The concept of a squarefull number can be generalized. Let $n \in \mathbb{Z}$ with $n>1$ Then $m \in \mathbb{Z}$ is $n$ full if, for every prime $p$ dividing $m$ $p^n$ divides $m$

Note that $m$ is $n$ full if and only if there exist $a_0, \dots, a_{n-1} \in \mathbb{Z}$ such that $\displaystyle m=\prod_{j=0}^{n-1} {a_j}^{n+j}$