A problem of Rankin on sets without geometric progressions

Abstract

A geometric progression of length k and integer ratio is a set of numbers of the form {a, ar, . . . , ark−1} for some positive real number a and integer r ≥ 2. For each integer k ≥ 3, a greedy algorithm is used to construct a strictly decreasing sequence (ai) ∞ i=1 of positive real numbers with a1 = 1 such that the set G = ∞ ⋃ i=1 (a2i, a2i−1] contains no geometric progression of length k and integer ratio. Moreover, G(k) is a maximal subset of (0, 1] that contains no geometric progression of length k and integer ratio. It is also proved that there is a strictly increasing sequence (Ai) ∞ i=1 of positive integers with A1 = 1 such that ai = 1/Ai for all i = 1, 2, 3, . . .. The set G(k) gives a new lower bound for the maximum cardinality of a subset of the set of integers {1, 2, . . . , n} that contains no geometric progression of length k and integer ratio. 1. Real and integral geometric progressions Let R denote the real numbers. For t ∈ R, let R>t denote the set of all real numbers x > t. Let [x] denote the integer part of the real number x. For real numbers u 0 with u 0 is the set q ∗X = {qx : x ∈ X}. The reciprocal of the set X is the set X = { x : x ∈ X } . For example, q ∗ (u, v] = (qu, qv] and (1/v, 1/u] = [u, v). If A = (a0, a1, . . . , ak−1) is a finite sequence of positive real numbers, then the dilation of the sequence A by q is the sequence q ∗ A = (qa0, qa1, . . . , qak−1) and the reciprocal of A is the sequence A = (1/a0, 1/a1, . . . , 1/ak−1). Let N denote the set of positive integers, and let N = N {1} denote the set of all integers r > 1. Let k ∈ N and let r, a ∈ R>0. A geometric progression of length k and ratio r with first term a is a sequence of the form (a, ar, ar, . . . , ar) = a ∗ (1, r, r, . . . , r). Date: August 14, 2014. 2010 Mathematics Subject Classification. 11B05 11B25, 11B75, 11B83, 05D10.