Fix a set $V$ of cardinality $n$. I'm trying to determine the number of possible DAGs (directed acyclic graphs) whose vertex set is $V$ with a longest path containing $k$ nodes, which I will write as $D^n_k$.
So far, I have
$$ D^n_k = {n \choose k} (\ldots) $$
The $n \choose k$ is required because we have to choose the vertices which form the longest path. I also know that the part in the parentheses needs to give the number of possible subgraphs of order $n - k$ which don't extend the path to more than $k$ vertices, but I don't know how to even describe this. Although I am after a closed form, a recurrence, or help in obtaining one, would be very helpful for me.
(Too long for a comment, not a full answer) If we forget about the $k$ parameter and just look at the number of DAGs on $n$ labeled nodes then the total number of DAGs is given by the recurrence $$ D_n = \sum_{k=1}^n(-1)^{k-1}\dbinom{n}k2^{k(n-k)}D_{n-k}.$$ If the nodes are not labeled, just divide by $n!$. This is also the OEIS sequence A003024. The first few numbers are $$1, 1, 3, 25, 543, 29281, 3781503, 1138779265, 783702329343, 1213442454842881, 4175098976430598143, 31603459396418917607425$$ so $D_n$ grows very quickly. As a very crude bound, if we only take the first term in the sum we get $$ D_n \approx 2^nD_{n-1}$$ so $D_n \approx 2^{n^2}$ (It grows even faster than this).
If we make the bad assumption that all longest path lengths are equally likely, then the number of DAGs on $n$ labeled nodes with longest path having length $k$ is roughly $\frac{D_n}k.$ However, some longest paths are much more likely than others. More precisely,
If we forget about labels, and also orientation of the edges, it is plausible to believe that the question is roughly the same as asking
This question has been answered here and the expected height is roughly $O(\sqrt{n}).$ Thus, for $k$ far away from $O(\sqrt{n})$, there might be much less than $D_n/n$ possibilities.