I was looking through some old notes of mine and I came across a couple lemmas/identities I wrote down in regards to a question I asked about four years ago. In particular I wrote that for an arbitrary fixed integer $ k>1$ we have:

$ $ \Psi_k(n)=\sum_{\substack{\mathbf{u}\cdot\mathbf{v}=n\(\mathbf{u},\mathbf{v})\in \mathbb{N}^k\times \mathbb{N}^k}}1=\left|\{(\mathbf{u},\mathbf{v})\in \mathbb{N}^k\times \mathbb{N}^k:\mathbf{u}\cdot\mathbf{v}=n\}\right|\sim \frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}$ $

Where $ \sigma_{k-1}(n)=\sum_{d\mid n}d^{k-1}$ and $ \zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}$ is the Riemann zeta function. In addition we can write out $ \Psi_k(n)=\left|\{(u_1,v_1,u_2,v_2,\ldots u_k,v_k)\in \mathbb{N}^{2k}:n=u_1v_1+u_2v_2+\cdots +u_kv_k\}\right|$ so that we have an equivalent representation $ \Psi_k(n)=\sum_{\substack{r_1+r_2+\cdots +r_k=n\(r_1,r_2,\ldots r_k)\in \mathbb{N}^k}}d(r_1)d(r_2)\cdots d(r_k)$ where $ d(n)=\sum_{d\mid n}1=\sigma_0(n)$ which unearths an ordinary generating function for $ \Psi_k$ as follows:

$ $ \sum_{n=1}^{\infty}\Psi_k(n)x^n=\left(\sum_{n=1}^{\infty}x^{n^2}\frac{x^n+1}{x^n-1}\right)^k$ $

Now with this I got that:

$ $ \Psi_k(1)+\Psi_k(2)+\Psi_k(3)+\cdots +\Psi_k(N)=\frac{N^k\text{log}(N)^{k}}{k!}+\mathcal{O}(N^k\text{log}(N)^{k-1})$ $

So using the same heuristic as before, it seems reasonable one should get $ \Psi_k(n)\sim \frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}$ which for the special at $ k=2$ would coincide with the answer to my previous question. Though yet again I’m unable to find a concrete proof of this and would thus appreciate any help in finding one.