A generalization of primitive sets and a conjecture of Erd\H{o}s

A set of integers greater than 1 is primitive if no element divides another. Erd\H{o}s proved in 1935 that the sum of $1/(n \log n)$ for $n$ running over a primitive set $A$ is universally bounded over all choices for $A$. In 1988 he asked if this universal bound is attained by the set of prime numbers. We answer the Erd\H{o}s question in the affirmative for 2-primitive sets. Here a set is 2-primitive if no element divides the product of 2 other elements.


Introduction and Statement of results
A set of integers greater than 1 is called primitive if no element divides any other. Erdős [4] showed that there is a constant K such that for any primitive set A, Further, in 1988 he asked if K can be taken as the sum of 1/(p log p), with p running over the primes. This is now referred to as the Erdős conjecture for primitive sets: where P is the set of prime numbers. By a simple argument, the Erdős conjecture is equivalent to the assertion that f (A) ≤ f (P(A)) for any primitive set A, where P(A) denotes the set of primes dividing some member of A.
One can strengthen the notion of primitivity as follows. We say that a set A of integers greater than 1 with |A| ≥ k + 1 is k-primitive if no element divides the product of k distinct other elements. Note that k-primitive implies j-primitive for all k ≥ j ≥ 1.
On the other hand, the primes are not optimal among primitive sets with respect to logarithmic density. Indeed, Erdős, Sárközy and Szemerédi [8] obtained the best possible upper bound for any primitive set A, while Erdős [7] showed that Nevertheless, one may wonder if the primes still maximize the logarithmic density among 2-primitive sets. Indeed, we prove this to be the case. We use this to deduce Theorem 2.
Proof of Theorem 2 given Proposition 1. For any 2-primitive set A, we have for all x ≥ 2 by Proposition 1. Then by partial summation, Hence taking x → ∞ gives f (P(A)) ≥ f (A) as desired. where t = τ 1 is the unique real solution to the equation The fact that τ 1 is markedly larger than 1 gives some indication as to why the full Erdős conjecture remains open.
In the setting of 2-primitive sets, we extend the range of valid exponents λ.
Theorem 3. For any λ ≥ 0.7983, x ≥ 2, and any 2-primitive set A, We remark it suffices to verify Theorem 3 with λ = 0.7983. Indeed, suppose that Then, by partial summation, for any t > λ, Hence we may define the critical exponent τ 2 for 2-primitive sets, as the infimum over all λ for which (1.2) holds. We also note that Theorem 3 with λ = 1 gives us Proposition 1. However, Theorem 3 does not hold for every positive value of λ. Indeed, in [6], Erdős showed that there is a 2-primitive set A in [1, x] of cardinality π(x) − π(x 1/3 ) + cx 2/3 /(log x) 2 . It consists of primes in (x 1/3 , x] and a subset of {p 1 p 2 p 3 : p i are primes ≤ x 1/3 } where the triples {p 1 , p 2 , p 3 } form a Steiner triple system. Thus, by the prime number theorem, when λ < 0.5 and x is sufficiently large. Hence the above argument and Theorem 3 together imply that the critical exponent lies in the interval In a sequel paper, we shall address the question of critical exponents for k-primitive sets, with k ≥ 3.

Combinatorial Lemmas
Before proving Theorem 3, we need lemmas in counting the maximal number of elements in a k-primitive set.
We first recall the following famous result due to Erdős and Szekeres [5], whose proof we provide for completeness.
Lemma 1 (Erdős-Szekeres). A sequence of (r − 1)(s − 1) + 1 real numbers has either a monotonic nondecreasing subsequence of length r or a monotonic nonincreasing subsequence of length s.
Proof. Say the sequence is a 1 , a 2 , . . . , a n , where n = (r−1)(s−1)+1. For each a i consider the ordered pair (b i , c i ), where b i is the length of the longest nondecreasing subsequence ending at a i and c i is the length of the longest nonincreasing subsequence ending at a i . Then no two pairs (b i , c i ) and (b j , c j ) can be equal, so for at least one choice of i we have b i ≥ r or c i ≥ s.
We next bound the size of a k-primitive set based on the number of prime factors used to generate its elements.
Let v 1 be such that e 1 is maximal. Then let v 2 be such that e 2 is maximal among the remaining vectors, and similarly define v 3 , . . . , v n . Thus, the chosen vectors are distinct.
Case n ≤ k: If |T | ≥ n+1 then T has some vector v = v i for all i. But then v ≤ v 1 +· · ·+ v n . This implies that T , and hence A, is not n-primitive, and since n ≤ k, it implies that A is not k-primitive, a contradiction. Hence we cannot have |T | ≥ k + 1 when n ≤ k.
Therefore, there can be at most 19 members in T .
Remark 2.1. We will not need it here, but by similar methods one can prove that if T is a 2-primitive set of positive integers with |P(T )| = n ≥ 3, then |T | ≤ 9 2 n−3 .

Proof of Theorem 3
Let A ⊂ (1, x] be a 2-primitive set. Let 0.79 ≤ λ < 1 be a parameter to be defined later. First, we partition A into primes S and composites T . Note S and P(T ) are disjoint since A is primitive. For a prime p, define then we replace the members of T p with the prime p (i.e., redefine A = (T \T p ) ∪ {p}). This would make Tp t −λ at least as big while keeping A 2-primitive. Repeat the process with each prime p ∈ P(T ) until no such prime satisfies (3.1). If T = ∅ after doing this, then A = S consists of primes so Proposition 1 follows. Otherwise T = ∅, so we may assume Consider the set We record some useful properties of T and D. (i) For each p ∈ P(T ), T p has at least 3 elements.
(ii) The map sending ordered pairs (t, p) with t ∈ T and p | t to t/p ∈ D is injective.
(iii) Each t ∈ T has at least 3 prime factors.
(iv) D is a primitive set of composite numbers.
(iii) If not, say t = pq. Since T p , T q each have at least 3 elements, there are some pm and qn other than t ∈ T . But then, t = pq | (pm)(qn) which contradicts T as 2-primitive.
(iv) If not, then (t/p) | (t 1 /p 1 ) for some t, t 1 ∈ T , p | t, p 1 | t 1 , and t/p = t 1 /p 1 . If p 1 = p, then t | t 1 which contradicts T as primitive. And if p 1 = p, then there is some pl ∈ T p other than t and t 1 . This implies t | t 1 · pl, and since t = t 1 (otherwise p = p 1 ), we have a contradiction to T being 2-primitive. Thus D is primitive, and also composite by (iii).
For Theorem 3, we must show Suppose P(T ) consists of primes q 1 < q 2 < · · · < q r . Let 2 = p 1 < p 2 < · · · < p r be the first r primes in P. We are going to modify the set T by the following process. First, if each q i = p i , we let T stand as it is. Otherwise, let i be the smallest index such that q i > p i . Then q j = p j for all j < i and we have p i ∤ t for all t ∈ T . Then replace each t ∈ T q i with p i /q i · t. This keeps T as 2-primitive, and by (3.2), So replacing each t ∈ T q i with p i /q i · t preserves (3.2). We repeat this process for each i with q i > p i and in the end we have P(T ) = {p 1 , p 2 , . . . , p r }. By showing (3.4) for this T it would follow that (3.2) fails for some p i , and this contradiction would prove the theorem.
As just noted, we may assume that P(T ) consists of the primes up to some Y , i.e., P(T ) = P ∩ (1, Y ], so (3.4) becomes For a parameter 0 < θ < 1 to be chosen later, we define λ as First consider those t ∈ T with greatest prime factor P (t) ≥ t θ . Then t 1−θ ≥ t/P (t) and so t −λ ≤ (t/P (t)) −λ/(1−θ) = (t/P (t)) −τ . Hence For a positive integer t, we consider the following unique factorization t = m(t)M(t) 6 into positive integers m(t) ≤ M(t) whose ratio M(t)/m(t) is minimal. Let We need two lemmas.
If m(t) < M(t), we can match t with M(t). Otherwise, we have t = m(t) 2 , and then m(t ′ ) < M(t ′ ). Let m ′ = t ′ /m(t). We would like to match t ′ with m ′ instead of m(t). Suppose this is blocked by some t ′′ different from t ′ (and necessarily different from t) with m ′ ∈ {m(t ′′ ), M(t ′′ )}. But then t ′ | tt ′′ , a violation of 2-primitivity. Thus, the matching can be completed.

Set
T p = {t ∈ T : P (t) = p}, so that T p ⊂ T p . We have the following variant of Lemma 5. Lemma 6. For any 2-primitive set T and prime p, let N p (z) denote the number of members t of T p with t ≤ z. With q running over the primes in I p := (max{p, z 1/4 }, z 1/2 ), we have Proof. Note that if T is 2-primitive, so too is T p /p = {t/p : t ∈ T p }. Thus, we may apply Lemma 4 to obtain a matching from T p /p into M(T p /p). The prime factors of each element t/p ∈ T p /p are at most p, so following the proof of Lemma 5, we have m(t/p), M(t/p) ∈ [t 1/2 /p, t 1/2 ). The lemma then follows in the same way as Lemma 5.