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 $\mathcal{P}([n]) \setminus \emptyset$?



Number of couples of sets with empty intersection in a separating union-closed family of sets
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 $n \le 4$.


