The number of representations of an integer as the inner product of integral lattice points

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 the following asymptotic expansion:

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

With $\sigma_{k-1}(n)=\sum_{d\mid n}d^{k-1}$ and $\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}$ the Riemann zeta function. Now alternatively if one expands the inner product of $\mathbf{u}\cdot\mathbf{v}$ we see $\Psi_k$ can be expressed in a number of other ways: $$\Psi_k(n)=\sum_{\substack{\mathbf{u}\cdot\mathbf{v}=n\\(\mathbf{u},\mathbf{v})\in \mathbb{N}^k\times \mathbb{N}^k}}1=\left|\{(u_1,v_1,\ldots u_k,v_k)\in \mathbb{N}^{2k}:n=u_1v_1+\cdots +u_kv_k\}\right|\\=\sum_{\substack{m_1+m_2+m_3+\cdots +m_k=n\\(m_1,m_2,m_3,\ldots ,m_k)\in \mathbb{N}^k}}d(m_1)d(m_2)d(m_3)\cdots d(m_k)$$

Where $d(m)=\sum_{d\mid m}1=\sigma_{0}(m)$ counts the divisors of any natural number $m$. Which unearths a number of $q$-series esque representation for the ordinary generating function of $\Psi_k$, for example:

$$\sum_{n=1}^{\infty}\Psi_k(n)q^n=\left(\sum_{n=1}^{\infty}q^{n^2}\frac{1+q^n}{1-q^n}\right)^k=\frac{\text{log}(1-q)^k}{\text{log}(q)^k}\sum_{j=0}^k\binom{k}{j}\left(\frac{\psi_{q}(1)}{\text{log}(1-q)}\right)^j$$

Where $\psi_{q}(z)=\frac{1}{\Gamma_{q}(z)}\frac{d}{dz}\Gamma_{q}(z)$ is the $q$-analog of the digamma function defined analogously in terms of the $q$-gamma function, expressible as $\Gamma_{q}(z)=(1-q)^{1-z}\prod_{n=0}^{\infty}\frac{1-q^{n+1}}{1-q^{n+z}}$ for $|q|<1$.

Now working with just some of these alternate representations, as well as fiddling with the order of the summands involved I was able to prove by induction on the integer $k>1$ that we have both:

$$\sum_{n\leq N}\Psi_k(n)=\frac{N^k\text{log}(N)^{k}}{k!}+\mathcal{O}(N^k\text{log}(N)^{k-1})$$ $$\sum_{n\leq N}\frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}=\frac{N^k\text{log}(N)^{k}}{k!}+\mathcal{O}(N^k\text{log}(N)^{k-1})\\$$

So using the same heuristics as before it seems reasonable that $\Psi_k(n)\sim \frac{\sigma_{k-1}(n)\text{log}(n)^{k}}{\zeta(k)(k-1)!}$ which at least for the case at $k=2$ would coincide with the answer to my previous question. However I'm unable to find a concrete proof of this result and would therefore appreciate any help in the matter.

• @Peter Heinig: it obviously does, since in dimension greater than two there are infinitely many possibilities if $0$ is allowed. – Igor Rivin Jan 5 '18 at 17:11
• These formulas remind me a bit of the expansion of Eisenstein series (except of the log) maybe there is some modification of your generating function that arrives at a modular form. – Rodrigo Jan 5 '18 at 19:36

The asymptotic formula is true for even dimensions $k\geq 2$. We can prove this by induction on $k$, inspired by Rodrigo's observation on Eisenstein series.
The case $k=2$ is classical and addressed in the OP's previous post that he linked. Now it suffices to show that if the formula is true for even dimensions $k,\ell\geq 2$, then it is also true for dimension $k+\ell$. Moreover, we can assume in this implication that either $k,\ell\geq 4$ or $k+\ell\in\{4,6\}$. Let us start from the obvious identity $$\Psi_{k+\ell}(n)=\sum_{r+s=n}\Psi_k(r)\Psi_\ell(s).$$ Using the fact that the logarithm is a slowly changing function, we are left with proving that $$\frac{\sigma_{k+\ell-1}(n)}{\zeta(k+\ell)\Gamma(k+\ell)}\sim\sum_{r+s=n}\frac{\sigma_{k-1}(r)}{\zeta(k)\Gamma(k)}\cdot\frac{\sigma_{\ell-1}(r)}{\zeta(\ell)\Gamma(\ell)}.$$ If $k,\ell\geq 4$, then the left hand side is $(2\pi i)^{-k-\ell}$ times the $n$-th Fourier coefficient of the standard holomorphic Eisenstein series $E_{k+\ell}$ of weight $k+\ell$ and full level, while the right hand side is $(2\pi i)^{-k-\ell}$ times the $n$-th Fourier coefficient of $E_kE_\ell$. As $E_{k+\ell}-E_kE_\ell$ is a cusp form, the result follows from standard (or even weaker) upper bounds for the Fourier coefficients of cusp forms.
If $k+\ell\in\{4,6\}$, then the result follows from the explicit identities $$5\sigma_3(n)+(1-6n)\sigma_1(n)=12\sum_{r+s=n}\sigma_1(r)\sigma_1(s),$$ $$21\sigma_5(n)+(10-30n)\sigma_3(n)-\sigma_1(n)=240\sum_{r+s=n}\sigma_1(r)\sigma_3(s).$$ These identities can also be proved with the help of Eisenstein series, or by elementary means, see e.g. (3.10) and (3.12) in the paper of Huard et al.
• @Ethan: I don't know about odd $k$. It is possible that some variant of the Eisenstein relations will do. Alternatively, a variant of the treatment of the binary additive divisor sum (e.g. by the circle method) will cover the general case of the reduced asymptotic for the additive convolution of $\sigma_{k-1}$ and $\sigma_{\ell-1}$. – GH from MO Jan 6 '18 at 0:28
• I really appreciate your time and think it was pretty clever how you worked in Eisenstein series at even integral weights to calculate the lower index terms in the recursion $\Psi_{l+j}(n)=\sum_{k=1}^{n-1}\Psi_l(n-k)\Psi_j(k)$ during the induction hypothesis. I upvoted your question and if no one can help with the odd index cases for the next day or two I will accept your answer. – Ethan Jan 6 '18 at 0:53
• @Ethan: Thank you. I hope though that someone can fill in the missing details for odd $k$. Naturally, one only needs to establish my second display for $l=2$ (and $k\geq 2$ arbitrary). – GH from MO Jan 6 '18 at 0:59