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I didn't find similar questions so decided to ask this one. Given positive integers $n$ and $d$ how can we efficiently estimate (or better calculate) cardinality of the set $~~ \{ \sigma^d ~~|~~ \sigma \in S_n \} ~~$? Here $S_n$ denotes the set of all permutations on the set $~~\{1, \dots, n\}$. Closed formula, recurrence relation or algorithmic approach are all wellcome.

Unfortunately i have no ideas how to approach this problem so any help will be highly appreciated. Thanks in advance.

Igor
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  • What is $\sigma$? – sinbadh Mar 16 '16 at 13:41
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    You mean $\sigma\in S_n$? – joriki Mar 16 '16 at 13:41
  • Yes, it was a typo – Igor Mar 16 '16 at 13:46
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    I think you can eventually get to an answer by looking at the prime factorization of $d$. – Jasper Mar 16 '16 at 13:47
  • The case $d=2$ is already not simple: http://math.stackexchange.com/questions/266569/how-to-find-the-root-of-permutation. – lhf Mar 16 '16 at 13:47
  • @lhf In that question a concrete permutation is checked for being a square. In my question for the case d = 2 i only wonder how many squares are there in S_n – Igor Mar 16 '16 at 13:48
  • I think this question is unrelated to that, @lhf. The question you are stating asks for the root of a specific $\sigma\in S_n$, whereas it's easy to find the cardinality of ${ \sigma^2 | \sigma\in S_n }$. Take some permutation $\sigma$. Then this is either even or odd. Then $\sigma ^2$ is even (or odd, this confuses me), so you count the number of even permutations, which is exactly the half of the set. Pardon me if I'm wrong, it'd been a long time I've done this the last time :) – Jasper Mar 16 '16 at 13:50
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    @Jasper: "All wood burns," states Sir Bedevere. "Therefore," he concludes, "all that burns is wood." – joriki Mar 16 '16 at 13:57
  • Indeed the question @lhf linked to is a slightly different question; but the exponential generating function for the case $d=2$ at Wikipedia does confirm that even this case is quite complicated. The corresponding OEIS entry doesn't give a closed form, so you're unlikely to find one. – joriki Mar 16 '16 at 14:01

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