# Number of $n$-permutations for which ${\tau}^k = id$

I am curious about the formula(any closed form) for the number of $n$-permutations $\tau$ such that ${\tau}^{n-1} = id$. How about for the case ${\tau}^n = id$ ?

• If $n$ is prime, there should be $(n-1)!+1$ of the form $\tau^n=\text{id}$, is that correct? Dec 31, 2014 at 0:06
• The OEIS entry A074759 is relevant. It was found with the exponential generating function $$n! [z^n]\exp\left(\sum_{d|n} \frac{z^d}{d}\right).$$ Dec 31, 2014 at 0:21
• You are right. I mean, a general formula. I know the first few numbers from A008307(OEIS), where diagonal elements are those.
– hkju
Dec 31, 2014 at 0:22
• Differentiate the generating function to obtain a recurrence relation. Dec 31, 2014 at 0:24
• What is the generating function for that?
– hkju
Dec 31, 2014 at 0:40