Originally posted on MathStackExchange but without answers.
Consider partitions $\mathcal{F}=\{\mathcal{A_1},\ldots,\mathcal{A_n} \}$ of the powerset without the empty set element $Q = \mathcal{P}([n]) \setminus \{\emptyset\}$. $[n]=\{1,2,\ldots,n\}$. The size of $\mathcal{F}$ is explicitly required to be $n$ too, that is not a printing error.
Let $\mathcal{F}_a = \{\mathcal{A} \in \mathcal{F} : \exists B \in \mathcal{A} : a \in B\}$, i.e. the number of families that have $a$ as element of at least one of their sets.
We know that for a given $m \le n$, $\vert \mathcal{F}_x \vert \le m, 1 \le x \le n$.
Let $G = \{\{B,C\} : \exists \mathcal{A} \in \mathcal{F} : (B \in \mathcal{A}) \land (C \in \mathcal{A}) \land (B \cap C = \emptyset)\}$, i.e. the unordered couples of non-empty subsets of $[n]$ such that the two subsets belong to the same family and are disjoint.
I would like to find a lower bound for $\vert G \vert$ over all possible partitions as a function of $m$:
$$\vert G \vert \ge f(m)$$
I think it is easier to start with higher values of $m$. For example for $m = n$, $f(m) = 0$ because we can choose $\mathcal{A_1} = \{A \in Q: 1 \in A \}, \mathcal{A_2} = \{A \in Q, A \not\in \mathcal{A_1}: 2 \in A \}, \mathcal{A_3} = \{A \in Q, A \not\in \mathcal{A_1},\mathcal{A_2}: 3 \in A \},\ldots,\mathcal{A_{n-1}} = \{A \in Q, A \not\in \mathcal{A_1},\ldots,\mathcal{A_{n-2}}: n-1 \in A \}=\{\{n-1\},\{n-1,n\}\}, \mathcal{A_n} = \{\{n\}\}$. For example, for $n=3$, we have $\mathcal{A_1} =\{\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}$, $\mathcal{A_2} =\{\{2\},\{2,3\}\}$, $\mathcal{A_3} =\{\{3\}\}$.
I believe $f(n-1) = 1$ although up to now I don't have a proof. Can we prove it and compute $f(m)$ for $m \le n-2$?
EDIT
I have posted a simplified version of the problem with a conjectured lower bound here.
