Let $f(G)$ give number of perfect matchings of a graph $G$.
Denote $\mathcal N_{2n}=\{0,1,2,\dots,n!-1,n!\}$.
Denote collection of all $2n$ vertex balanced bipartite graph to be $\mathcal G_{2n}$.
There are many $m\in\mathcal N_{2n}$ that do not have a $G\in \mathcal G_{2n}$ such that $f(G)=m$ (refer Are all numbers from $1$ to $n!$ the number of perfect matchings of some bipartite graph?).
How many numbers do we hit or miss in $\mathcal N_{2n}$?
Can $\bigg|{\big\{m\in\mathcal N_{2n}:f^{-1}(m)\cap\mathcal G_{2n}\neq\emptyset\big\}}\bigg|=O(2^{n^c})$ hold for some fixed $c\in(0,1)$?
Conjecture (Update):
$\bigg|{\big\{m\in\mathcal N_{2n}:f^{-1}(m)\cap\mathcal G_{2n}\neq\emptyset\big\}}\bigg|={n^{\Omega(n)}}$ holds.
