How many total orders consistent with a partial order?

Question

I have a finite set of objects $X$, whose power set is partially ordered by $\subseteq$.

1. Consider all possible total orderings of the power set $\mathscr{P}(X)$ which are compatible with the partial order $\subseteq$ in the sense that $A \subsetneq B \Rightarrow A \prec B$. How many compatible total orders are there?

2. Some orders $\prec$ have the special property that they can be concretely quantified by assigning numerical weights to each element in the set; then a subset has a smaller total weight than another subset if and only if the subsets are related by $\prec$.

   Specifically, this means that you can find a weight assignment function $f:\mathscr{P}(X)\rightarrow \mathbb{R}^+$ such that every subset's weight is the sum of its elements' weights: $$\forall S\subseteq X,\quad f(S) = \sum_{x\in S} f(\{x\})$$ and the weight respects order in that $f(S) < f(T) \iff S \prec T.$

   How many quantifiable total orders are there? (For my applications, I'm interested in weight assignments where $f(S) = 0 \iff S = \varnothing$.)

Answer 1

To question 2:

These are called coherent Boolean term orders by MacLagan (1999).

Definition 2.4. A Boolean term order is coherent if there is a weight vector $w = (w_1, w_2, \ldots, w_n) \in \mathbb N^n$ such that $$ \alpha \prec \beta \Leftrightarrow \sum_{i\in\alpha}w_i < \sum_{j\in\beta}w_j.$$

MacLagan uses integer weights, but it would not seem to make a difference; if you can assign positive real weights that quantify a total order, then you can just scale them up so that every pair of sets has weight difference of more than $1$, then round them down to integers, and the order is unchanged.

MacLagan counted them up to $n=|X|=7$ and the counts are in OEIS A009997: $$1, 1, 2, 14, 516, 124187, 214580603$$ The numbers in this sequence have been divided by $n!$ to account for the (somehow trivial) factor of $n!$ that ensues from permuting the $n$ elements of $X$. If you want to count such orders as distinct, just multiply the numbers by $n!$.

• MacLagan, Diane, Boolean term orders and the root system $B_n$, Order 15, No. 3, 279-295 (1999). ZBL0932.05093.