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Is there an asymptotic for: $$\sum_{n\leq x\atop n \not\equiv 0(mod\ 2)}d(n)\ \mathrm{\mathbf{and}}\ \sum_{n\leq x\atop n \not\equiv 0(mod\ 2,3)}d(n)$$

I need this for this https://mathoverflow.net/questions/286923/on-a-weak-version-of-the-grimms-conjecture question

Also, if there are any links, I would be only happy.

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  • $\begingroup$ Use $\sum_{n=1}^\infty d(n) \chi(n)n^{-s} = L(s,\chi)^2$ and the orthogonality $\frac{1}{\varphi(q)}\sum_{\chi \bmod q}\overline{ \chi(a) } \chi(n) = 1_{n \equiv a \bmod q}$. The same hyperbola method works for $\sum_{nm \le x} \chi(n) \chi(m)$. $\endgroup$
    – reuns
    Commented Nov 26, 2017 at 10:12
  • $\begingroup$ @reuns Is this for both expressions? $\endgroup$
    – user102007
    Commented Nov 26, 2017 at 11:29
  • $\begingroup$ Try and see ${}{}$ $\endgroup$
    – reuns
    Commented Nov 26, 2017 at 12:09

1 Answer 1

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For the first sum, you are counting pairs $x,y \in \mathbb{N}$ such that $x$ and $y$ are both odd, and $xy \leq X$. You can write this as a sum

$$\displaystyle \sum_{\substack{n \leq X \\ n \text{ odd}}} \sum_{\substack{m \leq X/n \\ m \text{ odd}}} 1.$$

The number of odd integers between 1 and a positive real number $Y$ is $\frac{Y}{2} + O(1)$. Therefore, the inner sum is equal to $\frac{X}{2n} + O(1)$. Now summing $\sum_{\substack{n \leq X \\ n \text{ odd}}}\left( \frac{X}{2n} + O(1)\right)$ gives the asymptotic

$$\displaystyle \sum_{\substack{n \leq X \\ n \text{ odd}}} \sum_{\substack{m \leq X/n \\ m \text{ odd}}} 1 \sim \frac{X \log X}{4}.$$

For the second case, we shall consider first the sum $\displaystyle \sum_{\substack{n \leq X \\ n \equiv 1 \pmod{6}}} d(n).$ Expanding, this can be written as

$$\displaystyle \sum_{\substack{n \leq X \\ n \equiv 1 \pmod{6}}} \sum_{\substack{m \leq X/n \\ m \equiv 1 \pmod{6}}} 1 + \sum_{\substack{n \leq X \\ n \equiv 5 \pmod{6}}} \sum_{\substack{m \leq X/n \\ m \equiv 5 \pmod{6}}} 1.$$

The evaluation of each sum is similar. In each case, the inner sum evaluates to $\frac{X}{6n} + O(1)$, and then the outer sum evaluates to $\sim \frac{X \log X}{36}$. Thus

$$\displaystyle \sum_{\substack{n \leq X \\ n \equiv 1 \pmod{6}}} d(n) \sim \frac{X \log X}{18}.$$

Similarly,

$$\displaystyle \sum_{\substack{n \leq X \\ n \equiv 5 \pmod{6}}} d(n) \sim \frac{X \log X}{18}.$$

Therefore the sum you ask for is asymptotic to $(X \log X)/9$.

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    $\begingroup$ @Retro I don't understand your question $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 26, 2017 at 11:00
  • $\begingroup$ Oh that is of course an error, I meant $X \log X$. My apologies $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 26, 2017 at 11:03
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    $\begingroup$ I don't understand your comment. 20 is not congruent to 1 mod 6 and is not congruent to 5 mod 6. If an integer $m \equiv 1 \pmod{6}$ is expressible as $m = ab$, then $a,b$ have to be both co-prime to 6 and $a \equiv b \pmod{6}$. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 26, 2017 at 11:26
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Stanley Yao Xiao
    Commented Nov 26, 2017 at 11:39

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