Ways to write $n=p^k$ as a product of integers Let's say that $F(n)$ is the number of ways to write $n$ as a product of integers greater than $1$. For example, $F(12)=4$ since $12=2\cdot 2 \cdot 3$, $12=2\cdot 6$, $12=3\cdot 4$ and $12=12$.
Given $n=p^k$ where $p$ is a prime number, what is the value of $F(n)$? I know how to manage this problem when $n=p_1\cdots p_k$ where $p_1,\ldots , p_k$ are different primes (the result is $B(k)$, where $B(n)$ is the Bell-Number), but how can I do it in this case?
Note: The order of the factors does not matter; that is, $a\cdot b$ and $b \cdot a$ do not count as different ways to write a number
 A: Clearly, any factor of $n$ will be of the form $p^m$ for some $m\le k$. The question thus is equivalent to asking how many ways are there to write $k$ as the sum of positive integers, where order doesn't matter. This is simply the number of partitions of $k$.
I'll admit that I'm not super familiar with partitions myself, so I'm not sure if there's a nice formula for this; I don't believe there is, though.
A: As already answered, the number is given by the number of partitions.
We add two more facts about it.
1 - By the fundamental theorem for finite abelian groups the number of abelian groups of order $n=p_1^{n_1}\dots p_k^{n_k}$ is the product of the partition numbers of $n_i$.
2- For the partition function $p(n)$ there are recursion formulas, asymptotic formulas and even an exact formula. The latter one is due to Rademacher (going back to work of Hardy and Ramanujan). We have
$$p(n) = \frac{1}{\pi \sqrt{2}} \sum_{k=1}^{\infty} \sqrt{k}\ A_k(n)\ F_k'(n),$$
with
$$A_k(n) = \sum_{0 \le m < k, \gcd(m, k) = 1} e^{i\pi\left(s(m, k) - 2nm/k\right)}$$
and
$$F_k(x) = \frac{1}{\sqrt{x - \frac{1}{24}}} \sinh\left(\frac{\pi}{k} \sqrt{\frac{2}{3}\left(x - \frac{1}{24}\right)}\right)$$
Here $s(m, k)$ is the Dedekind sum given by
$$s(m, k) = \sum_{n=1}^{k} \left(\left(\frac{n}{k}\right)\right)\left(\left(\frac{mn}{k}\right)\right)$$
where $((x))$ is the sawtooth function
$$((x)) = \begin{cases}
x - \lfloor x \rfloor - \frac{1}{2}, &\mbox{if } x \in \mathbb{R} \setminus \mathbb{Z}\\
0, &\mbox{if }x \in \mathbb{Z}
\end{cases}$$
