I have asked this question in MSE a few weeks back, but did not receive any responses. I have cross-posted it to MO, hoping that it is appropriate for this site.

Let $\sigma(x)$ be the (classical) sum of the divisors of $x$.

A number $N \in \mathbb{N}$ is called perfect if $\sigma(N)=2N$.

An even perfect number $U$ is said to be given in Euclidean form if $U=(2^p - 1){2^{p-1}}$ (where $2^p - 1$ is called the Mersenne prime). On the other hand, an odd perfect number $L$ is said to be given in Eulerian form if $L = {q^k}{n^2}$ (where $q$ is called the Euler prime).

Notice that for even perfect numbers, trivially we have $1 < 2^p - 1$ (where $1$ is the exponent of the Mersenne prime $2^p - 1$).

Does a similar statement hold for odd perfect numbers? That is, does $k < q$ always hold?

I have asked this question in MSE a few weeks back, but did not receive any responses. I have cross-posted it to MO, hoping that it is appropriate for this site.

Let $\sigma(x)$ be the (classical) sum of the divisors of $x$.

A number $N \in \mathbb{N}$ is called perfect if $\sigma(N)=2N$.

An even perfect number $U$ is said to be given in Euclidean form if $U=(2^p - 1){2^{p-1}}$ (where $2^p - 1$ is called the Mersenne prime). On the other hand, an odd perfect number $L$ is said to be given in Eulerian form if $L = {q^k}{n^2}$ (where $q$ is called the Euler prime).

Notice that for even perfect numbers, trivially we have $1 < 2^p - 1$ (where $1$ is the exponent of the Mersenne prime $2^p - 1$).

Does a similar statement hold for odd perfect numbers? That is, does $k < q$ always hold?