probability of non-catastrophicity, which is $1-t$; see [@ich].

$(n,1,1)$ convolutional codes for $n\geq 3$
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For codes with these parameters, we consider generator matrices of the form $G(z)=\sum_{i=0}^{\delta}g_iz^i=g_0+g_1z$ with $g_0,g_1\in\mathbb F^n$.

[@diss] \
For $n\geq 3$, the probability that $G(z)\in\mathbb F[z]^n$ with $\deg(G(z))=1$ generates an $(n,1,1)$ MDP convolutional code is $$(1-t^n)^{-1}(1-t)^{n+1}\prod_{i=2}^{n-1}(1-it).$$

 \
According to the proof of the preceding theorem, for $n\geq 3$, the number of $(n,1,1)$ MDP convolutional codes over $\mathbb F$ is $|\mathbb F|(|\mathbb F|-1)^{n+1}\prod_{i=2}^{n-1}(|\mathbb F|-i)$, i.e. such a code exists if and only if $|\mathbb F|\geq n$, and one can construct a generator matrix of such a code as follows: choose all entries of $g_0$ arbitrary but nonzero, choose the first entry of $g_1$ arbitrary and then choose entry $i$ of $g_1$ such that $\begin{pmatrix} g_{0,i}\\ g_{1,i}\end{pmatrix}$ is linear independent to $\begin{pmatrix} g_{0,j}\\ g_{1,j}\end{pmatrix}$ for $j\in\{1,\hdots,i-1\}$.\
One could see that all MDP codes with these parameters are reverse MDP: For $j=1,\hdots,n$, it holds $\overline{g_j}(z)=\begin{cases}g_j(z), & g_{1,j}=0\\
g_{1,j}+g_{0,j}z, & g_{1,j}\neq 0 \end{cases}$. Therefore, $\overline{g_{0,j}}\neq 0$ and it is easy to see that the other conditions are also fulfilled for $\overline{G}$.\
However, since $n-1\geq 2>\delta$ and thus $(n-k)\nmid\delta$ a complete MDP convolutional code with these parameters cannot exist [@cmdp].

