6
$\begingroup$

Let $c>1$, $c\not\in\mathbb{Z}$ and consider the sum $$ \sum_{n\leq x} \tau(\lfloor n^c \rfloor), $$ where $\tau(n)$ is the number of divisors of $n$. I'm almost certain I've seen an evaluation of this sum for an appropriate range of $c$, but do not know where. Does anyone happen to know where this sum has been studied? More general sums with $\tau$ replaced by a multiplicative function would also be of interest and seem like something that has been studied before. Any references are most appreciated.

Edit: Using a result of Jutila on exponential sums of the form $$ \sum_{N < n \leq 2N} \tau(n)e(f(n)), $$ I can show that $$ \sum_{n\leq x} \tau(\lfloor n^c \rfloor) = c x\log x + (2\gamma-c)x + O\left(x^{1-\frac{2c}{3}+\varepsilon}+x^{\frac{5c}{8}+\frac{1}{4}+\varepsilon}\right), $$ which gives an asymptotic formula for $c < \frac{6}{5}$. I'd love to know if this has been improved by other methods, which I suspect would be more intricate exponential sum estimates (perhaps in the spirit of Fouvry-Iwaniec's work on exponential sums with monomials).

$\endgroup$
1
  • $\begingroup$ For instance you may need to check this paper see theorem 2 and 3 $\endgroup$ Commented Oct 16, 2022 at 21:46

1 Answer 1

1
$\begingroup$

Since asking my question, I have stumbled upon the answer myself, so I post it here in case some future person finds this post.

It appears that the only paper that explicitly considers the problem above is the paper "On the number of divisors of $\lfloor n^c \rfloor$" of D. I. Tolev from 1990. The paper is 2 pages long and in Russian, and the MathSciNet review states that the result in the paper (which gives the appropriate asymptotic formula for $1 < c < \frac{12}{11}$) is merely sketched.

As indicated in the comments, this paper contains some improved multidimensional exponential sum estimates for monomials in the spirit of Fouvry-Iwaniec which are specifically applied to problems concerning Piatetski-Shapiro sequences. I fully expect that these yield an improvement, though I haven't worked through the details myself.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .