I recently discovered that the number of multiplicative partitions of some integer $n$ with $i$ prime factors is given by the Bell number $B_i$, provided that $n$ is a square-free integer. So, is there a method to finding the number of multiplicative partitions of a non-square-free integer?

I've searched for this online but it seems that I either can't word my search query well enough or what I'm looking for doesn't exist. Thanks in advance!

It seems my exact question was answered in How can we compute the multiplicative partition function. However, my mathematics education is at high school level and the explanation regarding symmetric groups confuses me greatly.

  • $\begingroup$ Computer algebra systems have this as a direct function call, so yes: there is an algorithm to find them. (reference.wolfram.com/language/ref/BellB.html) $\endgroup$ Aug 7, 2020 at 2:30
  • $\begingroup$ I have taken a look at the webpage but I am having trouble finding information regarding multiplicative partitions. I notice Bell number and Bell polynomial functions and I have having trouble finding a link to multiplicative partitions. Perhaps I'm looking in the wrong places? $\endgroup$
    – Scene
    Aug 7, 2020 at 2:39
  • $\begingroup$ This is Known as number of unordered factorisation of a positive integer rather than multiplicative partition. I have listed some known results regarding this function and this is an active area of research. $\endgroup$
    – ShBh
    Aug 9, 2020 at 15:34
  • $\begingroup$ @DavidG.Stork Are you sure? Mathematica's BellB function at most returns a univariate polynomial. $\endgroup$ Aug 11, 2020 at 1:07
  • $\begingroup$ I guess BellB is not as relevant as I thought. $\endgroup$ Aug 11, 2020 at 1:39

1 Answer 1


A General Discussion

In analytic number theory, the nature of an arithmetic function $f:\mathbb{N}\rightarrow\mathbb{R}$ is studied in many different ways, like

  • Studying the growth of the summatory function $F:\mathbb{R}_{\geq1}\rightarrow\mathbb{R}$ defined as follows $$F(x):=\sum_{n\leq x}f(n)$$
  • Studying frequency of values $f(n),n\in\mathbb{N}$ among the reals by considering the function $\mathcal{F}:\mathbb{R}_{\geq1}\rightarrow\mathbb{R}$ defined as follows$$\mathcal{F}(x):=|\{f(n):f(n)\leq x\}|$$
  • Finding functions $f_1:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(n)\sim f_1(n)$$ or $$f(n)=O(f_1(n))$$

Results regarding the number of unordered factorizations of an integer

It's an interesting question in analytic number theory to study the number of unordered factorizations of a positive integer. Many results regarding this are known and this is still an active field of research.

Let us define the function for formally stating some of the important results. For a positive integer, $n$ let $\psi(n)$ denote the number of tuples $(n_1,n_2,\ldots,n_k)$ such that $1<n_1\leq n_2\leq\cdots\leq n_k$ and $n=n_1n_2\cdots n_k$. This function was first studied comprehensively by A. Oppenheim. He proved that $$\sum_{n\leq x}\psi(n)\sim\frac{xe^{2\sqrt{\log(x)}}}{2\sqrt{\pi}(\log(x))^{3/4}}\tag{1}$$ For more details read the following paper

[Opp] A. Oppenheim, On an arithmetic function, J. London Math. Soc.1(1926), 205-211; part II in2(1927),123-130.

In some recent developments, the following function is studied,

$$\Psi(x):=|\{\psi(n):\psi(n)\leq x\}|$$ for $x\geq1$. R. Balasubramanian and Priyamvad Srivastav proved the following,

Theorem Let $C=2\pi\sqrt{2/3}$, then for sufficiently large $x$ we have $$\Psi(x)\leq\exp\left(C\sqrt{\frac{\log(x)}{\log(\log(x))}}\left(1+O\left(\frac{\log(\log(\log(x)))}{\log(\log(x))}\right)\right)\right)\tag{2}$$

For more details read the following paper

arXiv:1609.08602v1 [math.NT] 27 Sep 2016 ON THE NUMBER OF FACTORIZATIONS OF AN INTEGER

Some interesting special cases

For a prime $q$ and $n\geq1$, $\psi(q^n)$ is exactly equal to the number of unordered partitions of $n$ with positive integral parts. Therefore $$\psi(q^n)=p(n)$$ where $p(\cdot)$ is the well-known partition function.

For $n=p_1p_2\cdots p_r$ being a square-free positive integer with $r$ distinct prime factors, $\psi(n)$ is the same as the number of partitions of a set of $r$ distinct elements which is known as the $r^{th}$ Bell number and denoted as $B_r$. Therefore $$\psi(n)=B_r$$ for square-free $n$ with $r$ distinct prime factors.


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