Since a connected graph with $n$ vertices and $n$ edges is unicyclic, you just to subtract the ones whose cycle is a triangle (http://oeis.org/A053507) from the total (http://oeis.org/A057500).

Starting at $n=4$, I get 3, 72, 1500, 32280, 748440, 18898992, 520107840, 15555704400, 503580654720, 17569154733240, ... which is not in OEIS by itself.

Since a connected graph with $n$ vertices and $n$ edges is unicyclic, you just need to subtract the ones whose cycle is a triangle (http://oeis.org/A053507) from the total (http://oeis.org/A057500).

Starting at $n=4$, I get 3, 72, 1500, 32280, 748440, 18898992, 520107840, 15555704400, 503580654720, 17569154733240, ... which is not in OEIS by itself.

To the best of my knowledge, there is no theoretical formula which is suitable for computing these numbers. The value can of course be computed as far as one can generate the graphs up to isomorphism. As you know, the number of labelled graphs in the isomorphism class of a graph $G$ with $n$ vertices is $$\frac{n!}{|\mathrm{Aut}(G)|}.$$

Starting at $n=4$, I get 3, 72, 1500, 32280, 748440, 18898992, 520107840, 15555704400, 503580654720, 17569154733240, ... Not in OEIS. That takes us to 13 vertices in less than 1 second altogether. It would be possible to get $n$ into the 30s with special code.

Since a connected graph with $n$ vertices and $n$ edges is unicyclic, you just to subtract the ones whose cycle is a triangle (http://oeis.org/A053507) from the total (http://oeis.org/A057500).

Starting at $n=4$, I get 3, 72, 1500, 32280, 748440, 18898992, 520107840, 15555704400, 503580654720, 17569154733240, ... which is not in OEIS by itself.