For $n\in\mathbb{N}^{+}$, let $c_{n}$ denote the number of simple non-isomorphic cycle matroids of graphs on $n$ vertices. That is, let

$$A(n)=\{M(G)\;;\;G\text{ is a graph on }n\text{ vertices}\},$$

and let $B(n)$ be a largest subset of $A(n)$ such that no two elements of $B(n)$ are isomorphic (as matroids). Then

$$c_{n}\equiv|B(n)|.$$

(Here $M(G)$ denotes the cycle matroid of graph $G$. )

A simple upper bound for $c_{n}$ would be something like $2^{2^{\binom{n}{2}}}$, since the ground set of the cycle matroid of a graph on $n$ vertices is of size at most $\binom{n}{2}$.

Are there any better upper bounds known for $c_{n}$?

Edit: It looks like I am asking for the number of non-$2$-isomorphic graphs on $n$ vertices. Whitney's 2-isomorphism theorem (version of Oxley's *Matroid Theory*) states: "Let $G$ and $H$ be graphs having no isolated vertices. Then $M(G)$ and $M(H)$ are isomorphic if and only if $G$ and $H$ are $2$-isomorphic."

Intuitively, this number looks like it should be significantly smaller than the number of non-isomorphic graphs on $n$ vertices, since non-isomorphic graphs can be $2$-isomorphic.

Do we know of upper bounds on the number of non-$2$-isomorphic graphs on $n$ vertices? (Should I ask this as a separate question?)