Cycle generating function of permutations with only odd cycles Let $\mathrm{ODD}(n)$ be the set of permutations in $\mathfrak{S}_n$ whose cycle lengths are all odd. It is known that
$$ \#\mathrm{ODD}(n) = \begin{cases} ((n-1)!!)^2 &\textrm{ if $n$ is even}; \\ n\cdot((n-2)!!)^2 &\textrm{ if $n$ is odd}. \end{cases}$$
Set $\mathcal{C}_{\mathrm{ODD}}(n,t) = \sum_{\pi \in \mathrm{ODD}(n)} t^{\kappa(\pi)}$, where $\kappa(\pi)$ is the number of cycles of $\pi \in \mathfrak{S}_n$.
Then it is easy to show that
$$ \sum_{n\geq0}\mathcal{C}_{\mathrm{ODD}}(n,t) \cdot \frac{z^n}{n!} =  \left(\frac{\sqrt{1-z^2}}{1-z}\right)^t$$
From this one can show that
$$ \mathcal{C}_{\mathrm{ODD}}(n,t)= \sum_{i=0}^{\lfloor n/2\rfloor}\frac{1}{2^i\cdot i!}\prod_{j=0}^{n-1-2i}(t+j)\prod_{k=0}^{i-1}(t-2k)(n-2k)(n-2k-1)$$
In particular if we set $t:=2k$ to be an even integer, then
$$ \mathcal{C}_{\mathrm{ODD}}(n,2k)=\frac{(n-1)!}{(2k-1)!} \cdot \sum_{j=0}^{k}(-1)^j\binom{k}{j}\prod_{i=0}^{k-j-1}(n+2i)(n+2i+1)\prod_{i=0}^{j-1}(n-2i)(n-2i-1)$$
Thus, for example $\mathcal{C}_{\mathrm{ODD}}(n,2)=2\cdot n!$ and $\mathcal{C}_{\mathrm{ODD}}(n,4)=4n\cdot n!$.
Define the polynomial $P_k(x)$ by setting
$$ \mathcal{C}_{\mathrm{ODD}}(n,2k) = \frac{(n-1)!\cdot k! \cdot 2^k}{(2k-1)!}\cdot P_k(n)$$
For example,


*

*$P_1(x)=x$;

*$P_2(x)=3x^2$;

*$P_3(x)=10x^3+5x$;

*$P_4(x)=35x^4+70x^2$;

*$P_5(x)=7\cdot 3^2(2x^5+10x^3+3x)$;

*$P_6(x)=3\cdot 7\cdot 11(2x^6+20x^4+23x^2)$;

*et cetera


Conjecture 1: $P_k(x)$ is a polynomial of degree $k$ with nonnegative integer coefficients, with zero constant term, and which is odd if $k$ is odd and even if $k$ is even.
Conjecture 2: We have,
$$ \sum_{k \geq 0} \frac{P_k(x)}{(2k-1)!!} \cdot z^k = \frac{1}{2}\cdot\left( \frac{1+z}{1-z}\right)^x$$
(The constant term on the RHS is $1/2$ so take whatever convention for $P_0(x)$ or $(-1)!!$ for that to work.)
With regard to Conjecture 2, note that from the above we have
$$ \sum_{n\geq0}\mathcal{C}_{\mathrm{ODD}}(n,2k) \cdot \frac{z^n}{n!} =  \left(\frac{1+z}{1-z}\right)^k$$
Question: Are these conjectures correct? Are these cycle generating functions studied somewhere?
 A: Conjecture 2 follows from setting $t=2k$ in the formula
  $$ \sum_{n\geq 0}\mathcal{C}_{\mathrm{ODD}}(n,t)\cdot\frac{z^n}{n!}
      =\left(\frac{\sqrt{1-z^2}}{1-z}\right)^t. $$
It then follows easily that
  $$ \sum_{j\geq 0}P_k(j)x^j = \frac{(2k-1)!!\,x(1+x)^{k-1}}{(1-x)^{k+1}}.\
  \qquad (1) $$
By Theorem 3.2 of http://math.mit.edu/~rstan/papers/cycles.pdf we see
that all the zeros of $P_k(x)$ are purely imaginary, which implies
that $P_k(x)$ has nonnegative coefficients and is either even or odd,
depending on the parity of $k$. Clearly also from (1) $P_k(0)=0$. I
don't see immediately from (1) why $P_k(x)$ has integer coefficients.
Let me also remark that the polynomial $(P_k(x)+P_k(x+1))/(2k-1)!!$ is the
Ehrhart polynomial of the standard $k$-dimensional
cross-polytope. See Exercise 4.61 of Enumerative Combinatorics,
vol. 1, second edition.
A: I'll start by addressing conjecture 2. By summing your generating fun over all values of $k$ we obtain
$$F(z,w)=\sum_{n\geq 0}\sum_{k\geq 0}C_{\text{ODD}}(n,2k)\frac{z^n}{n!}w^k=\sum_{k\geq 0}w^k\left(\frac{1+z}{1-z}\right)^k=\frac{1-z}{1-w-z-wz}$$
From here we see that
$$\sum_{n\geq 1}\sum_{k\geq 1}\frac{P_k(n)}{(2k-1)!!}w^k \frac{z^{n-1}}{(n-1)!}=\frac{1}{2}\frac{d}{dz}\int\left(F(z,w)-1\right)dw$$
$$=\int \frac{w}{(1-w)^2}\frac{1}{\left(1-z(\frac{1+w}{1-w})\right)^2}dw$$
By extracting the coefficients of $z^{n-1}/(n-1)!$ on both sides we have
$$\sum_{k\geq 0}\frac{P_k(n)}{(2k-1)!!}w^k=\int \frac{nw}{(1-w)^2}\left(\frac{1+w}{1-w}\right)^{n-1} dw=\frac{1}{2}\left(\frac{1+w}{1-w}\right)^n+\text{constant}$$
which is what we wanted. 

I just realized that we can also answer conjecture 1 by making use of this identity. Start with the expansion
$$\left(\frac{1+w}{1-w}\right)^n=\left(1+\frac{2w}{1-w}\right)^n=1+\sum_{r\geq 1} \binom{n}{r}\left(\frac{2w}{1-w}\right)^r$$
$$=1+\sum_{k\geq 1}w^k\sum_{r\geq 1}2^r\binom{n}{r}\binom{k-1}{r-1}$$
which gives us an explicit formula for $P_k$
$$P_k(n)=\sum_{r\geq 1}2^{r-1}(2k-1)!!\binom{n}{r}\binom{k-1}{r-1}$$
Which immediately tells us that $P_k(n)$, as a linear combination of $\binom{n}{r}$ for $1\le r\le k$, is a polynomial of degree $k$ with no constant term. Combined with the fact that $\left(\frac{1+w}{1-w}\right)^n=\left(\frac{1-w}{1+w}\right)^{-n}$ we have $P_k(n)=(-1)^kP_k(-n)$ which tells us that $P_k$ has the same parity as $k$. Now it remains to establish integrality of the coefficients. The explicit formula for $P_k$ can be rearranged as
$$P_k(n)=\sum_{r\geq 1}\binom{k+r-1}{r,r-1,k-r}\frac{(2k-1)!}{2^{k-r}(k+r-1)!}(n)_r$$
and from here it is clear that the coefficients of $P_k$ have nonnegative $p$-adic valuation for any odd prime $p$. It remains to show the following lemma
Lemma: For any $k\geq r\geq 1$ we have
$$\nu_2\left(\frac{(2k-1)!}{r!(r-1)!(k-r)!}\right)\geq k-r.$$
Proof We make use of the fact that $\nu_2\left(\frac{s!}{\lfloor\frac{s}{2}\rfloor!}\right)=\lfloor\frac{s}{2}\rfloor$. Our expression can be written as
$$\nu_2\left(\frac{(2k-1)!}{r!(r-1)!(k-r)!}\right)=\nu_2 \left(\frac{(2k-1)!}{(k-1)!}\frac{\lfloor\frac{r}{2}\rfloor!\lfloor\frac{r-1}{2}\rfloor!}{r!(r-1)!}\right)+\nu_2\left(\binom{k-1}{\lfloor\frac{r}{2}\rfloor,\lfloor\frac{r-1}{2}\rfloor,k-r}\right)$$
the first term is equal to $k-1-\lfloor\frac{r}{2}\rfloor-\lfloor\frac{r-1}{2}\rfloor=k-r$ and the second term is clearly nonnegative. This completes the proof of integrality.
