Let $\sigma=(1\;2)(3\;4)\cdots (n-1\; n)$ be a fixed-point-free involution in $S_{2n}$. I want to count permutations $\pi$ such that the group $\langle \pi,\sigma\rangle$ generated by $\pi$ and $\sigma$ is transitive on $\{1,...,2n\}$.

I want to do this according to the coset type of $\pi$ with respect to the hyperoctahedron subgroup $H_n\subset S_{2n}$. These coset types are labeled by partitions of $n$.

The numbers I have obtained provide the following series (n=1,2,3,4 - partitions in lexicographic order):

$$2$$ $$4, 16$$ $$16, 192, 384$$ $$96, 2304, 3840, 9216, 18432$$

Clearly the first element in each row is $(n-1)!2^n$. Diving this out I get

$$1$$ $$1, 4$$ $$1, 12, 24$$ $$1, 24, 40, 96, 192$$

Now, second element is always $2n(n-1)$ and last element is always $n!2^{n-1}$

These numbers look very simple. Does anyone know of an explicit solution to this problem?