On maximal \((k,b)\)-linear-free sets of integers and its spectrum (Q2712524)

A set \(A\subseteq [1,n]\) is said to be \((k,b)\)-linear-free if for every \(a\in A\), \(ka+b\not \in A\). A \((k,b)\)-linear-free set \(A\) is maximal if \(A \cup {t}\) is not \((k,b)\)-linear-free for some \(t\in [1,n].\) Let \(\mathcal{M}\) be the set of all maximal \((k,b)\)-linear-free subsets of \([1,n]\) and let \(f(n,k,b)=\max \{|A|: A\in \mathcal{M}\}\) and \(g(n,k,b)=\min \{|A|: A\in \mathcal{M}\}\). The authors prove the following results: NEWLINENEWLINENEWLINE1. Assuming that \(k+b>2\), \(f(n,k,b)=g(n,k,b)\) if and only if \(n<k^2+kb+b.\) NEWLINENEWLINENEWLINE2. For every \(T\in [g(n,k,b),f(n,k,b)]\) there exists a maximal \((k,b)\)-linear-free subset of \([1,n]\) with cardinality \(T\). NEWLINENEWLINENEWLINE3. They prove \(f(n,k,b)=\sum_{p\in P}\lceil{n(p)+1\over 2}\rceil\) and \(g(n,k,b)=\sum_{p\in P}\lceil{n(p)+1\over 3}\rceil\), where \(P\) is the set defined by \(P=\{p:p\in [1,n]\) and \(p\neq km+b\) for any \(m\in \mathbb{N}\}\) and \(n(p)=\lfloor\log_{k}{n+b/(k-1)\over p+b/(k-1)}\rfloor.\)