number of permutations whose cycles all have odd length I found the number of permutations whose cycles all have odd length by using exponential generating function and found that it's 
n![1-sum(C(i-1)/2^(2i-1))]
where the sum is 1<=i<=n/2
and C is catalan number.
Now i wonder if i can find this number directly without using generating function but combinatorical way.
 A: What follows is more of a comment.
By classifying the odd-cycles  only permutations according to the size
of the cycle that $n+1$ is on we get the recurrence
$$Q_{n+1} = \sum_{k=0}^{\lfloor n/2\rfloor}
{n\choose 2k}\times (2k)!\times Q_{n-2k}$$
where $Q_0=1.$ We get the factor $(2k)!$ because placement of $n+1$ on the cycle transforms it into a sequence.
However as we solve this recurrence which is not difficult we invariably encounter the generating function of the species $\mathfrak{P}(\mathfrak{C}_{\mathrm{odd}}(\mathcal{Z}))$
$$G(z) = \exp\left(\sum_{q\ge 1} \frac{z^{2q+1}}{2q+1}\right)
= \exp\left(\sum_{q\ge 1} \frac{z^{q}}{q}
- \sum_{q\ge 1} \frac{z^{2q}}{2q}\right)
\\ = \frac{1}{1-z} \left(\frac{1}{1-z^2}\right)^{-1/2}
= \frac{1}{1-z} \sqrt{1-z^2}$$
or one of its guises. The closed form then follows by inspection using the generating function of the Catalan numbers.
For example, we may solve the recurrence by considering the exponential generating function $\sum_{n\ge 0}\frac{Q_n}{n!} z^n$  and transforming it into a differential equation but this EGF is precisely $G(z)$ which we may not use. 
