Given set $\mathcal T_n=\{0,1,3,4\dots,2^n-1\}$ (note there is no $2$) what is the minimum number of vertices $m$ needed in a planar graph such that at every $i\in\mathcal T_n$ there is a graph $G\in\mathcal G_{m}$ (set of all planar graphs on $m$ vertices) with Spanning tree count exactly $i$? Is there $m=O(poly(n))$?
Essentially I am asking whether for every $i$ in $\mathcal T_n$ we can construct a graph with spanning tree count exactly $i$ and the number of vertices is just $O(poly(n))$?
If $m=2^{O(n)}$ then regular $i$-agon suffices for $i=m$.
