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 $\{1,...,2n\}$ and I want to do this according to the coset type of $\pi$ with respect to the hyperoctahedron subgroup $H_n\subset S_{2n}$ (the centralizer of $\sigma$). These coset types are labeled by partitions of $n$.

So I want $N(\lambda)$, the number of permutations $\pi$ of coset type $\lambda\vdash n$ such that $\langle \pi,\sigma\rangle$ is transitive.

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

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

Clearly the first element in each row is $(n-1)!2^n$.

Edit: Proof of the above statement. Suppose $\pi$ has coset type $(1^n)$ and $\langle \pi,\sigma\rangle$ is transitive. Take some initial number $i_1$. Then a) $\pi$ cannot map $i_1$ to itself, or having coset type $(1^n)$ would imply that $\pi$ also maps $\sigma(i_1)$ to itself and then transitivity would not hold;

b) $\pi$ cannot map $i_1$ to $\sigma(i_1)$, or having coset type $(1^n)$ would imply that $\pi$ also maps $\sigma(i_1)$ to $i_1$ and then transitivity would not hold;

c) so there are $(2n-2)$ possibilities for the image of $i_1$ under $\pi$, call this $i_2$;

d) the image of $i_2$ under $\pi$ cannot belong to $\{i_2,\sigma(i_2),i_1,\sigma(i_1)\}$, for the same reasons as above. So there are $(2n-4)$ possibilities for $\pi(i_2)=i_3$. And so on.

e) There are thus $(2n-2)(2n-4)\cdots=2^{n-1}(n-1)!$ possibilities for the list $[i_1,...,i_n]$. Finally, the image of $i_n$ can be either $i_1$ or $\sigma(i_1)$, which gives an extra factor of $2$.

End edit

Dividing every row of the triangle by its first element, we get

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

Now, second element seems to be $2n(n-1)$ and last element seems to be $n!2^{n-1}$

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

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 $\{1,...,2n\}$ and I want to do this according to the coset type of $\pi$ with respect to the hyperoctahedron subgroup $H_n\subset S_{2n}$ (the centralizer of $\sigma$). These coset types are labeled by partitions of $n$.

So I want $N(\lambda)$, the number of permutations $\pi$ of coset type $\lambda\vdash n$ such that $\langle \pi,\sigma\rangle$ is transitive.

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

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

Clearly the first element in each row is $2^n(n-1)!=2(2n-2)!!$.

Edit: Proof of the above statement. Suppose $\pi$ has coset type $(1^n)$ and $\langle \pi,\sigma\rangle$ is transitive. Take some initial number $i_1$. Then a) $\pi$ cannot map $i_1$ to itself, or having coset type $(1^n)$ would imply that $\pi$ also maps $\sigma(i_1)$ to itself and then transitivity would not hold;

b) $\pi$ cannot map $i_1$ to $\sigma(i_1)$, or having coset type $(1^n)$ would imply that $\pi$ also maps $\sigma(i_1)$ to $i_1$ and then transitivity would not hold;

c) so there are $(2n-2)$ possibilities for the image of $i_1$ under $\pi$, call this $i_2$;

d) the image of $i_2$ under $\pi$ cannot belong to $\{i_2,\sigma(i_2),i_1,\sigma(i_1)\}$, for the same reasons as above. So there are $(2n-4)$ possibilities for $\pi(i_2)=i_3$. And so on.

e) There are thus $(2n-2)(2n-4)\cdots=2^{n-1}(n-1)!$ possibilities for the list $[i_1,...,i_n]$. Finally, the image of $i_n$ can be either $i_1$ or $\sigma(i_1)$, which gives an extra factor of $2$.

End edit

Dividing every row of the triangle by $(2n-2)!!$, we get

$$2$$ $$2, 8$$ $$2, 24, 48$$ $$2, 48, 80, 192, 384$$

Now, second element seems to be $(2n)(2n-2)$ and last element seems to be $(2n)!!$

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

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 $\{1,...,2n\}$ and I want to do this according to the coset type of $\pi$ with respect to the hyperoctahedron subgroup $H_n\subset S_{2n}$ (the centralizer of $\sigma$). These coset types are labeled by partitions of $n$.

So I want $N(\lambda)$, the number of permutations $\pi$ of coset type $\lambda\vdash n$ such that $\langle \pi,\sigma\rangle$ is transitive.

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

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

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?

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 $\{1,...,2n\}$ and I want to do this according to the coset type of $\pi$ with respect to the hyperoctahedron subgroup $H_n\subset S_{2n}$ (the centralizer of $\sigma$). These coset types are labeled by partitions of $n$.

So I want $N(\lambda)$, the number of permutations $\pi$ of coset type $\lambda\vdash n$ such that $\langle \pi,\sigma\rangle$ is transitive.

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

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

Clearly the first element in each row is $(n-1)!2^n$.

Edit: Proof of the above statement. Suppose $\pi$ has coset type $(1^n)$ and $\langle \pi,\sigma\rangle$ is transitive. Take some initial number $i_1$. Then a) $\pi$ cannot map $i_1$ to itself, or having coset type $(1^n)$ would imply that $\pi$ also maps $\sigma(i_1)$ to itself and then transitivity would not hold;

b) $\pi$ cannot map $i_1$ to $\sigma(i_1)$, or having coset type $(1^n)$ would imply that $\pi$ also maps $\sigma(i_1)$ to $i_1$ and then transitivity would not hold;

c) so there are $(2n-2)$ possibilities for the image of $i_1$ under $\pi$, call this $i_2$;

d) the image of $i_2$ under $\pi$ cannot belong to $\{i_2,\sigma(i_2),i_1,\sigma(i_1)\}$, for the same reasons as above. So there are $(2n-4)$ possibilities for $\pi(i_2)=i_3$. And so on.

e) There are thus $(2n-2)(2n-4)\cdots=2^{n-1}(n-1)!$ possibilities for the list $[i_1,...,i_n]$. Finally, the image of $i_n$ can be either $i_1$ or $\sigma(i_1)$, which gives an extra factor of $2$.

End edit

Dividing every row of the triangle by its first element, we get

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

Now, second element seems to be $2n(n-1)$ and last element seems to be $n!2^{n-1}$

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