G| = n, not on the isomorphism type of G; for example, all three abelian groups of order 8 (Z/8Z, Z/4Z X Z/2Z, and (Z/2Z)^3) give a(8) = 6. Hence for n <= 48, a(n) = S(Z/nZ, 2) is this common value. The invariance first fails at order 49: S(Z/49Z, 2) = 32 but S(Z/7Z X Z/7Z, 2) = 28. Indeed the argument above gives S(G,2) = 2*S(G,1) (the maximum size of a single sum-free subset) for every abelian group G of odd order, and by the Green-Ruzsa formula S(Z/49Z, 1) = 16 while S(Z/7Z X Z/7Z, 1) = 14. For three sum-free sets the invariance fails already at order 9, where Z/9Z gives 8 but Z/3Z X Z/3Z gives 7 (cf. A396816).

Links

Table of n, a(n) for n=2..40.

Ben Green and Imre Z. Ruzsa, Sum-free sets in abelian groups, arXiv:math/0307142 [math.CO], 2004.

Formula

a(n) = 2*A211316(n) for odd n (observed by Andrei Zabolotskii; see Comments for a proof).

Crossrefs

Cf. A211316 (k=1 case: maximal sum-free subset of Z/nZ), A396816 (k=3 case).

Sequence in context: A099777 A221917 A131798 * A206925 A339363 A377910

Adjacent sequences: A396812 A396813 A396814 * A396816 A396817 A396818

Keyword

nonn,more,new

Author

Alex Towell, Jun 07 2026

Status

approved