A Symmetric Form of the Mean Value Involving Non-Isomorphic Abelian Groups

Abstract

Let $a\left(n\right)$ be the number of non-isomorphic abelian groups of order n. In this paper, we study a symmetric form of the average value with respect to $a\left(n\right)$ and prove an asymptotic formula. Furthermore, we study an analogue of the well-known Titchmarsh divisor problem involving $a\left(n\right).$

1. Introduction

Let $a\left(n\right)$ denote the number of non-isomorphic abelian groups of order n. The Dirichlet series of $a\left(n\right)$ is

$$\sum _{n=1}^{\infty }a\left(n\right)n^{-s}=\zeta \left(s\right)\zeta \left(2s\right)\zeta \left(3s\right)\cdots (\mathfrak{R}s>1),$$

where $\zeta \left(s\right)$ is the Riemann zeta function. It is well-known that the arithmetical function $a\left(n\right)$ is multiplicative and satisfies the equality $a\left(p^{\alpha }\right)=P\left(\alpha \right)$ for any prime p and integer $\alpha \ge 1$, where $P\left(\alpha \right)$ is the number of partitions of $\alpha$. Hence, for each prime number $p,$ we have $a\left(p\right)=1,a\left(p^2\right)=2,a\left(p^3\right)=3,a\left(p^4\right)=5,a\left(p^5\right)=7.$

A vast amount of literature exists on the asymptotic properties of $a\left(n\right).$ See, e.g., refs. [1,2] for historical surveys. The classical problem is to study the summatory function

$$A\left(x\right):=\sum _{n\le x}a\left(n\right).$$

In 1935, Erdös and Szekeres [3] proved that

$$A\left(x\right)=A_{1}x+O\left(x^{1/2}\right),$$

(1)

where $A_1={\prod }_{v=2}^{\infty }\zeta \left(v\right).$ Schwarz [4] showed that

$$A\left(x\right)=A_{1}x+A_2{x}^{1/2}+A_3{x}^{1/3}+R\left(x\right),$$

with $R\left(x\right)\ll x^{\frac{3}{10}-\frac{7}{30·23}}{(logx)}^{21/23}$ and $A_j={\prod }_{v\ne j}\zeta (v/j)(j=1,2,3).$ Many authors have investigated the upper bound of $R\left(x\right)$. For later improvements, see [5,6,7]. The best result to date is

$$R\left(x\right)\ll x^{1/4+\epsilon }$$

(2)

for every $\epsilon >0,$ proved by O. Robert and P. Sargos [8].

For an arithmetic function $f:\mathbb{N}\to \mathbb{N},$ and any integer $r>1,$ one can define

$$f^{\left(r\right)}\left(n\right)=f\left(f(\cdots f\left(n\right)\cdots )\right)$$

as the r-th iterate of f. If $r\ge 2$ is fixed, then two among the most natural problems concerning $f^{\left(r\right)}\left(n\right)$ are an evaluation of the sums of $f^{\left(r\right)}\left(n\right)$ and the determination of the maximal order of $f^{\left(r\right)}\left(n\right).$ In the case of $f\left(n\right)=d\left(n\right)$, representing the Dirichlet divisor function, these problems were investigated by Erdös and Kátai [9,10]. In [10] it was shown that

$$\sum _{n\le x}{d}^{\left(r\right)}\left(n\right)=(1+o\left(1\right))D_{r}x{log}_{r}x(D_r>0,x\to \infty )$$

holds for $r=4,$ which was proved earlier by I. Kátai to also be true for $r=2,3$. Additionally, there has been work on the analogue of this problem for $a\left(n\right)$. A. Ivić [11] considered the 2nd iterate of $a\left(n\right)$ and proved that

$$\sum _{n\le x}a\left(a\left(n\right)\right)=Cx+O\left(x^{1/2}{log}^{4}x\right).$$

for a suitable $C>0.$

In 1986, C. Spiro [12] studied a new iteration problem involving the divisor function and obtained

$$\sum _{n\le x,d(n+d(n\left)\right)=d\left(n\right)}1\gg \frac{x}{{(logx)}^7}.$$

(3)

In view of the work of C. Spiro, one can conjecture that, for some $D>0,$

$$\sum _{n\le x}d(n+d\left(n\right))=Dxlogx+O\left(x\right).$$

(4)

However, it seems very difficult at present to determine the rationality of (4). A result analogous to (4) is much less difficult if $d\left(n\right)$ is replaced by $a\left(n\right)$, or a suitable prime-independent multiplicative function $f\left(n\right)$ such that $f\left(p\right)=1.$ This is roughly due to the fact that $d\left(p\right)=2$ and $a\left(p\right)=1.$

Inspired by (3), A. Ivić [13] pointed out an asymptotic formula for the symmetric sum

$$Q\left(x\right):=\sum _{n\le x}a(n+a\left(n\right)),$$

and derived that the result

$$Q\left(x\right)=C_{1}x+O\left(x^{11/12+\epsilon }\right)$$

(5)

holds, for a positive constant $C_1$. Recently, Fan and Zhai [14] improved Ivić’s result (5) and got

$$Q\left(x\right)=C_{1}x+O\left(x^{3/4+\epsilon }\right).$$

(6)

In this paper, we shall use a different approach to improve (6).

Let

$$D_k\left(x\right)=\sum _{p\le x}d(p-k),$$

(7)

where p runs through all prime numbers greater than $k,$ and $k\ge 1$ is a fixed integer. The Titchmarsh divisor problem is to understand the behavior of $D_k\left(x\right)$ as $x\to \infty .$ So far we know very little concerning the properties of $p-k,$ such as whether $p-2$ contains an infinity of primes; therefore, a problem regarding $p-k$ for which we can give some sort of answer makes some sense.

Assuming the generalized Riemann hypothesis, Titchmarsh [15] showed that

$$D_k\left(x\right)\sim E_{1}x$$

(8)

with

$$E_1=\frac{\zeta \left(2\right)\zeta \left(3\right)}{\zeta \left(6\right)}\prod _{p\mid k}\left(1-\frac{p}{p^2-p+1}\right).$$

In 1963, Linik [16] proved (8) unconditionally. Subsequently, Fouvry [17] and Bombieri et al. [18] gave a secondary term,

$$D_k\left(x\right)=E_{1}x+E_{2}Lix+O\left(\frac{x}{{(logx)}^c}\right),$$

(9)

for all $c>1$ and

$$E_2=E_1\left(\gamma -\sum _p\frac{logp}{p^2-p+1}+\sum _{p\mid a}\frac{p^{2}logp}{(p-1)(p^2-p+1)}\right),$$

where $\gamma$ denotes the Euler–Mascheroni constant and $Li\left(x\right)$ is the logarithmic integral function. Motivated by the above results, we shall study an analogue of the Titchmarsh divisor problem for the symmetric form with regard to $a\left(n\right)$.

Our main plan is as follows. In Section 2, we state some important lemmas, and in Section 3, we prove the symmetric form of the mean value concerning non-isomorphic abelian groups. The analogue of the Titchmarsh divisor problem for $a(n+a(n\left)\right)$ is given in Section 4, with the help of the well-known Bombieri–Vinogradov theorem. We note that the proofs of the two results are analogous; however, there are also differences in some details.

Notation. In this paper, $\mathcal{P}$ denotes the set of all prime numbers, $\epsilon$ always denotes a small enough positive constant. $\mu \left(n\right)$ denotes the Möbius function, $\phi \left(n\right)$ denotes Euler’s totient function, and $d\left(n\right)$ denotes the Dirichlet divisor function.

2. Some Preliminary Lemmas

In this section, we quote some lemmas used in this paper.

Lemma 1.

We have

$$\underset{n\to \infty }{lim\; sup}loga\left(n\right)·\frac{loglogn}{logn}=\frac{log5}{4}.$$

Proof. 

See, for example, Krätzel [2].  □

Lemma 2.

For a positive number $u>0,$ let $S\left(u\right)$ denote the number of square-full numbers not exceeding $u,$ then we have

$$S\left(u\right)\ll u^{1/2}.$$

Proof. 

P. T. Bateman and E. Grosswald [19] proved that

$$S\left(u\right)=\frac{\zeta \left(\frac{3}{2}\right)}{\zeta \left(3\right)}{u}^{\frac{1}{2}}+\frac{\zeta \left(\frac{2}{3}\right)}{\zeta \left(2\right)}{u}^{\frac{1}{3}}+O\left(u^{\frac{1}{6}}\right),$$

(10)

then Lemma 2 follows from (10) immediately.  □

Suppose $m\ge 1,(a,m)=1,$ and $1\le a<m.$ In the next Lemma, we care about the average distribution of primes in arithmetic progressions. Define

$$\pi (x;m,a)=\sum _{\stackrel{p\le x}{p\equiv a\left(modm\right)}}1$$

and

$$E(x;m,a)=\pi (x;m,a)-\frac{Lix}{\phi \left(m\right)}.$$

(11)

Lemma 3.

Suppose $x\ge 3.$ For any given positive number $A>1,$ we have the estimate

$$\sum _{m\le M}{d}^4\left(m\right)a\left(m\right)\underset{y\le x}{max}\underset{(a,m)=1}{max}\left|E(y;m,a)\right|\ll \frac{x}{{(logx)}^A},$$

where $M=x^{1/2}{(logx)}^{-B}$ with $B=2A+272,$ the implied constant depending on $A.$

Proof. 

Let $\lambda =A+257.$ We write

$$\sum _{m\le M}{d}^4\left(m\right)a\left(m\right)\underset{y\le x}{max}\underset{(a,m)=1}{max}\left|E(y;m,a)\right|=S_1+S_2,$$

(12)

where

$$\begin{array}{ccc}& & S_1:=\sum _{\stackrel{m\le M}{d^4\left(m\right)a\left(m\right)>{(logx)}^{\lambda }}}{d}^4\left(m\right)a\left(m\right)\underset{y\le x}{max}\underset{(a,m)=1}{max}\left|E(y;m,a)\right|,\hfill \\ & & S_2:=\sum _{\stackrel{m\le M}{d^4\left(m\right)a\left(m\right)\le {(logx)}^{\lambda }}}{d}^4\left(m\right)a\left(m\right)\underset{y\le x}{max}\underset{(a,m)=1}{max}\left|E(y;m,a)\right|.\hfill \end{array}$$

We estimate $S_1$ first. Trivially we have (note $y\le x$)

$$\begin{array}{c}\hfill \left|E(y;m,a)\right|\ll \frac{Lix}{\phi \left(m\right)}+\sum _{\stackrel{n\le x}{n\equiv a\left(modm\right)}}1\ll \frac{x}{\phi \left(m\right)logx}+\frac{x}{m}\ll \frac{x}{m},\end{array}$$

where we used the estimate $\phi \left(m\right)\gg m/logm.$ Inserting the above bound into $S_1,$ we see that

$$\begin{array}{c}\hfill S_1\ll \frac{x}{{(logx)}^{\lambda -1}}\sum _{m\le M}\frac{d^8\left(m\right)a^2\left(m\right)}{m}.\end{array}$$

(13)

Suppose $s=\sigma +it$ with $\sigma >1$. Since $d^8\left(m\right)a^2\left(m\right)$ is multiplicative, we have the following expression

$$\begin{array}{c}\hfill \sum _{m=1}^{\infty }\frac{d^8\left(m\right)a^2\left(m\right)}{m^s}=\prod _p\left(1+\frac{256}{p^s}+\frac{3^8·2^2}{p^{2s}}+\cdots \right)={\zeta }^{256}\left(s\right)G\left(s\right),\end{array}$$

(14)

where $G\left(s\right)$ can be written as an infinite product, which is absolutely convergent for $\sigma >1/2.$ By the standard method of analytic number theory, we can obtain from (14) that

$$\sum _{m\le M}\frac{d^8\left(m\right)a^2\left(m\right)}{m}\ll {(logM)}^{256}\ll {(logx)}^{256}.$$

(15)

From (13) and (15), we obtain

$$\begin{array}{c}\hfill S_1\ll \frac{x}{{(logx)}^{\lambda -257}}\ll \frac{x}{{(logx)}^A}.\end{array}$$

(16)

Now, we estimate $S_2.$ Let $A_1>1$ be any fixed real number. Then we have the estimate

$$\sum _{m\le x^{1/2}{(logx)}^{-B_1}}\underset{y\le x}{max}\underset{(a,m)=1}{max}\left|E(y;m,a)\right|\ll \frac{x}{{(logx)}^{A_1}},$$

(17)

where $B_1=A_1+15.$ This is the well-known Bombieri–Vinogradov theorem. See Theorem 8.1 of [20].

Take $A_1=2A+257$ and $B_1=B=2A+272$ in (17). We have

$$\begin{array}{ccc}\hfill S_2& & \ll {(logx)}^{\lambda }\sum _{m\le M}\underset{y\le x}{max}\underset{(a,m)=1}{max}\left|E(y;m,a)\right|\hfill \\ & & \ll \frac{x}{{(logx)}^{A_1-\lambda }}\ll \frac{x}{{(logx)}^A}.\hfill \end{array}$$

(18)

Now, Lemma 3 follows from (12), (16) and (18).  □

3. A Symmetric Form of Mean Value Concerning $a\left(n\right)$

In this section, we propose a symmetric form of mean value concerning $a\left(n\right)$. We have the following theorem.

Theorem 1.

For any $\epsilon >0,$ we have the asymptotic formula

$$Q\left(x\right)=C_{1}x+O\left(x^{2/3+\epsilon }\right),$$

(19)

where the O-constant relies only on $\epsilon .$

Proof. 

We begin by noting that each natural number n can be uniquely written as $n=qs$ such that $(q,s)=1.$ We use this fact to obtain

$$Q\left(x\right)=\sum _{k_1\le W\left(x\right)}\sum _{q_1{s}_1\le x,a\left(s_1\right)=k_1,(q_1,s_1)=1}a(q_1{s}_1+k_1),$$

(20)

where $q_1$ is square-free, $s_1$ is square-full, and $W\left(x\right):={max}_{n\le x}a\left(n\right);$ the property that $a\left(n\right)=a\left(s\right(n\left)\right)$ is also utilized, where $s\left(n\right)$ is the square-full part of $n.$ Functions with this property were named s-functions; one can see [21] for more details. Taking advantage of the fact again, we have

$$Q\left(x\right)=\sum _{k_1\le W\left(x\right)}\sum _{k_2\le W(x+W\left(x\right))}{k}_2\sum _{\stackrel{q_1{s}_1\le x,a\left(s_1\right)=k_1,(q_1,s_1)=1}{q_1{s}_1+k_1=q_2{s}_2,a\left(s_2\right)=k_2,(q_2,s_2)=1}}1$$

(21)

where $q_2$ is square-free and $s_2$ is square-full.

For convenience, we abbreviate the innermost sum of (21) as $S(k_1,k_2).$ Therefore the estimation of $Q\left(x\right)$ can be reduced to estimate $S(k_1,k_2).$

3.1. Evaluation of the Sum $S(k_1,k_2)$

In this subsection, we shall study the sum $S(k_1,k_2).$ From the elementary relations

$${\mu }^2\left(n\right)=\sum _{d^2|n}\mu \left(d\right),\sum _{d|n}\mu \left(d\right)=\left\{\begin{array}{cc}1\hfill & n=1\hfill \\ 0\hfill & n>1,\hfill \end{array}\right.$$

(22)

we have

$$\begin{array}{cc}\hfill S(k_1,k_2)& =\sum _{q_1{s}_1\le x,a\left(s_1\right)=k_1,(q_1,s_1)=1}1\sum _{q_1{s}_1+k_1=q_2{s}_2,a\left(s_2\right)=k_2,(q_2,s_2)=1}1\hfill \\ \hfill & =\sum _{\stackrel{d_1^2{n}_1{s}_1\le x,a\left(s_1\right)=k_1}{(d_1,s_1)=(n_1,s_1)=1}}\mu \left(d_1\right)\sum _{\stackrel{d_1^2{n}_1{s}_1+k_1=d_2^2{n}_2{s}_2,a\left(s_2\right)=k_2}{(d_2,s_2)=(n_2,s_2)=1}}\mu \left(d_2\right)\hfill \\ \hfill & =\sum _{\stackrel{d_1^2{d}_1^{*}{n}_1^{*}{s}_1\le x,a\left(s_1\right)=k_1}{(d_1,s_1)=1,d_1^{*}\mid s_1}}\mu \left(d_1\right)\mu \left(d_1^{*}\right)\sum _{\stackrel{d_1^2{d}_1^{*}{n}_1^{*}{s}_1+k_1=d_2^2{d}_2^{*}{n}_2^{*}{s}_2,a\left(s_2\right)=k_2}{(d_2,s_2)=1,d_2^{*}\mid s_2}}\mu \left(d_2\right)\mu \left(d_2^{*}\right).\hfill \end{array}$$

It follows that

$$S(k_1,k_2)=\sum _{m_1{n}_1^{*}\le x}{f}_1\left(m_1\right)\sum _{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}{f}_2\left(m_2\right),$$

where

$$f_1\left(m_1\right):=\sum _{\stackrel{m_1=d_1^2{d}_1^{*}{s}_1}{a\left(s_1\right)=k_1,(d_1,s_1)=1,d_1^{*}|s_1}}\mu \left(d_1\right)\mu \left(d_1^{*}\right),$$

(23)

and

$$f_2\left(m_2\right):=\sum _{\stackrel{m_2=d_2^2{d}_2^{*}{s}_2}{a\left(s_2\right)=k_2,(d_2,s_2)=1,d_2^{*}|s_2}}\mu \left(d_2\right)\mu \left(d_2^{*}\right).$$

(24)

Suppose $x^{\epsilon }\ll y\ll x$ is a parameter to be determined later, we split the sum $S(k_1,k_2)$ into four parts:

$$S(k_1,k_2)=S_1+S_2+S_3+S_4,$$

(25)

where

$$\begin{array}{cc}\hfill & S_1=\sum _{\stackrel{m_1\le y,m_2\le y}{m_1{n}_1^{*}\le x,m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right),\hfill \\ \hfill & S_2=\sum _{\stackrel{y<m_1\le \frac{x}{n_1^{*}},m_2\le y}{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right)\hfill \\ \hfill & S_3=\sum _{\stackrel{m_1\le y,y<m_2\le \frac{x+k_1}{n_2^{*}}}{m_1{n}_1^{*}\le x,m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right),\hfill \\ \hfill & S_4=\sum _{\stackrel{y<m_1\le \frac{x}{n_1^{*}},y<m_2\le \frac{x+k_1}{n_2^{*}}}{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right).\hfill \end{array}$$

Consider first the sum $S_2.$ It is obviously seen that in the sum $S(k_1,k_2)$$m_1$ and $m_2$ are both square-full; by noting that if $d_1^{*}$ and $d_2^{*}$ are not square-full, then $d_i^{*}{s}_i(i=1,2)$ is square-full due to $d_i^{*}|s_i(i=1,2)$. Let $d(a,b,c;n)$ denote the number of representations of an integer n in the form $n=n_1^a{n}_2^b{n}_3^c.$ We know from the property of the 3-dimensional divisor problem

$$d(a,b,c;n)\ll n^{\epsilon }.$$

(26)

Using (26) and the the definition of $f_i\left(m_i\right)(i=1,2)$, we obtain

$$f_i\left(m_i\right)\ll d(2,1,1;m_i)\ll m_i^{\epsilon }\ll x^{\epsilon }(i=1,2).$$

(27)

From (27), Lemma 2, and partial summation, the sum $S_2$ can be estimated by

$$\begin{array}{cc}\hfill & \ll x^{\epsilon }\sum _{\stackrel{y<m_1\le \frac{x}{n_1^{*}},m_2\le y}{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}1\hfill \\ \hfill & \ll \frac{x^{1+\epsilon }}{y^{1/2}}.\hfill \end{array}$$

(28)

Similar to the sum $S_2$, we can obtain

$$S_i\ll \frac{x^{1+\epsilon }}{y^{1/2}}(i=3,4).$$

(29)

Next, we evaluate the sum $S_1.$ Recalling the definition of $S_1$, we have

$$S_1=\sum _{\stackrel{m_1\le y}{m_2\le y}}{f}_1\left(m_1\right)f_2\left(m_2\right)\sum _{\stackrel{m_1{n}_1^{*}\le x}{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}1.$$

(30)

By observation, the innermost sum of (30) is equal to the solution of the following set of congruent equations

$$\left\{\begin{array}{c}n\equiv 0\left(mod{m}_1\right)\hfill \\ n\equiv -k_1\left(mod{m}_2\right)\hfill \end{array}\right.$$

(31)

for $n\le x.$ The Chinese remainder theorem (for example, see [22]) reveals that (31) has a solution

$$n\equiv l\left(mod[m_1,m_2]\right)$$

(32)

for some l if and only if $(m_1,m_2)|k_1.$ Let

$$\delta (k_1;m_1,m_2)=\left\{\begin{array}{c}1,(m_1,m_2)\mid k_1,\hfill \\ 0,(m_1,m_2)\nmid k_1.\hfill \end{array}\right.$$

From (30) to (32),

$$S_1=\sum _{\stackrel{m_1\le y}{m_2\le y}}{f}_1\left(m_1\right)f_2\left(m_2\right)\delta (k_1;m_1,m_2)\sum _{\stackrel{n\le x}{n\equiv l\left(mod[m_1,m_2]\right)}}1.$$

(33)

As for the innermost sum of (33), obviously we obtain

$$\sum _{\stackrel{n\le x}{n\equiv l\left(mod[m_1,m_2]\right)}}1=\frac{x}{[m_1,m_2]}+O\left(1\right).$$

(34)

From (33) and (34), we have by applying the fact both $m_1$ and $m_2$ are square-full and by using Lemma 2 that

$$S_1=x\sum _{\stackrel{m_1\le y}{m_2\le y}}\frac{f_1\left(m_1\right)f_2\left(m_2\right)}{[m_1,m_2]}\delta (k_1;m_1,m_2)+O\left(y{x}^{\epsilon }\right).$$

(35)

Substituting (28), (29) and (35) into (25), we obtain

$$\begin{array}{cc}\hfill S(k_1,k_2)& =x\sum _{\stackrel{m_1\le y}{m_2\le y}}\frac{f_1\left(m_1\right)f_2\left(m_2\right)}{[m_1,m_2]}\delta (k_1;m_1,m_2)+O\left(y{x}^{\epsilon }\right)+O\left(x^{1+\epsilon }{y}^{-1/2}\right)\hfill \\ \hfill & :=xB\left(y\right)+O\left(x^{2/3+\epsilon }\right),\hfill \end{array}$$

(36)

for $y=x^{2/3}$ with

$$B\left(y\right)=\sum _{\stackrel{m_1\le y}{m_2\le y}}\frac{f_1\left(m_1\right)f_2\left(m_2\right)}{[m_1,m_2]}\delta (k_1;m_1,m_2)$$

Now, we treat the sum $B\left(y\right).$ Unfolding variables, we obtain

$$\begin{array}{cc}\hfill B\left(y\right)& =\sum _{\stackrel{d_1^2{d}_1^{*}{s}_1\le y,a\left(s_1\right)=k_1}{(d_1,s_1)=1,d_1^{*}|s_1}}\sum _{\stackrel{d_2^2{d}_2^{*}{s}_2\le y,a\left(s_2\right)=k_2}{(d_2,s_2)=1,d_2^{*}|s_2}}\frac{\mu \left(d_1\right)\mu \left(d_1^{*}\right)\mu \left(d_2\right)\mu \left(d_2^{*}\right)\delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)}{[d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2]}\hfill \\ \hfill & =\sum _{\stackrel{d_1^2{d}_1^{*}{s}_1\le y,a\left(s_1\right)=k_1}{(d_1,s_1)=1,d_1^{*}|s_1}}\sum _{\stackrel{d_2^2{d}_2^{*}{s}_2\le y,a\left(s_2\right)=k_2}{(d_2,s_2)=1,d_2^{*}|s_2}}\frac{(d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\mu \left(d_1\right)\mu \left(d_1^{*}\right)\mu \left(d_2\right)\mu \left(d_2^{*}\right)}{d_1^2{d}_1^{*}{s}_1{d}_2^2{d}_2^{*}{s}_2}\hfill \\ \hfill & \times \delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\hfill \\ \hfill & =\sum _{\stackrel{d_1^{*}{s}_1\le y,a\left(s_1\right)=k_1}{d_1^{*}|s_1}}\frac{\mu \left(d_1^{*}\right)}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2\le y,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\frac{\mu \left(d_2^{*}\right)}{d_2^{*}{s}_2}\sum _{\stackrel{d_1\le \sqrt{\frac{y}{d_1^{*}{s}_1}}}{(d_1,s_1)=1}}\frac{\mu \left(d_1\right)}{d_1^2}\hfill \\ \hfill & \times \sum _{\stackrel{d_2\le \sqrt{\frac{y}{d_2^{*}{s}_2}}}{(d_2,s_2)=1}}\frac{(d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\mu \left(d_2\right)\delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)}{d_2^2}\hfill \\ \hfill & =B_1+O\left(B_2{x}^{\epsilon }\right)+O\left(B_3{x}^{\epsilon }\right),\hfill \end{array}$$

(37)

where

$$\begin{array}{cc}\hfill & B_1:=\sum _{\stackrel{d_1^{*}{s}_1\le y,a\left(s_1\right)=k_1}{d_1^{*}|s_1}}\frac{\mu \left(d_1^{*}\right)}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2\le y,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\frac{\mu \left(d_2^{*}\right)}{d_2^{*}{s}_2}\sum _{\stackrel{d_1=1}{(d_1,s_1)=1}}^{\infty }\frac{\mu \left(d_1\right)}{d_1^2}\hfill \\ \hfill & \times \sum _{\stackrel{d_2=1}{(d_2,s_2)=1}}^{\infty }\frac{(d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\mu \left(d_2\right)\delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)}{d_2^2}\hfill \\ \hfill & B_2:=\sum _{\stackrel{d_1^{*}{s}_1\le y,a\left(s_1\right)=k_1}{d_1^{*}|s_1}}\frac{\left|\mu \right(d_1^{*}\left)\right|}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2\le y,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\frac{\left|\mu \right(d_2^{*}\left)\right|}{d_2^{*}{s}_2}\sum _{d_1>\sqrt{\frac{y}{d_1^{*}{s}_1}}}\frac{\left|\mu \right(d_1\left)\right|}{d_1^2}\sum _{d_2=1}^{\infty }\frac{\left|\mu \right(d_2\left)\right|}{d_2^2}\hfill \\ \hfill & B_3:=\sum _{\stackrel{d_1^{*}{s}_1\le y,a\left(s_1\right)=k_1}{d_1^{*}|s_1}}\frac{\left|\mu \right(d_1^{*}\left)\right|}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2\le y,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\frac{\left|\mu \right(d_2^{*}\left)\right|}{d_2^{*}{s}_2}\sum _{d_1=1}^{\infty }\frac{\left|\mu \right(d_1\left)\right|}{d_1^2}\sum _{d_2>\sqrt{\frac{y}{d_2^{*}{s}_2}}}\frac{\left|\mu \right(d_2\left)\right|}{d_2^2},\hfill \end{array}$$

(38)

and where in the last sum in $B_2$ and $B_3,$ we use the fact $(d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)<x^{\epsilon },$ which follows from $(d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\mid k_1.$

Consider the sum $B_2.$ From Lemma 2 and partial summation, we deduce that

$$B_2\ll \sum _{d_1^{*}{s}_1\le y}\frac{1}{\sqrt{d_1^{*}{s}_1}\sqrt{y}}\sum _{d_2^{*}{s}_2\le y}\frac{1}{d_2^{*}{s}_2}\ll y^{-\frac{1}{2}+\epsilon }.$$

(39)

Consider $B_3.$ Following the same argument as (39), we obtain

$$B_3\ll y^{-\frac{1}{2}+\epsilon }.$$

(40)

We combine (37), (39) and (40), then

$$B\left(y\right)=B_1+O\left(y^{-\frac{1}{2}-\epsilon }\right).$$

(41)

On substituting (41) into (36), the result is

$$\begin{array}{cc}\hfill S(k_1,k_2)& =x{B}_1+O\left(x{y}^{-\frac{1}{2}+\epsilon }\right)+O\left(x^{\frac{2}{3}+\epsilon }\right)\hfill \\ \hfill & =x{B}_1+O\left(x^{\frac{2}{3}+\epsilon }\right).\hfill \end{array}$$

(42)

3.2. Proof of Theorem 1

It follows from (21) and (42) that

$$Q\left(x\right)=x{Q}_1(x,y)+O\left(x^{2/3+\epsilon }{Q}_2(x,y)\right),$$

(43)

where

$$\begin{array}{cc}\hfill & Q_1(x,y):=\sum _{k_1\le W\left(x\right)}\sum _{k_2\le W(x+W\left(x\right))}{k}_2{B}_1\hfill \\ \hfill & Q_2(x,y):=\sum _{k_1\le W\left(x\right)}\sum _{k_2\le W(x+W\left(x\right))}{k}_2.\hfill \end{array}$$

(44)

It is easy to prove

$$Q_2(x,y)\ll x^{\epsilon }.$$

(45)

Hence, it remains to estimate $Q_1(x,y).$ Since $d_i^{*}{s}_i\le y(i=1,2),$ it holds that $s_i\le y(i=1,2)$ and $k_i\le W\left(y\right)(i=1,2),$ from which and also from (44), $Q_1(x,y)$ can be rewritten as

$$Q_1(x,y):=\sum _{k_1\le W\left(y\right)}\sum _{k_2\le W\left(y\right)}{k}_2{B}_1.$$

(46)

Lemma 4.

Let y be a natural number, and $W\left(y\right)={max}_{n\le y}a\left(n\right).$ Denote by $y_0$ the smallest natural number not exceeding y such that $W\left(y\right)=a\left(y_0\right),$ then

$$y_0>y^{1-\epsilon }.$$

Proof. 

From Lemma 1, we know that for any small positive constant $\epsilon >0,$ the inequality

$$\begin{array}{c}\hfill (1-0.1\epsilon )\frac{log5}{4}·\frac{logy}{loglogy}<logW\left(y\right)<(1+0.1\epsilon )\frac{log5}{4}·\frac{logy}{loglogy}\end{array}$$

holds for $y_0>y^{1-\epsilon }.$ If $y_0\le y^{1-\epsilon },$ then

$$\begin{array}{cc}\hfill logW\left(y\right)& <(1+0.1\epsilon )\frac{log5}{4}\frac{log{y}^{1-\epsilon }}{loglog{y}^{1-\epsilon }}=(1-0.9\epsilon -0.1{\epsilon }^2)\frac{log5}{4}\frac{logy}{loglogy+log(1-\epsilon )}.\hfill \end{array}$$

The above two formulas imply that

$$(1-0.1\epsilon )\frac{log5}{4}·\frac{logy}{loglogy}<(1-0.9\epsilon -0.1{\epsilon }^2)\frac{log5}{4}\frac{logy}{loglogy}.$$

This is a contradiction if $\epsilon >0$ is small enough. So, we have $y_0>y^{1-\epsilon }$.  □

Inserting (38) into (44) and expanding the range of $k_1$ and $k_2$ to infinity, we obtain

$$Q_1(x,y)=C_1+O\left({\Sigma }_1\right)+O\left({\Sigma }_2\right),$$

(47)

where

$$\begin{array}{cc}\hfill C_1& =\sum _{k_1=1}^{\infty }\sum _{k_2=1}^{\infty }{k}_2\sum _{\stackrel{d_1^{*}{s}_1=1,a\left(s_1\right)=k_1}{d_1^{*}|s_1}}^{\infty }\frac{\mu \left(d_1^{*}\right)}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2=1,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}^{\infty }\frac{\mu \left(d_2^{*}\right)}{d_2^{*}{s}_2}\sum _{\stackrel{d_1=1}{(d_1,s_1)=1}}^{\infty }\frac{\mu \left(d_1\right)}{d_1^2}\hfill \\ \hfill & \times \sum _{\stackrel{d_2=1}{(d_2,s_2)=1}}^{\infty }\frac{(d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\mu \left(d_2\right)\delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)}{d_2^2}\hfill \\ \hfill {\Sigma }_1& =\sum _{k_1>W\left(y\right)}\sum _{k_2=1}^{\infty }{k}_2\sum _{d_1^{*}{s}_1>y_0}\frac{\left|\mu \right(d_1^{*}\left)\right|}{d_1^{*}{s}_1}\sum _{d_2^{*}{s}_2=1}^{\infty }\frac{\left|\mu \right(d_2^{*}\left)\right|}{d_2^{*}{s}_2}\sum _{d_1=1}^{\infty }\frac{\left|\mu \right(d_1\left)\right|}{d_1^2}\sum _{d_2=1}^{\infty }\frac{\left|\mu \right(d_2\left)\right|}{d_2^2},\hfill \\ \hfill {\Sigma }_2& =\sum _{k_1=1}^{\infty }\sum _{k_2>W\left(y\right)}{k}_2\sum _{d_1^{*}{s}_1=1}^{\infty }\frac{\left|\mu \right(d_1^{*}\left)\right|}{d_1^{*}{s}_1}\sum _{d_2^{*}{s}_2>y_0}\frac{\left|\mu \right(d_2^{*}\left)\right|}{d_2^{*}{s}_2}\sum _{d_1=1}^{\infty }\frac{\left|\mu \right(d_1\left)\right|}{d_1^2}\sum _{d_2=1}^{\infty }\frac{\left|\mu \right(d_2\left)\right|}{d_2^2}.\hfill \end{array}$$

(48)

We consider ${\Sigma }_1.$ By using Lemma 4, we have

$$\begin{array}{cc}\hfill {\Sigma }_1& \ll \sum _{k_1>W\left(y\right)}\sum _{k_2=1}^{\infty }{k}_2\sum _{d_1^{*}{s}_1>y^{1-\epsilon }}\frac{1}{d_1^{*}{s}_1}\sum _{d_2^{*}{s}_2=1}^{\infty }\frac{1}{d_2^{*}{s}_2}\hfill \\ \hfill & =\sum _{k_1>W\left(y\right)}\sum _{d_1^{*}{s}_1>y^{1-\epsilon }}\frac{1}{d_1^{*}{s}_1}·\sum _{k_2=1}^{\infty }{k}_2\sum _{d_2^{*}{s}_2=1}^{\infty }\frac{1}{d_2^{*}{s}_2}\hfill \\ \hfill & ={\Sigma }_{11}·{\Sigma }_{12},\hfill \end{array}$$

(49)

say. In view of the well-known upper bound $a\left(n\right)\ll n^{\epsilon }$ and recalling that $d_i^{*}{s}_i(i=1,2)$ is square-full, by partial summation, we have

$$\begin{array}{cc}\hfill {\Sigma }_{12}& \ll \sum _{d_2^{*}{s}_2=1}^{\infty }\frac{s_2^{\epsilon }}{d_2^{*}{s}_2}\ll 1.\hfill \end{array}$$

(50)

However, we have by applying summation by parts and by using Lemma 2 that

$${\Sigma }_{11}\ll y^{-\frac{1}{2}+\epsilon }.$$

(51)

Gathering the three estimates above, we arrive at

$${\Sigma }_1\ll y^{-\frac{1}{2}+\epsilon }.$$

(52)

As for ${\Sigma }_2,$ we repeat the above argument to obtain

$${\Sigma }_2\ll y^{-\frac{1}{2}+\epsilon }.$$

(53)

From (47), (52) and (53), we obtain

$$Q_1(x,y)=C_1+O\left(y^{-\frac{1}{2}+\epsilon }\right),$$

(54)

and by combining (43) and (45), we obtain

$$\begin{array}{cc}\hfill Q\left(x\right)& =C_{1}x+O\left(x{y}^{-\frac{1}{2}+\epsilon }\right)+O\left(x^{2/3+\epsilon }\right)\hfill \\ \hfill & =C_{1}x+O\left(x^{2/3+\epsilon }\right)\hfill \end{array}$$

(55)

on recalling $y=x^{2/3}.$ This completes the proof of Theorem 1.  □

4. An Analogue of the Titchmarsh Divisor Problem

Let

$$P\left(x\right):=\sum _{p\le x}a(p-1+a(p-1)).$$

Motivated by the work of [18], we shall study the asymptotic behavior of $P\left(x\right)$. As an analogue of (9), we have the following.

Theorem 2.

Let $A>1$ be any fixed constant. We have

$$P\left(x\right)=C_{2}Lix+O\left(\frac{x}{{(logx)}^A}\right),$$

(56)

where $C_2$ is a constant defined by (75)

Proof. 

We start from the definition of $P\left(x\right).$ Since $p-1$ can be uniquely written as $p-1=q_1{s}_1$ such that $(q_1,s_1)=1,$ we obtain

$$P\left(x\right)=\sum _{k_1\le W\left(x\right)}\sum _{\stackrel{q_1{s}_1\le x-1,a\left(s_1\right)=k_1,(q_1,s_1)=1}{q_1{s}_1+1\in \mathcal{P}}}a(q_1{s}_1+k_1),$$

(57)

where $q_1$ is square-free, $s_1$ is square-full, $W\left(x\right):={max}_{p\le x}a(p-1)$, and $\mathcal{P}$ denotes the set of all primes. Here, the important property that $a\left(n\right)$ is an s-function is used. Using the fact that $q_1{s}_1+k_1$ can be uniquely written as $q_1{s}_1+k_1=q_2{s}_2$ such that $(q_2,s_2)=1$, we obtain

$$P\left(x\right)=\sum _{k_1\le W\left(x\right)}\sum _{k_2\le W(x+W\left(x\right))}{k}_2\sum _{\stackrel{q_1{s}_1\le x-1,a\left(s_1\right)=k_1,(q_1,s_1)=1,q_1{s}_1+1\in \mathcal{P}}{q_1{s}_1+k_1=q_2{s}_2,a\left(s_2\right)=k_2,(q_2,s_2)=1}}1,$$

(58)

where $q_2$ is square-free and $s_2$ is square-full.

It suffices to consider the innermost sum of (58). For convenience, we abbreviate it as $T(k_1,k_2).$ By (22), we have

$$\begin{array}{cc}\hfill T(k_1,k_2)& =\sum _{\stackrel{q_1{s}_1\le x-1,a\left(s_1\right)=k_1,(q_1,s_1)=1}{q_1{s}_1+1\in \mathcal{P}}}1\sum _{q_1{s}_1+k_1=q_2{s}_2,a\left(s_2\right)=k_2,(q_2,s_2)=1}1\hfill \\ \hfill & =\sum _{\stackrel{d_1^2{n}_1{s}_1\le x-1,a\left(s_1\right)=k_1}{(d_1,s_1)=(n_1,s_1)=1,d_1^2{n}_1{s}_1+1\in \mathcal{P}}}\mu \left(d_1\right)\sum _{\stackrel{d_1^2{n}_1{s}_1+k_1=d_2^2{n}_2{s}_2,a\left(s_2\right)=k_2}{(d_2,s_2)=(n_2,s_2)=1}}\mu \left(d_2\right)\hfill \\ \hfill & =\sum _{\stackrel{d_1^2{d}_1^{*}{n}_1^{*}{s}_1\le x-1,a\left(s_1\right)=k_1}{(d_1,s_1)=1,d_1^{*}\mid s_1,d_1^2{d}_1^{*}{n}_1^{*}{s}_1+1\in \mathcal{P}}}\mu \left(d_1\right)\mu \left(d_1^{*}\right)\sum _{\stackrel{d_1^2{d}_1^{*}{n}_1^{*}{s}_1+k_1=d_2^2{d}_2^{*}{n}_2^{*}{s}_2,a\left(s_2\right)=k_2}{(d_2,s_2)=1,d_2^{*}\mid s_2}}\mu \left(d_2\right)\mu \left(d_2^{*}\right).\hfill \end{array}$$

(59)

There holds that

$$T(k_1,k_2)=\sum _{\stackrel{m_1{n}_1^{*}\le x-1}{m_1{n}_1^{*}+1\in \mathcal{P}}}{f}_1\left(m_1\right)\sum _{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}{f}_2\left(m_2\right).$$

(60)

where $f_1\left(m_1\right)$ and $f_2\left(m_2\right)$ were defined by (23) and (24), respectively.

Suppose $x^{\epsilon }\ll z\ll x$ is a parameter to be determined later. We can write $T(k_1,k_2)$ as

$$T(k_1,k_2)=T_1+T_2+T_3+T_4,$$

(61)

where

$$\begin{array}{cc}\hfill & T_1=\sum _{\stackrel{m_1\le z,m_2\le z,m_1{n}_1^{*}\le x-1}{m_1{n}_1^{*}+1\in \mathcal{P},m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right),\hfill \\ \hfill & T_2=\sum _{\stackrel{z<m_1\le \frac{x-1}{n_1^{*}},m_2\le z}{m_1{n}_1^{*}+1\in \mathcal{P},m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right),\hfill \\ \hfill & T_3=\sum _{\stackrel{m_1\le z,z<m_2\le \frac{x-1+k_1}{n_2^{*}},m_1{n}_1^{*}\le x-1}{m_1{n}_1^{*}+1\in \mathcal{P},m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right),\hfill \\ \hfill & T_4=\sum _{\stackrel{z<m_1\le \frac{x-1}{n_1^{*}},z<m_2\le \frac{x-1+k_1}{n_2^{*}}}{m_1{n}_1^{*}+1\in \mathcal{P},m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}{f}_1\left(m_1\right)f_2\left(m_2\right).\hfill \end{array}$$

By the same arguments as (28), we have

$$T_i\ll \frac{x^{1+\epsilon }}{z^{1/2}}(i=2,3,4).$$

(62)

From (61) and (62), we obtain

$$T(k_1,k_2)=T_1+O\left(\frac{x^{1+\epsilon }}{z^{1/2}}\right).$$

(63)

Now, we evaluate the sum $T_1.$ We have

$$T_1=\sum _{\stackrel{m_1\le z}{m_2\le z}}{f}_1\left(m_1\right)f_2\left(m_2\right)\sum _{\stackrel{m_1{n}_1^{*}\le x-1,m_1{n}_1^{*}+1\in \mathcal{P}}{m_1{n}_1^{*}+k_1=m_2{n}_2^{*}}}1.$$

(64)

Note that the innermost sum in (64) is equal to the number of solutions of the following congruence equations

$$\left\{\begin{array}{c}p-1\equiv 0\left(mod{m}_1\right)\hfill \\ p-1\equiv -k_1\left(mod{m}_2\right)\hfill \end{array}\right.$$

(65)

for $p\le x.$ By the Chinese remainder theorem, (65) has a solution

$$p\equiv t\left(mod[m_1,m_2]\right)$$

(66)

for some t satisfying $(t,[m_1,m_2])=1$ if and only if $(m_1,m_2)|k_1.$ Let

$$\delta (k_1;m_1,m_2)=\left\{\begin{array}{c}1,(m_1,m_2)\mid k_1\hfill \\ 0,(m_1,m_2)\nmid k_1.\hfill \end{array}\right.$$

and

$$\lambda (t;m_1,m_2)=\left\{\begin{array}{c}1,(t,[m_1,m_2])=1\hfill \\ 0,(t,[m_1,m_2])\ne 1.\hfill \end{array}\right.$$

Whence by (64)–(66) and (11), we have

$$\begin{array}{cc}\hfill T_1& =\sum _{\stackrel{m_1\le z}{m_2\le z}}{f}_1\left(m_1\right)f_2\left(m_2\right)\delta (k_1;m_1,m_2)\lambda (t;m_1,m_2)\sum _{\stackrel{p\le x}{p\equiv t\left(mod[m_1,m_2]\right)}}1\hfill \\ \hfill & =\sum _{\stackrel{m_1\le z}{m_2\le z}}{f}_1\left(m_1\right)f_2\left(m_2\right)\delta (k_1;m_1,m_2)\lambda (t;m_1,m_2)\pi (x;[m_1,m_2],t)\hfill \\ \hfill & =\sum _{\stackrel{m_1\le z}{m_2\le z}}{f}_1\left(m_1\right)f_2\left(m_2\right)\delta (k_1;m_1,m_2)\lambda (t;m_1,m_2)\frac{Lix}{\phi \left([m_1,m_2]\right)}\hfill \\ \hfill & +\sum _{\stackrel{m_1\le z}{m_2\le z}}{f}_1\left(m_1\right)f_2\left(m_2\right)\delta (k_1;m_1,m_2)\lambda (t;m_1,m_2)E(x;[m_1,m_2],t)\hfill \\ \hfill & :=T_{11}+T_{12},\hfill \end{array}$$

(67)

say.

First, we consider the contribution of $T_{11}.$ We can write

$$T_{11}=Lix\times G\left(z\right),$$

(68)

where

$$G\left(z\right)=\sum _{\stackrel{m_1\le z}{m_2\le z}}\frac{f_1\left(m_1\right)f_2\left(m_2\right)}{\phi \left([m_1,m_2]\right)}\delta (k_1;m_1,m_2)\lambda (t;m_1,m_2).$$

(69)

Using the familiar bound $\phi \left(q\right)\gg \frac{q}{logq}$ for any $q>0,$ we compare (69) and (37) to obtain

$$G\left(z\right)=G_1+O\left(G_2{x}^{\epsilon }\right)+O\left(G_3{x}^{\epsilon }\right),$$

(70)

where

$$\begin{array}{cc}\hfill & G_1:=\sum _{\stackrel{d_1^{*}{s}_1\le z,a\left(s_1\right)=k_1}{d_1^{*}|s_1,d_1^2{d}_1^{*}{s}_1{n}_1^{*}+1\in \mathcal{P}}}\mu \left(d_1^{*}\right)\sum _{\stackrel{d_2^{*}{s}_2\le z,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\mu \left(d_2^{*}\right)\sum _{\stackrel{d_1=1}{(d_1,s_1)=1}}^{\infty }\mu \left(d_1\right)\sum _{\stackrel{d_2=1}{(d_2,s_2)=1}}^{\infty }\frac{\mu \left(d_2\right)\delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)}{\phi \left([d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2]\right)}\hfill \\ \hfill & G_2:=\sum _{\stackrel{d_1^{*}{s}_1\le z,a\left(s_1\right)=k_1}{d_1^{*}|s_1,d_1^2{d}_1^{*}{s}_1{n}_1^{*}+1\in \mathcal{P}}}\frac{\left|\mu \right(d_1^{*}\left)\right|}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2\le z,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\frac{\left|\mu \right(d_2^{*}\left)\right|}{d_2^{*}{s}_2}\sum _{d_1>\sqrt{\frac{z}{d_1^{*}{s}_1}}}\frac{\left|\mu \right(d_1\left)\right|}{d_1^2}\sum _{d_2=1}^{\infty }\frac{\left|\mu \right(d_2\left)\right|}{d_2^2},\hfill \\ \hfill & G_3:=\sum _{\stackrel{d_1^{*}{s}_1\le z,a\left(s_1\right)=k_1}{d_1^{*}|s_1,d_1^2{d}_1^{*}{s}_1{n}_1^{*}+1\in \mathcal{P}}}\frac{\left|\mu \right(d_1^{*}\left)\right|}{d_1^{*}{s}_1}\sum _{\stackrel{d_2^{*}{s}_2\le z,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}\frac{\left|\mu \right(d_2^{*}\left)\right|}{d_2^{*}{s}_2}\sum _{d_1=1}^{\infty }\frac{\left|\mu \right(d_1\left)\right|}{d_1^2}\sum _{d_2>\sqrt{\frac{z}{d_2^{*}{s}_2}}}\frac{\left|\mu \right(d_2\left)\right|}{d_2^2}.\hfill \end{array}$$

(71)

We now have the following result by applying the same arguments as (39) and (40)

$$G_i\ll z^{-1/2+\epsilon }(i=2,3).$$

(72)

Combining (68), (70) and (72),

$$T_{11}=Lix\times G_1+O\left(x{z}^{-1/2+\epsilon }\right).$$

(73)

From (58), (63), (67) and (73), we see that the contribution of $T_{11}$ to $P\left(x\right)$ is

$$=Lix\sum _{k_1\le W\left(x\right)}\sum _{k_2\le W(x+W\left(x\right))}{k}_2{G}_1+O\left(x{z}^{-1/2+\epsilon }\right).$$

Note that the sum ${\sum }_{k_1\le W\left(x\right)}{\sum }_{k_2\le W(x+W\left(x\right))}{k}_2{G}_1$ can be treated similarly to $x{Q}_1(x,y)$ in Section 3 (see (47)–(54)). So, we obtain that the contribution of $T_{11}$ to $P\left(x\right)$ is

$$=C_{2}Lix+O\left(x{z}^{-1/2+\epsilon }\right),$$

(74)

which absorbed the effects of $T_i(i=2,3,4)$ and

$$\begin{array}{cc}\hfill C_2& =\sum _{k_1=1}^{\infty }\sum _{k_2=1}^{\infty }{k}_2\sum _{\stackrel{d_1^{*}{s}_1=1,a\left(s_1\right)=k_1}{d_1^{*}|s_1,d_1^2{d}_1^{*}{s}_1{n}_1^{*}+1\in \mathcal{P}}}^{\infty }\mu \left(d_1^{*}\right)\sum _{\stackrel{d_2^{*}{s}_2=1,a\left(s_2\right)=k_2}{d_2^{*}|s_2}}^{\infty }\mu \left(d_2^{*}\right)\sum _{\stackrel{d_1=1}{(d_1,s_1)=1}}^{\infty }\mu \left(d_1\right)\hfill \\ \hfill & \times \sum _{\stackrel{d_2=1}{(d_2,s_2)=1}}^{\infty }\frac{\mu \left(d_2\right)\delta (k_1;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)\lambda (t;d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2)}{\phi \left([d_1^2{d}_1^{*}{s}_1,d_2^2{d}_2^{*}{s}_2]\right)}.\hfill \end{array}$$

(75)

Now, we study the contribution of $T_{12}$. Let $z=x^{1/5}$ and let $m=[m_1,m_2]$. It is easy to see that $m\le z^2=x^{2/5}$ and

$$|f_1\left(m_1\right)|\le d_3\left(m_1\right)\le d^2\left(m_1\right)\le d^2\left(m\right),\left|f_2\left(m_2\right)\right|\le d_3\left(m_2\right)\le d^2\left(m_2\right)\le d^2\left(m\right).$$

So, by Lemma 3, we have

$$\begin{array}{cc}\hfill & |\sum _{k_1\le W\left(x\right)}\sum _{k_2\le W(x+W\left(x\right))}{k}_2{T}_{12}|\hfill \\ \hfill & \le \sum _{[m_1,m_2]\le z^2}{d}_3\left(m_1\right)d_3\left(m_2\right)a\left(m_2\right)\underset{y\le x}{max}\underset{(t,[m_1,m_2])=1}{max}\left|E(y;[m_1,m_2],t)\right|\hfill \\ \hfill & \le \sum _{m\le x^{2/5}}{d}^4\left(m\right)a\left(m\right)\underset{y\le x}{max}\underset{(t,m)=1}{max}\left|E(y;m,t)\right|\hfill \\ \hfill & \le x{(logx)}^{-A},\hfill \end{array}$$

(76)

where $A>1$ is any fixed number. Now, Theorem 2 follows from (61), (67), (74) and (76).  □

5. Conclusions

In this paper, we established a symmetric form of the average value with regard to the non-isomorphic abelian groups based on the arithmetic structure of natural numbers. In addition, we studied an analogue of the Titchmarsh divisor problem for the symmetric form of $a\left(n\right)$ with the help of the modified Bombieri–Vinogradov theorem. We can easily generalize the results obtained for $a\left(n\right)$ to a class of functions that are “prime-independent”.