Labelled unicyclic graph count
= Labelled unicyclic graph count
{title2=$C(n,n)=\tfrac12(n-1)!\sum_{j=0}^{n-3}n^j/j!$}
A labelled <unicyclic component> has a unique cycle of length $k\geq3$, with a rooted forest attached to its vertices. Choose its vertices in $\binom nk$ ways and its unoriented cycle in $(k-1)!/2$ ways. The <labelled forest count with prescribed roots> gives $kn^{n-k-1}$ attachments. Summing over $k$ and putting $j=n-k$ yields the displayed identity.