TITLE: Partial recurrence relation for the number of permutations in $S_n$ which have a square root.
QUESTION [6 upvotes]: I ran into this problem the other day.  The proof is supposed to be done by exhibiting an explicit bijection between two sets, without using induction, recurrence, or generating functions.

Denote by $\omega(n)$ the number of permutations $\sigma\in S_n$ so that $\sigma$ has a square root (that is, there exists $\tau\in S_n$ so that $\tau^2 = \sigma$).  Prove that $\omega(2n+1) = (2n+1)\omega(2n)$.

REPLY [2 votes]: The problem features as exercise 59 in chapter 3 of Miklos Bona's Combinatorics of permutations. 
The solution to the exercise is on page 337 of the book, which is not included in the Google preview but can be found here. The proof is due to Dennis White.