I am interested in estimating the number of non-isomorphic simple graphs on $n$ vertices with $O(n)$ edges. Specifically, I am wondering whether it is correct that the number of such graphs is at most $2^{O(n)}$.
The bound is correct for some sparse graphs like trees and planar graphs. However, I would like to understand whether this bound is correct for general sparse graphs having $O(n)$ edges.
Thank you for any insights!
No. Consider the graphs on, say, $k$ labelled vertices with $n$ edges, where $k=\lceil 2n^{\lambda}\rceil$ and $1/2<\lambda<1$ is fixed. If $n$ is large, then $k(k-1)/2>n^{2\lambda}+n$, and for the number of graphs we have the lower bound $${k(k-1)/2\choose n}=\frac{\prod_{j=0}^{n-1}(k(k-1)/2-j)}{n!}>\frac{n^{2\lambda n}}{n!}>n^{(2\lambda-1)n}$$ for large enough $n$. Each graph is counted at most $k!<n^k=e^{k\log n}=e^{o(n)}$ times. Thus, the number of distinct graphs is not less than $n^{(2\lambda-1)n-o(n)}$. Since $\lambda$ may be arbitrarily close to 1, we get the lower bound $n^{n-o(n)}$ for the number of distinct graphs on $n$ vertices with $n$ edges.
This property also fails for bounded expansion classes, as there are at least $n^{3n/2+o(n)}$ subcubic graphs (which are sparse and have bounded expansion). This example shows that you cannot even ask for small classes, namely with at most $c^nn!$ labelled graphs on $n$ vertices (which is $n^{n+o(n)}$ and fits @FedorPetrov's lower bound).
See Twin width II: small classes for a simple argument for this bound page 19.
