1 Introduction
As usual, denote by
$[t]$
the integral part of the real number t. Recently, Heyman [Reference Heyman2] quantified the cardinality of the set
$$ \begin{align*} \mathcal{S}(x) := \bigg\{\bigg[\frac{x}{n}\bigg] : 1\leqslant n\leqslant x\bigg\}. \end{align*} $$
By elementary arguments, he proved in [Reference Heyman2, Theorems 1 and 2] that
$$ \begin{align} S(x) := \lvert \mathcal{S}(x)\rvert = [\sqrt{4x+1}]-1 = 2\sqrt{x} + O(1) \end{align} $$
for
$x\to \infty $
. Subsequently [Reference Heyman3, Theorem 1], he also investigated the number of primes in the set
$\mathcal {S}(x)$
, showing that as
$x\rightarrow \infty $
,
$$ \begin{align} \pi_{\mathcal{S}}(x) := \bigg\lvert \bigg\{\bigg[\frac{x}{n}\bigg] : 1\leqslant n\leqslant x \;\,\text{and}\; \bigg[\frac{x}{n}\bigg] \; \text{is prime}\bigg\}\bigg\rvert \; = \frac{2\sqrt{x}}{\log\sqrt{x}} + O\bigg(\frac{\sqrt{x}}{(\log x)^{2}}\bigg). \end{align} $$
This can be considered an analogue of the prime number theorem for the set
$\mathcal {S}(x)$
. Very recently, Ma and Wu [Reference Ma and Wu4] sharpened this result by proving the strong form of the prime number theorem for
$\mathcal {S}(x)$
:
$$ \begin{align} \pi_{\mathcal{S}}(x) = {\textrm{ Li}}_{\mathcal{S}}(x) + O (\!\sqrt{x}\,e^{-c(\log x)^{3/5}(\log\log x)^{-1/5}}), \end{align} $$
where
$c>0$
is a positive constant and
$$ \begin{align*} {\textrm{ Li}}_{\mathcal{S}}(x) := \int_{2}^{\sqrt{x}} \frac{\,{\mathit{d}} t}{\log t} + \int_{2}^{\sqrt{x}} \frac{\,{\mathit{d}} t}{\log(x/t)}\cdot \end{align*} $$
Since
$\mathcal {S}(x)$
is a very sparse subset of
$[1, x]\cap {\mathbb N}$
, the results (1.2) and (1.3) seem rather interesting.
In [Reference Heyman2], Heyman also proposed a more general problem than (1.1), to study the asymptotic behaviour of the cardinality
$S(x; q)$
of the set
$$ \begin{align*} \mathcal{S}(x; q) := \bigg\{\bigg[\frac{x}{n}\bigg] : 1\leqslant n\leqslant x \;\, \text{and}\;\, q\,\bigg|\, \bigg[\frac{x}{n}\bigg]\bigg\} \end{align*} $$
for a fixed integer
$q\geqslant 1$
. Let
$\psi (t) := t-[t]-\tfrac 12$
and
$b:=(\!\sqrt {4x+1}-1)/2$
. Heyman first showed in [Reference Heyman2, Lemma 3] that
$$ \begin{align*} S(x; q) = \frac{4\sqrt{x}}{3q} + \sum_{r=1}^{[x/b]} \sum_{d=(x-r)/(qr)}^{x/(qr)} \bigg(\psi\bigg(\frac{x}{dq+1}\bigg)-\psi\bigg(\frac{x}{dq}\bigg)\bigg) + O(1) \end{align*} $$
for
$x\to \infty $
, and then wrote: ‘Calculating various sums using Maple suggests that the double sum cannot successfully be bounded. In fact Maple suggests that the double sum is asymptotically equivalent to
$2\sqrt {x}/(3q)$
. If this argument is correct then
$$ \begin{align} S(x; q)\sim \frac{2\sqrt{x}}{q} \qquad (x\to\infty), \end{align} $$
as one would expect heuristically.’
The aim of this short note is to prove (1.4). In fact, we can consider a more general problem: for
$1\leqslant a\leqslant q$
, study the distribution of elements of the set
$\mathcal {S}(x)$
in the arithmetical progression
$\{a+dq\}_{d\geqslant 0}$
. For this, we define
$$ \begin{align*} \mathcal{S}(x; q, a) := \bigg\{\bigg[\frac{x}{n}\bigg] : 1\leqslant n\leqslant x \;\, \text{and}\;\, \bigg[\frac{x}{n}\bigg]\equiv a \pmod q \bigg\} \end{align*} $$
and
$$ \begin{align*} S(x; q, a) := \lvert\mathcal{S}(x; q, a)\rvert. \end{align*} $$
Our result is as follows.
Theorem 1.1. Under the previous notation,
$$ \begin{align} S(x; q, a) = \frac{2\sqrt{x}}{q} + O((x/q)^{1/3}\log x) \end{align} $$
uniformly for
$x\geqslant 3$
,
$1\leqslant q\leqslant x^{1/4}/(\log x)^{3/2}$
and
$1\leqslant a\leqslant q$
, where the implied constant is absolute.
Since
$S(x; q, q) = S(x; q)$
, Heyman’s expected result (1.4) is a special case of our (1.5) with fixed q. The result of Theorem 1.1 implies
$$ \begin{align*} S(x; q, a)\sim \frac{2\sqrt{x}}{q} \end{align*} $$
uniformly for
$1\leqslant q = o(x^{1/4}/(\log x)^{3/2})$
and
$1\leqslant a\leqslant q$
. We did not try to get the best possible exponent and further improvements of the constant
$\tfrac 14$
are possible.
It seems interesting to establish analogues of the Dirichlet theorem or more generally the Siegel–Walfisz theorem, the Brun–Titchmarsh theorem and the Bombieri–Vinogradov theorem for the set
$\mathcal {S}(x)$
. We shall leave these problems to another occasion.
2 Preliminary lemmas
In this section, we shall cite two lemmas, which will be needed in the next section. The first one is due to Vaaler (see [Reference Graham and Kolesnik1, Theorem A.6]).
Lemma 2.1. Let
$\psi (t) := t - [t]-\tfrac 12$
. For
$x\geqslant 1$
and
$H\geqslant 1$
,
$$ \begin{align*} \psi(x) = - \sum_{1\leqslant \lvert h \rvert \leqslant H} \Phi\bigg(\frac{h}{H+1}\bigg) \frac{e(hx)}{2\pi{i} h} + R_{H}(x), \end{align*} $$
where
$e(t):=e^{2\pi {i}t}$
,
$\Phi (t):=\pi t (1- \lvert t \rvert )\cot (\pi t) + \lvert t\rvert $
and the error term
$R_{H}(x)$
satisfies
$$ \begin{align} \lvert R_{H}(x) \rvert \leqslant \frac{1}{2H+2} \sum_{0\leqslant \lvert h\rvert\leqslant H} \bigg(1-\frac{\lvert h\rvert}{H+1}\bigg) e(hx). \end{align} $$
Lemma 2.2. For
$1\leqslant a\leqslant q$
and
$\delta \in \{0, 1\}$
, define
$$ \begin{align*} \mathfrak{S}_{\delta}(D, D^{\prime}) := \sum_{D<d\leqslant D^{\prime}} \psi\bigg(\frac{x}{dq+a+\delta}\bigg). \end{align*} $$
If
$(\kappa , \lambda )$
is an exponent pair, then
$$ \begin{align} \mathfrak{S}_{\delta}(D, D^{\prime}) \ll (x^{\kappa} D^{-\kappa+\lambda} q^{-\kappa})^{1/(1+\kappa)} + x^{\kappa} D^{-2\kappa+\lambda} q^{-\kappa} + x^{-1} D^{2} q \end{align} $$
uniformly for
$1\leqslant a\leqslant q$
,
$(\!\sqrt {x}-a)/q<D\leqslant (x/q)^{2/3}$
and
$D<D^{\prime }\leqslant 2D$
, where the implied constant depends on
$(\kappa , \lambda )$
at most.
Proof. Using Lemma 2.1, we can write
$$ \begin{align} \mathfrak{S}_{\delta}(D, D^{\prime}) = - \frac{1}{2\pi{i}} (\mathfrak{S}_{\delta}^{\flat}(D, D^{\prime}) + \overline{\mathfrak{S}_{\delta}^{\flat}(D, D^{\prime})}) + \mathfrak{S}_{\delta}^{\dagger}(D, D^{\prime}), \end{align} $$
where
$H\leqslant D$
and
$$ \begin{align*} \mathfrak{S}_{\delta}^{\flat}(D, D^{\prime}) & := \sum_{h\leqslant H} \frac{1}{h}\Phi\bigg(\frac{h}{H+1}\bigg) \sum_{D<d\leqslant D^{\prime}} e\bigg(\frac{hx}{dq+a+\delta}\bigg), \\ \mathfrak{S}_{\delta}^{\dagger}(D, D^{\prime}) & := \sum_{D<d\leqslant D^{\prime}} R_{H}\bigg(\frac{x}{dq+a+\delta}\bigg). \end{align*} $$
Inverting the order of summations and applying the exponent pair
$(\kappa , \lambda )$
to the sum over d (see [Reference Graham and Kolesnik1, page 31, Definition]), it follows that
$$ \begin{align} \mathfrak{S}_{\delta}^{\flat}(D, D^{\prime}) & \ll \sum_{h\leqslant H} \frac{1}{h} \bigg(\bigg(\frac{xh}{D^{2}q}\bigg)^{\kappa} D^{\lambda} + \bigg(\frac{xh}{D^{2}q}\bigg)^{-1}\bigg) \notag \\[3pt] & \ll x^{\kappa} D^{-2\kappa+\lambda} q^{-\kappa} H^{\kappa} + x^{-1} D^{2}q. \end{align} $$
However, (2.1) of Lemma 2.1 allows us to derive
$$ \begin{align*} \lvert \mathfrak{S}_{\delta}^{\dagger}(D, D^{\prime}) \rvert & \leqslant \sum_{D<d\leqslant D^{\prime}} \bigg\lvert R_{H}\bigg(\frac{x}{dq+a+\delta}\bigg)\bigg\rvert \\ & \leqslant \frac{1}{2H+2} \sum_{0\leqslant \lvert h\rvert \leqslant H} \bigg(1-\frac{\lvert h\rvert}{H+1}\bigg) \sum_{D<d\leqslant D^{\prime}} e\bigg(\frac{xh}{dq+a+\delta}\bigg). \end{align*} $$
When
$h\not =0$
, as before, we apply the exponent pair
$(\kappa , \lambda )$
to the sum over d and obtain
$$ \begin{align} \mathfrak{S}_{\delta}^{\dagger}(D, D^{\prime}) \ll DH^{-1} + x^{\kappa} D^{-2\kappa+\lambda} q^{-\kappa} H^{\kappa} + x^{-1} D^{2}q. \end{align} $$
Inserting (2.4) and (2.5) into (2.3), it follows that
$$ \begin{align*} \mathfrak{S}_{\delta}(D, D^{\prime}) \ll DH^{-1} + x^{\kappa} D^{-2\kappa+\lambda} q^{-\kappa} H^{\kappa} + x^{-1} D^{2}q \end{align*} $$
for
$H\leqslant D$
. Optimising H on
$[1, D]$
, we obtain the required inequality (2.2).
3 Proof of Theorem 1.1
If
$[{x}/{n}] = m = dq+a$
with
$0\leqslant d\leqslant (x-a)/q$
, then
${x/(dq+a+1)<n\leqslant x/(dq+a)}$
. Thus we can write
$$ \begin{align} S(x; q, a) & = \sum_{d\leqslant (x-a)/q} \mathbb{1}\bigg(\bigg[\frac{x}{dq+a}\bigg]-\bigg[\frac{x}{dq+a+1}\bigg]>0\bigg) + O(1)\notag \\[3pt] & = S_{1}(x; q, a) + S_{2}(x; q, a) + O(1), \end{align} $$
where
$\mathbb {1}=1$
if the statement is true and 0 otherwise, and
$$ \begin{align*} S_{1}(x; q, a) & := \sum_{d\leqslant (\!\sqrt{x}-a)/q} \mathbb{1}\bigg(\bigg[\frac{x}{dq+a}\bigg]-\bigg[\frac{x}{dq+a+1}\bigg]>0\bigg), \\ S_{2}(x; q, a) & := \sum_{(\!\sqrt{x}-a)/q<d\leqslant (x-a)/q} \mathbb{1}\bigg(\bigg[\frac{x}{dq+a}\bigg]-\bigg[\frac{x}{dq+a+1}\bigg]>0\bigg). \end{align*} $$
For
$d\leqslant (\!\sqrt {x}-a-1)/q$
,
$$ \begin{align*} \bigg[\frac{x}{dq+a}\bigg]-\bigg[\frac{x}{dq+a+1}\bigg]>\frac{x}{(dq+a)(dq+a+1)}-1 >0. \end{align*} $$
Thus,
$$ \begin{align} S_{1}(x; q, a) = \frac{\sqrt{x}}{q} + O(1) \end{align} $$
for
$x\geqslant 3$
, where the implied constant is absolute.
Next, we treat
$S_{2}(x; q, a)$
. Since
$d>(\!\sqrt {x}-a)/q$
,
$$ \begin{align*} 0<\frac{x}{dq+a}-\frac{x}{dq+a+1}=\frac{x}{(dq+a)(dq+a+1)}<1. \end{align*} $$
Therefore, the quantity
$[{x}/{dq+a}]-[{x}/{dq+a+1}]$
can only equal 0 or 1. However, for
$d\geqslant (x/q)^{2/3}$
, we have
$dq+a=[{x}/{n}]$
for some
$n\leqslant (x/q)^{1/3}$
. Thus,
$$ \begin{align*} S_{2}(x; q, a) = \sum_{(\!\sqrt{x}-a)/q<d\leqslant (x/q)^{2/3}} \bigg(\bigg[\frac{x}{dq+a}\bigg]-\bigg[\frac{x}{dq+a+1}\bigg]\bigg) + O((x/q)^{1/3}). \end{align*} $$
Since
$$ \begin{align*} \bigg[\frac{x}{dq+a}\bigg]-\bigg[\frac{x}{dq+a+1}\bigg] = \frac{x}{dq+a}-\frac{x}{dq+a+1}-\psi\bigg(\frac{x}{dq+a}\bigg)+\psi\bigg(\frac{x}{dq+a+1}\bigg), \end{align*} $$
it follows that
$$ \begin{align} S_{2}(x; q, a) = S_{2, 1}(x; q, a) - S_{2, 2}^{\langle 0\rangle}(x; q, a) + S_{2, 2}^{\langle 1\rangle}(x; q, a) + O((x/q)^{1/3}), \end{align} $$
where
$$ \begin{align*} S_{2, 1}(x; q, a) & := \sum_{(\!\sqrt{x}-a)/q<p\leqslant (x/q)^{2/3}} \bigg(\frac{x}{dq+a}-\frac{x}{dq+a+1}\bigg), \\ S_{2, 2}^{\langle \delta\rangle}(x; q, a) & := \sum_{(\!\sqrt{x}-a)/q<d\leqslant (x/q)^{2/3}} \psi\bigg(\frac{x}{dq+a+\delta}\bigg). \end{align*} $$
An elementary computation shows that, as
$x\to \infty $
,
$$ \begin{align} S_{2, 1}(x; q, a) = \sum_{(\!\sqrt{x}-a)/q<d\leqslant (x/q)^{2/3}} \frac{x}{d^{2}q^{2}} + O(1) = \frac{\sqrt{x}}{q} + O ((x/q^{4})^{1/3}). \end{align} $$
It remains to bound
$S_{2, 2}^{\langle \delta \rangle }(x; q, a)$
. According to [Reference Graham and Kolesnik1, Theorem 3.10],
$(\tfrac 12, \tfrac 12)$
is an exponent pair (since
$(0, 1)$
is a trivial exponent pair). Thus we can take
$(\kappa , \lambda ) = (\tfrac 12, \tfrac 12)$
in (2.2) of Lemma 2.2 to get
$$ \begin{align*} \mathfrak{S}_{\delta}(D, D^{\prime}) \ll (x/q)^{1/3} + (xD^{-1}q^{-1})^{1/2} + x^{-1} D^{2} q \end{align*} $$
uniformly for
${1{{\kern-1pt}\leqslant{\kern-1pt}} a{{\kern-1pt}\leqslant{\kern-1pt}} q}$
,
${(\!\sqrt {x}-a)/q<D{{\kern-1pt}\leqslant{\kern-1pt}} (x/q)^{2/3}}$
and
${D<D^{\prime }{{\kern-1pt}\leqslant{\kern-1pt}} 2D}$
. Using the fact that
${(\!\sqrt {x}-a)/} q<D{{\kern-1pt}\leqslant{\kern-1pt}} (x/q)^{2/3}$
, we easily see that the preceding inequality implies
$$ \begin{align*} \mathfrak{S}_{\delta}(D, D^{\prime}) \ll (x/q)^{1/3} \end{align*} $$
uniformly for
$1\leqslant a\leqslant q$
,
$(\!\sqrt {x}-a)/q<D\leqslant (x/q)^{2/3}$
and
$D<D^{\prime }\leqslant 2D$
. From this,
$$ \begin{align} S_{2, 2}^{\langle \delta\rangle}(x; q) \ll (\log x) \max_{(\!\sqrt{x}-a)/q<D\leqslant (x/q)^{2/3}} \lvert\mathfrak{S}_{\delta}(D, 2D)\rvert \ll (x/q)^{1/3}\log x. \end{align} $$
Inserting (3.4) and (3.5) into (3.3),
$$ \begin{align} S_{2}(x; q, a) = \frac{\sqrt{x}}{q} + O((x/q)^{1/3}\log x) \end{align} $$
for
$x\to \infty $
. Now the required result follows from (3.1), (3.2) and (3.6).
Acknowledgement
We began to work on this project when the second author visited Luoyang Institute of Science and Technology in the summer of 2021, despite the difficulty caused by the coronavirus. It is a pleasure to record his gratitude to this institution for their hospitality and support.
