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I am trying to see how I can find the number of permutations pi - lets say $a(n)$ - of $n$ objects with pi^3=id or pi^4=id. for example - $a(4)=24$, $a(5)=76$, $a(6)=336$. Is it something that can be found? I was trying to use generating functions to define $\sum_{n\ge 0} a(n) x^n$ and find this function. Can someone please guide me.

SHAR HASSIN
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  • It can be done. I would probably use recurrences rather than generating functions, but that is a matter of habit. – André Nicolas Oct 09 '15 at 21:22
  • If you have a three-cycle and apply it four times you will definitely not get the identity so you need to clarify the question. The generating function here is $$G(z) =\exp\left(\sum_{d|k} \frac{z^d}{d}\right)\quad \text{and the species is}\quad \mathfrak{P}\left(\sum_{d|k} \mathfrak{C}_{=d}(\mathcal{Z})\right).$$ Differentiate to obtain a recurrence. – Marko Riedel Oct 09 '15 at 21:48
  • a(n) is simply all the permuations on n objects with order 2 or 3 or 4 - so i dont see how there is a problem here. – SHAR HASSIN Oct 09 '15 at 21:54
  • I see you have an "or" where I thought you had an "and." Since we have $(3,4)=1$ you can just add the two generating functions as Andre Nicolas pointed out, taking into account the identity permutation which is the only shared permutation. – Marko Riedel Oct 09 '15 at 21:59

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We count the permutations of order that divides $3$, count the permutations of order that divides $4$, add, and subtract $1$ for the identity permutation that has been counted twice.

Permutations whose order divides $3$: Let $a_n$ be the number of such permutations of the $n$-element set $1,2,\dots,n$. Now look at $a_{n+1}$. Maybe $n+1$ is sent to itself. There are $a_n$ such permutations of order that divides $3$. Maybe $n+1$ is part of a $3$-cycle. Then there are $n$ choices for $\sigma(n+1)$, and then $n-1$ choices for $\sigma(\sigma(n))$. That yields the recurrence $$a_{n+1}=a_n+n(n-1)a_{n-2}.$$ Let $b_n$ be the number of permutations whose order divides $4$. One uses the same idea to get a recurrence for $b_n$. Either $n+1$ is sent to itself, or it is part of a $2$-cycle, or it is part of a $4$-cycle.

André Nicolas
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  • thanks Andre. i know this method (same thing as for order 2) but is there a recurrence for both 3 or 4? – SHAR HASSIN Oct 09 '15 at 22:02
  • I think it will be awkward to do it simultaneously, since then we have to avoid orders $6$ and $12$. – André Nicolas Oct 09 '15 at 22:05
  • There is a derivation of the recurrence at this MSE link. – Marko Riedel Oct 09 '15 at 22:10
  • thanks. do you know a way to find this recurrence for permutations of order that divides some odd k (instead of 3 or 4)? – SHAR HASSIN Oct 09 '15 at 22:11
  • You are welcome. Suspect there may be an explicit formula, perhaps involving Stirling numbers. But I have not written down anything, and don't remember seeing the problem. – André Nicolas Oct 09 '15 at 22:17
  • I have added some Maple code to the MSE link that should make it easy to write down the recurrence for fixed order $k.$ For $k=24$ we get $$1, 2, 6, 24, 96, 576, 3312, 26496, 198144, 1691136,\ldots$$ which is OEIS A053504 where no explicit formula other than the EGF is given. – Marko Riedel Oct 09 '15 at 22:48
  • thanks Marko that is very helpful. as i can see - the number of permutation on n objects of order that divides some r is b(n)=C(r)(n+1). if i want a power function generator of sum(b(n)x^n) - can i find clear formula of this function? – SHAR HASSIN Oct 09 '15 at 23:01