How to prove there are $\frac{3^n-1}{2}$ couples $A,B \in \mathcal{P}([n])$ such that $A \cup B = [n]$, $A \neq B$. Consider the unordered couple of sets $A,B \in \mathcal{P}([n])$ such that $A \cup B = [n]$, $A \neq B$. I would like to prove that the number of such couples is:
$$\frac{3^n-1}{2}$$
I derived that expression by enumerating the solutions for the first values of $n$, and also considering the sum of the number of those couples for all different values of $n$ over $\mathcal{P}([m])$:
$${2^m \choose 2} = \sum_{n=0}^m {m \choose n}\frac{3^n-1}{2}$$
(this is easy to show using the binomial expansion).
I have checked OEIS A003462 where they count the number of couples $A,B \in \mathcal{P}([n])$ with $A \cap B = \emptyset$ and $A \neq \emptyset$ or $B \neq \emptyset$, with the same $\frac{3^n-1}{2}$ result, but it's not the same thing and I was not able to adapt that reasoning.

Note that $A\cup B=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)$ is a disjoint union. Assign to each unordered couple $\{A,B\}$ the $n$-length ternary string $(a_1,a_2,\ldots,a_n)$ defined by $$a_j=\begin{cases}0\quad\text{if $a_j\in A\setminus B$}\\1\quad\text{if $a_j\in A\cap B$}\\2\quad\text{if $a_j\in B\setminus A$}\end{cases}$$
