Im looking for an algorithm that does the following in a quick way:

Input: Natural number $r \geq 2$, natural number $s \geq 3$, prime power $q$. Output:

Finds all two-sided ideals in $J^2/J^s \subseteq \mathbb{F}_q \langle x_1,...,x_r\rangle/J^s$ ,where $A=\mathbb{F}_q \langle x_1,...,x_r\rangle $ is the noncommutative polynomial ring in r variables over the finite field with q elements and $J=(x_1,...,x_r)$ is the ideal generated by $x_1, ... , x_r$.

How many ideals are there?

And how many up to isomorphism? (Two ideals $I_1$ and $I_2$ are isomorphic iff $A/ I_1$ and $A/I_2$ are isomorphic as $K$-algebras)

It might be also interesting to replace the noncommutative polynomial ring by a commutative one.

Note that the problem can also be formulated as to find all admissible ideals containing $J^2$ of a quiver algebra with one point and $r$ loops over a finite field.

As a motivation I offer a 100 Euro reward if someone can make a quick programm which works for small r and s (lets say $r=2,3$ and $s=2,3,4$ and choosing $q=2,3,5$). It should be programmed with the GAP-packet qpa.

edit: I rewrote the question to make it shorter.Easiest special case is a question on stackexchange: https://math.stackexchange.com/questions/2322214/number-of-ideals-with-gap .