3
$\begingroup$

I would like to estimate from above the following sum
$$ \sum_{1 \leq x_1 < X} .. . \sum_{1 \leq x_n < X} \frac{\prod_{1 \leq i \leq n } \phi(x_i)}{\mathrm{lcm}(x_1, .., x_n)^a}. $$ $\phi$ is the Euler totient function and $a$ is a positive integer less than $2n$. A trivial estimate would be $\ll X^{2n - a}$. Is there a way to get a better bound? Thank you!

$\endgroup$
1

1 Answer 1

1
$\begingroup$

One can improve on $X^{2n-a}$ as long as $(a,n)\neq(1,1)$ (for $a=n=1$, the sum grows like $X/\zeta(2)$ so there's no room for improvement). Let us introduce $$f_n(m) := \# \{ (x_1,\ldots,x_n) : \mathrm{lcm}(x_1,\ldots,x_n) = m\},$$ which satisfies $f_n(m) \le \tau(m)^n \ll_{n,\varepsilon} m^{\varepsilon}$ where $\tau$ is the usual divisor function. First suppose that $a >n$. We have $$\frac{\prod_{i=1}^{n} \phi(x_i)}{\mathrm{lcm}(x_1,\ldots,x_n)^a} \le \frac{\prod_{i=1}^{n} \phi(x_i)}{\max_{1\le i \le n} x_i^n} \frac{1}{\mathrm{lcm}(x_1,\ldots,x_n)} \le \frac{1}{\mathrm{lcm}(x_1,\ldots,x_n)}$$ which gives the upper bound $$< \sum_{1 \le m < X^{na}} \frac{f_n(m)}{m} = X^{o(1)}.$$ This is optimal since we have the lower bound $\ge 1$. We may assume $n \ge a$ from now on. Your sum is $$< X^n \sum_{1 \le m < X^{na}} \frac{f_n(m)}{m^a} \ll_{a,n} X^{n+o(1)}.$$ If $n\neq a$ this already beats $X^{2n-a}$. We now sketch how one can do better than $X^{n+o(1)}$. In section 3 of R. R. Hall's ``The distribution of squarefree numbers'' (Reine Angew. Math. 394 (1989), 107–117), the author introduces `total decomposition sets', which help him study a sum related to yours with $a=2$ and $n \ge 2$ (see his Lemma 3). Modifying the proof of Lemma 3 slightly, we obtain the bound $$\ll_{a,n} \begin{cases} X^{n-\frac{n(a-1)}{n-1}+o(1)} & \text{if }n > a,\\ X^{1+o(1)} & \text{if }n=a,\end{cases}$$ which beats $2n-a$ as long as $(a,n) \neq (1,1)$. The dependence on $a,n$ can be made explicit but is quite horrible. To see that $n=a$ and $a=1$ are optimal consider the contribution of $x_1=x_2=\ldots=x_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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