$n$-full number

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 $a_0,\mathrm{\dots },a_{n-1}\in \mathbb{Z}$ such that $m={\displaystyle \prod _{j=0}^{n-1}}a_j{}^{n+j}$.