Number of couples of sets with empty intersection in a separating union-closed family of sets

Question

The ratio between the number of unordered couples of sets, with empty intersection between the two sets, and the total number of unordered couples of sets, for a powerset on $n$ elements without the empty set, $\mathcal{P}([n]) \setminus \emptyset$, is:

$$\frac{{n+1 \brace 3}}{{2^n-1 \choose 2}}=\frac{(1 + 3^n - 2^{n+1})}{(2^n-1)(2^n-2)}$$

Where ${n+1 \brace 3}$ denotes a Stirling number of the second kind. Is it possible to find a finite separating union closed family $\mathcal{F}$, $\emptyset \notin \mathcal{F}$, with size of the universe $|U(\mathcal{F})| = n$, with a biconnected Hasse diagram graph (without articulation vertices), and with the ratio defined as above higher than the value for $\mathcal{P}([n]) \setminus \emptyset$?

Answer 1

The numerator grows faster than the denominator, so we can do better by making a minimal extension of a previous powerset: $2^{[k]} \cup \{[m] : k+1 \le m \le n \} \setminus \emptyset$ gives $$\frac{(1 + 3^k - 2^{k+1})}{(2^k+n-k-1)(2^n+n-k-2)}$$ and for $n \ge 5$ the optimal value of $k$ is less than $n$ and grows logarithmically.

k    First n for which k is optimal
2    2
3    3
4    4
5    7
6    12
7    24
8    46
9    91
10   182
11   366

By brute force enumeration, $2^{[n]} \setminus \emptyset$ is optimal for $n \le 4$.