• 1 Institut de Mathématiques de Bordeaux

Using a method due to G. J. Rieger, we show that for $1 < c < 2 $ one has, as $x$ tends to infinity $$\textrm{Card}{n \leq x : \lfloor{n^c}\rfloor} \ \textrm{ is cube-free} } = \frac{x}{\zeta(3)} + O (x^{ (c+1)/3} \log x)$$ , thus improving on a recent result by Zhang Min and Li Jinjiang.