In the question Explicit proof that algebra is derived wild it was asked whether there are examples of algebras $A$ where it is possible to show explicitly that $A$ is derived wild by finding an appropiate strict family.
I am interested in weakening this question for $A = \mathbb{k}[x]/(x^n)$, $n \geq 3$ as follows: From On derived tame algebras (Bautista, 2007) it follows that since $\mathbb{k}[x]/(x^n)$ is self-injective, for it to be derived wild means not being derived discrete. So to show the derived wildness of this algebra one only needs to show that it is not derived discrete. This means finding for a vector $(d_j)_{j \in \mathbb{Z}}$ of non-negative numbers with finitely many entries an infinite family of pairwise non-isomorphic, indecomposable chain complexes $(P^{\bullet}_n)_{n \in \mathbb{N}}$ of finite-dimensional $\mathbb{k}[x]/(x^n)$-complexes with $\dim(\operatorname{H}^j(P^{\bullet}_n)) = d_j$.
Question: Is there an explicit example of such a $(P^{\bullet}_n)_{n \in \mathbb{N}}$ known? I have tried coming up with one, but have not been successful. Since $\mathbb{k}[x]/(x^n)$ is derived tame for $n=2$ and derived wild for $n=3$, one then should be explicitly be able to see "why this jump happens". In particular, over $\mathbb{k}[x]/(x^3)$ one can write $P/S \cong S$ for the simple module $S$ and the two-dimensional indecomposable module $P$. If $n=2$, there is only one indecomposable module $S$, so the construction should probably be based on this fact.
