Variations around a problem of Mahler and Mendès France (Q2898884)

Let \(2\leq b\leq b_1<b_2\) be given positive integers and let \((\varepsilon_k)\) be a sequence of integers with \(0\leq\varepsilon_k<b\). Consider the numbers \(\xi_1=\sum_{k\geq1}\frac{\epsilon_k}{b_1^k}\) and \(\xi_2=\sum_{k\geq1}\frac{\varepsilon_k}{b_2^k}\). The author proposes several problems concerning simultaneous behaviour of the numbers \(\xi_1\) and \(\xi_2\). For instance he proposes a problem whether the numbers \(\xi_1\) and \(\xi_2\) are algebraically independent if the sequence \((\epsilon_k)\) is not ultimately periodic.NEWLINENEWLINEThe main result of the paper is the following. Let \(b\) and \(b_1\) be positive integers with \(2\leq b<b_1\neq b^2\). Let \(a\) be a real number and \(w\) an integer with \(3\leq a\leq \frac{w}{3}\). The author gives an explicit construction of the sequence \((\varepsilon_k)\) of integers with \(\varepsilon_k\in\{0,1,b\}\) such that the numbers \(\xi_1=\sum_{k\geq1}\frac{\epsilon_k}{b_1^k}\) and \(\xi_2=\sum_{k\geq1}\frac{\varepsilon_k}{(b^2)^k}\) have the irrationality exponents equal to \(\mu(\xi_1)=\frac{a(2w+1)}{2a+2}\) and \(\mu(\xi_2)=\frac{a(2w+1)}{a+2}\).