Recently I wondered whether there might be a natural topological complexity measure for convex polytopes embedded in $\mathbb{R}^N$. After some reflection it occurred to me that the number of distinct Hamiltonian cycles on a convex polytope may be a useful proxy measure. Now, let's suppose that the asymptotic formula for the maximum number of Hamiltonian cycles on an $n$-vertex convex polytope embedded in $\mathbb{R}^N$ is given by: 


\begin{equation}
f\_N(n) \tag{1}
\end{equation}


What I'm curious about is whether there exists a polynomial $P(n)$ such that:


\begin{equation}
\forall N \in \mathbb{N},\frac{f\_{N+1}(n)}{f\_{N}(n)} \leq P(n) \tag{2}
\end{equation}


For concreteness, in the case $(N=2,n=4)$ we have a square with 8 H-cycles and in the case $(N=3,n=4)$ we have a tetrahedron with 24 H-cycles. 


**Note:** After doing several Google searches it's still unknown to me whether this problem has already been solved. 



The number of Hamiltonian circuits on a convex polytope embedded in $\mathbb{R}^N$
For $N\ge 4$ and arbitrary $n\ge N+1$ there is a [polytope with edge graph $K\_n$](https://en.wikipedia.org/wiki/Neighborly_polytope), hence $f\_N(n)=n!$. So if $n\ge N+2$ then


$$\frac{f\_{N+1}(n)}{f\_N(n)} = 1.$$


Note also, that $f\_N(n)$ is not well-defined for $n\le N$. So talking about *arbitrarily large* $N$ for fixed $n$ might be problematic.


