# A sum of divisor functions

Let $d(n)$ denote the number of positive divisors of $n$. Is it known how to evaluate the sum

$$\displaystyle \sum_{1 \leq m < n \leq X} d(m) d(n) d(n-m)?$$

A slightly more difficult question is if we change the height condition in the summation, to obtain the sum

$$\displaystyle \sum_{\substack{1 \leq mn(n-m) \leq X \\ 1 \leq m < n}} d(m) d(n) d(n-m).$$

This is a generalization of the single variable case, where it is known how to evaluate sums of the form

$$\displaystyle \sum_{1 \leq n \leq X} d(an + b) d(cn+d)$$

for fixed positive integers $a,b,c,d$.

• What do you mean by "evaluate"? – Will Sawin Jun 7 '18 at 22:37
• Opening up the last factor $d(n-m)$ yields $$\sum_{c\le X} \sum_{m\le X} d(m) \sum_{\substack{m<n\le X \\ n \equiv m \pmod c}} d(n),$$ which might be tractable.... – Greg Martin Jun 8 '18 at 0:48
• @WillSawin: I mean an asymptotic formula, or lower and upper bounds of the right order of magnitude – Stanley Yao Xiao Jun 8 '18 at 16:36