The $\mathbb{Z}$-extensions of $\mathbb{Z}/n\mathbb{Z}$
$$
0\longrightarrow \mathbb{Z}\longrightarrow E\longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow0
$$
are classifiied by $\operatorname{Ext}\_{\mathbb{Z}}^1(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z})$. But one can compute $\operatorname{Ext}\_{\mathbb{Z}}^1(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z})=\mathbb{Z}/n\mathbb{Z}$. Hence there are $n$ such extensions. But is there any way to explicitly write down those extensions?



$\mathbb{Z}$-extensions of $\mathbb{Z}/n\mathbb{Z}$
Yes. May be a general remark: the group $\mathrm{Ext}^1$ classifies the short exact sequences, which is more structure than remembering $E$ or even $E$ with the $2$-term filtration. In any case, $0\rightarrow \mathbb Z \xrightarrow{\cdot n} \mathbb Z \rightarrow \mathbb Z/n\mathbb Z \rightarrow 0$ is a generator of $\mathrm{Ext}^1\_{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z)$, let's call it $\alpha$. Then $k\cdot\alpha\in\mathrm{Ext}^1\_{\mathbb Z}(\mathbb Z/n\mathbb Z,\mathbb Z)$ is obtained as pull-back of $\alpha$ via the multiplication by $k$: $\mathbb Z/n\mathbb Z \xrightarrow{\cdot k} \mathbb Z/n\mathbb Z$. The fiber product $\mathbb Z\times\_{\mathbb Z/n\mathbb Z,\cdot k} \mathbb Z/n\mathbb Z$ is computed as $\mathbb Z$ plus the kernel of $\ker(\cdot k)=\frac{n}{(n,k)}\mathbb Z/n\mathbb Z $ and the extension we get looks as follows:
$$
0\rightarrow \mathbb Z \xrightarrow{} \mathbb Z\oplus\frac{n}{(n,k)}\mathbb Z/n\mathbb Z \rightarrow \mathbb Z/n\mathbb Z \rightarrow 0
$$
(where the map $\mathbb Z\oplus\frac{n}{(n,k)}\mathbb Z/n\mathbb Z \rightarrow \mathbb Z/n\mathbb Z$ is identified with the projection $\mathbb Z\times\_{\mathbb Z/n\mathbb Z,\cdot k} \mathbb Z/n\mathbb Z\twoheadrightarrow \mathbb Z/n\mathbb Z$). For example if $k\in (\mathbb Z/n\mathbb Z)^\times$ we get the usual sequence $0\rightarrow \mathbb Z \xrightarrow{} \mathbb Z \rightarrow \mathbb Z/n\mathbb Z \rightarrow 0$ but where the second map sends $1$ to $k$ instead of $1$.


