Ordered analogue of the chromatic polynomial Let $G=(V,E), E\subseteq V^2$ be a finite directed graph. For $n\in\mathbb{N}$, consider
$$\chi_{G}^{\leqslant}(n)=\#\Big\{f : V \to \{1, \ldots, n\}\ \Big|\ \big(\forall (u,v)\in E\big)\big(f(u) \leqslant f(v)\big)\Big\}.$$
It can be shown that this is a polynomial in $n$ of degree $\leqslant|V|$ with rational coefficients.
We could take $<$ or $\neq$ in place of $\leqslant$. Thus $\chi_{G}^{\neq}$ is the chromatic polynomial of $G$.

Do $\chi_{G}^{\leqslant}$ and/or $\chi_{G}^{<}$ have a name? How to compute $\chi_{G}^{\leqslant}$ for moderate-sized $G$?

For computing, something similar to this would be sufficient to me. I don't see it myself yet. Something along these lines might also help, but currently I'm feeling stuck.

For extending deletion-contraction: even if we assume $G$ pruned (with cycles contracted, loops thrown away, etc.), then for $e\in E$ we only have $\color{blue}{\chi_{G}^{\leqslant}+\chi_{G\downarrow e}^{\leqslant}=\chi_{G\setminus e}^{\leqslant}+\chi_{G/e}^{\leqslant}}$, where $G\downarrow e$ is $G$ with $e$ reversed, $G\setminus e$ is $G$ with $e$ removed, and $G/e$ is $G$ with $e$ contracted. This does not provide a reduction (even for a simple $G=(\{u,v\},\{(u,v)\})$, for which $\chi_{G}^{\leqslant}=\chi_{G\downarrow e}^{\leqslant}$ however)...
 A: These are called the weak and strong chromatic polynomial of $G$, respectively; see this paper, for example.
Regarding the deletion-contraction rule, this paper suggests using it to obtain strongly connected components in $G$ by reversing arrows, then contracting those components to a point. I'm not sure how efficient this is when all cycles are long cycles.
Doing this repeatedly will eventually leave us with a graph which is the orientation of a 
tree. I don't know what the best thing to do here is. One possibility is to keep flipping edges until all edges are oriented away from a single root vertex. Then we have a recursion for the number of colorings, based on how we color the root and what we do with the subtrees.
Computing $\chi^<_G$ is related to counting the topological sorts of an acyclic graph; see this MSE answer. Specifically, a topological sort is a $<$-coloring using exactly $N:=|V(G)|$ colors, which is equal to $\chi_G^{<}(N) - N \chi_G^{<}(N-1)$. I say this to argue that computing $\chi_G^<$ is computationally difficult, because counting the topological sorts is computationally difficult. I expect that something similar holds for $\chi_G^{\leqslant}$ but don't have a proof.
We can give $\chi_G^<(n)$ a recursive formula based on the set $S$ of vertices of degree $0$:
$$
   \chi_G^{<}(n) = \sum_{A \subseteq S} \chi_{G-A}^{<}(n-1).
$$
This is probably reasonable for evaluating $\chi_G^<$ at small values of $n$, but will be obnoxious for actually computing the polynomial. I'm not sure if it's better than brute force in that case.
