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The problem is:

For $f:X\rightarrow X$, where $X$ has $n$ distinct elements, how many different $f$ exist that meet $f(f(f(x)))=x$ for all $x\in X$?

I can calculate for small number of $n$'s by the following way (for example, $n=8$):

case-1) $f$ is an identity function ($f(x)=x$) for all $x$.

case-2) $f$ is an identity function for 5 elements in $X$ and for the rest of 3, it is cyclic, i.e. $f(a)=b,f(b)=c,f(c)=a$.

case-3) $f$ is an identity function for 2 elements in $X$ and for the rest of 6, it can be divided into two groups of each having 3, and each groups are cyclic.

As there are only two different ways of being cyclic (1,2,3 or 1,3,2), the number of functions for $n=8$ can be calculated as:

$$1+\binom85\cdot2+\binom82\binom63\frac12\cdot2^2=1233$$

I could also generalize it for any $n$ as:

$$n=3m+p\quad (p=0,1,2)$$ $$a_n=\sum_{k=0}^m\left(\binom{n}{n-3k}\cdot2^k\cdot\frac1{k!}\prod_{i=0}^{k-1}\binom{n-3k}{3}\right)$$

And then I found that $a_n$ are $1,1,3,9,21,81,351,1233,\dots$ that is the sequence in this link https://oeis.org/A001470/internal. In the link, it shows a simple recurrence equation of

$$a_n=a_{n-1}+(n-1)(n-2)a_{n-3}$$

But I can't figure out how one can reach to that recurrence equation.

So my question is "how one can reach to the above recurrence equation"?

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    For continuous complex functions, consider $f(x)=e^{2k\pi i/3}x$ – 2017-01-08
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    @SimpleArt From the question body, this is quite clearly about permutations of finite sets. – 2017-01-08
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    Obviously $f$ is one-to-one. I think you're looking for permutations of $\{1,...,n\}$ that are products of disjoint $3$-cycles. – 2017-01-08
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    @LeGrandDODOM This is all explained in the question, if you just read it. – 2017-01-08
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    @LeGrandDODOM: I believe OP understands this and is just looking to justify the recurrence s/he found. – 2017-01-08
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    I think the question caption needs editing, otherwise it is confusing. – 2017-01-09
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    @Anixx, I agree. If someone has a better suggestion, please edit it. – 2017-01-09

1 Answers 1

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To understand the recurrence, note that a function on a set of $n$ members can either fix the last member or not. If the last member is fixed, you have a function on the first $n-1$ members. If the last member is not fixed, you have $n-1$ choices of the next element in its cycle, $n-2$ choices for the third element in its cycle, and a function on the remaining $n-3$ elements.

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    Knowing what the answer is makes it a lot easier to come up with this kind of reasoning. – 2017-01-08
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    @martycohen: yes, it does. That is what makes the question so good. OP gave it a lot of thought, found the OEIS entry so we had data to work with, and asked a clear question. It seemed like there ought to be a bijection approach to the answer. The $n-3$ term was a strong clue. – 2017-01-08
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    @RossMillikan Thanks a lot. I fixed the equation. I am still trying to understand what you wrote. – 2017-01-08
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    @KayK.: I would leave what you had as it is an accurate quote from OEIS. When $n$ appears in the recurrence a shift in the base of the list changes the recurrence. You could add in the new version as a PS edit if you want. As an example, you found there were three functions for $n=3$, which is correct. Now for $n=4$ we can either have $f(4)=4$ in which case you just have one of the previous functions on the first three elements, or you have $f(4)=$ something else. There are $3$ choices for $f(4)$ and $2$ choices for $f(that)$, then you have to finish the cycle. That leaves just one – 2017-01-09
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    element over. You need a function on that one element, and there is only one of those. – 2017-01-09
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    Hi @RossMillikan, thanks. That's what I understood so far. The only thing that is not clear to me is, (let's assume $n=5$ instead), I am not sure why it can't be $a_5 = a_4 + 4\cdot3\cdot a_3$, instead of $a_5 = a_4 + 4\cdot3\cdot a_2$.. (or, why it can't be $a_4 = a_3 + 3\cdot2\cdot a_2$, instead of $a_4 = a_3 + 3\cdot2\cdot a_1$) – 2017-01-09
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    Assuming $n=5$ you either have $f(5)=5$, in which case you just extend some function from $n=4$ by adding in that case, which gives you the $a_4$ term, or you need $5$ to be in a three cycle. There are four choices for $f(5)$ because we can't have $f(5)=5$ and three choices for $f(f(5))$ because that can't be $f(5)$ or $5$. That accounts for three of the elements, so you need a function on the other two that you are extending by this three cycle. That is why it is $4 \cdot 3 \cdot a_2$ – 2017-01-09
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    @RossMillikan I completely understood now. I appreciate your kind explanation. – 2017-01-09