Is there a good upper bound for the number of ways $n$ can be expressed as the product of distinct positive integers?

Question

I've been pondering this problem as it relates to the convergence of an infinite sum, and I haven't been able to bound it at all. I've tried calculating small values and looking in the OEIS, but I've come up empty. I thought I had a way to bound it by $n$ (by removing the distinct condition and disallowing $1$ and using some properties of partitions) but it turns out I was mistaken.

Ideally, I'd be looking to prove that it is asymptotically less than any power $n^\epsilon$, but I'm not even sure if that's true. Thoughts?

Answer 1

I assume you mean this sequence. http://oeis.org/A045778

The $k$th Bell number is the number of ways to partition a set of $k$ elements. If $n$ is square-free and has $k$ prime factors, the number of ways to partition that set of prime factors is identical to the number of ways to write $n$ as a product of positive integers. If $n$ is not square-free and has $k$ prime factors (in total, counting repetitions), then $B_k$ gives an upper bound for $a_n$ since some partitions will yield identical products.

Hence, we can bound $a_n$ (the number of ways to factor $n$ into distinct factors greater than $1$) by $$ B_{\omega(n)} \le a_n \le B_{\Omega(n)} $$ where $B_k$ is the $k$th Bell number.

When $n$ is square-free, both bounds are achieved.

Using $\Omega(n)\le \frac{\log n}{\log 2}$, and an upper bound on the $k$-th Bell number, you can get an upper bound on $a_n$.

Calculations with primorials up to $\prod_{i=1}^{1000} p_i$ suggest that $a_n > n^{1/2}$ for primorials.

If we let $f(n)$ be the sought number of such factorizations of $n$, then we have $\log(f(5040))/\log(5040)=0.649...$, $\log(f(10080))/\log(10080) = 0.65678...$, and $\log(f(20160))/\log(20160) = 0.65877...$ (these are record-setters for this ratio).

Answer 2

For products of two distinct integers, you can use facts about the divisor function $$\tau(n) = \sum_{d|n} 1.$$ It is true that $\tau(n) \ll_\epsilon n^\epsilon$ for every $\epsilon > 0$, and in fact it is known that $$\tau(n) \leq n^{1.538 \log 2/\log \log n}, \ \ \ \ \ n \geq 3,$$ so $\tau(n) \ll n^{o(1)}$.