MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a fixed natural number $n$ is it possible to obtain an asymptotic for the number of solutions $(a,x,b,y)\in\mathbb{N^4}$ to $ax+by=n$ or equivalently an asymptotic for $\sum_{k=1}^{n-1}d(k)d(n-k)$?

Using some heuristics I got that possibly:

$$\sum_{k=1}^{n-1}d(k)d(n-k)\sim \frac{3}{\pi^2}\sigma(n)\ln(n)^2$$

I wrote:

$$\sum\limits_{k=1}^{n-1}d(k)d(n-k)=\sum\limits_{{(a,x,b,y)\in \mathbb{N^4}}\atop{ax+by=n} }1=\sum\limits_{d\mid n}\sum\limits_{{(a,x,b,y)\in \mathbb{N^4}}\atop{(a,b)=d,ax+by=n}}1 =\sum\limits_{d\mid n}\sum_{{(a,x,b,y)\in \mathbb{N^4}}\atop(a,b)=1,{ax+by=\frac{n}{d}}}1$$ $$=\sum\limits_{d\mid n}\sum\limits_{{(a,b)\in \mathbb{N^2}}\atop{(a,b)=1}}\sum\limits_{{(x,y)\in \mathbb{N^2}}\atop{ax+by=\frac{n}{d}}}1$$

Now I considered a fixed natural number $m$ and looked at the solutions $(x,y)$ to $ax+by=m$

Then noted that $x\leq \frac{m-b}{a}$ and that the solutions $x$ would be of the form: $x\equiv c+a,c+2b,c+3b,c+4b,...c+kb$, so I wrote $c+kb\leq\frac{m-b}{a}\implies k\leq \frac{m-b}{ab}-\frac{c}{b}$

So the largest such $k$ that should satisfy this is $\lfloor{\frac{m-b}{ab}-\frac{c}{a}}\rfloor\approx \lfloor{\frac{m}{ab}}\rfloor$

Then went back and wrote:

$$\sum\limits_{k=1}^{n-1}d(k)d(n-k)\approx \sum\limits_{d\mid n}\sum\limits_{{(a,b)\in \mathbb{N^2}}\atop(a,b)=1}\lfloor{\frac{n/d}{ab}}\rfloor\approx \sum\limits_{d\mid n}\frac{n}{d}\sum\limits_{{ab\leq n,(a,b)=1}\atop(a,b)\in \mathbb{N^2}}\frac{1}{ab}$$ $$=\sigma(n)\sum\limits_{ab\leq n\atop(a,b)=1}\frac{1}{ab}=\sigma(n)\sum\limits_{k\leq \sqrt{n}}\frac{\mu(k)}{k^2}\sum\limits_{ab\leq \frac{n}{k^2}}\frac{1}{ab}=\sigma(n)\sum\limits_{k\leq \sqrt{n}}\frac{\mu(k)}{k^2}\sum\limits_{j\leq \frac{n}{k^2}}\frac{d(j)}{j}$$ $$\approx\sigma(n)\sum\limits_{k\leq \sqrt{n}}\frac{\mu(k)}{k^2}\frac{\ln(n/k^2)^2}{2}\sim\frac{\sigma(n)\ln(n)^2}{2}\sum_{k\ge 1}\frac{\mu(k)}{k^2}=\frac{3}{\pi^2}\sigma(n)\ln(n)^2$$

But this doesn't seem right either because I know that,

$$\sum_{ax+by\leq m\atop(a,x,b,y)\in \mathbb{N^4}}1=\frac{m^2\ln(m)^2}{2}+(2\gamma-\frac{3}{2})m^2\ln(m)+(2\gamma^2-\frac{5}{2}\gamma-\frac{\pi^2}{12}+\frac{3}{2})m^2+O(m^{1+\theta}\ln(m))$$

Where $\theta$ is the same $\theta$ in Dirichlet's divisor problem.

Thus I would expect that if $\sum\limits_{k=1}^{n-1}d(k)d(n-k)\sim C'\sigma(n)\ln(n)^2$ for some constant $C'$ that we would then have that $C'=\frac{6}{\pi^2}$, which it isn't according to the last heuristic.

So I'm not really sure how to proceed, and would really appreciate any help in finding an asymptotic expansion for $\sum\limits_{k=1}^{n-1}d(k)d(n-k)$.

share|cite|improve this question
I think the problem in your initial computation is that $ab$ can be as large as $n^2$, and your approximation $\lfloor \frac{m}{ab} \rfloor \approx \frac{m}{ab}$ is terrible in that regime. But note that one of $ab$ or $xy$ has to be bounded by $n$, so by paying a factor of two one can basically restrict to the regime $ab \leq n$ where your heuristic argument is reasonably accurate, explaining the loss of 2 in your initial prediction. – Terry Tao Feb 7 '15 at 21:00
up vote 13 down vote accepted

This is a binary additive divisor problem (also known as the shifted convolution problem) and related questions are to understand asymptotics for $\sum_{n\le x} d(n) d(n+k)$, or $\sum_{n\le x} \lambda_f(n) \lambda_f(n+k)$ where $\lambda_f$ denotes the Fourier coefficients of a cusp form. For your particular problem, Ingham first showed the asymptotic formula $$ \sum_{k=1}^{n-1} d(k) d(n-k) \sim \frac{6}{\pi^2} \sigma(n) (\log n)^2. $$ See for example Motohashi's paper which discusses this asymptotic formula, with strong error terms.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.