 1 solved out of 3 shown
$500 Let$A\subset \mathbb{R}$be a countably infinite set (with$1\not\in A$) such that for all$x\neq y\in A$and integers$k\geq 1$we have $\lvert kx -y\rvert \geq 1.$ Does this imply that $\sum_{x\in A}\frac{1}{x\log x}<\infty$ or $\sum_{x <n}\frac{1}{x}=o(\log n)?$ Note that if$A$is a set of integers then the condition implies that$A$is a primitive set (that is, no element of$A$is divisible by any other), for which the convergence of$\sum_{n\in A}\frac{1}{n\log n}$was proved by Erdős [Er35], and that$\sum_{n<x}\frac{1}{n}=o(\log x)$was proved by Behrend [Be35]. A set$A\subset \mathbb{N}$is primitive if no member of$A$divides another. Is the sum $\sum_{n\in A}\frac{1}{n\log n}$ maximised over all primitive sets when$A$is the set of primes? Erdős [Er35] proved that this sum always converges for a primitive set. Solved by Lichtman [Li23]. Let$A\subseteq \mathbb{N}$, and for each$n\in A$choose some$X_n\subseteq \mathbb{Z}/n\mathbb{Z}$. Let $B = \{ m\in \mathbb{N} : m\not\in X_n\pmod{n}\textrm{ for all }n\in A\}.$ Must$B$have a logarithmic density, i.e. is it true that $\lim_{x\to \infty} \frac{1}{\log x}\sum_{\substack{m\in B\\ m<x}}\frac{1}{m}$ exists? Davenport and Erdős [DaEr37] proved that the answer is yes when$X_n=\{0\}$for all$n\in A$. The problem considers logarithmic density since Besicovitch [Be34] showed examples exist without a natural density, even when$X_n=\{0\}$for all$n\in A\$.