On sequences without geometric progressions

An improved upper bound is obtained for the density of sequences of positive integers that contain no k-term geometric progression.


A problem of Rankin
Let k ≥ 3 be an integer. Let r = 0, ±1 be a real number. A geometric progression of length k with common ratio r is a sequence (a 0 , a 1 , a 2 , . . . , a k−1 ) of nonzero real numbers such that r = a i a i−1 for 1, 2, . . . , k − 1. For example, (3/4, 3/2, 3, 6) and (8, 12, 18, 27) are geometric progressions of length 4 with common ratios 2 and 3/2, respectively. A k-geometric progression is a geometric progression of length k with common ratio r for some r. If the sequence (a 0 , a 1 , a 2 , . . . , a k−1 ) is a k-geometric progression, then a i = a j for 0 ≤ i < j ≤ k − 1.
A finite or infinite set of real numbers is k-geometric progression free if the set does not contain numbers a 0 , a 1 , . . . , a k−1 such that the sequence (a 0 , a 1 , . . . , a k−1 ) is a k-geometric progression. Rankin [3] introduced k-geometric progression free sets, and proved that there exist infinite k-geometric progression free sets with positive asymptotic density. 1 For example, the set Q of square-free positive integers, with asymptotic density d(Q ) = π 2 /6, contains no k-term geometric progression for k ≥ 3.
Let A be a set of positive integers that contains no k-term geometric progression. Brown and Gordon [2] proved 2 that the upper asymptotic density of A, denoted d U (A), has the following upper bound: Riddell [4] and Beiglböck, Bergelson, Hindman, and Strauss [1] proved that The purpose of this note is to improve these results. Key words and phrases. Geometric progression-free sequences, Ramsey theory. 1 If A(n) denotes the number of positive integers a ∈ A with a ≤ n, then the upper asymptotic density of A is d U (A) = lim sup n→∞ A(n)/n, and the asymptotic density of A is d(A) = limn→∞ A(n)/n, if this limit exists. 2 Brown and Gordon claimed a slightly stronger result, but their proof contains an (easily corrected) error. 1 2. An upper bound for sets with no k-term geometric progression Proof. Let For 1 ≤ ℓ ≤ L we have 2 ℓk−1 ≤ n. Let a be an odd positive integer such that is a geometric progression of length k with common ratio 2. If A ∈ GP F k (n), then A does not contain this geometric progression, and so at least one element in the set is not an element of A. Because every nonzero integer has a unique representation as the product of an odd integer and a power of 2, it follows that, for integers ℓ = 1, . . . , L and odd positive integers a ≤ 2 1−ℓk n, the sets X ℓ (a) are pairwise disjoint subsets of {1, 2, . . . , n}.
For every real number t ≥ 1, the number of odd positive integers not exceeding t is strictly greater than (t − 1)/2. It follows that the cardinality of the set {1, 2, . . . , n} \ A is strictly greater than L ℓ=1 1 2 Note that if r is an odd integer and r ∈ X ℓ (a), then ℓ = 1 and r = a. Let b be an odd integer such that (1) n 6 k−1 < b ≤ n 5 k−1 and b is not divisible by 5, that is, (2) b ≡ 1, 3, 7, or 9 (mod 10).
We consider the following geometric progression of length k with ratio 5/3: Every integer in this progression is odd, and n 2 k−1 < 3 k−1 b < · · · < 5 k−1 b ≤ n.
It follows that X ℓ (a) ∩ Z(c) = ∅ for all ℓ, a, and c. If c and c ′ satisfy (3) and (4) with c < c ′ and if Z(c) ∩ Z(c ′ ) = ∅, then there exist integers i, j ∈ {0, 1, 2, . . . , k − 1} such that 5 k−1−i 7 i c = 5 k−1−j 7 j c ′ or, equivalently, The inequality c < c ′ implies that 0 ≤ j < i ≤ k − 1 and so c ≡ 0 (mod 5), which contradicts (4). Therefore, the sets Z(c) are pairwise disjoint. If b and c satisfy inequalities (1) and (3), respectively, then c < b. If Y (b)∩Z(c) = ∅, then there exist integers i, j ∈ {0, 1, . . . , k − 1} such that Because bc ≡ 0 (mod 5), it follows that i = j and so Because c < b, we must have i ≥ 1 and so c ≡ 0 (mod 3), which contradicts congruence (4). Therefore, Y (b)∩Z(c) = ∅ and the sets X ℓ (a), Y (b), and Z(c) are pairwise disjoint. The number of integers c satisfying inequality (3) and congruence (4) is 4 15 Because A contains no k-term geometic progression, at least one element from each of the sets X ℓ (a), Y (b), and Z(c) is not in A. This completes the proof. Corollary 1. If A k is a set of positive integers that contains no k-term geometric progression, then Here is a

Open problems
For every integer k ≥ 3, let GP F k denote the set of sets of positive integers that contain no k-term geometric progression. It would interesting to determine precisely sup{d U (A) : A ∈ GP F k } and sup{d(A) : A has asymptotic density and A ∈ GP F k }. In the special case k = 3, Riddell [4, p. 145] claimed that if A ∈ GP F 3 , then d U (A) < 0.8339, but wrote, "The details are too lengthy to be included here." An infinite sequence A = (a i ) ∞ i=1 of positive integers is syndetic if it is strictly increasing with bounded gaps. Equivalently, A is syndetic if there is a number c such that 1 ≤ a i+1 − a i ≤ c for all positive integers i. Beiglböck, Bergelson, Hindman, and Strauss [1] asked if every syndetic sequence must contain arbitrarily long finite geometric progressions.