Let $f(n):=$min{ $ k \in N$: There is no graph with $n$ nodes and $k$ hamilton-cycles} for $n\ge 3$
The values I found out so far :
$$f(3)=2$$ $$f(4)=2$$ $$f(5)=3$$ $$f(6)=9$$ $$f(7)=13$$ $$f(8)=89$$ $$f(9)=485$$
It is clear that $f(n)\le \frac{(n-1)!}{2}+1$ because the $K_n$ has $\frac{(n-1)!}{2}$ hamilton-cycles and more than that is imposssible for graphs with $n$ nodes.
My questions :
How can $f(n)$ be calculated efficiently ? Is there an algorithm to decide efficiently if there is a graph with $n$ nodes and $k$ hamilton-cycles for a given pair $(n,k)$ ?
Is $f(n)$ a strictly motonon increasing function ?
Especially, I am interested in the value of $f(10)$.