1 Introduction
The distribution of prime numbers is one of the most important problems in number theory. Denote by
$\pi (x)$
the number of primes
$p\leqslant x$
. The prime number theorem states that
$$ \begin{align*} \pi(x) = \frac{x}{\log x} + O\bigg(\frac{x}{(\log x)^2}\bigg) \quad (x\to\infty). \end{align*} $$
A strong form of this theorem is
$$ \begin{align} \pi(x) = {\mathrm{Li}}(x) + O(x \exp(-c(\log x)^{3/5}(\log_2x)^{-1/5})) \quad (x\to\infty), \end{align} $$
where c is a positive constant,
$\log _2$
denotes the iterated logarithm function and
$$ \begin{align*} {\mathrm{Li}}(x) := \int_2^x \frac{d t}{\log t}. \end{align*} $$
The Riemann hypothesis is equivalent to the asymptotic formula
$$ \begin{align} \pi(x) = {\mathrm{Li}}(x) + O_{\varepsilon}(x^{1/2+\varepsilon}) \quad\; (x\to\infty), \end{align} $$
where
$\varepsilon $
is an arbitrarily small positive number. More generally, let
$\mathcal {N}(x)$
be a set of integers of
$[1, x]$
and let
$\mathcal {N}_{{\mathbb P}}(x)$
be the set of prime numbers in
$\mathcal {N}(x)$
. We expect that
$$ \begin{align} |\mathcal{N}_{{\mathbb P}}(x)|\sim \frac{|\mathcal{N}(x)|}{\log |\mathcal{N}(x)|} \quad (x\to \infty), \end{align} $$
provided
$\mathcal {N}(x)$
is rather regular and is not too sparse. Some well-known examples are
$$ \begin{align*} \{qn+a\leqslant x\}, \;\; \{[n^c]\leqslant x\}, \;\; \{m^2+n^4\leqslant x\}, \;\; \{m^3+2n^3\leqslant x\}, \;\; \{x<n\leqslant x+x^{7/12+\varepsilon}\}, \end{align*} $$
the respective densities for which are
$$ \begin{align*} x/\varphi(q) \quad &(q\leqslant (\log x)^A, \text{Walfisz--Siegel } [3]), \\ x^{1/c} \quad &\bigg(c\leqslant \dfrac{2817}{2426}, \text{Rivat--Sargos } [12]\bigg), \\ x^{3/4} \quad &(\text{Friedlander--Iwaniec } [4]), \\ x^{2/3} \quad &(\text{Heath-Brown } [5]), \\ x^{7/12+\varepsilon} \quad &(\text{Huxley } [8]), \end{align*} $$
where
$[t]$
is the integral part of the real number t,
$\varphi (q)$
is the Euler function, A is any positive constant and
$\varepsilon>0$
is an arbitrarily small positive number.
Recently, Bordellès et al. [Reference Bordellès, Dai, Heyman, Pan and Shparlinski2] investigated the asymptotic behaviour of the summative function
$$ \begin{align*} S_f(x) := \sum_{n\leqslant x} f\bigg(\bigg[\frac{x}{n}\bigg]\bigg) \end{align*} $$
under some simple hypothesis on the growth of f and there are a number of further developments on this theme. If we use
$\Lambda (n)$
to denote the von Mangoldt function, then [Reference Wu13, Theorem 1.2(i)] or [Reference Zhai15, Theorem 1] give us immediately
$$ \begin{align} S_{\Lambda}(x) = C_{\Lambda} x + O_{\varepsilon}(x^{1/2+\varepsilon}), \end{align} $$
for any
$\varepsilon>0$
and
$x\to \infty $
, where
$C_{\Lambda } := \sum _{n\geqslant 1} {\Lambda (n)}/{n(n+1)}$
. Ma and Wu [Reference Ma and Wu11] applied the Vaughan identity and the technique of one-dimensional exponential sums to break the
$\tfrac 12$
-barrier by establishing
$$ \begin{align} S_{\Lambda}(x) = C_{\Lambda} x + O_{\varepsilon}(x^{35/71+\varepsilon}). \end{align} $$
This result seems rather interesting if we compare it with (1.2). The exponent
${35}/{71}$
has been improved to
${97}/{203}$
by Bordellès [Reference Bordellès1] and
${9}/{19}$
by Liu et al. [Reference Liu, Wu and Yang10], using more sophisticated techniques of multiple exponential sums. Obviously, (1.5) is the prime number theorem for the floor function set
$$ \begin{align*} \mathcal{S}(x) := \bigg\{\bigg[\frac{x}{n}\bigg] : 1\leqslant n\leqslant x\bigg\} \end{align*} $$
considered as the weighted count of prime powers. Very recently, Heyman [Reference Heyman7] examined the number of primes in the floor function set
$\mathcal {S}(x)$
without the multiplicity. The principal result of Heyman [Reference Heyman7, Theorem 1] is the asymptotic formula
$$ \begin{align} \pi_{\mathcal{S}}(x) := \sum_{\substack{p\leqslant x\\ \exists\,n\,\text{such that}\; [x/n]=p}} 1 = \frac{4\sqrt{x}}{\log x} + O\bigg(\frac{\sqrt{x}}{(\log x)^2}\bigg) \quad(x\to\infty). \end{align} $$
Since Heyman [Reference Heyman6, Theorems 1 and 2] proved that
$$ \begin{align} |\mathcal{S}(x)| = 2\sqrt{x} + O(1) \quad(x\to\infty). \end{align} $$
it follows from (1.6) that (1.3) holds for this sparse set
$\mathcal {S}(x)$
. This may be the first example of such a sparse subset of
$[1, x]\cap {\mathbb N}$
(of density
$x^{1/2}$
) for which the prime number theorem holds.
It seems natural and interesting to establish an analogue of the strong form of the prime number theorem in (1.1) for the set
$\mathcal {S}(x)$
. We prove such a result.
Theorem 1.1. (i) For
$x\to \infty $
,
$$ \begin{align} \pi_{\mathcal{S}}(x) = {\mathrm{Li}}_{\mathcal{S}}(x) + O(\sqrt{x} \exp(-c'(\log x)^{3/5}(\log_2x)^{-1/5})), \end{align} $$
where
$c'>0$
is a positive constant and
$$ \begin{align*} {\mathrm{Li}}_{\mathcal{S}}(x) := \int_2^{\sqrt{x}} \frac{d t}{\log t} + \int_2^{\sqrt{x}} \frac{d t}{\log(x/t)}\cdot \end{align*} $$
(ii) There is a real sequence
$\{a_n\}_{n\geqslant 1}$
with
$a_1=4$
such that for any positive integer
$N\geqslant 1$
,
$$ \begin{align*} \pi_{\mathcal{S}}(x) = \sqrt{x} \sum_{n=1}^{N} \frac{a_n}{(\log x)^n} + O_N\bigg(\frac{\sqrt{x}}{(\log x)^{N+1}}\bigg) \quad (x\to\infty). \end{align*} $$
Let
${\mathbb P}$
be the set of all primes and let
${{\mathbb P}}_{\mathrm {ower}}$
be the set of all prime powers. Denote by
$\mathbb {1}_{{\mathbb P}}$
and
$\mathbb {1}_{{{\mathbb P}}_{\mathrm {ower}}}$
their characteristic functions. Define
$$ \begin{align*} S_{\mathbb{1}_{{\mathbb P}}}(x) := \sum_{n\leqslant x} \mathbb{1}_{{\mathbb P}}\bigg(\bigg[\frac{x}{n}\bigg]\bigg), \quad S_{\mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}}(x) := \sum_{n\leqslant x} \mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}\bigg(\bigg[\frac{x}{n}\bigg]\bigg). \end{align*} $$
Theorems 5 and 7 of [Reference Heyman7] can be stated as follows:
$$ \begin{align} S_{\mathbb{1}_{{\mathbb P}}}(x) & = C_{\mathbb{1}_{{\mathbb P}}} x + O(x^{1/2}), \end{align} $$
$$ \begin{align} S_{\mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}}(x) & = C_{\mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}} x + O(x^{1/2}), \end{align} $$
where
$C_{\mathbb {1}_{{\mathbb P}}} := \sum _{p} {1}/{p(p+1)}$
and
$C_{\mathbb {1}_{{{\mathbb P}}_{\mathrm {ower}}}} := \sum _{p, \, \nu \geqslant 1} {1}/{p^{\nu }(p^{\nu }+1)}$
. Similar to (1.4), these are immediate consequences of [Reference Wu13, Theorem 1.2(i)] or [Reference Zhai15, Theorem 1]. Heyman [Reference Heyman7, Theorem 6] also proved that there is a positive constant
$B>0$
such that the inequality
$$ \begin{align} S_{\mathbb{1}_{{\mathbb P}}}(x) \geqslant C_{\mathbb{1}_{{\mathbb P}}} x - \frac{Bx^{1/2}}{\log x} \end{align} $$
for
$x\geqslant 2$
. We improve these results by breaking the
$\tfrac 12$
-barrier in the error terms of (1.9), (1.10) and (1.11).
Theorem 1.2. For any
$\varepsilon>0$
,
$$ \begin{align} S_{\mathbb{1}_{{\mathbb P}}}(x) & = C_{\mathbb{1}_{{\mathbb P}}} x + O_{\varepsilon}(x^{9/19+\varepsilon}), \end{align} $$
$$ \begin{align} S_{\mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}}(x) & = C_{\mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}} x + O_{\varepsilon}(x^{9/19+\varepsilon}), \end{align} $$
as
$x\to \infty $
, where the implied constants depend on
$\varepsilon $
.
Remark 1.3. It is possible to improve the error terms in (1.12) and (1.13). It seems interesting to prove
$\Omega $
-results for the error terms in (1.8), (1.12) and (1.13). We shall return to this problem in forthcoming work.
Very recently, Yu and Wu [Reference Yu and Wu14] generalised Heyman’s (1.7) by showing
$$ \begin{align*} \mathcal{S}(x; q, a) := \sum_{\substack{m\in \mathcal{S}(x)\\ m\equiv a \,(\mathrm{mod}\,q)}} 1 = \frac{2\sqrt{x}}{q} + O((x/q)^{1/3}\log x) \end{align*} $$
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. This confirms a numerical test of Heyman.
2 Proof of Theorem 1.1
We begin by following the argument of [Reference Heyman7]. First, we note that
$$ \begin{align*} \mathcal{S}(x) = \bigg\{p\in \mathbb{P} : \exists \; n\in [1, x] \;\text{such that}\; \bigg[\frac{x}{n}\bigg]=p\bigg\}. \end{align*} $$
Further, if
$[{x}/{n}] = p\in \mathbb {P}$
, then
$x/(p+1)<n\leqslant x/p$
. Thus, we can write
$$ \begin{align} \pi_{\mathcal{S}}(x) = \sum_{p\leqslant x} \mathbb{1}\bigg(\bigg[\frac{x}{p}\bigg]-\bigg[\frac{x}{p+1}\bigg]>0\bigg) = G_1(x) + G_2(x), \end{align} $$
where
$\mathbb {1}(Q)=1$
if the statement Q is true and 0 otherwise, and
$$ \begin{align*} G_1(x) & := \sum_{p\leqslant \sqrt{x}} \mathbb{1}\bigg(\bigg[\frac{x}{p}\bigg]-\bigg[\frac{x}{p+1}\bigg]>0\bigg), \\ G_2(x) & := \sum_{\sqrt{x}<p\leqslant x} \mathbb{1}\bigg(\bigg[\frac{x}{p}\bigg]-\bigg[\frac{x}{p+1}\bigg]>0\bigg). \end{align*} $$
For
$p\leqslant \sqrt {x}-1$
,
$$ \begin{align*} \bigg[\frac{x}{p}\bigg]-\bigg[\frac{x}{p+1}\bigg]>\frac{x}{p(p+1)}-1 >0. \end{align*} $$
Thus, the prime number theorem (1.1) gives us
$$ \begin{align} G_1(x) = \pi(\sqrt{x}) + O(1) = {\mathrm{Li}}(\sqrt{x}) + O(\sqrt{x} \exp(-c'(\log x)^{3/5}(\log_2x)^{-1/5})) \end{align} $$
for
$x\geqslant 3$
, where
$c'>0$
is a positive constant.
Next, we treat
$G_2(x)$
. Noticing that
$$ \begin{align*} 0<\frac{x}{p}-\frac{x}{p+1}=\frac{x}{p(p+1)}<1 \end{align*} $$
for
$p>\sqrt {x}$
, the quantity
$[{x}/{p}]-[{x}/{p+1}]$
can only equal 0 or 1. However, for
$p>x^{10/19}$
, we have
$p=[{x}/{n}]$
for some
$n\leqslant x^{9/19}$
. Thus, we can write
$$ \begin{align} \begin{aligned} G_2(x) & = \sum_{x^{1/2}<p\leqslant x^{10/19}} \bigg(\bigg[\frac{x}{p}\bigg]-\bigg[\frac{x}{p+1}\bigg]\bigg) + O(x^{9/19}) \\ & = \sum_{x^{1/2}<p\leqslant x^{10/19}} \bigg(\frac{x}{p}-\frac{x}{p+1}-\psi\bigg(\frac{x}{p}\bigg)+\psi\bigg(\frac{x}{p+1}\bigg)\bigg) + O(x^{9/19}) \\ & = G_{2, 1}(x) - G_{2, 2}^{\langle 0\rangle}(x) + G_{2, 2}^{\langle 1\rangle}(x) + O(x^{9/19}), \end{aligned} \end{align} $$
where
$\psi (t):=t-[t]-\tfrac 12$
and
$$ \begin{align*} G_{2, 1}(x) & := \sum_{x^{1/2}<p\leqslant x^{10/19}} \bigg(\frac{x}{p}-\frac{x}{p+1}\bigg), \\ G_{2, 2}^{\langle\delta\rangle}(x) & := \sum_{x^{1/2}<p\leqslant x^{10/19}} \psi\bigg(\frac{x}{p+\delta}\bigg) \quad (\delta=0, 1). \end{align*} $$
With the help of the prime number theorem (1.1), a simple partial integration gives
$$ \begin{align*} G_{2, 1}(x) & = \sum_{x^{1/2}<p\leqslant x/2} \frac{x}{p^2} + O(x^{9/19}) = x \int_{\sqrt{x}}^{x/2} \frac{d \pi(t)}{t^2} + O(x^{9/19}) \\ & = x \int_{\sqrt{x}}^{x/2} \frac{d t}{t^2\log t} + O(\sqrt{x}\exp(-c'(\log x)^{3/5}(\log_2x)^{-1/5}), \end{align*} $$
where
$c'>0$
is a positive constant. Making the change of variables
$t\to x/t$
in the last integral, it follows that
$$ \begin{align} G_{2, 1}(x) = \int_2^{\sqrt{x}} \frac{d t}{\log(x/t)} + O(\sqrt{x}\exp(-c'(\log x)^{3/5}(\log_2x)^{-1/5}) \end{align} $$
for
$x\to \infty $
.
It remains to bound
$G_{2, 2}^{\langle \delta \rangle }(x)$
. Similar to [Reference Liu, Wu and Yang10], define
$$ \begin{align*} \mathfrak{S}_{\delta}(x; D, D') := \sum_{D<d\leqslant D'} \Lambda(d) \psi\bigg(\frac{x}{d+\delta}\bigg). \end{align*} $$
According to [Reference Liu, Wu and Yang10, (4.3)], for any
$\varepsilon>0$
,
$$ \begin{align*} \mathfrak{S}_{\delta}(x; D, 2D) \ll_{\varepsilon} (x^2 D^7)^{1/12} x^{\varepsilon} \end{align*} $$
uniformly for
$x\geqslant 3$
and
$x^{6/13}\leqslant D\leqslant x^{2/3}$
. The same proof shows that for any
$\varepsilon>0$
,
$$ \begin{align} \mathfrak{S}_{\delta}(x; D, D') \ll_{\varepsilon} (x^2 D^7)^{1/12} x^{\varepsilon} \end{align} $$
uniformly for
$x\geqslant 3$
,
$x^{6/13}\leqslant D\leqslant x^{2/3}$
and
$D<D'\leqslant 2D$
. Since we have trivially
$$ \begin{align*} \sum_{D<p^{\nu}\leqslant D', \, \nu\geqslant 2} \Lambda(p^{\nu}) \psi\bigg(\frac{x}{p^{\nu}+\delta}\bigg) \ll \sum_{p\leqslant (2D)^{1/2}} \sum_{\nu\leqslant (\log 2D)/\log p} \log p \ll D^{1/2}, \end{align*} $$
the inequality (2.5) implies that the bound
$$ \begin{align} \sum_{D<p\leqslant D'} (\log p) \psi\bigg(\frac{x}{p+\delta}\bigg) \ll_{\varepsilon} (x^2 D^7)^{1/12} x^{\varepsilon} \end{align} $$
holds uniformly for
$x\geqslant 3$
,
$x^{6/13}\leqslant D\leqslant x^{2/3}$
and
$D<D'\leqslant 2D$
. Using (2.6),
$$ \begin{align} \begin{aligned} G_{2, 2}^{\langle\delta\rangle}(x) & \ll \max_{x^{1/2}<D\leqslant x^{10/19}} \sum_{D<p\leqslant 2D} \psi\bigg(\frac{x}{p+\delta}\bigg) \\ & \ll \max_{x^{1/2}<D\leqslant x^{10/19}} \int_D^{2D} \frac{1}{\log t} \,d \bigg(\sum_{D<p\leqslant t} (\log p) \psi\bigg(\frac{x}{p+\delta}\bigg)\bigg) \\ & \ll_{\varepsilon} \max_{x^{1/2}<D\leqslant x^{10/19}} (x^2 D^7)^{1/12} x^{\varepsilon} \\ & \ll_{\varepsilon} x^{9/19+\varepsilon}. \end{aligned} \end{align} $$
Inserting (2.4) and (2.7) into (2.3), we find that
$$ \begin{align} G_2(x) = \int_2^{\sqrt{x}} \frac{d t}{\log(x/t)} + O(\sqrt{x}\exp(-c'(\log x)^{3/5}(\log_2x)^{-1/5}). \end{align} $$
Now the required result (1.8) follows from (2.1), (2.2) and (2.8).
The second assertion is an immediate consequence of the first one thanks to a simple partial integration.
3 Proof of Theorem 1.2
We begin by following the argument of [Reference Liu, Wu and Yang9]. Let
$f=\mathbb {1}_{{\mathbb P}}$
or
$\mathbb {1}_{{{\mathbb P}}_{\mathrm {ower}}}$
and let
$N\in [x^{1/3}, x^{1/2})$
be a parameter which can be chosen later. First, we write
$$ \begin{align} S_f(x) = \sum_{n\leqslant x} f\bigg(\bigg[\frac{x}{n}\bigg]\bigg) = S_f^{\dagger}(x)+S_f^{\sharp}(x) \end{align} $$
with
$$ \begin{align*} S_f^{\dagger}(x):=\sum_{n\leqslant N} f\bigg(\bigg[\frac{x}{n}\bigg]\bigg), \quad S_f^{\sharp}(x):=\sum_{N<n\leqslant x} f\bigg(\bigg[\frac{x}{n}\bigg]\bigg). \end{align*} $$
We have trivially
$$ \begin{align} S_f^{\dagger}(x) \ll N. \end{align} $$
To bound
$S_f^{\sharp }(x)$
, we put
$d=[x/n]$
. Noticing that
$$ \begin{align*} x/n-1<d\leqslant x/n \Leftrightarrow x/(d+1)<n\leqslant x/d, \end{align*} $$
we see that
$$ \begin{align} \begin{aligned} S_f^{\sharp}(x) & = \sum_{d\leqslant x/N} f(d) \sum_{x/(d+1)<n\leqslant x/d} 1 \\ & = \sum_{d\leqslant x/N} f(d) \bigg(\frac{x}{d}-\psi\bigg(\frac{x}{d}\bigg)-\frac{x}{d+1}+\psi\bigg(\frac{x}{d+1} \bigg) \bigg) \\ & = x \sum_{d\geqslant 1} \frac{f(d)}{d(d+1)} + \mathcal{R}_1^{f}(x, N) - \mathcal{R}_0^{f}(x, N) + O(N), \end{aligned} \end{align} $$
where we have used the bounds
$$ \begin{align*} x \sum_{d>x/N} \frac{f(d)}{d(d+1)}\ll N, \quad \sum_{d\leqslant N} f(d)\bigg(\psi\bigg(\frac{x}{d+1}\bigg) - \psi\bigg(\frac{x}{d}\bigg)\bigg) \ll N \end{align*} $$
and
$$ \begin{align*} \mathcal{R}_{\delta}^{f}(x, N) = \sum_{N<d\leqslant x/N} f(d) \psi\bigg(\frac{x}{d+\delta}\bigg). \end{align*} $$
Combining (3.1), (3.2) and (3.3), it follows that
$$ \begin{align*} S_f(x) = x \sum_{d\geqslant 1} \frac{f(d)}{d(d+1)} + O_{\varepsilon}(|\mathcal{R}_1^{f}(x, N)| + |\mathcal{R}_0^{f}(x, N)| + N). \end{align*} $$
However,
$$ \begin{align*} \mathcal{R}_{\delta}^{\mathbb{1}_{{{\mathbb P}}_{\mathrm{ower}}}}(x, N) = \sum_{N<p^{\nu}\leqslant x/N} \psi\bigg(\frac{x}{p^{\nu}+\delta}\bigg) = \mathcal{R}_{\delta}^{\mathbb{1}_{{\mathbb P}}}(x, N) + O((x/N)^{1/2}). \end{align*} $$
Thus, to prove Theorem 1.2, it suffices to show that
$$ \begin{align*} \mathcal{R}_{\delta}^{\mathbb{1}_{{\mathbb P}}}(x, N)\ll_{\varepsilon} N x^{\varepsilon} \quad (x\geqslant 1) \end{align*} $$
for
$N=x^{9/19}$
. This can be done exactly as for (2.7) by using (2.6):
$$ \begin{align*} \mathcal{R}_{\delta}^{\mathbb{1}_{{\mathbb P}}}(x, N) & \ll x^{\varepsilon} \max_{x^{9/19}<D\leqslant x^{10/19}} \sum_{D<p\leqslant 2D} \psi\bigg(\frac{x}{p+\delta}\bigg) \\ & \ll x^{\varepsilon} \max_{x^{9/19}<D\leqslant x^{10/19}} \int_D^{2D} \frac{1}{\log t} \,d \bigg(\sum_{D<p\leqslant t} (\log p) \psi\bigg(\frac{x}{p+\delta}\bigg)\bigg) \\ & \ll_{\varepsilon} \max_{x^{9/19}<D\leqslant x^{10/19}} (x^2 D^7)^{1/12} x^{\varepsilon} \\ & \ll_{\varepsilon} x^{9/19+\varepsilon}. \end{align*} $$
This completes the proof.
