Let $\mathcal F_n$ be the set of suitable partitions of $ \{1,...,n\}$ and $f_n=|\mathcal F_n|$. ($n\ge3$) We'll prove by induction $f_n=2^{n-2}-1.$ For $n=3$, there is only one way, $\mathcal F_n= \{\{1\},\{2\},\{3\}\}$. For each member of $\mathcal F_n$, we can add $n+1$ to the 2 sets of it that don't contain $n$ to obtain a member of $\mathcal F_{n+1}$. Another member of $\mathcal F_{n+1}$ is obtained by taking as its three sets, $\{n+1\}$, the set of all even numbers $\le n$ and the set of all odd numbers $\le n$. Obviously that is all, so $f_{n+1}=2f_n+1$ and the result follows.