I would like to know more about divisibility among $\sigma_k(n)$ the power divisor function , Let $\sigma_x(n)$ , $\sigma_y(n)$ be a two power divisor function where $x, y$ are two positive integers and $x \neq y$ ,

My question here is : for which values or forms of $x$ and $y$ :$\sigma_x(n) $ devides $\sigma_y(n) $ if we take $x<y$ and $n>1$ ?

Note: $\sigma_x(n) = \sum_{d \mid n}{d^x}$ and $\sigma_y(n) = \sum_{d \mid n}{d^y}$

EDIT01 : for instance i find for $y=11$ and $x=1$ this result from $n=2$ to $1000$, may the closed form will :$y=11k$ and ,$x=k$ and $k$ is positive integer . could be this a suitable form ?

End EDIT01:

Thank you for your help !!!

I would like to know more about divisibility among power-divisor functions. Put $\sigma_k(n) = \sum_{d \mid n} d^k$ for all positive integers $k$ and $n$.

My question here is : for which positive integers $x$ and $y$ do we have that $\sigma_x(n) $ divides $\sigma_y(n) $ for all $n$?

EDIT01 : For instance, this is true for $y = 11$ and $x = 1$, at least for $n$ from $2$ to $1000$ (see this Wolfram|Alpha calculation). Could it be that the solutions are always of the form $y=11x$?