Are all known $k$-multiperfect numbers (for $k > 2$) *not* squarefree?

Question

A positive integer $N$ is said to be $k$-multiperfect if

$$\sigma(N) = kN$$

where $\sigma(x)$ is the sum of the divisors of $x$ and $k$ is a positive integer.

(The case $k = 2$ reduces to the original notion of perfect numbers.)

Now my question is the following: Are all known $k$-multiperfect numbers (for $k > 2$) not squarefree?

For the case $k = 2$, the only known exception is $N = 6 = 2\cdot3$.

Update [October 06 2013 - Manila time] :: This question has been cross-posted to MathOverflow here.

Answer 1

This question has been answered in MathOverflow here.

In particular, the answer confirms that all known $k$-multiperfect numbers (for $k > 2$) are not squarefree.

Additionally, $1$ and $6$ are the only squarefree $k$-perfect numbers (allowing for $k = 1$, and where $k = 2$ corresponds to the classical notion of perfect numbers).

These results were further validated by Flammenkamp (in a response to an e-mail inquiry), who maintains a database of $k$-multiperfect numbers. Here is a copy of Flammenkamp's e-mail:

"Well, from memory I know that the two smallest MPNs, namely $1$ and $6$ are not divisible by $4$. And yes, I think all greater known MPNs have a two-power exponent of at least $2$, i.e. are not squarefree."