On the Generalized Antiaverage Problem

Abstract

For integers \(k, \theta \leq 3\) and \(\beta \geq 1\), an integer \(k\)-set \(S\) with the smallest element \(0\) is a \((k; \beta, \theta)\)-free set if it does not contain distinct elements \(a_{i,j}\) (\(1 \leq i \leq j \leq \theta\)) such that \(\sum_{j=1}^{\theta -1}a_{i ,j} = \beta a_{i_\theta}\). The largest integer of \(S\) is denoted by \(\max(S)\). The generalized antiaverage number \(\lambda(k; \beta, \theta)\) is equal to \(\min\{\max(S) : S \text{ is a } (k^0; \delta, 0)\text{-free set}\}\). We obtain:(1) If \(\beta \notin \{\theta-2, \theta-1, \theta\}\), then \(\lambda(m; \beta, \theta) \leq (\theta-1)(m-2) + 1\); (2) If \(\beta \geq {\theta-1}\), then \(\lambda(k; \beta, \theta) \leq \min\limits_{k=m+n}\{\lambda(m;\beta,\theta)+\beta \lambda (n;\beta,\theta)+1\}\), where \(k =m+n \) with \(n>m\geq 3\) and \(\lambda(2n;\beta,\theta)\leq \lambda(n;\beta,\theta)(\beta+1)+\varepsilon\), for \(\varepsilon=1\) for \(\theta=3\) and \(\varepsilon=0\) otherwise.