All of the graphs considered in this question are connected. We can find the number of spanning trees $t(G)$ of $G$ using Kirchhoff's matrix-tree theorem or the deletion-contraction method. I'm interested in the converse of this problem. That is, given $n$, find $G$ such that $t(G)=n$.
Without some restrictions on $G$ this problem becomes almost trivial. For example, the cycle $C_n$ has $n$ spanning trees. Additionally, if we identify a vertex of graphs $G$ and $H$, the resulting graph has $t(G) \cdot t(H)$ spanning trees. Likewise if we join $G$ and $H$ with any path. So if we could factor $n$ we would just be dealing with the same question for smaller numbers. Thus I would like to restrict $G$ from being a cycle or a graph with a cut-vertex or cut-edge.
My question is then, given $n$, can we always find a graph $G$ which is not a cycle and which has no cut-vertices or cut-edges such that $t(G)=n$? If it is possible to find one, is it possible to find many, or all graphs that satisfy those properties? If it is not possible in general, are we able to determine which $n$ it is true for?
I've gone through all of the relevant graphs up to six vertices from http://www.graphclasses.org/smallgraphs.html and calculated the number of spanning trees for each. The smallest graph which satisfies all of the criteria is $K_4-e$, which has 8 spanning trees. The first ten $n>8$ for which I haven't found a respective $G$ are 9, 10, 13, 18, 22, 25, 33, 37, 42, and 46. All of these numbers are one less than a prime or one less than 2 times a prime.