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 ${\tau}^n = id$ ?
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Ben Grossmann
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hkju
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If $n$ is prime, there should be $(n-1)!+1$ of the form $\tau^n=\text{id}$, is that correct? – user2345215 Dec 31 '14 at 00:06
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4The OEIS entry A074759 is relevant. It was found with the exponential generating function $$n! [z^n]\exp\left(\sum_{d|n} \frac{z^d}{d}\right).$$ – Marko Riedel Dec 31 '14 at 00:21
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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 '14 at 00:22
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1Differentiate the generating function to obtain a recurrence relation. – Marko Riedel Dec 31 '14 at 00:24
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What is the generating function for that? – hkju Dec 31 '14 at 00:40
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egf for the sequence $a_n = n![z^n]exp(...)$ – hkju Dec 31 '14 at 00:53
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@Marko Are you sure that differentiating is helpful here? The sum in the generating function depends on $n$ (and the divisors of $n$), so the desired value cannot (obviously) be expressed as $f^{(n)}(0)$ for some known $f$. – Milo Brandt Dec 31 '14 at 00:54
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I understand it. Thanks! – hkju Dec 31 '14 at 00:57
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This problem appeared by way of a question that I didn't understand properly at this MSE link. – Marko Riedel Dec 31 '14 at 01:23
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So the answer for the first question is $$n! [z^n]\exp\left(\sum_{d|(n-1)} \frac{z^d}{d}\right)?$$ – hkju Dec 31 '14 at 04:49