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There is a bijective function defined as $f: A \rightarrow A$, and $n (A) =m$ and what's the number of bijective functions $a_ {m, n} $ which is become an identity function when composite n times? I think I heard it's related to Mobius function, Can anyone help me to solve this problem?

kyj519
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    Does $n(A)$ mean that $A$ has $n$ elements? – Stefan Mesken Mar 04 '16 at 02:54
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    A bijective function from a set to itself is a permutation. You ask how many permutations of an $m$ element set have order $n$ or some divisor of $n$. Consider the prime decomposition of $n$ and go from there. – JMoravitz Mar 04 '16 at 03:09
  • Note that the order of a permutation broken into cycles of length $k_1,k_2,\dots,k_j$ respectively where $k_1+k_2+\dots+k_j=m$ will have order $\text{lcm}(k_1,k_2,\dots,k_j)$. Note further that if picking which elements belong to a cycle of length $k_1$, and there is another set of cycles of the same length, you will need to account for symmetry. For example, the number of permutations of $[6]$ with three two-cycles is $\binom{6}{2,2,2}\frac{1}{3!} = \frac{6!}{2!2!2!3!}=15$. Range over all possible ways to partition $m$ where $\text{lcm}(k_1,\dots,k_j)\mid n$. – JMoravitz Mar 04 '16 at 03:17
  • @Stefan yes, $n(A)=m$ means A has m elements. – kyj519 Mar 04 '16 at 04:14
  • @JMoravitz Thank you for an answer, I have to Generalize your idea and get an equation written in m and n, or recurrence relation, can you give me more hints? – kyj519 Mar 04 '16 at 04:18
  • For $\text{lcm}(k_1,\dots,k_j)\mid n$ it would suffice to just guarantee that every $k_i\mid n$. One could describe the way I describe using partition notation, and a sum ranging over all partitions satisfying that every part divides $n$ where the contribution depending on the partition, you calculate as described here. It will be a messy notation, which is more cleanly written as you already did. I cant personally think of a closed form atm. – JMoravitz Mar 04 '16 at 04:30

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To expand on the comments I made above, I would think to try to write as:

$$\underbrace{\sum\limits_{\lambda\in \Lambda}}_{\text{sum over all partitions in}~\Lambda}\left(\underbrace{\frac{m!}{\prod_i (\lambda(i)!)}}_{\text{split into parts}}\underbrace{\prod_i(\lambda(i)-1)!}_{\text{arrange within parts}}\underbrace{\prod_j\frac{1}{|\{k~:~\lambda(k)=j\}|!}}_{\text{account for symmetry}}\right)$$

where $\Lambda = \{\lambda \vdash m~:~\forall i~\lambda(i)\mid n\}$ is the set of all partitions where the size of every part divides $n$.

This can be simplified a bit by combining the first two product symbols and cancelling, and by giving a name to the set used in the third product symbol.

Feel free to continue working on the problem searching for a more elegant way to write it using recurrence methods etc.

JMoravitz
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  • I'm little unfamiliar with math symbols, so can you explain me with little easier expression? Thank you very much for the answer. – kyj519 Mar 04 '16 at 06:29
  • @Yeon which symbols? $\lambda \vdash m$ means that $\lambda$ is a partition of $m$, $\lambda(i)\mid n$ means that $\lambda(i)$ divides evenly into $n$, $\prod$ and $\sum$ are for repeated multiplication or repeated addition of what is inside it ranging over an indexing variable within an indexing set, $!$ denotes the factorial,... – JMoravitz Mar 04 '16 at 06:32
  • I got it, Thanks, I think recurrence equation is $a_{m,n}=\sum\limits_{k|n}a_{m-k,n}P(n-1,k-1)$. thanks again for the great answer. – kyj519 Mar 04 '16 at 16:57
  • @YeonJoonKim It might be worth checking your formula against some easily known results. For example, $a_{4,4!} = 4!$ (by Lagrange: every element of a group has order dividing the order of the group). – JMoravitz Mar 04 '16 at 20:18
  • I test the fomula with computer, using brute-force argorithm, I think it works. Thanks. – kyj519 Mar 06 '16 at 03:20
  • The equation in the previous comment is wrong, not $P (n-1, k-1) $ it should be $P (m-1, k-1) $ – kyj519 Mar 08 '16 at 04:13