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I am trying to find an approximation formula for the number of multiplicative partitions of $n$ with $k$ parts. I found that an approximation formula for the number of multiplicative partitions with no constraints with regard to $k$, i.e., the number of parts, was found initially by Oppenheim in "On an arithmetic function, J. London Math. Soc.1(1926), 205-211; part II in2(1927),123-130":

$$\sum_{n\leq x}\psi(n)\sim\frac{xe^{2\sqrt{\log(x)}}}{2\sqrt{\pi}(\log(x))^{3/4}}\tag{1}$$

More details can be found in this post:

Number of unordered factorizations of a non-square-free positive integer

Recent research on the problem is available here:

https://arxiv.org/abs/1907.07364

But, I think that no closed expression was found for calculation of the number of $k$-factorizations of positive integer $n$, only recursive formulas.

To further clarify, I am interested in function $f_k(n)$ as defined in https://arxiv.org/abs/1907.07364, which denotes the number of factorizations of $n$ with exactly $k$ parts $\geq 2$. For instance, the $3$-factorizations of $36$ are:

2 x 3 x 6

2 x 2 x 9

4 x 3 x 3

thus, $f_3(36) = 3$. $2$-factorizations of $36$, i.e., would be:

4 x 9

6 x 6

2 x 18

12 x 3

thus, $f_2(36) = 4$.

I'm interested in the approximations for summatory function $\sum_{n \leq x} f_k(n)$ as well. Note that this is somewhat different from Piltz divisor functions, e.g., http://oeis.org/A007425, which allow that a factor in factorization equals 1, and in the function defined above factors are greater than 1. Moreover, Piltz functions count ordered factorizations, and I am interested in unordered.

Any hints or suggestions are greatly appreciated.

dumpram
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    For $k=2$, the number of $2$-factorizations of $n$ is just the number of divisors of $n$, often denoted $d(n)$. What kind of "approximate formula" do you expect to find for $d(n)$? There's an exact formula, in terms of the prime factorization of $n$, but I don't think that's what you want. There are good asymptotics for $\sum_{n\le x}d(n)$, but that's the summatory function of $d(n)$, not $d(n)$ itself. So, what do you want? – Gerry Myerson Jun 08 '21 at 12:41
  • @GerryMyerson I addressed the comment in the question. – dumpram Jun 09 '21 at 07:11
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    OK, so, $f_2(n)$ is essentially $(d(n)-2)/2$ (a little different if $n$ is a square). You won't find a closed form formula for it except in terms of the prime factorization of $n$, and you will find good formulas for the summatory function in any good intro number theory text, or any of a hundred websites. Similar comments apply to other values of $k$. – Gerry Myerson Jun 09 '21 at 08:19
  • Does that help? – Gerry Myerson Jun 10 '21 at 13:11
  • @GerryMyerson I have looked in a lot of places, but I have not found anything so far. If you can please suggest a textbook, website or paper I would be grateful. I am not in the field so I have a hard time parsing the literature. Can you clarify what is "formula in terms of the prime factorizations of $n$"? Thank you in advance. – dumpram Jun 11 '21 at 06:53
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    If $n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$ where $p_1,p_2,\dots,p_r$ are distinct primes, then $d(n)=(a_1+1)(a_2+1)\cdots(a_r+1)$. https://oeis.org/A006218 will get you started on the summatory function for the divisor function. – Gerry Myerson Jun 11 '21 at 07:26
  • So, is that OK? – Gerry Myerson Jun 12 '21 at 12:39
  • Thank you for the reference. I think that the problem I described in the question is in line with the Piltz divisor functions, i.e., finding the k-th Piltz divisor function (e.g., https://oeis.org/A007425). However, as in Dirichlet problem 1 is allowed as a factor, and I am interested in factorization where parts are larger than 1. – dumpram Jun 14 '21 at 07:09
  • So, you count the number of factorizations using $1$, and subtract them. – Gerry Myerson Jun 14 '21 at 13:26

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