(I posted this on MathOverflow, but I haven't gotten a response yet, so I've reposted it here.)
Let $x\in\left(0,1\right)$. For each integer $n\geq2$, let $\Omega\left(n\right)$ denote the number of prime factors of $n$, counted according to multiplicities; thus $\Omega\left(2\right)=1$, $\Omega\left(4\right)=2$, etc. I am looking for asymptotics for the quantity:
$$\left|\sum_{k=1}^{\Omega\left(n\right)}\left(-\frac{1}{x!}\right)^{k}\sum_{\begin{array}{c} d_{1}\cdots d_{k}=n\\ d_{1},\ldots,d_{k}\geq2 \end{array}}\prod_{j=1}^{k}\frac{1}{\left(d_{j}+1\right)^{1-x}}\right|$$ for large $n$. The central sum is taken over all factorizations of $n$ into $k$ possibly non-distinct integers, all of which are greater than or equal to $2$. This takes me into the realm of multiplicative partitions, but, unfortunately, the literature on them seems to be much less rich than that for their additive counterparts. The best (though, admittedly, wasteful) estimate I've been able to obtain is: $$\left|\sum_{k=1}^{\Omega\left(n\right)}\left(-\frac{1}{x!}\right)^{k}\sum_{\begin{array}{c} d_{1}\cdots d_{k}=n\\ d_{1},\ldots,d_{k}\geq2 \end{array}}\prod_{j=1}^{k}\frac{1}{\left(d_{j}+1\right)^{1-x}}\right|\ll\frac{n^{1-o\left(1\right)}}{x!\left(M\left(n\right)+1\right)^{1-x}-1}\textrm{ as }n\rightarrow\infty$$ where $M\left(n\right)$ is the smallest prime number that divides $n$. This is not very satisfactory, seeing as the upper bound doesn't even decrease to $0$ as we let $n$ tend to $\infty$ along the primes ($n=2,3,5,7,11,\ldots)$; the quantity being bounded tends to $0$ in this case (and, in general, seems to be decreasing, with the decrease getting more intense the more prime factors $n$ has), even though the upper bound diverges to $\infty$.