Sum of $\frac{1}{(\delta_1,\delta_2)}$ with congruence restrictions

Question

In the course of my work, I encountered the following sum ($(x,y)$ stands for the GCD of $x$ and $y$):

$$L(Q)=\sum_{\substack{\delta_1,\delta_2\leq Q\\\delta_1\equiv0\ (a)\\\delta_2\equiv0\ (b)}}\frac{1}{(\delta_1,\delta_2)},$$

where $a$ and $b$ may get as large as $Q^{\frac{1}{2}}$. Heuristics suggest that it may be $O(\frac{Q^{2}}{ab})$, but I am worried this might be horribly wrong. In any case, what I would like the know is an asymptotic for this sum. Any help would be much appreciated.

Answer 1

Your sum is \begin{align*} L(Q)&=\sum _{d\leq Q}\frac {1}{d}\sum _{\delta _1,\delta _2\leq Q/d\atop {a|d\delta _1\atop {b|d\delta _2\atop {(\delta _1,\delta _2)=1}}}}1=\sum _{dh\leq Q}\frac {\mu (h)}{d}\sum _{\delta _1,\delta _2\leq Q/dh\atop {a|dh\delta _1\atop {b|dh\delta _2}}}1\\ &=\sum _{dh\leq Q}\frac {\mu (h)}{d}\sum _{\delta _1\leq Q/[a,dh]\atop {\delta _2\leq Q/[b,dh]}}1=Q^2\sum _{dh\leq Q}\frac {1}{d[a,dh][b,dh]}+\mathcal O\left (Q\sum _{dh\leq Q}\frac {\mu (h)}{d^2h}\right )\\ &=\frac {Q^2}{ab}\sum _{dh\leq Q}\frac {\mu (h)(a,dh)(b,dh)}{d^3h^2}+\mathcal O\left (Q\right )\\ &=\frac {Q^2}{ab}\sum _{dh=1}^\infty \frac {\mu (h)(a,dh)(b,dh)}{d^3h^2}+\mathcal O\left (Q\right )=:\frac {c_{a,b}Q^2}{ab} \end{align*}

Answer 2

Define $f(n)=\frac{1}{n}\prod_{p\mid n}(1-p)$ now $\sum_{d\mid n}f(d)=\frac{1}{n}$ thus $\sum_{d=1}^{\infty}f(d)[d\mid \delta_1][d\mid \delta_2]=\frac{1}{(\delta_1,\delta_2)}$ which means $\sum_{d=1}^{\infty}f(d)[d\mid \delta_1][d\mid \delta_2][a\mid \delta_1][b\mid \delta_2]=\frac{1}{(\delta_1,\delta_2)}[a\mid \delta_1][b\mid \delta_2]$ using inversion brackets and therefore we have $L(Q)=\sum_{\delta_2\leq \frac{Q}{b}}\sum_{\delta_1\leq \frac{Q}{a}}\sum_{d=1}^{\infty}f(d)[d\mid a\delta_1][d\mid b\delta_2]$ thus if we let $d_m=d/(m,d)$ for all $m\in\mathbb{N}$ then because $d\mid mk$ if and only if $d_m\mid k$ we must have that $\small L(Q)=\sum_{\delta_2\leq \frac{Q}{b}}\sum_{\delta_1\leq \frac{Q}{a}}\sum_{d=1}^{\infty}f(d)[d_a\mid \delta_1][d_b\mid \delta_2]=\sum_{d=1}^{\infty}f(d)\left(\sum_{\delta_1\leq \frac{Q}{a}}[d_a\mid \delta_1]\right)\left(\sum_{\delta_2\leq \frac{Q}{b}}[d_b\mid \delta_2]\right)$ which means $L(Q)=\sum_{d=1}^{\infty}f(d)\left(\sum_{\delta_1\leq \frac{Q}{d_a a}}1\right)\left(\sum_{\delta_2\leq \frac{Q}{d_bb}}1\right)=\sum_{d=1}^{\infty}f(d)\lfloor \frac{Q}{d_a a}\rfloor\lfloor \frac{Q}{d_a b}\rfloor$ or equivalently stated $L(Q)=\sum_{d=1}^{\infty}f(d)\left\lfloor\frac{(a,d)}{a}\frac{Q}{d}\right\rfloor \left\lfloor\frac{(b,d)}{b}\frac{Q}{d}\right\rfloor=\sum_{d\leq Q}f(d)\left\lfloor\frac{(a,d)}{a}\frac{Q}{d}\right\rfloor \left\lfloor\frac{(b,d)}{b}\frac{Q}{d}\right\rfloor$ which means if we let $\alpha(d)=\left\{\frac{(a,d)}{a}\frac{Q}{d}\right\}$ and $\beta(d)=\left\{\frac{(b,d)}{b}\frac{Q}{d}\right\}$ then $\alpha(d),\beta(d)\in [0,1]$ thus:

$$L(Q)=\sum_{d\leq Q}f(d)\left(\frac{(a,d)}{a}\frac{Q}{d}-\alpha(d)\right)\left(\frac{(b,d)}{b}\frac{Q}{d}-\beta(d)\right)\\=\sum_{d\leq Q}f(d)\frac{(a,d)(b,d)}{ab}\frac{Q^2}{d^2}-\sum_{d\leq Q}f(d)\frac{Q}{d}(\frac{(a,d)\beta(d)}{a}+\frac{(b,d)\alpha(d)}{b})+\sum_{d\leq Q}f(d)\alpha(d)\beta(d)\\\small=\frac{Q^2}{ab}\sum_{d=1}^{\infty}\frac{f(d)(a,d)(b,d)}{d^2}+\mathcal{O}\left(\frac{Q}{ab}(a+b)\log(Q)\right)=\frac{Q^2}{ab}\sum_{n=1}^{\infty}\frac{f(n)(a,n)(b,n)}{n^2}+\mathcal{O}\left(\frac{Q^{2}}{ab}\frac{\log(Q)}{\sqrt{Q}}\right)$$