1 Introduction and statement of results
Let
$\mu (n)$
be the Möbius function defined by
$$\begin{align*}\mu(n)=\begin{cases} 1 & \text{if } n=1,\\ 0 & \text{if } p^2\mid n \text{ for some prime } p,\\ (-1)^{r} & \text{if } n=p_{1} \dots p_{r}, \text{ where the } p_i \text{ are distinct primes}.\\ \end{cases} \end{align*}$$
Let
$k\geq 1$
be an integer. In 1970, Apostol [Reference Apostol2] introduced the Möbius function
$\mu _k(n)$
of order k, which vanishes if
$p^{k+1}\mid n$
for some prime p and equals to
$(-1)^r$
if n contains the kth powers of exactly r distinct primes. That is,
$$\begin{align*}\mu_k(n)=\begin{cases} 1 & \text{if } n=1,\\ 0 & \text{if } p^{k+1}\mid n \text{ for some prime } p,\\ (-1)^{r} & \text{if } n=p_{1}^k \dots p_{r}^k \prod\limits_{i>r}p_i^{a_i}, \text{ with } 0\leq a_i<k,\\ 1& \text{otherwise}. \end{cases} \end{align*}$$
Clearly,
$\mu _1(n)=\mu (n)$
. Let
$m\geq k \geq 1$
be an integer. In [Reference Bege4], Bege generalized Apostol’s
$\mu _{k}(n)$
to
$\mu _{k,m}(n)$
as follows:
$$\begin{align*}\mu_{k,m}(n)=\begin{cases} 1 & \text{if } n=1,\\ 1 & \text{if } p^{k}\nmid n \text{ for each prime } p,\\ (-1)^{r} & \text{if } n=p_{1}^{m}\dots p_{r}^{m}\prod\limits_{i>r}p_{i}^{\alpha_{i}}, \text{ with } 0\leq\alpha_{i}<k,\\ 0 & \text{otherwise.} \end{cases} \end{align*}$$
Clearly,
$\mu _{k,k}(n)=\mu _k(n)$
for all
$k\ge 1$
. Moreover, by [Reference Bege4, Lemma 2.1], Bege’s generalization
$\mu _{k,m}$
of the Möbius function satisfies the following relation to
$\mu $
:
$$ \begin{align} \mu_{k,m}(n) =\sum_{\substack{d^m\delta=n \\ (d,\delta)=1}} \mu(d)q_k(\delta) \end{align} $$
for any
$m\ge k\ge 1$
, where
$q_k(n)$
is the characteristic function of the k-free integers:
$$\begin{align*}q_{k}(n)=\begin{cases} 0 & \text{if } p^k\mid n \text{ for some prime } p;\\ 1& \text{if } p^\alpha \mid n \text{ implies } \alpha < k. \end{cases} \end{align*}$$
Using (1.1), Bege obtained the asymptotic formula for the partial sum of
$\mu _{k,m}$
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n) = A_{k,m} x+O\Big(x^{\frac1k}\exp\left\{-c \log^{\frac{3}{5}} x (\log \log x)^{-\frac{1}{5}}\right\}\Big) \end{align} $$
for some absolute constant
$c>0$
, where
$$ \begin{align} A_{k,m} = \frac1{\zeta(k)}\prod_{p}\Big(1-\frac1{p^{m-k+1} + p^{m-k+2} + \cdots + p^m} \Big), \end{align} $$
the product runs over all primes p, and
$\zeta (s)=\sum _{n=1}^\infty \frac 1{n^s}$
is the Riemann zeta function for
$\Re s>1$
. In this article, based on (1.1), we will establish three kinds of asymptotic formulas related to
$\mu _{k,m}$
for
$m\geq k\geq 2$
, using a unified preliminary estimate in Proposition 3.1.
First, we consider the shifted convolution sums of Fourier coefficients of holomorphic cusp forms. Let
$H_{\kappa }$
denote the set of normalized primitive holomorphic cusp forms of even integral weight
$\kappa $
for the full modular group
$\Gamma =SL(2,\mathbb {Z})$
, and suppose
$f\in H_{\kappa }$
. Let
$\lambda _f(n)$
be the nth normalized Fourier coefficient of
$f(z)$
. Let
$a(n)$
be a function with squarefull kernel introduced by Ivić and Tenenbaum [Reference Ivić and Tenenbaum13], which means that
$a(n)=a(s)$
for
$n=qs, (q,s)=1$
, where q is squarefree and s is squarefull, and
$a(n)\ll n^\varepsilon $
. Let
$\tau (n)=\sum _{d\mid n}1$
be the divisor function, and let
$\omega (n)=\sum _{p\mid n}1$
be the number of distinct prime factors of n. In [Reference Erdős and Ivić9, equations (1.5) and (1.6)], Erdős and Ivić established the asymptotic formulas for the following shifted convolution sums on
$\tau (n)$
and
$\omega (n)$
:
$$ \begin{align} \sum_{n\leq x} a(n)\tau(n+1) & = C_1 x\log x + C_2x+O(x^{\frac89+\varepsilon}),\qquad \end{align} $$
$$ \begin{align} \,\sum_{n\leq x} a(n)\omega(n+1) &= D_1 x\log \log x + D_2 x + O\left(\frac{x}{\log x}\right), \end{align} $$
where
$C_1, D_1>0, C_2, D_2$
are constants that can be explicitly evaluated. Inspired by their work, in 2020, Lü and Wang [Reference Lü and Wang16, Theorem 1] obtained an asymptotic estimate for the shifted convolution sum of
$a(n)$
with
$\lambda _f^2(n+1)$
: for any
$\varepsilon>0$
, we have
$$ \begin{align} \sum_{n \leq x} a(n)\lambda_f^2(n + 1) = C x + O(x^{13/14+\varepsilon}), \end{align} $$
where C is a constant that can be explicitly evaluated. Based on the work of Lü and Wang, we will establish an asymptotic formula for the shifted convolution sum of
$\mu _{k,m}(n)$
with
$\lambda _f^2(n+1)$
in our first main theorem. Moreover, for any integer
$j\ge 2$
, based on the results in [Reference Hua12, Reference Venkatasubbareddy and Sankaranarayanan22, Reference Wang27], using the method in Theorem 1.1, one can establish asymptotic formulas for the shifted convolution sum of
$\mu _{k,m}(n)$
with higher power moments
$\lambda _f^{2j} (n+1)$
of Fourier coefficients as well.
Theorem 1.1 Suppose
$f\in H_{\kappa }$
and
$m\geq k\ge 2$
. Then for any
$\varepsilon>0$
, we have
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n) \lambda_f^2(n+1) = A_{k,m, f} \cdot x + O\Big(x^{1-\frac{k-1}{4(m+k+1)}+\varepsilon}\Big), \end{align} $$
where
$A_{k,m, f}$
is a constant defined by (4.5).
Second, we consider a dynamical generalization of the prime number theorem (PNT) with respect to
$\mu _{k,m}(n)$
. For any
$n\in \mathbb {N}$
, let
$\Omega (n)$
be the number of prime divisors of n counted with multiplicity. Let
$\lambda (n)=(-1)^{\Omega (n)}$
be the Liouville function. In 2022, Bergelson and Richter [Reference Bergelson and Richter5, Theorem A] established a new dynamical generalization of the PNT, showing that given any uniquely ergodic topological dynamical system
$(X, \nu , T)$
,
$$ \begin{align} \lim_{N\to\infty}\frac1N\sum_{n=1}^N g(T^{\Omega(n)}x_0)=\int_X g \,d\nu \end{align} $$
holds for any
$g\in C(X)$
and
$x_0\in X$
, where
$C(X)$
denotes the set of continuous functions on X. They also proved that (1.8) holds over squarefree numbers:
$$ \begin{align} \lim_{N\to\infty}\frac1N\sum_{n=1}^N \mu^2(n)g(T^{\Omega(n)}x_0)=\frac{6}{\pi^2}\cdot\int_X g \,d\nu \end{align} $$
for any
$g\in C(X)$
and
$x_0\in X$
. This is another dynamical generalization of the PNT. In [Reference Loyd15, Theorem 1.3], Loyd discovered that the sequences capturing the behaviors of the Erdős–Kac theorem and Bergelson–Richter’s theorem are asymptotically uncorrelated, that is,
$$ \begin{align*} \lim_{N \to \infty} \frac1N\sum_{n=1}^N F \Big( \frac{\Omega(n) - \log \log N }{\sqrt{\log \log N}} \Big) g(T^{\Omega(n) }x_0) =\Big(\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(t) e^{-t^2/2} \, dt\Big)\Big( \int_X g \, d\nu \Big) \end{align*} $$
holds for any compactly supported continuous function F on
$\mathbb {R}$
,
$g\in C(X)$
and
$x_0\in X$
. In [Reference Wang24–Reference Wang and Yi26], the authors also established some analogs of (1.9) related to
$\mu ^2(n^2+n)$
,
$\mu ^2(n^2+1),$
and
$\Omega (n)-\omega (n)$
, respectively.
Inspired by these works, we will establish an analog of (1.9) with respect to
$\mu _{k,m}$
in our second main result. Similar to [Reference Wang24], we consider an arithmetic function
${b: \mathbb {N}\to \mathbb {C}}$
of invariant average under multiplications, which means that the average of
$b(n)$
exists and satisfies
$$ \begin{align*} \lim_{N\to\infty}\frac1N\sum_{n=1}^N b(mn)= \lim_{N\to\infty}\frac1N\sum_{n=1}^N b(n) \end{align*} $$
for all
$m\in \mathbb {N}$
. For any
$g\in C(X)$
and
$x_0\in X$
in (1.8), the orbit
$\{g(T^{\Omega (n)}x_0)\}_{n\in \mathbb {N}}$
is a typical example of a sequence of invariant average
$\int _X g \,d\nu $
under multiplications. We refer readers to [Reference Wang24] for more interesting examples.
Theorem 1.2 Let
$b(n)$
be a bounded arithmetic function of invariant average
$\beta $
under multiplications. Then we have
$$ \begin{align} \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N \mu_{k,m}(n) b(n) = A_{k,m}\cdot\beta \end{align} $$
for
$m\geq k\ge 2$
, where
$A_{k,m}$
is defined by (1.3).
Taking
$b(n)=g(T^{\Omega (n)}x_0)$
in Theorem 1.2, we obtain the following analog of (1.9) with respect to
$\mu _{k,m}$
, immediately.
Corollary 1.3 Let
$(X, \nu , T)$
be a uniquely ergodic topological dynamical system. Then we have
$$ \begin{align} \lim_{N\to\infty}\frac1N \sum_{n=1}^N \mu_{k,m}(n) g(T^{\Omega(n)}x_0)= A_{k,m}\cdot \int_X g \,d\nu \end{align} $$
for any
$m\geq k\ge 2$
,
$g\in C(X),$
and
$x_0\in X$
, where
$A_{k,m}$
is defined by (1.3).
Third, we consider an analog of the Titchmarsh divisor problem with respect to
$\mu _{k,m}$
. In 1930, under the Generalized Riemann Hypothesis, Titchmarsh [Reference Titchmarsh21, Theorem 7] proved that
$$ \begin{align} \sum_{p\leq x} \tau(p-a)= \frac{\zeta(2)\zeta(3)}{\zeta(6)}\prod_{p\mid a}\left(1-\frac{p}{p^2-p+1}\right)\cdot x+O\left(\frac{x\log\log x}{\log x}\right) \end{align} $$
for any fixed nonzero integer a. In 1961, Linnik [Reference Linnik14, Chapter VIII] proved (1.12) unconditionally by applying his dispersion method. Later, Halberstam [Reference Halberstam10] gave a very short proof using the Bombieri–Vinogradov theorem. A number of generalizations and analogs of the Titchmarsh divisor problem have been established in the literature (see [Reference Akbary and Ghioca1, Reference Assing, Blomer and Li3, Reference Bombieri, Friedlander and Iwaniec6, Reference Drappeau and Topacogullari8, Reference Pitt19, Reference Pollack20, Reference Virdol23]). In our third main theorem, we will use the Siegel–Walfisz theorem to show an analog of (1.12) for
$\mu _{k,m}$
.
Theorem 1.4 Let
$m\geq k\ge 2$
. For any
$A>0$
, we have
$$ \begin{align} \sum_{p\leq x} \mu_{k,m}(p-1) = C_{k,m}\operatorname{li}(x) + O\left(\frac{x}{(\log x)^A}\right), \end{align} $$
where
$\operatorname {li}(x)=\int _2^x\frac 1{\log t}dt$
and
$$\begin{align*}C_{k,m}=\prod_{p}\Big(1-\frac{1}{p^{k-1}(p-1)}-\frac1{p^m}\Big). \end{align*}$$
The implied constant depends on
$k,m,$
and A only.
In Section 2, we will introduce some basic knowledge on holomorphic cusp forms and uniquely ergodic topological dynamical systems. Then in Section 3, we will use (1.1) to build a relationship between the partial sum with weight
$\mu _{k,m}$
and the standard partial sum. In the end, in Sections 4–6, we will use this relationship to show Theorems 1.1, 1.2, and 1.4.
Notations: We always use letter p to denote primes. The notations
$f(n)=O_{k,m,\varepsilon , \dots }(g(n))$
or
$f(n)\ll _{k,m,\varepsilon ,\dots }g(n)$
mean that there is a positive constant
$C=C(k,m,\varepsilon ,\dots )>0$
such that
$|f(n)|\le C|g(n)|$
for all
$n\ge 1$
. Here, C is called the implied constant of this O-term. Sometimes we write
$f=O(g)$
or
$f\ll g$
for simplicity, if the implied constant depends at most on some fixed positives.
2 Preliminaries
2.1 Holomorphic cusp forms
Let
$f \in H_\kappa $
. The Fourier expansion of
$f(z)$
at cusp
$\infty $
is given by
$$ \begin{align*} f(z)=\sum_{n=1}^\infty \lambda_f(n)n^{\frac{\kappa-1}{2}}e(nz), \end{align*} $$
where
$e(z)=e^{2\pi i z}$
. The Fourier coefficients
$\lambda _f(n)$
are real and satisfy the following multiplicative property:
$$ \begin{align*} \lambda_f(m)\lambda_f(n)=\sum_{d\mid (m,n)} \lambda_f\big(\frac{mn}{d^2}\big) \end{align*} $$
for any integers
$m,n\ge 1$
. By the celebrated work of Deligne [Reference Deligne7, Theorem (8.2)], the Ramanujan–Petersson conjecture
$$ \begin{align} |\lambda_f(n)|\leq \tau(n) \end{align} $$
holds for all
$n\ge 1$
. Moreover, the L-function associated with
$f(z)$
is given by
$$\begin{align*}L(s, f) = \sum_{n=1}^{\infty} \frac{\lambda_f(n)}{n^s} = \prod_p \left( 1 - \frac{\alpha_p}{p^s} \right)^{-1} \left( 1 - \frac{\beta_p}{p^s} \right)^{-1}, \end{align*}$$
where the local parameters
$\alpha _p$
and
$\beta _p$
satisfy
$\alpha _p + \beta _p = \lambda _f(p)$
and
$\alpha _p \beta _p = 1$
, and
$L(s, f)$
is absolutely convergent for
$\Re s> 1$
.
For any integer
$m\ge 1$
, the mth symmetric power L-function is defined by the following degree m + 1 Euler product:
$$\begin{align*}L(s, \text{sym}^m f) = \prod_p \prod_{0 \leq j \leq m} \left( 1 - \frac{\alpha_p^{m-j} \beta_p^j}{p^s} \right)^{-1}. \end{align*}$$
By the result of Newton and Thorne [Reference Newton and Thorne18, Theorem A], it can be continued analytically over all of ℂ for m ≥ 1. Let χ be a primitive Dirichlet character modulo q. By the Rankin–Selberg theory, the twisted mth symmetric power L-function, which is defined by
$$\begin{align*}L(s, \text{sym}^m f \otimes \chi) = \prod_p \prod_{0 \leq j \leq m} \left( 1 - \frac{\alpha_p^{m-2j} \chi(p)}{p^s} \right)^{-1} \end{align*}$$
for
$\Re s> 1$
, can be analytically continued over all of
$\mathbb {C}$
for m ≥ 1 as well.
2.2 Uniquely ergodic topological dynamical systems
A topological dynamical system is a pair
$(X, T)$
for a compact metric space X and a continuous map
$T\colon X\to X$
. A Borel probability measure
$\nu $
on X is said to be T-invariant if
$\nu (T^{-1}A)=\mu (A)$
for all Borel measurable subsets
$A\subset X$
. By the Bogolyubov–Krylov Theorem, every topological dynamical system
$(X, T)$
possesses at least one T-invariant Borel probability measure. If
$(X, T)$
admits only one such measure, then it is called uniquely ergodic. It is well-known that
$(X, \nu , T)$
is uniquely ergodic if and only if
$$ \begin{align} \lim_{N \to \infty} \frac1N\sum_{n=1}^N f(T^n x) = \int_X f \, d\nu \end{align} $$
holds for all
$x \in X$
and
$f\in C(X)$
. For example, let
$X=\{0,1\}$
be a rotation on two points with
$\nu (\left \{0\right \})=\nu (\left \{1\right \})=1/2$
and
$T: x\mapsto x+1 \mod {2}$
. Then
$(X, \nu , T)$
is a uniquely ergodic topological dynamical system, and it is the simplest one.
In Bergelson and Richter’s theorem (1.8), taking
$(X,T)$
as a rotation on two points, we get the following equivalent form of the PNT:
$$ \begin{align} \lim_{N\to\infty} \frac1N\sum_{n=1}^N \lambda(n) = 0. \end{align} $$
If we take
$(X,T)$
as a rotation on two points in (1.9), then we obtain another well-known equivalent form of the PNT:
$$ \begin{align} \lim_{N\to\infty} \frac1N\sum_{n=1}^N \mu(n) = 0. \end{align} $$
Thus, either (1.8) or (1.9) provides a dynamical generalization of the PNT. Bergelson and Richter’s theorem also unifies some other number-theoretic results from a dynamical point of view. We refer readers to [Reference Bergelson and Richter5] for more interesting applications of (1.8) and (1.9).
3 Partial sum with weight
$\mu _{k,m}$
In the following sections, we always let
$m\ge k \ge 2$
be integers. To prove Theorems 1.1, 1.2, and 1.4, we build a relationship between the partial sum with weight
$\mu _{k,m}$
and the standard partial sum.
Proposition 3.1 Let
$g: \mathbb {N}\to \mathbb {C}$
be an arithmetic function such that
$g(n)\ll _\varepsilon n^\varepsilon $
for any
$\varepsilon>0$
. Then for any
$\varepsilon>0$
and
$H\leq x^{\frac 1{m+k}}$
, we have
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n)g(n) = \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{t\leq \frac{x}{d^ml^kr}} g(d^ml^krt) + O\left(\frac{x^{1+\varepsilon}}{H^{k-1}}\right), \end{align} $$
where the implied constant depends on
$\varepsilon $
only.
Moreover, if
$g: \mathbb {N}\to \mathbb {C}$
is bounded, then
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n)g(n) = \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{t\leq \frac{x}{d^ml^kr}} g(d^ml^krt) + O\left(\frac{x}{H^{k-1}}\right), \end{align} $$
where the implied constant depends on
$k,m, $
and
$\Vert g\Vert _{\infty }$
only.
Proof Since the proof of (3.2) is similar to that of (3.1), we will only show that (3.1) holds. First, using the relation (1.1), we observe that
$$ \begin{align*} \sum_{n\leq x} \mu_{k,m}(n)g(n) &= \sum_{n\leq x} \sum_{\substack{d^m\delta=n \\ (d,\delta)=1}} \mu(d)q_k(\delta)g(n) \nonumber\\ &= \sum_{\substack{d^m\delta \leq x \\ (d,\delta)=1}} \mu(d)q_k(\delta)g(d^m\delta) \nonumber\\ &= \sum_{d\leq x^{\frac1m}} \mu(d) \sum_{ \substack{\delta\leq \frac{x}{d^m} \nonumber\\ (\delta,d)=1}} q_k(\delta)g(d^m\delta). \end{align*} $$
Then for
$H\leq x^{\frac 1{m+k}}$
, we split up the summation into two parts according to the size of the modulus d:
To estimate
$S_2$
, by
$|\mu (d)|\leq 1, g(n)\ll n^\varepsilon $
, and
$0\leq q_k(\delta ) \leq 1$
,
$$ \begin{align*} S_2 &\ll_\varepsilon \sum_{H< d\leq x^{\frac1m}} \sum_{ \substack{\delta\leq \frac{x}{d^m} \\ (\delta,d)=1}} (d^m\delta)^\varepsilon \leq x^\varepsilon \sum_{H< d\leq x^{\frac1m}} \sum_{\delta\leq \frac{x}{d^m}} 1 \\ &\leq x^{1+\varepsilon} \sum_{H< d\leq x^{\frac1m}} \frac{1}{d^m} \ll \frac{x^{1+\varepsilon}}{(m-1)H^{m-1}}, \end{align*} $$
which implies that
$$ \begin{align} S_2=O_{\varepsilon}\left(\frac{x^{1+\varepsilon}}{H^{m-1}}\right). \end{align} $$
Now, we estimate
$S_1$
as follows. For
$q_k(n)$
, it is well-known that
$$\begin{align*}q_k(n) = \sum_{d^k\mid n} \mu(d). \end{align*}$$
This implies that
$$\begin{align*}S_1 = \sum_{d\leq H} \mu(d) \sum_{ \substack{\delta\leq \frac{x}{d^m} \\ (\delta,d)=1}} \Big(\sum_{l^k\mid \delta} \mu(l) \Big)g(d^m\delta). \end{align*}$$
Writing
$\delta =l^kt$
, then by
$(\delta , d)=1,$
we have
$(l,d)=1$
and
$(t,d)=1$
. So
$$\begin{align*}S_1 =\sum_{d\leq H} \mu(d) \sum_{\substack{ l\leq \frac{x^{1/k}}{d^{m/k}}\\ (l,d)=1}} \mu(l) \sum_{\substack{ t\leq \frac{x}{d^ml^k}\\ (t,d)=1}} g( d^ml^kt). \end{align*}$$
By
$$\begin{align*}\sum_{d\mid n} \mu(d) = \begin{cases} 1 & \text{ if } n=1 \\ 0 & \text{ if } n>1, \end{cases} \end{align*}$$
the condition
$(t,d)=1$
can be by replaced by
$\sum _{r\mid (t,d)}\mu (r)$
. We get that
$$ \begin{align*} S_1 &= \sum_{d\leq H} \mu(d) \sum_{\substack{ l\leq \frac{x^{1/k}}{d^{m/k}}\\ (l,d)=1}} \mu(l) \sum_{ t\leq \frac{x}{d^ml^k}} g( d^ml^kt) \sum_{r\mid (t,d)}\mu(r) \nonumber\\ &= \sum_{d\leq H} \mu(d) \sum_{\substack{ l\leq \frac{x^{1/k}}{d^{m/k}}\\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{ t\leq \frac{x}{d^ml^kr}} g( d^ml^krt). \end{align*} $$
Similar to (3.3), we divide the summation into two parts according to the size of the modulus l:
For
$S_4$
, similar to the estimation of
$S_2$
, we have
$$ \begin{align*} S_4 & \ll_\varepsilon x^\varepsilon \sum_{d\leq H} \sum_{\substack{ H<l\leq \frac{x^{1/k}}{d^{m/k}}\\ (l,d)=1}} \sum_{r\mid d} \sum_{ t\leq \frac{x}{d^ml^kr}} 1 \\ &\leq x^{1+\varepsilon} \sum_{d\leq H} \frac1{d^m}\sum_{\substack{ H<l\leq \frac{x^{1/k}}{d^{m/k}}\\ (l,d)=1}} \frac{1}{l^k} \sum_{r\mid d} \frac1r \\ &\leq x^{1+\varepsilon} \sum_{d\leq H} \frac{\sigma_{-1}(d)}{d^{m}} \sum_{l>H} \frac{1}{l^k} \\ &\ll \frac{x^{1+\varepsilon}}{(k-1)H^{k-1}}, \end{align*} $$
where
$\sigma _s(n)=\sum _{d\mid n} d^{s}$
. This gives us that
$$ \begin{align} S_1 = \sum_{d\leq H} \mu(d) \sum_{\substack{ l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{ t\leq \frac{x}{d^ml^kr}} g( d^ml^krt) + O_{\varepsilon}\left(\frac{x^{1+\varepsilon}}{H^{k-1}}\right). \end{align} $$
As a direct application of Proposition 3.1, taking
$g(n)=1$
for all
$n\geq 1$
in (3.2), we obtain an estimate for the partial sum of
$\mu _{k,m}$
with a weaker error term than Bege’s result (1.2).
Theorem 3.2 For
$m\ge k\ge 2$
, we have
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n) =A_{k,m} x + O\big(x^{\frac{m+1}{m+k}} \big). \end{align} $$
Proof By (3.2) of Proposition 3.1, for
$H\leq x^{\frac 1{m+k}}$
, we have
$$ \begin{align*} \sum_{n\leq x} \mu_{k,m}(n) &= \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{t\leq \frac{x}{d^ml^kr}} 1 + O\left(\frac{x}{H^{k-1}}\right) \nonumber \\ & = \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \Big(\frac{x}{d^ml^kr} +O(1) \Big) + O\left(\frac{x}{H^{k-1}}\right) \nonumber \\ &= x \sum_{d\leq H} \frac{\mu(d)}{d^{m}} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{l^k} \sum_{r\mid d} \frac{\mu(r)}{r} \\ & \qquad + O\left(\sum_{d\leq H} \tau(d) \sum_{\substack{l\leq H \\ (l,d)=1}}1\right) + O\left(\frac{x}{H^{k-1}}\right). \nonumber \end{align*} $$
By the well-known facts that
$$\begin{align*}\sum_{d\mid n}\frac{\mu(d)}{d}=\frac{\varphi(n)}{n}, \end{align*}$$
where
$\varphi (n)$
is the Euler totient function, and
$\sum _{d\leq H} \tau (d) \ll H\log H$
, we get that
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n) &= x \sum_{d\leq H} \frac{\mu(d)\varphi(d)}{d^{m+1}} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{l^k} + O(H^2\log H) + O\left(\frac{x}{H^{k-1}}\right) \nonumber \\ &= x \sum_{d=1}^\infty \frac{\mu(d)\varphi(d)}{d^{m+1}} \sum_{\substack{l=1 \\ (l,d)=1}}^\infty \frac{\mu(l)}{l^k} + O(H^2\log H) + O\left(\frac{x}{H^{k-1}}\right). \end{align} $$
Here, we have used the fact that the tail of the first term of (3.7) is bounded by
$$ \begin{align*}\Big| x \sum_{d\leq H} \frac{\mu(d)\varphi(d)}{d^{m+1}} \sum_{\substack{l> H \\ (l,d)=1}} \frac{\mu(l)}{l^k} \Big| \leq x \sum_{d=1}^\infty \frac{\varphi(d)}{d^{m+1}} \sum_{d>H} \frac{1}{l^k} \ll \frac{x}{H^{k-1}}.\end{align*} $$
For the coefficient of the first term of (3.7),
$$ \begin{align} \sum_{d=1}^\infty \frac{\mu(d)\varphi(d)}{d^{m+1}} \sum_{\substack{l=1 \\ (l,d)=1}}^\infty \frac{\mu(l)}{l^k} & = \sum_{d=1}^\infty \frac{\mu(d)\varphi(d)}{d^{m+1}} \cdot \prod_{p\nmid d} \Big(1-\frac1{p^{k}}\Big) \nonumber\\ &= \frac1{\zeta(k)} \sum_{d=1}^\infty \frac{\mu(d)\varphi(d)}{d^{m+1}} \cdot \prod_{p\mid d} \Big(1-\frac1{p^{k}}\Big)^{-1} \nonumber\\ &= \frac1{\zeta(k)} \prod_{p} \Big(1- \frac{1}{p^m}\cdot\frac{1-\frac1p}{1-\frac1{p^{k}}}\Big)\nonumber\\ &= \frac1{\zeta(k)} \prod_{p} \Big(1- \frac{1}{p^m}\cdot\frac{1}{1+\frac1p+\cdots+\frac1{p^{k-1}}}\Big), \end{align} $$
which is equal to
$A_{k,m}$
by the definition (1.3). By (3.7), this implies that
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n)= A_{k,m} x + O(H^2\log H) + O\left(\frac{x}{H^{k-1}}\right). \end{align} $$
Thus, (3.6) follows by taking
$H=x^{\frac 1{m+k}}$
in (3.9).
In the following three sections, we will use Proposition 3.1 to prove Theorems 1.1, 1.2, and 1.4.
4 Proof of Theorem 1.1
To prove Theorem 1.1, we cite a proposition in the work [Reference Lü and Wang16] of Lü and Wang. This is also the main tool they used to establish the asymptotic formula (1.6) on the shifted convolution sums of Fourier coefficients with squarefull kernel functions.
Proposition 4.1 [Reference Lü and Wang16, Proposition 1]
Let
$f \in H_{\kappa }$
. Then for any
$\epsilon> 0$
and
$q < x^{1/4}$
, we have
$$\begin{align*}\sum_{\substack{n \leq x+1 \\ n\equiv1\mod{q}}} \lambda_f^2(n) =\frac{x}{q} L(1, \operatorname{sym}^2f \otimes \chi_0) L(2,\chi_0^2)^{-1} + O\left(\frac{x^{3/5+\varepsilon}}{\varphi(q)}\right) + O(x^{1/2+\varepsilon}), \end{align*}$$
where
$\chi _0$
is the principal Dirichlet character modulo q.
Proof of Theorem 1.1
For any
$\varepsilon>0$
and
$H\leq x^{\frac 1{m+k}}$
, taking
$g(n) =\lambda _f^2(n+1)$
in (3.1) of Proposition 3.1,
$$ \begin{align} &\quad\sum_{n\leq x} \mu_{k,m}(n) \lambda_f^2(n+1) \nonumber\\ &= \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{t\leq \frac{x}{d^ml^kr}} \lambda_f^2(d^ml^krt+1) + O\left(\frac{x^{1+\varepsilon}}{H^{k-1}}\right) \nonumber\\ &=\sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{\substack{n \leq x+1 \\ n\equiv1\mod{d^ml^kr}}} \lambda_f^2(n) + O\left(\frac{x^{1+\varepsilon}}{H^{k-1}}\right). \end{align} $$
Suppose
$H\leq x^{\frac 1{4(m+k+1)}-\varepsilon }$
, then
$d^ml^kr\leq H^{m+k+1} < x^{1/4}$
. Set
$A_f= L(1, \operatorname {sym}^2f \otimes \chi _0) L(2,\chi _0^2)^{-1}$
. By Proposition 4.1, we have
$$ \begin{align} &\quad\sum_{n\leq x} \mu_{k,m}(n) \lambda_f^2(n+1) \nonumber\\ &=\sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \cdot \nonumber \\ &\quad\qquad\Big( \frac{A_f}{d^ml^kr} x + O\left(\frac{x^{3/5+\varepsilon}}{\varphi(d^ml^kr)}\right) + O(x^{1/2+\varepsilon}) \Big) + O\left(\frac{x^{1+\varepsilon}}{H^{k-1}}\right) \nonumber\\ &= A_fA_{k,m } \cdot x + O(x^{3/5+\varepsilon}) + O(x^{1/2+\varepsilon}H^2\log H) + O\left(\frac{x^{1+\varepsilon}}{H^{k-1}}\right). \end{align} $$
Here, we have used the following estimates for the first three terms in (4.2), respectively:
$$\begin{align*}\sum_{d\leq H}\frac{\mu(d)}{d^m}\sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{l^k} \sum_{r\mid d} \frac{\mu(r)}{r} = A_{k,m} +O\left(\frac1{H^{k-1}}\right), \end{align*}$$
$$\begin{align*}\sum_{d\leq H} \sum_{\substack{l\leq H \\ (l,d)=1}} \sum_{r\mid d} \frac1{\varphi(d^ml^kr)} < \sum_{d=1}^\infty \sum_{r\mid d} \frac1{\varphi(d^mr)} \sum_{l=1}^\infty \frac1{\varphi(l^k)} <\infty, \end{align*}$$
and
$$ \begin{align} \Big|\sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \Big| \leq \sum_{d\leq H} \tau(d) \sum_{l\leq H}1\ll H^2\log H. \end{align} $$
Thus, taking
$H=x^{\frac 1{4(m+k+1)}-\varepsilon }$
in (4.2), we conclude that
$$ \begin{align} \sum_{n\leq x} \mu_{k,m}(n) \lambda_f^2(n+1) = A_{k,m,f} \cdot x + O\Big(x^{1-\frac{k-1}{4(m+k+1)}+\varepsilon}\Big), \end{align} $$
where
This completes the proof of Theorem 1.1.
5 Proof of Theorem 1.2
In this section, we use (3.2) of Proposition 3.1 and the definition of invariant average under multiplications to prove Theorem 1.2.
Proof of Theorem 1.2
Since
$b(n)$
is of invariant average
$\beta $
under multiplications, we have
$$ \begin{align} \lim_{N\to\infty}\frac1N\sum_{n=1}^N b(mn)= \beta \end{align} $$
for all
$m\in \mathbb {N}$
. Then by (3.2) of Proposition 3.1, for
$H\leq N^{\frac 1{m+k}}$
, we find that
$$ \begin{align*} &\quad \sum_{n\leq N} \mu_{k,m}(n)b(n) \nonumber\\ & = \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{t\leq \frac{N}{d^ml^kr}} b(d^ml^krt) + O\left(\frac{N}{H^{k-1}}\right) \nonumber\\ & = \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \Big(\frac{N}{d^ml^kr} +O(1) \Big) \frac1{\lfloor\frac{N}{d^ml^kr}\rfloor}\sum_{t\leq \frac{N}{d^ml^kr}} b(d^ml^krt) + O\left(\frac{N}{H^{k-1}}\right) \nonumber \\ & = N \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{l^k} \sum_{r\mid d} \frac{\mu(r)}{r} \frac1{\lfloor\frac{N}{d^ml^kr}\rfloor}\sum_{t\leq \frac{N}{d^ml^kr}} b(d^ml^krt) + O(H^2\log H) + O\left(\frac{N}{H^{k-1}}\right),\nonumber \end{align*} $$
where
$\lfloor x \rfloor := \max \{n\in \mathbb {Z}: n\leq x\}$
denotes the integral part of a real number x. It follows that
$$ \begin{align} \frac1N\sum_{n\leq N} \mu_{k,m}(n)b(n)= & \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{l^k} \sum_{r\mid d} \frac{\mu(r)}{r} \frac1{\lfloor\frac{N}{d^ml^kr}\rfloor}\sum_{t\leq \frac{N}{d^ml^kr}} b(d^ml^krt) \nonumber\\ &\qquad\quad + O\left(\frac{H^2\log H}{N}\right) + O\left(\frac{1}{H^{k-1}}\right). \end{align} $$
Fix H, taking
$N\to \infty $
in (5.2), by (5.1), we get that
$$ \begin{align} \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N \mu_{k,m}(n) b(n) = \beta \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{l^k} \sum_{r\mid d} \frac{\mu(r)}{r} + O\left(\frac{1}{H^{k-1}}\right). \end{align} $$
Then taking
$H\to \infty $
gives us that
$$ \begin{align} \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N \mu_{k,m}(n) b(n) = \beta \sum_{d=1}^\infty \frac{\mu(d)}{d^m} \sum_{\substack{l=1\\ (l,d)=1}}^\infty \frac{\mu(l)}{l^k} \sum_{r\mid d} \frac{\mu(r)}{r}, \end{align} $$
which is equal to
$A_{k,m}\cdot \beta $
by (3.8), as desired. This completes the proof of Theorem 1.2.
6 Proof of Theorem 1.4
In this section, we use (3.2) of Proposition 3.1 and the Siegel–Walfisz theorem to prove Theorem 1.4. For integers
$1\leq a\leq q$
with
$(a,q)=1$
, let 
be the prime-counting function in arithmetic progressions.
Theorem 6.1 (The Siegel–Walfisz theorem, [Reference Montgomery and Vaughan17, Corollary 11.21])
There is a constant
$c_1>0$
such that for any given
$A>0$
, if
$q\leq (\log x)^A$
and
$(a,q)=1$
, then
$$ \begin{align} \pi(x; q, a)= \frac{\operatorname{li}(x)}{\varphi(q)} + O_A\left(x\exp(-c_1\sqrt{\log x})\right). \end{align} $$
Proof of Theorem 1.4
Let
$\mathbb {P}$
be the set of all primes. Denote by
$1_{\mathbb {P}}$
the characteristic function of primes defined by
$$\begin{align*}1_{\mathbb{P}}(n) = \begin{cases} 1 & \text{ if } n\in \mathbb{P},\\ 0 & \text{ if } n\notin \mathbb{P}. \end{cases} \end{align*}$$
Then
$$\begin{align*}\sum_{p\leq x} \mu_{k,m}(p-1) = \sum_{n\leq x-1} \mu_{k,m}(n) 1_{\mathbb{P}}(n+1) \text{ and } \sum_{t\leq \frac{x}{q}} 1_{\mathbb{P}} (qt+1)= \pi(x+1; q,1). \end{align*}$$
Taking
$g(n)=1_{\mathbb {P}}(n+1)$
in (3.2) of Proposition 3.1, for
$H\leq (x-1)^{\frac {1}{m+k}}$
, we have
$$ \begin{align} \sum_{p\leq x} \mu_{k,m}(p-1) &= \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \sum_{t\leq \frac{x}{d^ml^kr}} g(d^ml^krt) + O\left(\frac{x}{H^{k-1}}\right) \nonumber\\ &= \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \pi(x+1; d^ml^kr,1 ) + O\left(\frac{x}{H^{k-1}}\right). \end{align} $$
Since
$r\mid d$
, we have
$\varphi (d^mr)=r\varphi (d^m)$
. Moreover, by
$(l,d)=1$
, we get that
$\varphi (d^ml^kr)=\varphi (d^mr)\varphi (l^k)=r\varphi (d^m)\varphi (l^k)$
. Clearly,
$\varphi (d^m)=d^{m-1}\varphi (d)$
. Take
$H=(\log x)^{\frac {A}{k-3/2}}$
, then
$d^ml^kr\leq H^{m+k+1}$
. We invoke the Siegel–Walfisz theorem in (6.2) and the upper bound in (4.3) to deduce that
$$ \begin{align*} &\quad \sum_{p\leq x} \mu_{k,m}(p-1) \nonumber\\ &= \sum_{d\leq H} \mu(d) \sum_{\substack{l\leq H \\ (l,d)=1}} \mu(l) \sum_{r\mid d} \mu(r) \Big(\frac{\operatorname{li}(x)}{\varphi(d^ml^kr)} +O\big(x\exp(-c_1\sqrt{\log x})\big)\Big) + O\left(\frac{x}{H^{k-1}}\right)\nonumber\\ &=\operatorname{li}(x) \sum_{d\leq H} \frac{\mu(d)}{\varphi(d^m)} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{\varphi(l^k)} \sum_{r\mid d} \frac{\mu(r)}{r} +O\Big(x\exp(-c_1\sqrt{\log x}) H^2\log H\Big) \nonumber\\ &\qquad + O\left(\frac{x}{H^{k-1}}\right). \nonumber \end{align*} $$
Since
$c_1>0$
, we have
$\exp (c_1\sqrt {\log x})\gg (\log x)^{AC/(k-3/2)}=H^C$
for any positive constant
$C>0$
. So the error term
$O\Big (x\exp (-c_1\sqrt {\log x}) H^2\log H\Big )$
can be absorbed in
$O\left ({x}/{H^{k-1}}\right )$
, and
$$ \begin{align} \sum_{p\leq x} \mu_{k,m}(p-1)=\operatorname{li}(x) \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{\varphi(l^k)} + O\left(\frac{x}{H^{k-1}}\right). \end{align} $$
Now, we estimate the first term of (6.3). By [Reference Hardy and Wright11, Theorem 327], we have
$\frac {l}{\varphi (l)}\ll l^{1/2}$
, so
$$ \begin{align} \Big|\sum_{\substack{l> H \\ (l,d)=1}} \frac{\mu(l)}{\varphi(l^k)}\Big| \leq \sum_{l > H} \frac{1}{l^{k-1}\varphi(l)} \ll\sum_{l> H}\frac{1}{l^{k-\frac12}} \ll \frac1{H^{k-\frac32}}. \end{align} $$
Then
$$ \begin{align} \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{\varphi(l^k)} &= \sum_{d\leq H} \frac{\mu(d)}{d^m} \Big( \sum_{\substack{l=1\\ (l,d)=1}}^\infty\frac{\mu(l)}{\varphi(l^k)} -\sum_{\substack{l> H \\ (l,d)=1}} \frac{\mu(l)}{\varphi(l^k)} \Big)\nonumber\\ &= \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l=1\\ (l,d)=1}}^\infty\frac{\mu(l)}{\varphi(l^k)} +O\left(\frac1{H^{k-\frac32}}\right) \nonumber \\ &= \sum_{d=1}^\infty \frac{\mu(d)}{d^m} \sum_{\substack{l=1\\ (l,d)=1}}^\infty\frac{\mu(l)}{\varphi(l^k)} +O\left(\sum_{d>H} \frac{1}{d^m}\right) +O\left(\frac1{H^{k-\frac32}}\right) \nonumber \\ &= \sum_{d=1}^\infty \frac{\mu(d)}{d^m} \sum_{\substack{l=1\\ (l,d)=1}}^\infty\frac{\mu(l)}{\varphi(l^k)} +O\left(\frac1{H^{k-\frac32}}\right) \end{align} $$
due to the fact that
$\sum _{d>H} {1}/{d^m} \ll 1/H^{m-1} < 1/ H^{k-3/2}$
for
$m\geq k\ge 2$
.
As regards the double summation over d and l in (6.5), we have
$$ \begin{align} &\quad\sum_{d=1}^\infty \frac{\mu(d)}{d^m} \sum_{\substack{l=1\\ (l,d)=1}}^\infty\frac{\mu(l)}{\varphi(l^k)} \nonumber\\ &= \sum_{d=1}^\infty \frac{\mu(d)}{d^m} \prod_{p\nmid d} \Big(1-\frac{1}{p^{k-1}(p-1)}\Big) \nonumber\\&= \prod_{p} \Big(1-\frac{1}{p^{k-1}(p-1)}\Big) \cdot \sum_{d=1}^\infty \frac{\mu(d)}{d^m} \prod_{p\mid d} \Big(1-\frac{1}{p^{k-1}(p-1)}\Big)^{-1} \nonumber\\ &= \prod_{p} \Big(1-\frac{1}{p^{k-1}(p-1)}\Big) \cdot \prod_p \Big(1-\frac1{p^m}\Big(1-\frac{1}{p^{k-1}(p-1)}\Big)^{-1}\Big)\nonumber\\ &=\prod_{p}\Big(1-\frac{1}{p^{k-1}(p-1)}-\frac1{p^m}\Big), \end{align} $$
which is equal to
$C_{k,m}$
. It follows by the trivial estimate
$\operatorname {li}(x)\ll x$
that
$$ \begin{align} \operatorname{li}(x) \sum_{d\leq H} \frac{\mu(d)}{d^m} \sum_{\substack{l\leq H \\ (l,d)=1}} \frac{\mu(l)}{\varphi(l^k)} = C_{k,m} \operatorname{li}(x) + O\left(\frac{x}{H^{k-\frac32}}\right). \end{align} $$
Thus, (1.13) follows by plugging
$H=(\log x)^{\frac {A}{k-3/2}}$
into (6.3) and (6.7). This completes the proof of Theorem 1.4.
Acknowledgements
The author is deeply grateful to the referees for a very careful review and a number of precise comments and helpful suggestions, which have improved this article a lot.
