CAMERON-ERDŐS MODULO A PRIME

VSEVOLOD F. LEV AND TOMASZ SCHOEN

Abstract. We prove that for p prime and sufficiently large, the number of ℤ_{p}-subsets free of solutions of the equation x + y = z (that is, free of Schur triples) satisfies

2^{⌊(p−2)/3⌋} (p − 1) (1 + O(2^{−ε_{1}p})) ≤ |SF[ℤ_{p}]| ≤ 2^{p/2−ε_{2}p},

where ε_{1} and ε_{2} are positive absolute constants.

1. Introduction

For a subset A ⊆ G of the set of elements of an Abelian group G we denote by SF[A] the family of all sum-free subsets A ⊆ A, that is subsets with the property that no two elements of A sum to another; in other words, a_{1} + a_{2} = a_{3} has no solutions in a_{i} ∈ A. An open conjecture proposed by Cameron and Erdős in [CE90] is that

|SF[n]| ≪ 2^{n/2},  (1)

where SF[n] abbreviates SF[[1, n] ∩ ℤ]. (If true, this is best possible since SF[n] obviously contains all subsets of

{1, 3, …, 2⌈n/2⌉ − 1}  (2)

as well as all subsets of [⌊n/2⌋+1, n].) Numerous partial and related results are known. We quote here one that will be used in the sequel; it was obtained independently by Alon and Calkin. Theorem A (Alon [A91, Proposition 4.1], Calkin [C90, Theorem 1]). We have:

|SF[n]| ≤ 2^{n/2+o(n)}.

We would like to mention also another result of Alon ([A91, Corollary 1.3]):

|SF[G]| ≤ 2^{|G|/2+o(|G|)}

for any G, and even a non-abelian one.

As a “modular” version of the question of Cameron and Erdős, one can ask about the number of sum-free subsets of ℤ_{n} = ℤ/nℤ. In view of the natural inclusion SF[ℤ_{n}] ↪ SF[n], an estimate of the sort

|SF[ℤ_{n}]| ≪ 2^{n/2}

can be viewed as a weak form of (1).

The first author is partially supported by the Edmund Landau Center for Research in Mathematical Analysis and Related Areas, sponsored by the Minerva Foundation (Germany).

For n even this would be best possible — consider the reduction modulo n of the set (2); some results for this case are established in [LS??]. In the present paper we investigate the case n = p, a prime number, and prove the following. Theorem 1. For p prime and sufficiently large we have

2^{⌊^{\frac{p−2}{3}}⌋} (p − 1) (1 + O(2^{−εp})) ≤ |SF[ℤ_{p}]| ≤ 2^{0.498p},

where ε > 0 is an absolute constant.

The expression at the right-hand side may be somewhat deceptive; it is worth pointing out that we know no simple way to prove that |SF[ℤ_{p}]| ≪ 2^{p/2}. We could have reduced the constant 0.498 somewhat, but we cannot get reasonably close to the lower bound which, we believe, gives the correct order of magnitude of |SF[ℤ_{p}]|. Notice that this lower bound is sharp: all multiplicative shifts of subsets of I_{0} := [⌊p/3⌋ + 1, ⌊2p/3⌋] ⊆ ℤ_{p} are sum-free, and we will show that the number of such shifts is given by the expression at the left-hand side of the inequality of Theorem 1. (By a multiplicative shift of a set A ⊆ ℤ_{p} by an element λ ∈ ℤ_{p}^{×} we mean the set λA := {λa: a ∈ A}; here and below ℤ_{p}^{×} denotes the set of non-zero elements of ℤ_{p}.) Indeed, the number of subsets A ⊆ I_{0} is

2^{⌊^{\frac{2p}{3}}⌋−⌊\frac{p}{3}⌋} = 2^{⌊^{\frac{p+1}{3}}⌋}

and the number of possible multipliers λ ∈ ℤ_{p}^{×} is p − 1. A naïve multiplication of the two quantities doesn’t give the total number of multiplicative shifts, as different pairs (λ, A) can yield the same shift λA; for instance, this always happens with the pairs (λ, A) and (−λ, −A). We show, however, that this “rarely” happens if the values of λ are restricted to [1, (p − 1)/2](mod p), and apply an inclusion-exclusion argument.

The proof of the lower-bound estimate of Theorem 1 is given in Section 3. In fact, we establish there a much more general result. Theorem 2. Let I be a block of consecutive residue classes modulo a prime p such that |I| < p/3 + 1. Then the number of sets A ⊆ ℤ_{p} which can be represented as a multiplicative shift of a subset of I is

2^{|I|} (1 − 2^{−|I\(−I)|−1}) (p − 1) + O (2^{5|I|/6}p^{2})

with an absolute constant in the O-sign.

Taking I = I_{0} gives the lower bound in Theorem 1.

The proof of the upper bound includes several steps. First, using exponential sums we show that for any A ∈ SF[ℤ_{p}] there exists an arithmetic progression P_{A} ⊆ ℤ_{p} of the cardinality |P_{A}| = (p + 1)/2 on which A is unproportionally dense. Replacing A with a suitable multiplicative shift, we can assume that the difference of P_{A} is one; that is, P_{A} is an interval modulo p. (This adds a factor of at most p − 1 to our estimate, which is negligible compared to the exponential reduction in 2^{−0.002p}.) Next, we consider the interval I of length δp centered the same as P_{A} + P_{A}, where δ > 0 is a suitably chosen constant. If A is sufficiently dense on this interval, then by Szemerédi’s theorem the intersection A ∩ I contains a “long” arithmetic progression, which substantially reduces the number of ways to choose A ∩ P_{A}. Similarly, if A is dense on the appropriately centered interval of length δp, then the number of ways to choose (ℤ_{p} \ P_{A}) ∩ A is small. Finally, we use Theorem A to show that there is only a small number of A ∈ SF[ℤ_{p}] which are not dense on some interval of length δp. Details are provided in the next section.

2. The upper-bound estimate

Lemma 1. For any A ∈ SF[ℤ_{p}] there is an arithmetic progression P_{A} ⊆ ℤ_{p} of cardinality |P_{A}| = (p+1)/2 such that |A\P_{A}| < γp, where γ = (3−2\sqrt{2})/2 ≈ 0.0857.

Proof. We write

S_{A}(z) = ∑_{a∈A} e^{2πi \frac{az}{p}} (z ∈ ℤ_{p})

so that S_{A}(0) = |A| and

∑_{z=0}^{p−1} |S_{A}(z)|^{2} = |A|p.

Since A is sum-free we have

∑_{z=0}^{p−1} S_{A}^{2}(z)S_{A}(−z) = 0,

whence

|A|^{3} = − ∑_{z=1}^{p−1} S_{A}^{2}(z)S_{A}(−z)

≤ max_{z∈ℤ_{p}^{×}}|S_{A}(z)| ∑_{z=1}^{p−1} |S_{A}(z)|^{2}

= max_{z∈ℤ_{p}^{×}}|S_{A}(z)||A|(p − |A|)

and

max_{z∈ℤ_{p}^{×}}|S_{A}(z)| ≥ \frac{|A|^{2}}{p − |A|} = \frac{α}{1 − α} |A|; α = |A|/p.

By a well-known lemma of Freiman ([F66, Lemma 2.2]), there is an arithmetic progression P_{A} ⊆ ℤ_{p} of |P_{A}| = (p + 1)/2 terms such that

|A ∩ P_{A}| > \frac{1 + α/(1 − α)}{2} |A|,

and then

|A \ P_{A}| < \frac{1 − α/(1 − α)}{2} |A| = \frac{α − 2α^{2}}{2 − 2α} p.

It remains to observe that the right-hand side, considered as a function of α ∈ [0, 1), attains its maximum value at α_{0} = (2 − \sqrt{2})/2 and this value is γ = (3 − 2\sqrt{2})/2. □

The next lemma we need is proved in [S??] and implements a version of the argument, first used by Calkin in [C90]. Lemma 2. Let p, u, d, k, v, w, and t be integer numbers such that d, t > 0, 1 ≤ k < p/2, (d, p) = 1, 0 ≤ w − v < p, and v + w ≡ u(mod p). Define

Q := {u − kd, u − (k − 1)d, ..., u + kd} (mod p).

Then the number of sets A ⊆ [v, w] mod p such that

(A + A) ∩ Q = ∅ and |A| = t

is at most 2^{kd+l/k}\binom{ℓ}{t}, where ℓ = ⌈(w − v)/2⌉. Corollary 1. For any ε > 0 there exists μ_{0} = μ_{0}(ε) with the following property. Let p be a prime number and let u, μ, v, w, and t > 0 be integer numbers such that 0 < w − v < p, v + w ≡ u(mod p), and μ_{0} < μ < p/2. Suppose that B ⊆ [u − μ, u + μ] (mod p) is a subset of ℤ_{p} of cardinality |B| ≥ εμ. Then the number of sets A ⊆ [v, w] (mod p) such that

(A + A) ∩ B = ∅ and |A| ≤ t

is at most 2^{1+μ+εℓ} ∑_{τ=0}^{t} \binom{ℓ}{τ}, where ℓ = ⌈(w − v)/2⌉.

Proof. Fix an integer k > ε^{−1}. By Szemerédi’s density theorem, if μ_{0} is large enough then the canonical pre-image of B in [u − μ, u + μ] contains an arithmetic progression

{u_{0} − kd, u_{0} − (k − 1)d, …, u_{0} + kd},  (3)

where d ≥ 1 and u_{0} are integers. Plainly, kd ≤ μ and |u_{0} − u| ≤ μ − kd. We define Q to be the image of progression (3) in ℤ_{p}, and put

v_{0} := v − ⌊\frac{u − u_{0}}{2} ⌋ , w_{0} := w − ⌈\frac{u − u_{0}}{2} ⌉ .

Evidently, we have then 0 ≤ w_{0} − v_{0} ≤ w − v and

v_{0} + w_{0} = v + w − (u − u_{0}) ≡ u_{0} (mod p).

Given a set A satisfying the conditions of the corollary we write

A_{0} := A ∩ [v_{0}, w_{0}] (mod p), A_{1} := A \ A_{0}.

As |A_{1}| ≤ |⌈(u − u_{0})/2⌉| ≤ (1 + μ − kd)/2, there are at most 2^{(1+μ−kd)/2} possible sets A_{1}. On the other hand, by Lemma 2 there are at most

∑_{τ=0}^{t} 2^{kd+εℓ_{0}} \binom{ℓ_{0}}{τ}

possible sets A_{0}, where ℓ_{0} = ⌈(w_{0} − v_{0})/2⌉ ≤ ℓ. Therefore, the number of A possible is at most

∑_{τ=0}^{t} 2^{(1+μ+kd)/2+εℓ}\binom{ℓ}{τ} ≤ 2^{1+μ+εℓ} ∑_{τ=0}^{t} \binom{ℓ}{τ}.

□

Now we are in a position to prove the upper bound of Theorem 1. Set δ = 0.008 + \sqrt{ε}, where ε > 0 is small enough. (It will be seen in the course of the proof what is “enough” for our purposes.) It suffices to count A ∈ SF[ℤ_{p}] such that the corresponding progression P_{A} (recall Lemma 1) has difference one: all other A can be obtained as multiplicative shifts. We first consider only sets A which are evenly distributed in ℤ_{p} in the sense that |A ∩ I| ≥ ε|I| for any interval I ⊆ ℤ_{p} of cardinality |I| > δp.

We write P_{A} = [v, w](mod p) with an integer v and w = v + (p − 1)/2, and we set μ = ⌈δp/2⌉ and B = A ∩ [u − μ, u + μ](mod p), where u = v + w. There are p possible values of u(mod p) and for each value there are at most O(2^{δp}) possible sets B. Moreover, if p is sufficiently large then by Corollary 1 for each B the number of possible sets A ∩ P_{A} is

≪ 2^{\frac{δp}{2} + \frac{εp}{4} + \frac{p}{4}} .

Furthermore, we have A \ P_{A} ⊆ [v′, w′](mod p), where v′ = v + (p + 1)/2, w′ = w + (p − 1)/2, and we notice that v′ + w′ ≡ v + w (mod p). By Corollary 1 to each B there correspond at most

≪ 2^{\frac{δp}{2} + \frac{εp}{4}} p \binom{(p + 3)/4}{γp}

possible sets A \ P_{A}. Putting everything together, we see that the number of A ∈ SF[ℤ_{p}] such that all multiplicative shifts of A are evenly distributed is

≪ 2^{2δp+ \frac{εp}{2} +p/4+0.23195p}p^{3} < 2^{0.48196p+2δp} < 2^{0.498p}.

(We omit here a boring exercise in Stirling formula which shows that \binom{(p+3)/4}{γp} ≪ 2^{0.23195p}. The crucial point, of course, is that γ < 1/8.)

To complete the proof we have to estimate the number of A ∈ SF[ℤ_{p}] with the property that there exists an interval I ⊆ ℤ_{p} such that |I| > δp, while |λA ∩ I| < ε|I| for some λ ∈ ℤ_{p}^{×}.

As above, we can concentrate on the case λ = 1 and we let I = [m, M](mod p). Suppose first that 0 ∉ I. Then m and M can be chosen so that 1 ≤ m < M ≤ p − 1, and we write

A_{1} = A ∩ [1, m − 1] (mod p), A_{3} = A ∪ [M + 1, p − 1] (mod p),
A_{2} = A ∩ [m, M] (mod p).

As A_{1} induces a sum-free subset of [1, m−1], by Theorem A the number of possible sets A_{1} is O(2^{m/2+εm}). As −A_{3} ⊆ [1, p−M −1](mod p) is sum-free in ℤ_{p}, it induces a sumfree subset of [1, p−M −1], hence the number of possible sets A_{3} is O(2^{(p−M)/2+ε(p−M)}). As |A_{2}| ≤ ε|I|, the total number of possible sets A_{2} is at most

∑_{0≤j≤ε|I|} \binom{|I|}{j} < p\binom{|I|}{ε|I|} < 2^{3|I|ε ln ε^{−1}}.

Therefore, the total number of A corresponding to a given interval I is

≪ 2^{(1/2+ε)(p−M+m)+3|I|ε ln ε^{−1}} ≪ 2^{p/2−(1/2−4ε ln ε^{−1})δp} ≪ 2^{0.496p}.

Similarly, if 0 ∈ I then the number of A corresponding to I is

≪ 2^{(1/2+ε)(p−|I|/2)+3|I|ε ln ε^{−1}} ≪ 2^{p/2−(1/4−4ε ln ε^{−1})δp} ≪ 2^{0.498p−\sqrt{ε}p/8}.

Finally, we notice that the number of intervals I is O(p^{2}).

3. Proof of Theorem 2

The number of subsets S ⊆ I ∩ (−I) is 2^{|I∩(−I)|} and the number of multiplicative shifts of these S is at most

\frac{1}{2} (p − 1) 2^{|I∩(−I)|},

as the multipliers λ and −λ yield the same system of shifts. Furthermore, the number of S ⊆ I which are not subsets of −I is 2^{|I|}−2^{|I∩(−I)|} and the number of multiplicative shifts of these S is at most

(p − 1) (2^{|I|} − 2^{|I∩(−I)|}) .

Therefore, the total number of shifts of all subsets S ⊆ I is at most

(2^{|I|} − 2^{|I∩(−I)|−1}) (p − 1) = 2^{|I|} (1 − 2^{|I∩(−I)|−|I|−1}) (p − 1)
= 2^{|I|} (1 − 2^{−|I\(−I)|−1}) (p − 1).

We proceed to prove the lower bound. Let I = [m, M](mod p), so that |I| = M − m + 1. For A ⊆ ℤ_{p}, define k(A) to be the number of representations of A as a multiplicative shift of a subset S ⊆ I with the agreement that the two representations A = λS and A = −λ(−S) are considered the same. That is, k(A) is the number of pairs (λ, S) such that A = λS, λ ∈ ℤ_{p}^{×}, S ⊆ I, and moreover, if S ⊆ −I, then λ ∈ [1, (p − 1)/2](mod p). We seek to estimate N := #{A ⊆ ℤ_{p}: k(A) > 0}. Clearly,

N ≥ ∑_{A⊆ℤ_{p}} (k(A) − 2\binom{k(A)}{2}) = N_{1} − N_{2},  (4)

where

N_{1} = ∑_{A⊆ℤ_{p}} k(A) and N_{2} = 2 ∑_{A⊆ℤ_{p}} \binom{k(A)}{2};

indeed, as k − 2\binom{k}{2} ≤ 1 for k ≥ 1 and k − 2\binom{k}{2} = 0 for k = 0, each A with k(A) ≥ 1 is counted by the sum in (4) at most once, and no other A is counted. The quantity N_{1} was evaluated at the beginning of this section and gives the main term. We have, therefore, to show that N_{2} = O(p^{2} 2^{5|I|/6}). For this we observe that

N_{2} ≤ #{(λ_{1}, S_{1}, λ_{2}, S_{2}): λ_{i} ∈ ℤ_{p}^{×}, S_{i} ⊆ I, λ_{1}S_{1} = λ_{2}S_{2}, (λ_{1}, S_{1}) ≠ ±(λ_{2}, S_{2})}

= #{(λ_{1}, S_{1}, λ_{2}, S_{2}): λ_{i} ∈ ℤ_{p}^{×}, S_{i} ⊆ I, λ_{1}S_{1} = λ_{2}S_{2}, λ_{1} ≠ ±λ_{2}}

= #{(λ_{1}, S_{1}, λ_{2}, S_{2}): λ_{i} ∈ ℤ_{p}^{×}, S_{i} ⊆ I, S_{2} = (λ_{2}^{−1}λ_{1})S_{1}, λ_{2}^{−1}λ_{1} ≠ ±1}

= (p − 1)#{(λ, S_{1}, S_{2}): λ ∈ ℤ_{p}^{×} \ {±1}, S_{i} ⊆ I, S_{2} = λS_{1}}

= (p − 1) ∑_{λ∈ℤ_{p}\{0,±1}}#{(S_{1}, S_{2}): S_{i} ⊆ I, S_{2} = λS_{1}}

= (p − 1) ∑_{λ∈ℤ_{p}\{0,±1}}#{S ⊆ I: λS ⊆ I}

= (p − 1) ∑_{λ∈ℤ_{p}\{0,±1}}2^{N(λ)},

where N(λ) = #{s ∈ I: λs ∈ I}.

We now prove that N(λ) ≤ 5|I|/6 + O(1) for λ ≠ 0, ±1. It is convenient to treat λ as an integer number satisfying 2 ≤ |λ| ≤ (p − 1)/2. Plainly, we have then

N(λ) = #{(x, y, k) ∈ ℤ^{3} : m ≤ x, y ≤ M, y = λx + kp}.  (5)

Suppose that (x, y, k) is a triple of integers satisfying the conditions of (5). Then k = p^{−1}(y − λx), whence

p^{−1}(m − λM) ≤ k ≤ p^{−1}(M − λm),

if λ > 0, and

p^{−1}(m − λm) ≤ k ≤ p^{−1}(M − λM),

if λ < 0. In any case, the number of k that can appear in (5) is at most

p^{−1}(M − m)(|λ| + 1) + 1 = p^{−1}(|I| − 1)(|λ| + 1) + 1.  (6)

Furthermore, for any given value of k the number of y that can appear in (5) is at most

⌈\frac{|I|}{|λ|}⌉ < \frac{|I|}{|λ|} + 1,

and it follows that

N(λ) < (\frac{|I|}{|λ|} + 1) (p^{−1}(|I| − 1)(|λ| + 1) + 1) .

The right-hand side is a convex function of |λ| and denoting it by f we have

f(3) = (\frac{1}{3} |I| + 1) (\frac{4}{p} (|I| − 1) + 1)

= \frac{4}{3} |I|^{2} p^{−1} + \frac{1}{3} |I| + O(1)

≤ \frac{7}{9} |I| + O(1),

f(p/2) = (\frac{2}{p} |I| + 1) ((\frac{1}{2} + \frac{1}{p}) (|I| − 1) + 1)

= \frac{1}{2} |I| (\frac{2}{p} |I| + 1) + O(1)

≤ \frac{5}{6} |I| + O(1).

Therefore, N(λ) ≤ 5|I|/6 + O(1) for |λ| ∈ [3, (p − 1)/2], and it remains to estimate N(±2). Here we use (6) in a subtler way: in view of |I| < p/3 + 1, it implies that for λ = ±2 the number of k that can appear in (5) is less than p^{−1}(p/3)(|±2|+1)+1 = 2; thus, there is at most one value of k possible! As above, the number of y possible is then less than |I|/|λ| + 1 = |I|/2 + 1, which means that N(±2) < |I|/2 + 1. This completes the proof.

Appendix: Arithmetic progressions in ℤ_{p}

Theorem 2 suggests the question about the number of multiplicative shifts of the whole interval I, and since all these shifts are arithmetic progressions of cardinality |I|, the next natural question is about the total number of arithmetic progressions in ℤ_{p} of given cardinality. The answer to these questions is in sharp contrast with Theorem 2: it turns out that any progression determines its first term and difference essentially uniquely. More precisely, there is at most one representation of a given set in the form {a, a + d, …, a + (n − 1)d} with a, d ∈ ℤ_{p} and d ∈ [1, (p − 1)/2](mod p). In particular, λ_{1}I_{1} ≠ λ_{2}I_{2}, unless (λ_{1}, I_{1}) = ±(λ_{2}, I_{2}).

Theorem 3. Let p > 3 be a prime number and suppose that A ⊆ ℤ_{p} is a set of n = |A| residues, where 2 ≤ n ≤ p − 2. Then there is at most one pair (a, d) ∈ ℤ_{p}^{2} such that d ∈ [1, (p − 1)/2](mod p) and A = {a, a + d, …, a + (n − 1)d}.

Proof. Let a_{1}, …, a_{n} be the elements of A and put

σ_{1} := a_{1} + ⋯ + a_{n},
σ_{2} := ∑_{i,j=1}^{n}(a_{i} − a_{j})^{2}.

Suppose that our set can be represented as A = {a, a + d, …, a + (n − 1)d}. Then

σ_{1} = na + \frac{1}{2} n(n − 1) d,  (7)
σ_{2} = d^{2} ∑_{i,j=0}^{n−1}(i − j)^{2}

= 2d^{2} (n ∑_{i=0}^{n−1} i^{2} − (∑_{i=0}^{n−1} i)^{2})

= 2d^{2} (\frac{n^{2}(n − 1)(2n − 1)}{6} − \frac{n^{2}(n − 1)^{2}}{4} )

= \frac{1}{6} d^{2}n^{2}(n − 1)(n + 1).

It follows that σ_{2} uniquely determines d, and then σ_{1} uniquely determines a. □

We conclude our paper with a description of a very effective algorithm, based on the approach we used in the proof of Theorem 3. Given a set A = {a_{1}, …, a_{n}} ⊆ ℤ_{p}, the algorithm determines in O(n+ln p) operations whether A is an arithmetic progression. We break the algorithm into 4 steps and accompany each with necessary explanations. To avoid trivialities, we assume p > 3 and n ∈ [2, p − 2].

i) Compute σ_{1} := a_{1}+⋯+a_{n} and σ_{2} := a_{1}^{2}+⋯+a_{n}^{2}. (Notice, that the definition of σ_{2} is somewhat different from one we used above.) If A = {a, a + d, …, a + (n − 1)d}, then σ_{1} satisfies (7) and

σ_{2} = na^{2} + n(n − 1)ad + \frac{1}{6} n(n − 1)(2n − 1)d^{2}

= na^{2} + 2(σ_{1} − na)a + \frac{1}{3} (σ_{1} − na)(2n − 1)d,

3n(n − 1)σ_{2} = 6n(n − 1)aσ_{1} − 3n^{2}(n − 1)a^{2} + 2(2n − 1)(σ_{1} − na)^{2}

= (n + 1)(an − σ_{1})^{2} + 3(n − 1)σ_{1}^{2},

whence

(n + 1)(an − σ_{1})^{2} = 3(n − 1)(σ_{2}n − σ_{1}^{2}).  (8)

ii) For each a_{i} ∈ A check whether a = a_{i} satisfies equality (8). If A is an

arithmetic progression, precisely two a_{i} will be found. iii) Let a_{i} be an element of A satisfying (8). For each a_{j} ∈ A (j ≠ i) check whether

d = a_{j} − a_{i} satisfies (7). If A is an arithmetic progression, such d exists. iv) It remains to check whether A is an arithmetic progression with the first

term a and difference d, where a and d are the values, found at steps ii) and iii). For this, compute d′, the inverse of d in ℤ_{p} (this requires O(ln p) operations) and loop over all a_{i} ∈ A, for each a_{i} computing the least nonnegative representative of the residue class (a_{i}−a)d′. The set A is a progression in ℤ_{p} if and only if all representatives belong to [0, n − 1].

References

[A91] N. Alon, Independent sets in regular graphs and sum-free subsets of finite groups, Israel J. Math. 73 (1991), 247–256.

[C90] N. J. Calkin, On the number of sum-free sets, Bull. London Math. Soc. 22 (1990), 141–144.

[CE90] P.J. Cameron and P. Erdős, On the number of sets of integers with various properties, in “Number Theory” (R.A. Mollin, Ed.), 61–79, de Gruyter, Berlin 1990.

[F66] G. Freiman, Foundations of a structural theory of set addition, Kazan 1966 [Russian]. English translation in: Translations of Math. Monographs 37 (1973), American Math. Soc., Providence.

[LS??] V. Lev and T. Schoen, On the number of sum-free subsets of ℤ_{n}, in preparation. [S??] T. Schoen, The number of (2, 3) sum-free sets, in preparation.

Institute of Mathematics, Hebrew University, Jerusalem 91904, Israel.

  E-mail address: seva@math.huji.ac.il

Department of Mathematics, Christian Albrecht University, Kiel, Germany, and Department of Discrete Mathematics, Adam Mickiewicz University, Poznań, Poland.

  E-mail address: tos@numerik.uni-kiel.de