Permutations with odd-length cycles I need to find - as a homework problem - the exponential generating function for the number of permutations of $n$ consisting of an even number of odd-length cycles. I can retrieve the exponential generating function for the number of permutations with just odd-length cycles, by
$$
E(z)=\exp(\sum_{k\geq 0}\frac{z^{2k+1}}{2k+1})=\exp(\sum_{k\geq 1}\frac{z^{k}}{k})/\exp(\sum_{k\geq 1}\frac{z^{2k}}{2k})=\frac{1}{1-z}(\sqrt{1-z^2})=\sqrt{\frac{1+z}{1-z}}.
$$
Since $n$ has permutations of an even number of odd-length cycles if and only if it is even, in which case all permutations with odd-length cycles have an even number of cycles, I believe that I have to find the power series obtained by $E(z)$ by just selecting the terms in $z^n$ with $n$ even. Is this correct? In any case, I don't know how to proceed.

There are  two possible interpretations here,  the first, permutations
consisting of  an even number of  odd cycles and some  even cycles and
second, permutations consisting of an even number of odd cycles only.
First interpretation.
Observe that  the generating function of permutations  with odd cycles
marked is
$$G(z, u) =
\exp\left(\sum_{k\ge 1} \frac{z^{2k}}{2k}
+ u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$
This is
$$G(z, u) =
\exp\left((1-u)\sum_{k\ge 1} \frac{z^{2k}}{2k}
+ u \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
To get the permutations with an even number of odd cycles use
$$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$
which yields
$$\frac{1}{2}\exp\left(\sum_{k\ge 1}\frac{z^{k}}{k}\right)
+ \frac{1}{2} \exp\left(2\sum_{k\ge 1} \frac{z^{2k}}{2k}
- \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
This simplifies to
$$\frac{1}{2} \frac{1}{1-z}
+ \frac{1}{2} (1-z) \frac{1}{1-z^2}$$
which is
$$\frac{1}{2} \frac{1}{1-z}
+ \frac{1}{2} \frac{1}{1+z}.$$

This simply  says that  when $n$ is  even then  there must be  an even
number  of odd cycles  and when  $n$ is  odd there  cannot be  an even
number of odd cycles, which follows by inspection (parity).
Second interpretation.
Here we have
$$G(z, u) =
\exp\left(u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$
This is
$$G(z, u) =
\exp\left(u \sum_{k\ge 0}\frac{z^{k}}{k}
- u \sum_{k\ge 1}\frac{z^{2k}}{2k}\right).$$
To get the permutations with an even number of odd cycles use
$$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$
which yields
$$\frac{1}{2}\frac{1}{1-z}
\left(\frac{1}{1-z^2}\right)^{-1/2}
+ \frac{1}{2} (1-z)
\left(\frac{1}{1-z^2}\right)^{1/2}.$$
This gives the sequence
$$0, 1, 0, 9, 0, 225, 0, 11025, 0, 893025, 0, 
108056025, 0, 18261468225,\ldots$$
which points us to
OEIS A177145, 
where we find this computation confirmed.

This generating function maybe written as
$$\frac{1}{2}\frac{1}{1-z}
\sqrt{1-z^2}
+ \frac{1}{2} (1-z)
\frac{1}{\sqrt{1-z^2}}$$
or
$$\frac{1}{2}\sqrt{\frac{1+z}{1-z}}
+ \frac{1}{2}\sqrt{\frac{1-z}{1+z}}.$$
