3
$\begingroup$

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 ${\tau}^n = id$ ?

$\endgroup$
10
  • $\begingroup$ If $n$ is prime, there should be $(n-1)!+1$ of the form $\tau^n=\text{id}$, is that correct? $\endgroup$ Dec 31, 2014 at 0:06
  • 4
    $\begingroup$ The OEIS entry A074759 is relevant. It was found with the exponential generating function $$n! [z^n]\exp\left(\sum_{d|n} \frac{z^d}{d}\right).$$ $\endgroup$ Dec 31, 2014 at 0:21
  • $\begingroup$ You are right. I mean, a general formula. I know the first few numbers from A008307(OEIS), where diagonal elements are those. $\endgroup$
    – hkju
    Dec 31, 2014 at 0:22
  • 1
    $\begingroup$ Differentiate the generating function to obtain a recurrence relation. $\endgroup$ Dec 31, 2014 at 0:24
  • $\begingroup$ What is the generating function for that? $\endgroup$
    – hkju
    Dec 31, 2014 at 0:40

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.