If $q^k n^2$ is an odd perfect number with Euler prime $q$, then

$$k = 1 \Longleftrightarrow D(n^2) \mid n^2,$$

where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of $n^2$.

In particular, this shows that the Descartes-Frenicle-Sorli conjecture for odd perfect numbers is true if and only if the non-Euler part $n^2$ is deficient-perfect.

(I have a new paper out there that discusses the proof for the biconditional

$$k = 1 \Longleftrightarrow D(n^2) \mid n^2,$$

if $q^k n^2$ is an odd perfect number with Euler prime $q$.)

(I have a new paper out there that discusses the proof for the biconditional

$$k = 1 \Longleftrightarrow D(n^2) \mid n^2,$$

if $q^k n^2$ is an odd perfect number with Euler prime $q$.)