A new arithmetical function and its mean value properties

Abstract

In this article, we use elementary methods and the estimate for character sums to study the mean value properties of a new arithmetical function, and obtain a sharp asymptotic formula for it.

1 Introduction

Let p and q be two fixed odd primes, ${\mathbb {A}}(p)$ denotes the set of all primitive roots modulo p. As for the definition of the primitive roots and its properties, refer to references [3, 8, 11, 21] and [22], which will not be stated here. For any positive integer n, we define a new arithmetical function d(n; p, q) as follows:

$$\begin{aligned} d(n;p,q)=\mathop {\sum _{ab=n}}_{a\in {\mathbb {A}}(p),\ b\in {\mathbb {A}}(q) }1. \end{aligned}$$

If $p=q$, then we write $d(n;p,p)=d(n;p)$ for convenience. For example, $d(4; 3,5)=1$, since $4=2\times 2$, $2\in {\mathbb {A}}(3)$ and $2\in {\mathbb {A}}(5)$; $d(6; 3,5)=1$, since $6=2\times 3$, $2\in {\mathbb {A}}(3)$ and $3\in {\mathbb {A}}(5)$. If n is any prime, then we have $d(n; 3,5)=0$. As we can see from these examples, the function d(n; p, q) has an irregular distribution of values, and its value is zero on many integers n. Since this function is new, we know very little about its arithmetical properties. From the form, d(n; p, q) is a bit like the divisor function d(n), but there are fundamental differences between them. Because if n is any prime, then we have $d(n;p,q)=0$ and $d(n)=2$. And, of course, there are similarities between d(n; p, q) and d(n). For example, their mean values, which have almost similar asymptotic properties. The main purpose of this article is to illustrate this point. Why do we want to study the mean value properties of d(n; p, q)? The problem begins with a recent paper by the second author Teerapat Srichan. In fact, Teerapat Srichan [14] used the analytic methods to study the estimate problem of one kind of character sums:

$$\begin{aligned} S_{\chi _1,\chi _2}(x)=\sum _{ab\le x}\chi _1\left( a\right) \chi _2\left( b\right) , \end{aligned}$$

and obtained a sharp upper bound estimate for it (see Lemma 2 below), where $\chi _i$ is a primitive character modulo $p_i$, $i=1$, 2, $p_1$ and $p_2$ are two positive integers with $\sqrt{x}\ge \max \{p_1,p_2\}$.

The main purpose of this article is to give a specific application of [14]. Based on this idea, we studied the mean value properties of d(n; p, q), and proved the following main result:

Theorem 1

Let p and q be two fixed odd primes. Then for any positive number $\sqrt{x}\ge \max \{p, q\}$, we have the asymptotic formula

$$\begin{aligned}&\sum _{n\le x}d(n; p,q)=\frac{\phi (p-1)\phi (q-1)}{pq}\cdot x\cdot \left( \ln x + 2\gamma -1+ \frac{\ln p}{p-1}+ \frac{\ln q}{q-1}\right) \\&\quad +\frac{\phi (p-1)\phi (q-1)}{q(p-1)}\cdot x \cdot \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum }~'}_{\chi ^h=\chi _p^0}L\left( 1,\chi \right) \\&\quad +\frac{\phi (p-1)\phi (q-1)}{p(q-1)}\cdot x \cdot \mathop {\sum _{k\mid q-1}}_{k>1}\frac{\mu (k)}{\phi (k)}\mathop {{\sum }~'}_{\chi ^k=\chi _q^0}L\left( 1,\chi \right) \\&\quad + O\left( x^{\frac{1}{3}} p^{\frac{5}{9}} q^{\frac{7}{9}} 2^{\omega (p-1)+ \omega (q-1)} \ln p\right) +O\left( x^{\frac{1}{3}} q^{\frac{5}{9}} p^{\frac{7}{9}} 2^{\omega (p-1)+ \omega (q-1)} \ln q\right) , \end{aligned}$$

where $\phi (n)$ and $\gamma$ denote the Euler function and Euler constant, respectively, $\mu (n)$ is the Möbius function, $\omega (n)$ denotes the number of all distinct prime divisors of n, $L(s,\chi _p)$ denotes the Dirichlet L-function corresponding to character $\chi _p$ modulo p, $\displaystyle \mathop {{\sum }~'}_{\chi ^h=\chi _p^0}$ denotes the summation over all h-order characters modulo p, and $\chi _p^0$ denotes the principal character modulo p.

If $p=q$, then from this theorem we may immediately deduce the following:

Corollary 1

Let p be fixed odd prime. Then for any $x>p^2$, we have the asymptotic formula:

$$\begin{aligned}&\sum _{n\le x}d(n; p)=\frac{\phi ^2(p-1)}{p^2}\cdot x\cdot \left( \ln x + 2\gamma -1+ \frac{2\ln p}{p-1}\right) \\&\quad +2\cdot \frac{\phi ^2(p-1)}{p(p-1)}\cdot x \cdot \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum }~'}_{\chi ^h=\chi _p^0}L\left( 1,\chi \right) + O\left( x^{\frac{1}{3}} p^{\frac{4}{3}} 4^{\omega (p-1)}\ln p\right) . \end{aligned}$$

Some notes: It is clear that if $\sqrt{x}> \max \{p, q\}$, then our asymptotic formula is nontrivial. In addition, if one can give a valid calculation formula for the mean value of the Dirichlet L-functions

$$\begin{aligned} \mathop {{\sum }~'}_{\chi ^h=\chi _p^0}L\left( 1,\chi \right) \ \text {and}\ \mathop {{\sum }~'}_{\chi ^k=\chi _q^0}L\left( 1,\chi \right) , \end{aligned}$$

that will make our theorem more concise and beautiful.

2 Several lemmas

To complete the proof of our main result, we need following several simple lemmas. Of course, some of the lemmas can be found in the references, and the proofs of several other lemmas require some elementary and analytic number theory knowledge. In particular, the contents of primitive roots and Dirichlet characters modulo p are required. All these can be found in references [3, 11] and [21]. We also provide several relevant references [1, 2, 4, 6, 7, 10, 13] and [15-20] for interested readers. First we have the following:

Lemma 1

Let p be an odd prime, then for any integer a coprime to p (i.e., $(a,p)=1$), we have the identity

$$\begin{aligned} \frac{\phi (p-1)}{p-1}\sum _{k\mid p-1}\frac{\mu (k)}{\phi (k)}\mathop {{\sum _{r=1}^{k}}\ '} \ e\left( \frac{r\cdot \text {ind}(a)}{k}\right) =\left\{ \begin{array}{ll} 1 &{}\ \text { if }a\text { is a primitive root }\bmod ~p\text {;}\\ 0 &{}\ \text { if }a\text { is not a primitive root }\bmod ~p\text {,} \end{array}\right. \end{aligned}$$

where $e(y)=e^{2\pi iy}$, $\displaystyle \mathop {{\sum _{r=1}^{k}}\ '}$ denotes the summation over all integers $1\le r\le k$ such that r is coprime to k, $\mu (n)$ is the Möbius function, and $\text {ind}(a)$ denotes the index of a relative to some fixed primitive root $g\bmod p$.

Proof

See Proposition 2.2 in reference [11] . $\square$

Lemma 2

For $x>1$ we have

$$\begin{aligned} \sum _{n\le x} {1\over n}=\ln x +\gamma -\psi (x) x^{-1}+O(x^{-2}). \end{aligned}$$

Proof

See Equation 14.41 in reference [9]. $\square$

Lemma 3

Let l and k be positive integers. For $x>1$ we have

$$\begin{aligned} \sum _{\begin{array}{c}n\le x \\ n\equiv l \bmod k\end{array}} {1\over n}={\ln x\over k} +{\gamma (l,k)\over k} -\psi \Big ({x-l\over k}\Big ) x^{-1}+O(kx^{-2}), \end{aligned}$$

where

$$\begin{aligned} \gamma (l,k)=\lim _{x\rightarrow \infty }\left (\sum _{\begin{array}{c}n\le x \\ n\equiv l \bmod k\end{array}} {1\over n}-{\ln x\over k}\right ). \end{aligned}$$

Proof

It follows form using Euler’s summation formula. $\square$

Lemma 4

For any non-principal character $\chi$ modulo k. Define for $0<a\le 1$

$$\begin{aligned} \gamma (a)=\lim _{N\rightarrow \infty }\Big (\sum _{n=0}^\infty {1\over n+a}-\ln (N+a)\Big ). \end{aligned}$$

Then

$$\begin{aligned} L(1,\chi )={1\over k} \sum _{n=1}^{k-1}\chi (n)\gamma (n/k). \end{aligned}$$

Proof

See Lemma 5 in reference [5]. $\square$

Lemma 5

I. Let $\omega \in {\mathbb {R}},\, \omega >0,\ q\in {\mathbb {N}},\,q\ge 2$. If f(n) is an arithmetic function, then

$$\begin{aligned} \sum _{n\le \omega ,\, (n,q)=1}f(n)=\sum _{d\vert q}\mu (d)\sum _{m\le \omega /d}f(md). \end{aligned}$$

Proof

See Lemma 13 in reference [12]. $\square$

II. For real $x>1$ and any non-principal character $\chi$ modulo q, we have

$$\begin{aligned} \sum _{k\le x}\chi (k)&=\sum _{j\le q}\chi (j)\left\lfloor {x\over q}-{j\over q}+1 \right\rfloor , \end{aligned}$$

(1)

and for any arithmetical function f, 

$$\begin{aligned} \sum _{k\le x}\chi (k)\cdot f(k)=\sum _{j\le q}\chi (j)\sum _{\begin{array}{c}k\le x\\ k\equiv j \bmod q\end{array}} f(k). \end{aligned}$$

(2)

Proof

The equations (1) and (2) follow from the periodicity of the primitive character modulo q. Namely,

$$\begin{aligned} \sum _{a\le z}\chi (a)&=\sum _{j\le q}\sum _{\begin{subarray}{c}a\le z\\ {a\equiv j \, \bmod \, q}\end{subarray}}\chi (a)=\sum _{j\le q}\sum _{\begin{subarray}{c}a\le z\\ {a\equiv j \,\bmod \, q}\end{subarray}}\chi (j)\\&=\sum _{j\le q}\chi (j)\sum _{\begin{subarray}{c}a\le z\\ {a\equiv j \,\bmod \, q}\end{subarray}}1 =\sum _{j\le q}\chi (j)\left\lfloor {z\over q}-{j\over q}+1\right\rfloor . \end{aligned}$$

$\square$

III. Let $x,\eta ,\alpha ,\omega$ be real numbers with $x\ge 1, \alpha >0, \eta \ge 1$, let j and q be positive integers with $1\le j\le q$, let $(k,\ell )$ be an exponent pair with $k>0$ and let

$$\begin{aligned} R(x,\eta ,\alpha ;q,j;\omega )=\sum _{\begin{array}{c}n\le \eta \\ n\equiv j\bmod {q}\end{array}}\psi \left( \frac{x}{n^\alpha }+\omega \right) , \end{aligned}$$

where $\omega$ is independent of n. Then

$$\begin{aligned}&R(x,\eta ,\alpha ;q,j;\omega )=O(1)+O(x^{-1/2}\eta ^{1+\alpha /2}q^{-1})\\&\quad + {\left\{ \begin{array}{ll} O\left( x^\frac{k}{k+1}\eta ^\frac{\ell -\alpha k}{k+1}q^\frac{-\ell }{k+1}\right) &{}\hbox { for}\ \ell >\alpha k,\\ O\left( x^\frac{k}{k+1}\ln \eta \, q^\frac{-\alpha k}{k+1}\right) &{}\hbox { for}\ \ell =\alpha k,\\ O\left( (xq^{-\alpha })^\frac{k}{1+(1+\alpha )k-\ell }\right) &{}\hbox { for}\ \ell <\alpha k,\\ \end{array}\right. } \end{aligned}$$

where the constants in the O-symbols depend on only $\alpha$.

Proof

See Lemma 17 in reference [12]. $\square$

Lemma 6

Let p and q be two fixed primes, $\chi _p^0$ be the principal character modulo p and $\chi _q^0$ be the principal character modulo q. Then for any real number $x>\max \{p, q\}$, we have the asymptotic formula

$$\begin{aligned} \sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)&=\frac{(p-1)(q-1)}{pq}\cdot x\cdot \left( \ln x + 2\gamma -1+ \frac{\ln p}{p-1}+ \frac{\ln q}{q-1}\right) \\&\quad +O(x^{1/3}q^{7/9}p^{5/9}+x^{1/3}p^{7/9}q^{5/9}). \end{aligned}$$

Proof

For any $x>1$,

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)\\&\quad =\sum _{a\le \sqrt{x}}\chi _p^0(a)\sum _{b\le {x\over a}}\chi _q^0(b)+\sum _{b\le \sqrt{x}}\chi _q^0(b)\sum _{a\le {x\over b}}\chi _p^0(a)-\sum _{a\le \sqrt{x}}\chi _p^0(a)\sum _{b\le \sqrt{x}}\chi _q^0(b)\\&\quad =\sum _{\begin{array}{c}a\le \sqrt{x}\\ (a,p)=1\end{array}}\sum _{\begin{array}{c}b\le {x\over a}\\ (b,q)=1\end{array}}1+\sum _{\begin{array}{c}b\le \sqrt{x}\\ (b,q)=1\end{array}}\sum _{\begin{array}{c}a\le {x\over b}\\ (a,p)=1\end{array}}1-\sum _{\begin{array}{c}a\le \sqrt{x}\\ (a,p)=1\end{array}}1\sum _{\begin{array}{c}b\le \sqrt{x}\\ (b,q)=1\end{array}}1. \end{aligned}$$

In view of I in Lemma 5, we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)\\&\quad =\sum _{\begin{array}{c}a\le \sqrt{x}\\ (a,p)=1\end{array}}\sum _{d\mid q} \mu (d)\left\lfloor x\over ad\right\rfloor +\sum _{\begin{array}{c}b\le \sqrt{x}\\ (b,q)=1\end{array}}\sum _{t\mid p} \mu (t)\left\lfloor x\over bt\right\rfloor \\&\qquad -\Big (\sum _{t\mid p}\mu (t)\left\lfloor \sqrt{x}\over t\right\rfloor \Big )\Big (\sum _{d\mid q}\mu (d)\left\lfloor \sqrt{x}\over d\right\rfloor \Big )\\&\quad =\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\left\lfloor x\over atd\right\rfloor +\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{b\le \sqrt{x}/d}\left\lfloor x\over bdt\right\rfloor \\&\qquad -\Big (\sum _{t\mid p}\mu (t)\left\lfloor \sqrt{x}\over t\right\rfloor \Big )\Big (\sum _{d\mid q}\mu (d)\left\lfloor \sqrt{x}\over d\right\rfloor \Big ). \end{aligned}$$

In view of $\left\lfloor z\right\rfloor =z-\psi (z)-{1\over 2}$, we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)\\&\quad =\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\left( {x\over atd}-\psi \left( x\over atd\right) -{1\over 2}\right) \\&\qquad +\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{b\le \sqrt{x}/d}\left( {x\over bdt}-\psi \left( x\over bdt\right) -{1\over 2}\right) \\&\qquad -\Big (\sum _{t\mid p}\mu (t)\left( {\sqrt{x}\over t}-\psi \left( \sqrt{x}\over t\right) -{1\over 2}\right) \Big )\Big (\sum _{d\mid q}\mu (d)\left( {\sqrt{x}\over d}-\psi \left( \sqrt{x}\over d\right) -{1\over 2}\right) \Big ). \end{aligned}$$

In view of $\sum _{d\mid q}\mu (d)=0$, when $q>1,$ we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)\\&\quad =\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\left( {x\over atd}-\psi \left( x\over atd\right) \right) +\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\\&\qquad \sum _{b\le \sqrt{x}/d}\left( {x\over bdt}-\psi \left( x\over bdt\right) \right) \\&\qquad -\Big (\sum _{t\mid p}\mu (t)\left( {\sqrt{x}\over t}-\psi \left( \sqrt{x}\over t\right) \right) \Big )\Big (\sum _{d\mid q}\mu (d)\left( {\sqrt{x}\over d}-\psi \left( \sqrt{x}\over d\right) \right) \Big )\\&\quad =x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\sum _{a\le \sqrt{x}/t}{1\over a}-\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\psi \left( x\over atd\right) \\&\qquad +x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\sum _{b\le \sqrt{x}/d}{1\over b}-\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{b\le \sqrt{x}/d}\psi \left( x\over btd\right) \\&\qquad -x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}+\sqrt{x}\sum _{d\mid q}\sum _{t\mid p}{\mu (d)\over d}\mu (t)\psi \left( \sqrt{x}\over t\right) \\&\qquad +\sqrt{x}\sum _{d\mid q}\sum _{t\mid p}{\mu (t)\over t}\mu (d)\psi \left( \sqrt{x}\over d\right) +O((pq)^{\epsilon }). \end{aligned}$$

In view of Lemma 2, we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)\\&\quad =x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\Big (\ln \sqrt{x}/t +\gamma -\psi (\sqrt{x}/t) (\sqrt{x}/t)^{-1}+O((\sqrt{x}/t)^{-2})\Big )\\&\qquad -\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\psi \left( x\over atd\right) \\&\qquad +x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\Big (\ln \sqrt{x}/d +\gamma -\psi (\sqrt{x}/d) (\sqrt{x}/d)^{-1}+O((\sqrt{x}/d)^{-2})\Big )\\&\qquad -\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{b\le \sqrt{x}/d}\psi \left( x\over btd\right) \\&\qquad -x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}+\sqrt{x}\sum _{d\mid q}\sum _{t\mid p}{\mu (d)\over d}\mu (t)\psi \left( \sqrt{x}\over t\right) \\&\qquad +\sqrt{x}\sum _{d\mid q}\sum _{t\mid p}{\mu (t)\over t}\mu (d)\psi \left( \sqrt{x}\over d\right) +O((pq)^{\epsilon })\\&\quad =x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\Big (\ln \sqrt{x}/t +\gamma \Big )-\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\psi \left( x\over atd\right) \\&\qquad +x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\Big (\ln \sqrt{x}/d +\gamma \Big )-\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{b\le \sqrt{x}/d}\psi \left( x\over btd\right) \\&\qquad -x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}+O((pq)^{\epsilon }+p\ln q+q\ln p).\\ \end{aligned}$$

Now we use III in Lemma 2 with the exponent pair (2/7, 4/7) to bound the two sums involving function $\psi (.)$. Thus, we have

$$\begin{aligned}&\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{a\le \sqrt{x}/t}\psi \left( x\over atd\right) \\&\quad =\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)R\left ({x\over td}, {x^{1/2}\over d}, 1,1,1,0\right )\\&\quad \ll \sum _{d\mid q}\sum _{t\mid p}\left (1+\left({x\over td}\right)^{-1/2}\left({x^{1/2}\over d}\right)^{3/2}+\left({x\over td}\right)^{2/9}\left({x^{1/2}\over d}\right)^{2/9}\right )\\&\quad \ll \sum _{d\mid q}\sum _{t\mid p}\left (1+x^{1/4}t^{1/2}d^{-1}+x^{1/3}t^{-2/9}d^{-4/9}\right )\\&\quad =O((pq^{\epsilon })+x^{1/4}p^{3/2}\log q+x^{1/3}p^{7/9}q^{5/9}). \end{aligned}$$

In the same way, we also have

$$\begin{aligned}&\sum _{d\mid q}\sum _{t\mid p} \mu (d)\mu (t)\sum _{b\le \sqrt{x}/d}\psi \left( x\over btd\right) \\&\quad =O((pq^{\epsilon })+x^{1/4}q^{3/2}\ln p+x^{1/3}q^{7/9}p^{5/9}). \end{aligned}$$

Thus, we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q^0(b)\\&\quad =x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\Big (\ln \sqrt{x}/t +\gamma \Big )\\&\qquad +x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}\Big (\ln \sqrt{x}/d +\gamma \Big )\\&\qquad -x\sum _{d\mid q}\sum _{t\mid p} {\mu (d)\over d}{\mu (t)\over t}+O((pq)^{\epsilon }\\&\qquad +p\ln q+q\ln p +x^{1/3}q^{7/9}p^{5/9}+x^{1/3}p^{7/9}q^{5/9}). \end{aligned}$$

The complete proof follows from the identity $\displaystyle \sum _{d\mid q} {\mu (d)\ln d \over d}=-{\phi (q)\over q}{\ln q\over q-1}.$ $\square$

Lemma 7

Let p and q be two fixed primes, $\chi _p^0$ be the principal character modulo p, $\chi _q$ be the non-principal character modulo q. Then for any real number $x>\max \{p, q\}$, we have the asymptotic formula

$$\begin{aligned} \sum _{ab\le x}\chi _p^0(a)\chi _q(b)=\frac{p-1}{p}\cdot x\cdot L(1,\chi _q)+ O\left( x^{1/3}q^{7/9}p^{5/9}+x^{1/3}p^{7/9}q^{5/9}\right) , \end{aligned}$$

where $L(1,\chi _q)$ denotes the Dirichlet L-function corresponding to $\chi _q$.

Proof

For any $x>1$,

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q(b) =\sum _{a\le \sqrt{x}}\chi _p^0(a)\sum _{b\le {x\over a}}\chi _q(b)\\&\qquad +\sum _{b\le \sqrt{x}}\chi _q(b)\sum _{a\le {x\over b}}\chi _p^0(a) -\sum _{a\le \sqrt{x}}\chi _p^0(a)\sum _{b\le \sqrt{x}}\chi _q(b)\\&\quad =\sum _{\begin{array}{c}a\le \sqrt{x}\\ (a,p)=1\end{array}}\sum _{b\le {x\over a}}\chi _q(b)+\sum _{b\le \sqrt{x}}\chi _q(b)\sum _{\begin{array}{c}a\le {x\over b}\\ (a,p)=1\end{array}}1-\sum _{\begin{array}{c}a\le \sqrt{x}\\ (a,p)=1\end{array}}1\sum _{b\le \sqrt{x}}\chi _q(b). \end{aligned}$$

In view of I in Lemma 2, we have

$$\begin{aligned} \sum _{ab\le x} \chi _p^0(a)\chi _q(b)&=\sum _{d\mid p}\mu (d)\sum _{a\le {\sqrt{x}\over d}}\sum _{b\le {x\over ad}}\chi _q(b)+\sum _{d\mid p}\mu (d)\sum _{b\le \sqrt{x}}\chi _q(b)\left\lfloor {x\over bd} \right\rfloor \\&-\left( \sum _{d\mid p}\mu (d)\left\lfloor {\sqrt{x}\over d}\right\rfloor \right) \sum _{b\le \sqrt{x}}\chi _q(b). \end{aligned}$$

From (1) and (2) in Lemma 2, we have

$$\begin{aligned} \sum _{ab\le x} \chi _p^0(a)\chi _q(b)&=\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{a\le {\sqrt{x}\over d}}\left\lfloor {x\over aqd}-{j\over q}+1 \right\rfloor \\&\quad +\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}\left\lfloor {x\over bd} \right\rfloor \\&\quad -\left( \sum _{d\mid p}\mu (d)\left\lfloor {\sqrt{x}\over d}\right\rfloor \right) \Big (\sum _{j\le q}\chi _q(j)\left\lfloor {\sqrt{x}\over q}-{j\over q}+1\right\rfloor \Big ).\\ \end{aligned}$$

From $\left\lfloor z \right\rfloor =z-\psi (z)-{1\over 2}$ and $\psi (z+1)=\psi (z)$, we have

$$\begin{aligned} \sum _{ab\le x} \chi _p^0(a)\chi _q(b)&=\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{a\le {\sqrt{x}\over d}}\left( {x\over aqd}-{j\over q}+{1\over 2}-\psi \left( {x\over aqd}-{j\over q} \right) \right) \\&\quad +\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}\left( {x\over bd}-{1\over 2}-\psi \left( {x\over bd} \right) \right) \\&\quad -\left( \sum _{d\mid p}\mu (d)\left( {\sqrt{x}\over d}-{1\over 2}\right. \right. \left. \left. -\psi \left( {\sqrt{x}\over d}\right) \right) \right)\\&\qquad \Big (\sum _{j\le q}\chi _q(j)\left( {\sqrt{x}\over q}-{j\over q}+{1\over 2}-\psi \left( {\sqrt{x}\over q}-{j\over q}\right) \right) \Big ).\\ \end{aligned}$$

In view of $\sum _{j\le q}\chi _q(j)=0$ and $\sum _{d\mid p}\mu (d)=0$, we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q(b)\\&\quad =\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{a\le {\sqrt{x}\over d}}\left( -{j\over q}-\psi \left( {x\over aqd}-{j\over q} \right) \right) \\&\qquad +\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}\left( {x\over bd}-\psi \left( {x\over bd} \right) \right) \\&\qquad -\left( \sum _{d\mid p}\mu (d)\left( {\sqrt{x}\over d}-\psi \left( {\sqrt{x}\over d}\right) \right) \right) \Big (\sum _{j\le q}\chi _q(j)\left( -{j\over q}-\psi \left( {\sqrt{x}\over q}-{j\over q}\right) \right) \Big )\\&\quad =-{1\over q}\sum _{d\mid p}\mu (d)j\sum _{j\le q}\chi _q(j)\left\lfloor {\sqrt{x}\over d}\right\rfloor -\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{a\le {\sqrt{x}\over d}}\psi \left( {x\over aqd}-{j\over q} \right) \\&\qquad +x{\phi (p)\over p}\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}{1\over b}-\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}\psi \left( {x\over bd} \right) \\&\qquad +{1\over q}\sum _{d\mid p}\mu (d)\sum _{j\le q}j\chi _q(j)\left( {\sqrt{x}\over d}-\psi \left( {\sqrt{x}\over d}\right) \right) \\&\qquad +\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\left( {\sqrt{x}\over d}-\psi \left( {\sqrt{x}\over d}\right) \right) \psi \left( {\sqrt{x}\over q}-{j\over q}\right) +O(p^\epsilon q)\\&\quad =x{\phi (p)\over p}\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}{1\over b}-S_1-S_2\\&\qquad +\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\left( {\sqrt{x}\over d}-\psi \left( {\sqrt{x}\over d}\right) \right) \psi \left( {\sqrt{x}\over q}-{j\over q}\right) +O(p^\epsilon q), \end{aligned}$$

where

$$\begin{aligned} S_1=\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{a\le {\sqrt{x}\over d}}\psi \left( {x\over aqd}-{j\over q} \right) \end{aligned}$$

and

$$\begin{aligned} S_2=\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\sum _{\begin{array}{c}b\le \sqrt{x}\\ b\equiv j \bmod q\end{array}}\psi \left( {x\over bd} \right) . \end{aligned}$$

In view of Lemma 3, we have

$$\begin{aligned}&\sum _{ab\le x} \chi _p^0(a)\chi _q(b)\\&\quad =x{\phi (p)\over p}\sum _{j\le q}\chi _q(j)\Big ({\ln x\over 2q} +{\gamma (j,q)\over q}-\psi \Big ({\sqrt{x}-j\over q}\Big )x^{-1/2}+O(qx^{-1})\Big )-S_1-S_2\\&\qquad +\sum _{d\mid p}\mu (d)\sum _{j\le q}\chi _q(j)\left( {\sqrt{x}\over d}-\psi \left( {\sqrt{x}\over d}\right) \right) \psi \left( {\sqrt{x}\over q}-{j\over q}\right) +O(p^\epsilon q)\\&\quad =x{\phi (p)\over pq}\sum _{j\le q}\chi _q(j)\gamma (j,q)-S_1-S_2+O(p^\epsilon q^2). \end{aligned}$$

In view of Lemma 4, we have

$$\begin{aligned} \sum _{ab\le x} \chi _p^0(a)\chi _q(b)&=x{\phi (p)\over p}L(1,\chi _q)-S_1-S_2+O(p^\epsilon q^2). \end{aligned}$$

(3)

Now, we use III in Lemma 2 with exponent pair (2/7, 4/7) to bound $S_1$ and $S_2$. We have

$$\begin{aligned} S_1&=\sum _{d\mid p}\sum _{j\le q} \mu (d)\chi _q(j)R\Big ({x\over dq}, {x^{1/2}\over d}, 1,1,0,{-j\over q}\Big )\\&\ll \sum _{d\mid p}\sum _{j\le q} \left (1+\left({x\over dq}\right)^{-1/2}\left({x^{1/2}\over d}\right)^{3/2}+\left({x\over dq}\right)^{2/9}\left({x^{1/2}\over d}\right)^{2/9}\right )\\&\ll \sum _{d\mid p}\sum _{j\le q} \Big (1+x^{1/4}q^{1/2}d^{-1}+x^{1/3}q^{-2/9}d^{-4/9}\Big )\\&=O((qp^{\epsilon })+x^{1/4}q^{3/2}p^\epsilon +x^{1/3}q^{7/9}p^{\epsilon }). \end{aligned}$$

In the same way, we also have

$$\begin{aligned} S_2&=\sum _{d\mid p}\sum _{j\le q} \mu (d)\chi _q(j)R\Big ({x\over d}, \sqrt{x}, 1, q, j, 0\Big )\\&\ll \sum _{d\mid p}\sum _{j\le q} \left (1+\left({x\over d}\right)^{-1/2}\left(x^{1/2}\right)^{3/2}q^{-1}+\left({x\over d}\right)^{2/9}\left(x^{1/2}\right)^{2/9}q^{-4/9}\right )\\&\ll \sum _{d\mid p}\sum _{j\le q} \Big (1+x^{1/4}q^{-1}d^{1/2}+x^{1/3}q^{-4/9}d^{-2/9}\Big )\\&=O((qp^{\epsilon })+x^{1/4}p^{1/2}+x^{1/3}q^{5/9}p^{\epsilon }). \end{aligned}$$

We put these bounds of $S_1$ and $S_2$ in to (3), we have

$$\begin{aligned} \sum _{ab\le x} \chi _p^0(a)\chi _q(b)&=x{\phi (p)\over p}L(1,\chi _q)+O(x^{1/3}q^{7/9}p^{\epsilon }+x^{1/3}q^{5/9}p^{\epsilon }). \end{aligned}$$

This proves Lemma 7. $\square$

Lemma 8

Let $p_1$ and $p_2$ be two odd primes, $\chi _i$ be a non-principal Dirichlet character modulo $p_i$, $i=1$, 2. Then for any real number $x\ge p^2_i$, $i=1$ and 2, we have the estimate

$$\begin{aligned} \sum _{ab\le x}\chi _1\left( a\right) \chi _2\left( b\right) =O\left( x^{\frac{1}{3}}\cdot p_1^{\frac{5}{9}}\cdot p_2^{\frac{7}{9}}\cdot \ln p_1\right) . \end{aligned}$$

Proof

See Theorem 1.1 in reference [14]. $\square$

3 Proof of the Theorem 1

In this part, we shall complete the proof of our main result. For any fixed primes p and q, from Lemma 1 and the definition of d(n; p, q) we have

$$\begin{aligned}&\sum _{n\le x}d(n; p,q)=\sum _{n\le x}\mathop {\sum _{ab=n}}_{{a\in {\mathbb {A}}(p), b\in {\mathbb {A}}(q)}}1 =\mathop {\sum _{ab\le x}}_{a\in {\mathbb {A}}(p), b\in {\mathbb {A}}(q)}1\nonumber \\&\quad =\frac{\phi (p-1)}{p-1}\cdot \frac{\phi (q-1)}{q-1}\cdot \sum _{ab\le x}\sum _{h\mid p-1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum _{r=1}^{h}}\ '} \ e\left( \frac{r\cdot \text {ind}(a)}{h}\right) \nonumber \\&\qquad \times \sum _{k\mid q-1}\frac{\mu (k)}{\phi (k)}\mathop {{\sum _{s=1}^{k}}\ '} \ e\left( \frac{s\cdot \text {ind}(b)}{k}\right) . \end{aligned}$$

(4)

For any integer $1\le r\le h\le p-1$ with $h\mid p-1$ and $(r,h)=1$, we write: $e\left( \frac{r\cdot \text {ind}(a)}{h}\right) =\chi _{r,h}(a)$, and $\chi _{r,h}(a)=0$, if $p\mid a$. It is clear that $\chi _{r,h}(a)$ is a Dirichlet character modulo p, $\chi _{1,1}=\chi _p^0$ denotes the principal character modulo p. So from (4) and the above notations we have

$$\begin{aligned}&\sum _{n\le x}d(n; p,q)=\frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\nonumber \\&\qquad \cdot \sum _{h\mid p-1}\sum _{k\mid q-1}\frac{\mu (h)\mu (k)}{\phi (h)\phi (k)}\mathop {{\sum _{r=1}^{h}}\ '}\mathop {{\sum _{s=1}^{k}}\ '} \sum _{ab\le x} \chi _{r,h}(a)\chi _{s,k}(b) \nonumber \\&\quad =\frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \mathop {\sum _{h\mid p-1}}_{h>1}\mathop {\sum _{k\mid q-1}}_{k>1}\frac{\mu (h)\mu (k)}{\phi (h)\phi (k)}\mathop {{\sum _{r=1}^{h}}\ '}\mathop {{\sum _{s=1}^{k}}\ '} \sum _{ab\le x} \chi _{r,h}(a)\chi _{s,k}(b)\nonumber \\&\qquad + \frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum _{r=1}^{h}}\ '} \sum _{ab\le x} \chi _{r,h}(a)\chi _q^0(b)\nonumber \\&\qquad +\frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \mathop {\sum _{k\mid p-1}}_{k>1}\frac{\mu (k)}{\phi (k)}\mathop {{\sum _{s=1}^{k}}\ '} \sum _{ab\le x} \chi _p^0(a)\chi _{s,k}(b)\nonumber \\&\qquad +\frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \sum _{ab\le x}\chi _p^0(a)\chi _q^0(b)\equiv W_1+W_2+W_3+W_4. \end{aligned}$$

(5)

Now we estimate $W_1$, $W_2$, $W_3$ and $W_4$ in (5), respectively. Note that the estimate

$$\begin{aligned} \mathop {\sum _{h\mid p-1}}_{h>1}|\mu (h)|= 2^{\omega (p-1)}-1\ \ \text { and }\ \ \mathop {\sum _{k\mid q-1}}_{k>1}|\mu (k)|= 2^{\omega (q-1)}-1, \end{aligned}$$

from Lemma 8 we have the estimate

$$\begin{aligned} W_1\ll & {} \frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \mathop {\sum _{h\mid p-1}}_{h>1}|\mu (h)|\mathop {\sum _{k\mid q-1}}_{k>1}|\mu (k)|\cdot x^{\frac{1}{3}}\cdot p^{\frac{5}{9}}\cdot q^{\frac{7}{9}}\cdot \ln p\nonumber \\\ll & {} x^{\frac{1}{3}}\cdot p^{\frac{5}{9}}\cdot q^{\frac{7}{9}}\cdot 2^{\omega (p-1)+ \omega (q-1)}\cdot \ln p. \end{aligned}$$

(6)

Applying Lemma 7 we have

$$\begin{aligned} W_2= & {} \frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum _{r=1}^{h}}\ '} \left( \frac{q-1}{q}\right. \nonumber \left. \cdot x\cdot L(1,\chi _{r,h})+ O\left( x^{\frac{1}{3}}\cdot p^{\frac{7}{9}}\cdot q^{\epsilon }\right) \right) \\& \nonumber \\= & {} \frac{\phi (p-1)\phi (q-1)}{q(p-1)} x \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum _{r=1}^{h}}\ '} L\left( 1,\chi _{r,h}\right) \nonumber \\&+O\left( x^{\frac{1}{3}}\cdot p^{\frac{7}{9}}\cdot q^{\epsilon } 2^{\omega (p-1)} \right) . \end{aligned}$$

(7)

Similarly, we also have

$$\begin{aligned} W_3=\frac{\phi (p-1)\phi (q-1)}{p(q-1)} x \mathop {\sum _{k\mid q-1}}_{k>1}\frac{\mu (k)}{\phi (k)}\mathop {{\sum _{s=1}^{k}}\ '} L\left( 1,\chi _{s,k}\right) +O\left( x^{\frac{1}{3}}\cdot q^{\frac{7}{9}}\cdot p^{\epsilon }2^{\omega (q-1)} \right) . \end{aligned}$$

(8)

From Lemma 6, we have the asymptotic formula

$$\begin{aligned} W_4= & {} \frac{\phi (p-1)\phi (q-1)}{(p-1)(q-1)}\cdot \sum _{ab\le x}\chi _p^0(a)\chi _q^0(b)\nonumber \\= & {} \frac{\phi (p-1)\phi (q-1)}{pq}\cdot x\cdot \left( \ln x + 2\gamma -1+ \frac{\ln p}{p-1}+ \frac{\ln q}{q-1}\right) \nonumber \\&+ O\left( x^{\frac{1}{3}}\cdot p^{\frac{5}{9}}\cdot q^{\frac{7}{9}}+x^{\frac{1}{3}}\cdot q^{\frac{5}{9}}\cdot p^{\frac{7}{9}}\right) . \end{aligned}$$

(9)

Combining (5)–(9), we have the asymptotic formula

$$\begin{aligned}&\sum _{n\le x}d(n; p,q)=\frac{\phi (p-1)\phi (q-1)}{pq}\cdot x\cdot \left( \ln x + 2\gamma -1+ \frac{\ln p}{p-1}+ \frac{\ln q}{q-1}\right) \\&\quad +\frac{\phi (p-1)\phi (q-1)}{q(p-1)}\cdot x \cdot \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum _{r=1}^{h}}\ '} L\left( 1,\chi _{r,h}\right) \\&\quad +\frac{\phi (p-1)\phi (q-1)}{p(q-1)}\cdot x \cdot \mathop {\sum _{k\mid q-1}}_{k>1}\frac{\mu (k)}{\phi (k)}\mathop {{\sum _{s=1}^{k}}\ '} L\left( 1,\chi _{s,k}\right) \\&\quad + O\left( x^{\frac{1}{3}} p^{\frac{5}{9}} q^{\frac{7}{9}} 2^{\omega (p-1)+ \omega (q-1)} \ln p\right) +O\left( x^{\frac{1}{3}} q^{\frac{5}{9}} p^{\frac{7}{9}} 2^{\omega (p-1)+ \omega (q-1)} \ln q\right) . \end{aligned}$$

It is clear that $\chi ^h_{r,h}=\chi _p^0$ and $\chi ^j_{r,h}\ne \chi _p^0$ for all $1\le j<h$. So $\chi _{r,h}$ is a h-order character modulo p. Therefore, we have the identity

$$\begin{aligned} \mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum _{r=1}^{h}}\ '} L\left( 1,\chi _{r,h}\right) =\mathop {\sum _{h\mid p-1}}_{h>1}\frac{\mu (h)}{\phi (h)}\mathop {{\sum }~'}_{\chi ^h=\chi _p^0}L\left( 1,\chi \right) . \end{aligned}$$

This completes the proof of our theorem.

4 Conclusion

In this article, we first introduced a new arithmetical function d(n; p, q), then we using the analytic methods and Teerapat Srichan’s work [14] to study the mean value properties of d(n; p, q), and prove a strong asymptotic formula for it. Especially for any fixed primes p and q, we have the following simple form

$$\begin{aligned} \sum _{n\le x}d(n; p,q)=\frac{\phi (p-1)\phi (q-1)}{pq}\cdot x\cdot \ln x + O\left( x\right) , \ \ x\rightarrow +\infty . \end{aligned}$$