Higher Moments of the n^th Fourier Coefficients of j^th Symmetric Power L-Functions on Certain Sequence

1. Introduction

Within the study of number theory, the Fourier coefficients derived from modular forms serve as pivotal and deeply intriguing mathematical entities. Let $L(s,f)$ be the L-function attached with the primitive holomorphic cusp form f of weight k for the full modular group $S{L}_2\left(\mathbb{Z}\right)$. Let ${\lambda }_f\left(n\right)$ be the $n^{th}$ normalized Fourier coefficient of the Fourier expansion of $f\left(z\right)$ at the cusp ∞, that is,

$$\begin{array}{c}\hfill f\left(z\right)=\sum _{n=1}^{\infty }{\lambda }_f\left(n\right)n^{{\displaystyle \frac{k-1}{2}}}{e}^{2\pi inz},\Im \left(z\right)>0.\end{array}$$

Then for $\Re \left(s\right)>1$, the L-function attached to ${\lambda }_f\left(n\right)$ is defined as

$$L(s,f)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\lambda }_f\left(n\right)}{n^s}},$$

where ${\lambda }_f\left(n\right)$ are Hecke eigenvalues of all Hecke operators $T_n$.

The associated L-function is given by

$$\begin{array}{c}\hfill L(s,f)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\lambda }_f\left(n\right)}{n^s}}=\prod _p{\left(1-{\displaystyle \frac{\alpha \left(p\right)}{p^s}}\right)}^{-1}{\left(1-{\displaystyle \frac{\beta \left(p\right)}{p^s}}\right)}^{-1},\end{array}$$

which converges absolutely for $\sigma =\Re \left(s\right)>1$, where $\alpha \left(p\right)$ and $\beta \left(p\right)$ are related to the normalized Fourier coefficients in the following way

$$\begin{array}{c}\hfill \alpha \left(p\right)+\beta \left(p\right)={\lambda }_f\left(p\right),\left|\alpha \left(p\right)\right|=\left|\beta \left(p\right)\right|=\alpha \left(p\right)\beta \left(p\right)=1.\end{array}$$

From Ramanujan-Petersson conjecture,

$$\begin{array}{c}\hfill |{\lambda }_f\left(n\right)|\le d\left(n\right),\end{array}$$

where $d\left(n\right)$ is the divisor function. Studying the properties and average behaviors of various sums concerning ${\lambda }_f\left(n\right)$ is an interesting problem. In number theory, classical problems are investigate mean value estimates of these Fourier coefficients and related problems with the corresponding automorphic L-functions (for examples, see [2,5,6,21] etc.)

Let $j\in \mathbb{N}$. The $j^{th}$ symmetric power L-function associated with f is defined as

$$\begin{array}{c}\hfill L(s,{\mathrm{sym}}^{j}f)=\prod _p\prod _{m=0}^j{\left(1-{\alpha }^{j-m}\left(p\right){\beta }^m\left(p\right)p^{-s}\right)}^{-1}\end{array}$$

(1)

for $\Re \left(s\right)>1$, which can also be written as the following Dirichlet series

$$\begin{array}{c}\hfill L(s,{\mathrm{sym}}^{j}f)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)}{n^s}}=\prod _p\left(1+\sum _{i\ge 1}{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}\left(p^i\right)}{p^{is}}}\right),\end{array}$$

here ${\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)$ is real valued and multiplicative. In particular,

$$L(s,{\mathrm{sym}}^{0}f)=\zeta \left(s\right),L(s,{\mathrm{sym}}^{1}f)=L(s,f).$$

The $j^{th}$ symmetric power L-function twisted by ${\chi }^{\prime }$ is defined as

$$\begin{array}{cc}\hfill L(s,{\mathrm{sym}}^{j}f\times {\chi }^{\prime }):=& \prod _p\prod _{u=0}^j{\left(1-{\displaystyle \frac{{\alpha }^{j-m}\left(p\right){\beta }^m\left(p\right){\chi }^{\prime }\left(p\right)}{p^s}}\right)}^{-1}\hfill \\ \hfill =& \sum _{n=1}^{\infty }{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right){\chi }^{\prime }\left(n\right)}{n^s}}\hfill \end{array}$$

for $\Re \left(s\right)>1$. We will take ${\chi }^{\prime }$ as the specific $\chi$ or ${\tilde{\chi }}_0$ in this paper.

Several authors have considered the average behaviors of the Fourier coefficients of the $j^{th}$ symmetric power L-function $L(s,{\mathrm{sym}}^{j}f)$. In [3], Fomenko showed that

$$\sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)\ll x^{1/2}{log}^{2}x,$$

and he further established that

$$\sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^2\left(n\right)=cx+O\left(x^{\theta }\right)$$

in [4], where $\theta <1$. In addition, many scholars have studied related problem, see [8–10,12,18].

In [20], Zhai gave asymptotic formulas for

$$\sum _{{\displaystyle \genfrac{}{}{0pt}{}{a^2+b^2\le x}{(a,b)\in {\mathbb{Z}}^2}}}{\lambda }_f^{\ell }(a^2+b^2)$$

for $x\ge 1$ and $3\le l\le 8$. Afterwards, Xu [19] and Liu [11] improved Zhai’s result. For results related of the Fourier coefficients of symmetric square L-functions on a certain sequence of positive integers, see [15–17].

In 2023, Sharma and Sankaranarayanan considered some higher moments of these $n^{th}$ normalized Fourier coefficients and established the asymptotic formulas

$$\sum _{{\displaystyle \genfrac{}{}{0pt}{}{a^2+b^2+c^2+d^2\le x}{(a,b,s,d)\in {\mathbb{Z}}^4}}}{\lambda }_{{\mathrm{sym}}^{2}f}^3(a^2+b^2+c^2+d^2)=c_1{x}^2+O\left(x^{27/14+\epsilon }\right)$$

and

$$\sum _{{\displaystyle \genfrac{}{}{0pt}{}{a^2+b^2+c^2+d^2\le x}{(a,b,s,d)\in {\mathbb{Z}}^4}}}{\lambda }_{{\mathrm{sym}}^{2}f}^4(a^2+b^2+c^2+d^2)=c_2{x}^2+O\left(x^{160/81+\epsilon }\right)$$

for a sufficiently large x, where $c_1$ and $c_2$ are effective constants.

In this paper, we consider some higher moments of the $n^{th}$ Fourier coefficients of $j^{th}$ symmetric power L-functions on certain sequence. The main results are as follows.

Theorem 1.

Let $j\ge 2$ and $j\in \mathbb{N}$. For a sufficiently large x and any $\epsilon >0$, we have

$$\sum _{{\displaystyle \genfrac{}{}{0pt}{}{a^2+b^2+c^2+d^2\le x}{(a,b,s,d)\in {\mathbb{Z}}^4}}}{\lambda }_{{\mathrm{sym}}^{j}f}^3(a^2+b^2+c^2+d^2)=c_1{x}^2+O\left(x^{2-{\displaystyle \frac{84}{63j^2+105j+26}}+2\epsilon }\right),$$

(2)

where $c_1$ is an effective constant defined as

$$c_1=-2\zeta \left(2\right)\prod _{n=1}^{j}L(2,{\mathrm{sym}}^{3n}f)L(1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0)H_3\left(2\right),$$

$H_3\left(2\right)$ is a Dirichlet series that converges uniformly, and absolutely in the half plane $\Re \left(s\right)>{\displaystyle \frac{3}{2}}$, and $H_3\left(s\right)\ne 0$ on $\Re \left(s\right)=2$, and ${\tilde{\chi }}_0$ is a character modulo 4.

Remark. 

When $j=2$, the O-term in (2) is $O\left(x^{{\displaystyle \frac{223}{122}}+2\epsilon }\right)$, which is better than the Theorem 1 in [15].

Theorem 2.

Let $j\ge 2$ and $j\in \mathbb{N}$. For a sufficiently large x and any $\epsilon >0$, we have

$$\sum _{{\displaystyle \genfrac{}{}{0pt}{}{a^2+b^2+c^2+d^2\le x}{(a,b,s,d)\in {\mathbb{Z}}^4}}}{\lambda }_{{\mathrm{sym}}^{j}f}^4(a^2+b^2+c^2+d^2)=c_2{x}^2+O\left(x^{2-{\displaystyle \frac{42}{42j^2+63j+13}}+2\epsilon }\right),$$

(3)

where $c_2$ is an effective constant defined as

$$c_2=-2\zeta \left(2\right)\prod _{n=1}^{j}L(2,{\mathrm{sym}}^{4n}f)L(1,{\mathrm{sym}}^{4n}f\otimes {\tilde{\chi }}_0)H_4\left(2\right),$$

$H_4\left(2\right)$ is a Dirichlet series that converges uniformly, and absolutely in the half plane $\Re \left(s\right)>{\displaystyle \frac{3}{2}}$, and $H_4\left(s\right)\ne 0$ on $\Re \left(s\right)=2$, and ${\tilde{\chi }}_0$ is a character modulo 4.

Remark. 

When $j=2$, the O-term in (3) is $O\left(x^{{\displaystyle \frac{538}{290}}+2\epsilon }\right)$, which is better than the Theorem 2 in [15].

The organization of this paper is as follows. In Section 2, we introduce some preliminaries and also give some useful lemmas. In Section 3 and Section 4, we are give the proof of Theorems 1 and 2, respectively.

2. Some Preliminary Lemmas

In this section, we will establish some lemmas and preliminary results which are used to prove the theorems. Let $r_k\left(n\right):=♯\{(n_1,n_2,\dots ,n_k)\in {\mathbb{Z}}^k:n_1^2+n_2^2+\dots +n_k^2=n\}$ allowing zeros, distinguishing signs and order. We are interested in the function $r_4\left(n\right)$.

Lemma 1.

For any positive integer n, we have

$$r_4\left(n\right)=8\sum _{d|n,4\nmid d}d.$$

(1)

Moreover, $r_4\left(n\right)$ is multiplicative function.

Proof. 

This is [16, Lemma 2.1]. □

We note that,

$$\begin{array}{cc}\hfill \sum _{{\displaystyle \genfrac{}{}{0pt}{}{a^2+b^2+c^2+d^2\le x}{(a,b,c,d)\in {\mathbb{Z}}^4}}}{\lambda }_{{\mathrm{sym}}^{j}f}^{\eta }(a^2+b^2+c^2+d^2)=& \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^{\eta }\left(n\right)\sum _{{\displaystyle \genfrac{}{}{0pt}{}{n=a^2+b^2+c^2+d^2}{(a,b,c,d)\in {\mathbb{Z}}^4}}}1\hfill \\ \hfill =& \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^{\eta }\left(n\right)r_4\left(n\right)\hfill \\ \hfill =& 8\sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^{\eta }\left(n\right)r\left(n\right),\hfill \end{array}$$

where $r\left(n\right)={\sum }_{d|n,4\nmid d}d$ with $\eta =3$ or 4.

We observe that $r\left(n\right)$ possesses a multiplicative property and is defined as follows:

$$r\left(p^u\right):=\left\{\begin{array}{cc}{\displaystyle \frac{1-p^{u+1}}{1-p}},\hfill & p>2,\hfill \\ 3,\hfill & p=2.\hfill \end{array}\right.$$

We express $r_4\left(n\right)=8{\sum }_{d|n}{\tilde{\chi }}_0\left(d\right)d$, where ${\tilde{\chi }}_0$ represents a character modulo 4, defined by:

$${\tilde{\chi }}_0\left(p^u\right):=\left\{\begin{array}{cc}{\tilde{\chi }}_0\left(p^u\right),\hfill & p>2,\hfill \\ 3,\hfill & p=2,\hfill \end{array}\right.$$

and ${\tilde{\chi }}_0$ is the principal character modula 4.

Note that, $r\left(p\right)={\sum }_{d|p}{\tilde{\chi }}_0\left(d\right)d=1_p{\tilde{\chi }}_0\left(p\right)$.

Lemma 2.

For any $\epsilon >0$, we have

$${\int }_1^T{\left|\zeta \left({\displaystyle \frac{5}{7}}+it\right)\right|}^{12}dt\ll T^{1+\epsilon }$$

(2)

uniformly for $T\ge 1$, and

$$\zeta (\sigma +it)\ll {(1+|t\left|\right)}^{max\{{\displaystyle \frac{13}{42}}(1-\sigma ),0\}+\epsilon }$$

(3)

uniformly for ${\displaystyle \frac{1}{2}}\le \sigma \le 2$ and $\left|t\right|\ge 1$.

Proof. 

The first result can be founded in [7], the second result can be founded in [1]. □

Lemma 3.

For any $\epsilon >0$, we have

$${\int }_1^T{\left|\zeta \left({\displaystyle \frac{5}{7}}+it\right)\right|}^{12}dt\ll T^{1+\epsilon }$$

uniformly for $T\ge 1$, and

$$\zeta (\sigma +it)\ll {(1+|t\left|\right)}^{max\{{\displaystyle \frac{13}{42}}(1-\sigma ),0\}+\epsilon }$$

uniformly for ${\displaystyle \frac{1}{2}}\le \sigma \le 2$ and $\left|t\right|\ge 1$.

Proof. 

□

For a prime p, $0\le j\le 4$ and $\Re \left(s\right)>1$, we know that, the $p^{th}$ Fourier coefficient of $j^{th}$ symmetric power L-function of f can be written as

$$\begin{array}{c}\hfill {\lambda }_{{\mathrm{sym}}^{j}f}\left(p\right)=\sum _{m=0}^j{\alpha }^{j-m}\left(p\right){\beta }^m\left(p\right).\end{array}$$

(4)

For $0\le j\le 4$ and $\Re \left(s\right)>1$, Rankin-Selberg L-function attached to ${\mathrm{sym}}^{i}f$ and ${\mathrm{sym}}^{j}f$ has the following equation

$$\begin{array}{cc}\hfill {\lambda }_{{\mathrm{sym}}^{i}f\times {\mathrm{sym}}^{j}f}\left(p\right)=& \sum _{m=0}^i\sum _{u=0}^j{\alpha }^{i-m}\left(p\right){\beta }^m\left(p\right){\alpha }^{j-u}\left(p\right){\beta }^u\left(p\right)\hfill \\ \hfill =& \left(\sum _{m=0}^i{\alpha }^{i-m}\left(p\right){\beta }^m\left(p\right)\right)\left(\sum _{u=0}^j{\alpha }^{j-u}\left(p\right){\beta }^u\left(p\right)\right)\hfill \\ \hfill =& {\lambda }_{{\mathrm{sym}}^{i}f}\left(p\right){\lambda }_{{\mathrm{sym}}^{j}f}\left(p\right).\hfill \end{array}$$

Since ${\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)$ is a multiplicative function and $|{\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)|\le d_{j+1}\left(n\right)$, here $d_{j+1}\left(n\right)$ is the number of ways of expressing n as a product of $j+1$ factors, we can write the Euler product of $L(s,{\mathrm{sym}}^{j}f)$ as

$$\begin{array}{c}\hfill \prod _p\left(1+{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}\left(p\right)}{p^s}}+\dots +{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}\left(p^l\right)}{p^{ls}}}+\dots \right).\end{array}$$

(5)

We can obtain (4) from (1) and (5).

Moreover, according to Hecke

$${\lambda }_{{\mathrm{sym}}^{j}f}\left(p\right)={\lambda }_f\left(p^j\right).$$

Also, observe that

$${\lambda }_f^2\left(p^j\right)=1+\sum _{l=1}^j{\lambda }_f\left(p^{2l}\right).$$

Since

$$\begin{array}{cc}\hfill {\lambda }_f^2\left(p^j\right)=& {\left(\sum _{m=0}^j{\alpha }^{j-m}\left(p\right){\beta }^m\left(p\right)\right)}^2=\left(\sum _{m=0}^j{\alpha }^{j-m}\left(p\right){\beta }^m\left(p\right)\right)\left(\sum _{m^{\prime }=0}^j{\alpha }^{j-m^{\prime }}\left(p\right){\beta }^{m^{\prime }}\left(p\right)\right)\hfill \\ \hfill =& \sum _{m=0}^j\sum _{m^{\prime }=0}^j\left({\alpha }^{2j-(m+m^{\prime })}\left(p\right)\right)\left({\beta }^{(m+m^{\prime })}\right)=\sum _{l=0}^j\left(\sum _{t=0}^{2l}{\alpha }^{2j-t}\left(p\right){\beta }^t\left(p\right)\right)\hfill \\ \hfill =& 1+\sum _{l=1}^j\left(\sum _{t=0}^{2l}{\alpha }^{2j-t}\left(p\right){\beta }^t\left(p\right)\right)=1+\sum _{l=1}^j{\lambda }_f\left(p^{2l}\right).\hfill \end{array}$$

Lemma 4.

Suppose that f is a normalized primitive holomorphic cusp form of weight k for $S{L}_2\left(\mathbb{Z}\right)$ and ${\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)$ is the $n^{th}$ normalized Fourier coefficients of the $j^{th}$ symmetric power L-function related to f. For $\Re \left(s\right)>2$, if

$$F_3\left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r\left(n\right)}{n^s}},$$

we have

$$F_3\left(s\right)=G_3\left(s\right)H_3\left(s\right),$$

where

$$\begin{array}{c}\hfill G_3\left(s\right):=\zeta \left(s\right)L(s-1,{\tilde{\chi }}_0)\prod _{n=1}^{j}L(s,{\mathrm{sym}}^{3n}f)L(s-1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0),\end{array}$$

here ${\tilde{\chi }}_0$ is a character modulo 4, and $H_3\left(s\right)$ is a Dirichlet series which converges uniformly and absolutely in the half plane $\Re \left(s\right)>{\displaystyle \frac{3}{2}}$ and $H_3\left(s\right)\ne 0$ on $\Re \left(s\right)=2$.

Proof. 

Note that

$$\begin{array}{cc}\hfill {\lambda }_{{\mathrm{sym}}^{j}f}^3\left(p\right)r\left(p\right)=& {\left(\sum _{m=0}^j{\alpha }^{j-m}\left(p\right){\beta }^m\left(p\right)\right)}^3(1+{\tilde{\chi }}_0\left(p\right)p)\hfill \\ \hfill =& {\lambda }_f^3\left(p^j\right)(1+{\tilde{\chi }}_0\left(p\right)p)=\left(1+\sum _{m=1}^j{\lambda }_f\left(p^{3m}\right)\right)(1+{\tilde{\chi }}_0\left(p\right)p)\hfill \\ \hfill =& \left(1+\sum _{m=1}^j{\lambda }_{{\mathrm{sym}}^{3m}f}\left(p\right)\right)(1+{\tilde{\chi }}_0\left(p\right)p)\hfill \\ \hfill =& 1+{\tilde{\chi }}_0\left(p\right)p+\sum _{m=1}^j{\lambda }_{{\mathrm{sym}}^{3m}f}\left(p\right)+\sum _{m=1}^j{\lambda }_{{\mathrm{sym}}^{3m}f}\left(p\right){\tilde{\chi }}_0\left(p\right)p\hfill \\ \hfill =& :b\left(p\right).\hfill \end{array}$$

From the structure of $b\left(p\right)$, we define the coefficients $b\left(n\right)$ as

$$\sum _{n=1}^{\infty }{\displaystyle \frac{b\left(n\right)}{n^s}}=\zeta \left(s\right)L(s-1,{\tilde{\chi }}_0)\prod _{n=1}^{j}L(s,{\mathrm{sym}}^{3n}f)L(s-1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0),$$

which is absolutely convergent in $\Re \left(s\right)>3$. We also note that, for $\Re \left(s\right)>2$,

$$\begin{array}{cc}\hfill & \prod _p\left(1+{\displaystyle \frac{b\left(p\right)}{p^s}}+\dots +{\displaystyle \frac{b\left(p^m\right)}{p^{ms}}}+\dots \right)\hfill \\ \hfill & =\zeta \left(s\right)L(s-1,{\tilde{\chi }}_0)\prod _{n=1}^{j}L(s,{\mathrm{sym}}^{3n}f)L(s-1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0)\hfill \\ \hfill & =:G_3\left(s\right).\hfill \end{array}$$

Observe that $b\left(n\right){\ll }_{\epsilon }{n}^{1+\epsilon }$ for any small positive constant $\epsilon$.

We remark that

$$\begin{array}{cc}\hfill & \left|{\displaystyle \frac{b\left(p\right)}{p^s}}+{\displaystyle \frac{b\left(p^2\right)}{p^{2s}}}+\dots +{\displaystyle \frac{b\left(p^m\right)}{p^{ms}+\dots }}\right|\hfill \\ \hfill & \ll \sum _{m=1}^{\infty }{\displaystyle \frac{p^{(1+\epsilon )l}}{p^{m\sigma }}}\le \sum _{m=1}^{\infty }{\displaystyle \frac{p^{(1+\epsilon )m}}{p^{(2+2\epsilon )m}}}(\mathrm{in}\Re \left(s\right)\ge 2+2\epsilon )\hfill \\ \hfill & =\sum _{m=1}^{\infty }{\displaystyle \frac{1}{p^{(1+\epsilon )m}}}={\displaystyle \frac{1/p^{1+\epsilon }}{1-1/p^{1+\epsilon }}}={\displaystyle \frac{1}{p^{1+\epsilon }-1}}<1.\hfill \end{array}$$

Let us write

$$\begin{array}{c}\hfill A={\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(p\right)r\left(p\right)}{p^s}}+\dots +{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(p\right)r\left(p\right)}{p^s}}+\dots \end{array}$$

and

$$\begin{array}{c}\hfill B={\displaystyle \frac{b\left(p\right)}{p^s}}+\dots +{\displaystyle \frac{b\left(p^m\right)}{p^{ms}}}+\dots .\end{array}$$

From the above calculations we observe that $\left|B\right|<1$ in $\Re \left(s\right)\ge 2+2\epsilon$.

Take note that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1+A}{1+B}}=& (1+A)(1-B+B^2-B^3+cdots)(\mathrm{in}\Re \left(s\right)\ge 2+2\epsilon )\hfill \\ \hfill =& 1+A-B-AB+\mathrm{higher}\mathrm{terms}\hfill \\ \hfill =& 1+{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(p^2\right)r\left(p^2\right)-b\left(p^2\right)}{p^{2s}}}+\dots +{\displaystyle \frac{c_m\left(p^m\right)}{p^{ms}}}+\dots ,\hfill \end{array}$$

where $c_m\left(n\right){\ll }_{\epsilon }{n}^{1+\epsilon }$. Thus,

$$\begin{array}{cc}\hfill \prod _p\left({\displaystyle \frac{1+A}{1+B}}\right)=& \prod _p\left(1+{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(p^2\right)r\left(p^2\right)-b\left(p^2\right)}{p^{2s}}}+\dots +{\displaystyle \frac{c_m\left(p^m\right)}{p^{ms}}}+\dots \right)\hfill \\ \hfill {\ll }_{\epsilon }1\left(\mathrm{in}\Re \left(s\right)>{\displaystyle \frac{3}{2}}\right).\end{array}$$

So we have

$$\begin{array}{cc}\hfill H_3\left(s\right):=& {\displaystyle \frac{F_3\left(s\right)}{G_3\left(s\right)}}\hfill \\ \hfill =& \prod _p\left({\displaystyle \frac{1+A}{1+B}}\right)\hfill \\ \hfill {\ll }_{\epsilon }& 1\left(\mathrm{in}\Re \left(s\right)>{\displaystyle \frac{3}{2}}\right).\hfill \end{array}$$

Furthermore, $H_3\left(s\right)\ne 0$ on $\Re \left(s\right)=2$. □

Lemma 5.

Suppose that f is a normalized primitive holomorphic cusp form of weight k for $S{L}_2\left(\mathbb{Z}\right)$ and ${\lambda }_{{\mathrm{sym}}^{j}f}\left(n\right)$ is the $n^{th}$ normalized Fourier coefficients of the $j^{th}$ symmetric power L-function related to f. For $\Re \left(s\right)>2$, if

$$F_4\left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{\lambda }_{{\mathrm{sym}}^{j}f}^4\left(n\right)r\left(n\right)}{n^s}},$$

we have

$$F_4\left(s\right)=G_4\left(s\right)H_4\left(s\right),$$

where

$$\begin{array}{c}\hfill G_4\left(s\right):=\zeta \left(s\right)L(s-1,{\tilde{\chi }}_0)\prod _{n=1}^{j}L(s,{\mathrm{sym}}^{4n}f)L(s-1,{\mathrm{sym}}^{4n}f\otimes {\tilde{\chi }}_0),\end{array}$$

here ${\tilde{\chi }}_0$ is a character modulo 4, and $H_4\left(s\right)$ is a Dirichlet series which converges uniformly and absolutely in the half plane $\Re \left(s\right)>{\displaystyle \frac{3}{2}}$ and $H_4\left(s\right)\ne 0$ on $\Re \left(s\right)=2$.

Proof. 

The proof of this lemma is similar to the proof of Lemma 5. We omit the detail. □

Lemma 6.

Suppose that $L(s,f)$ is a Dirichlet series with Euler product of degree $m\ge 2$, which is defined as

$$\begin{array}{c}\hfill L(s,f)=\prod _{p<\infty }\prod _{i=0}^l{\left(1-{\displaystyle \frac{\alpha (p,i)}{p^s}}\right)}^{-1},\end{array}$$

where $\alpha (p,i)$ are local parameters of $L(s,f)$ at prime p. Suppose that:

(i) Its Euler product representation converges absolutely in the half-plane $\Re \left(s\right)>1$,

(ii) It admits a meromorphic continuation to the entire complex plane $\mathbb{C}$,

(iii) It satisfies a degree m functional equation of Riemann type.

Then we have

$$\begin{array}{c}\hfill {\int }_T^{2T}{\left|L\left({\displaystyle \frac{1}{2}}+it+\epsilon ,f\right)\right|}^{2}dt\ll T^{{\displaystyle \frac{m}{2}}+\epsilon }\end{array}$$

(6)

for $T\ge 1$, and we have

$$\begin{array}{c}\hfill \left|L(\sigma +it,f)\right|\ll {(1+|t\left|\right)}^{{\displaystyle \frac{m}{2}}(1-\sigma +it)+\epsilon }\end{array}$$

(7)

for $0\le \sigma \le 1+\epsilon$.

Proof. 

This two results follow from Perelli [14] and Matsumoto [13]. □

3. Proof of Theorem 1

By applying Perron’s formula with $\xi =2+\epsilon$ and $10\le T\le x$ to $F_3\left(s\right)$, we can get

$$\begin{array}{cc}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r_4\left(n\right)=& 8\sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r\left(n\right)\hfill \\ \hfill =& {\displaystyle \frac{8}{2\pi i}}{\int }_{\xi -iT}^{\xi +iT}{F}_3\left(s\right){\displaystyle \frac{x^s}{s}}ds+O\left({\displaystyle \frac{x^{2+2\epsilon }}{T}}\right).\hfill \end{array}$$

Note that

$$\begin{array}{cc}\hfill L(s-1,{\tilde{\chi }}_0)=& \sum _{n=1}^{\infty }{\displaystyle \frac{{\tilde{\chi }}_0\left(n\right)}{n^{s-1}}}\hfill \\ \hfill =& \prod _p{\left(1-{\displaystyle \frac{{\tilde{\chi }}_0\left(p\right)}{p^{s-1}}}\right)}^{-1}\hfill \\ \hfill =& {\left(1-{\displaystyle \frac{{\tilde{\chi }}_0\left(2\right)}{2^{s-1}}}\right)}^{-1}\prod _{p>2}\left(1-{\displaystyle \frac{{\tilde{\chi }}_0\left(p\right)}{p^{s-1}}}\right)\hfill \\ \hfill =& {\left(1-{\displaystyle \frac{3}{2^{s-1}}}\right)}^{-1}\left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)\prod _p{\left(1-{\displaystyle \frac{{\chi }_0\left(p\right)}{p^{s-1}}}\right)}^{-1}\hfill \\ \hfill =& {\left(1-{\displaystyle \frac{3}{2^{s-1}}}\right)}^{-1}\left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)L(s-1,{\chi }_0)\hfill \\ \hfill =& {\left(1-{\displaystyle \frac{3}{2^{s-1}}}\right)}^{-1}{\left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)}^2\zeta (s-1),\left(1\right)\hfill \end{array}$$

where we used

$$\begin{array}{cc}\hfill L(s-1,{\tilde{\chi }}_0)=& \sum _{n=1}^{\infty }{\displaystyle \frac{{\tilde{\chi }}_0\left(n\right)}{n^{s-1}}}\hfill \\ \hfill =& \prod _{{\displaystyle \genfrac{}{}{0pt}{}{p}{(p,4)=1}}}{\left(1-{\displaystyle \frac{1}{p^{s-1}}}\right)}^{-1}\hfill \\ \hfill =& \left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)\prod _p{\left(1-{\displaystyle \frac{1}{p^{s-1}}}\right)}^{-1}\hfill \\ \hfill =& \left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)\zeta (s-1).\hfill \end{array}$$

So we have

$$\begin{array}{cc}\hfill F_3\left(s\right):=& \zeta \left(s\right)L(s-1,{\tilde{\chi }}_0)\prod _{n=1}^{j}L(s,{\mathrm{sym}}^{3n}f)L(s-1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0)H_3\left(s\right)\hfill \\ \hfill =& \zeta \left(s\right){\left(1-{\displaystyle \frac{3}{2^{s-1}}}\right)}^{-1}{\left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)}^2\zeta (s-1)\hfill \\ \hfill & \times \prod _{n=1}^{j}L(s,{\mathrm{sym}}^{3n}f)L(s-1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0)H_3\left(s\right).\hfill \end{array}$$

Shifting the line of integration to $\Re \left(s\right)={\displaystyle \frac{12}{7}}+\epsilon$ and applying Cauchy’s residue theorem, we can get there is only one simple pole at $s=2$, which is coming from factor $\zeta (s-1)$, contributes the residue $c_1{x}^2$, where $c_1$ is an effective constant depending on the values of various L-functions appearing in $G_3\left(s\right)$ at $s=2$. To be more precise,

$$\begin{array}{cc}\hfill c_1=& 8\underset{s\to 2}{lim}(s-2){\displaystyle \frac{F_3\left(s\right)}{s}}\hfill \\ \hfill =& 8\zeta \left(2\right)\left(-{\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\prod _{n=1}^{j}L(2,{\mathrm{sym}}^{3n}f)L(1,{\mathrm{sym}}^{3n}f\otimes {\tilde{\chi }}_0)H_3\left(2\right).\hfill \end{array}$$

Thus we can get

$$\begin{array}{cc}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r_4\left(n\right)=& c_1{x}^2+{\displaystyle \frac{8}{2\pi i}}\left\{{\int }_{{\displaystyle \frac{12}{7}}-iT+\epsilon }^{{\displaystyle \frac{12}{7}}+iT+\epsilon }+{\int }_{2-iT+\epsilon }^{{\displaystyle \frac{12}{7}}-iT+\epsilon }+{\int }_{{\displaystyle \frac{12}{7}}+iT+\epsilon }^{2+iT+\epsilon }\right\}\hfill \\ \hfill & \times F_3\left(s\right){\displaystyle \frac{x^s}{s}}ds+O\left({\displaystyle \frac{x^{2+2\epsilon }}{T}}\right)\hfill \\ \hfill =& c_1{x}^2+{\displaystyle \frac{8}{2\pi i}}(U_1+U_2+U_3)+O\left({\displaystyle \frac{x^{2+2\epsilon }}{T}}\right).\hfill \end{array}$$

By Lemmas 3, 4 and 6, we can obtain that the contribution of the horizontal line integrals ($U_2$ and $U_3$) in absolute value is

$$\begin{array}{cc}\hfill & \ll {\int }_{{\displaystyle \frac{12}{7}}+\epsilon }^{2+\epsilon }{\displaystyle \frac{\left|\zeta (\sigma -1+iT)L(\sigma -1+iT,{\mathrm{sym}}^{3}f\otimes {\tilde{\chi }}_0)\dots L(\sigma -1+iT,{\mathrm{sym}}^{3j}f\otimes {\tilde{\chi }}_0)\right|}{T}}{x}^{\sigma }d\sigma \hfill \\ \hfill & \ll {\int }_{{\displaystyle \frac{5}{7}}+\epsilon }^{1+\epsilon }{\displaystyle \frac{\left|\zeta (\sigma +iT)L(\sigma +iT,{\mathrm{sym}}^{3}f\otimes {\tilde{\chi }}_0)\dots L(\sigma +iT,{\mathrm{sym}}^{3j}f\otimes {\tilde{\chi }}_0)\right|}{T}}{x}^{\sigma +1}d\sigma \hfill \\ \hfill & \ll \left({\displaystyle \frac{x}{T}}\right)\underset{{\displaystyle \frac{5}{7}}+\epsilon \le \sigma \le 1+\epsilon }{max}{x}^{\sigma }{T}^{{\displaystyle \frac{j(3j+5)}{4}}(1-\sigma +\epsilon )}{T}^{{\displaystyle \frac{13}{42}}(1-\sigma +\epsilon )}\hfill \\ \hfill & \ll \left({\displaystyle \frac{x^{1+\epsilon }}{T}}\right)\underset{{\displaystyle \frac{5}{7}}+\epsilon \le \sigma \le 1+\epsilon }{max}{\left({\displaystyle \frac{x}{T^{\frac{63j^2+105j+26}{84}}}}\right)}^{\sigma }{T}^{{\displaystyle \frac{63j^2+105j+26}{84}}}\hfill \\ \hfill & \ll x^{1+\epsilon }\left(x^{{\displaystyle \frac{5}{7}}+\epsilon }{T}^{{\displaystyle \frac{63j^2+105j+26}{294}}-1+\epsilon }+x^{1+\epsilon }{T}^{-1}\right)\hfill \\ \hfill & \ll x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{3j^2+5j}{14}}-{\displaystyle \frac{134}{147}}+\epsilon }+x^{2+2\epsilon }{T}^{-1}.\hfill \end{array}$$

By using Hölder’s inequality and Lemmas 2, 4, 6, we can get the contribution of left vertical line integral $U_1$ in absolute value is

$$\begin{array}{cc}\hfill \ll & {\int }_{{\displaystyle \frac{12}{7}}-iT+\epsilon }^{{\displaystyle \frac{12}{7}}+iT+\epsilon }{\displaystyle \frac{\left|\zeta (\frac{5}{7}+it+\epsilon )L(\frac{5}{7}+it+\epsilon ,{\mathrm{sym}}^{3}f\otimes {\tilde{\chi }}_0)\dots L(\frac{5}{7}+it+\epsilon ,{\mathrm{sym}}^{3j}f\otimes {\tilde{\chi }}_0)\right|}{|\frac{12}{7}+it+\epsilon |}}{x}^{{\displaystyle \frac{12}{7}}+\epsilon }dt\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}+\epsilon }+{\displaystyle \frac{x^{\frac{12}{7}+\epsilon }}{T}}{\left({\int }_{10\le \left|t\right|\le T}{\left|\zeta \left({\displaystyle \frac{5}{7}}+it+\epsilon \right)\right|}^{12}\right)}^{{\displaystyle \frac{1}{12}}}{\left({\int }_{10\le \left|t\right|\le T}{\left|L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{3}f\otimes {\tilde{\chi }}_0\right)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \times {\left({\int }_{10\le \left|t\right|\le T}{\left|L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{6}f\otimes {\tilde{\chi }}_0\right)\right|}^2\right)}^{{\displaystyle \frac{5}{12}}}\underset{10\le \left|t\right|\le T}{max}\left|L{\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{6}f\otimes {\tilde{\chi }}_0\right)}^{{\displaystyle \frac{1}{6}}}\right|\hfill \\ \hfill & \times \underset{10\le \left|t\right|\le T}{max}\left|L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{9}f\otimes {\tilde{\chi }}_0\right)\dots L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{3j}f\otimes {\tilde{\chi }}_0\right)\right|\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}+\epsilon }+{\displaystyle \frac{x^{\frac{12}{7}+\epsilon }}{T}}\left(T^{{\displaystyle \frac{1}{12}}+\epsilon }{T}^{{\displaystyle \frac{4}{7}}+\epsilon }{T}^{{\displaystyle \frac{5}{6}}+\epsilon }{T}^{{\displaystyle \frac{1}{6}}+\epsilon }{T}^{{\displaystyle \frac{10+13+\dots +(3j+1)}{7}}+\epsilon }\right)\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}+\epsilon }+x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{3j^2+5j}{14}}-{\displaystyle \frac{11}{12}}+\epsilon }\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{3j^2+5j}{14}}-{\displaystyle \frac{11}{12}}+\epsilon }.\hfill \end{array}$$

Therefore, we can get

$$\begin{array}{c}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r_4\left(n\right)=c_1{x}^2+O\left(x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{3j^2+5j}{14}}-{\displaystyle \frac{134}{147}}+\epsilon }+x^{2+2\epsilon }{T}^{-1}\right)\end{array}$$

Taking $T=x^{{\displaystyle \frac{84}{63j^2+105j+26}}}$, then we have

$$\begin{array}{c}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r_4\left(n\right)=c_1{x}^2+O\left(x^{2-{\displaystyle \frac{84}{63j^2+105j+26}}+2\epsilon }\right).\end{array}$$

Thus we complete the proof of Theorem 1.

4. Proof of Theorem 2

By applying Perron’s formula with $\xi =2+\epsilon$ and $10\le T\le x$ to $F_4\left(s\right)$, we can get

$$\begin{array}{cc}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^4\left(n\right)r_4\left(n\right)=& 8\sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r\left(n\right)\hfill \\ \hfill =& {\displaystyle \frac{8}{2\pi i}}{\int }_{\xi -iT}^{\xi +iT}{F}_4\left(s\right){\displaystyle \frac{x^s}{s}}ds+O\left({\displaystyle \frac{x^{2+2\epsilon }}{T}}\right).\hfill \end{array}$$

By (1), we have

$$\begin{array}{cc}\hfill F_4\left(s\right):=& \zeta \left(s\right)L(s-1,{\tilde{\chi }}_0)\prod _{n=1}^{j}L(s,{\mathrm{sym}}^{4n}f)L(s-1,{\mathrm{sym}}^{4n}f\otimes {\tilde{\chi }}_0)H_4\left(s\right)\hfill \\ \hfill =& \zeta \left(s\right){\left(1-{\displaystyle \frac{3}{2^{s-1}}}\right)}^{-1}{\left(1-{\displaystyle \frac{1}{2^{s-1}}}\right)}^2\zeta (s-1)\hfill \\ \hfill & \times \prod _{n=1}^{j}L(s,{\mathrm{sym}}^{4n}f)L(s-1,{\mathrm{sym}}^{4n}f\otimes {\tilde{\chi }}_0)H_4\left(s\right).\hfill \end{array}$$

Shifting the line of integration to $\Re \left(s\right)={\displaystyle \frac{12}{7}}+\epsilon$ and applying Cauchy’s residue theorem, we can get there is only one simple pole at $s=2$, which is coming from factor $\zeta (s-1)$, contributes the residue $c_2{x}^2$, where $c_2$ is an effective constant depending on the values of various L-functions appearing in $G_4\left(s\right)$ at $s=2$. To be more precise,

$$\begin{array}{cc}\hfill c_2=& 8\underset{s\to 2}{lim}(s-2){\displaystyle \frac{F_4\left(s\right)}{s}}\hfill \\ \hfill =& 8\zeta \left(2\right)\left(-{\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\prod _{n=1}^{j}L(2,{\mathrm{sym}}^{4n}f)L(1,{\mathrm{sym}}^{4n}f\otimes {\tilde{\chi }}_0)H_4\left(2\right).\hfill \end{array}$$

Thus we can get

$$\begin{array}{cc}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^4\left(n\right)r_4\left(n\right)=& c_2{x}^2+{\displaystyle \frac{8}{2\pi i}}\left\{{\int }_{{\displaystyle \frac{12}{7}}-iT+\epsilon }^{{\displaystyle \frac{12}{7}}+iT+\epsilon }+{\int }_{2-iT+\epsilon }^{{\displaystyle \frac{12}{7}}-iT+\epsilon }+{\int }_{{\displaystyle \frac{12}{7}}+iT+\epsilon }^{2+iT+\epsilon }\right\}\hfill \\ \hfill & \times F_4\left(s\right){\displaystyle \frac{x^s}{s}}ds+O\left({\displaystyle \frac{x^{2+2\epsilon }}{T}}\right)\hfill \\ \hfill =& c_1{x}^2+{\displaystyle \frac{8}{2\pi i}}(D_1+D_2+D_3)+O\left({\displaystyle \frac{x^{2+2\epsilon }}{T}}\right).\hfill \end{array}$$

By Lemmas 3, 4 and 6, we can obtain that the contribution of the horizontal line integrals ($D_2$ and $D_3$) in absolute value is

$$\begin{array}{cc}\hfill & \ll {\int }_{{\displaystyle \frac{12}{7}}+\epsilon }^{2+\epsilon }{\displaystyle \frac{\left|\zeta (\sigma -1+iT)L(\sigma -1+iT,{\mathrm{sym}}^{4}f\otimes {\tilde{\chi }}_0)\dots L(\sigma -1+iT,{\mathrm{sym}}^{4j}f\otimes {\tilde{\chi }}_0)\right|}{T}}{x}^{\sigma }d\sigma \hfill \\ \hfill & \ll {\int }_{{\displaystyle \frac{5}{7}}+\epsilon }^{1+\epsilon }{\displaystyle \frac{\left|\zeta (\sigma +iT)L(\sigma +iT,{\mathrm{sym}}^{4}f\otimes {\tilde{\chi }}_0)\dots L(\sigma +iT,{\mathrm{sym}}^{4j}f\otimes {\tilde{\chi }}_0)\right|}{T}}{x}^{\sigma +1}d\sigma \hfill \end{array}$$

$$\begin{array}{cc}\hfill & \ll \left({\displaystyle \frac{x}{T}}\right)\underset{{\displaystyle \frac{5}{7}}+\epsilon \le \sigma \le 1+\epsilon }{max}{x}^{\sigma }{T}^{{\displaystyle \frac{j(2j+3)}{2}}(1-\sigma +\epsilon )}{T}^{{\displaystyle \frac{13}{42}}(1-\sigma +\epsilon )}\hfill \\ \hfill & \ll \left({\displaystyle \frac{x^{1+\epsilon }}{T}}\right)\underset{{\displaystyle \frac{5}{7}}+\epsilon \le \sigma \le 1+\epsilon }{max}{\left({\displaystyle \frac{x}{T^{\frac{42j^2+63j+13}{42}}}}\right)}^{\sigma }{T}^{{\displaystyle \frac{42j^2+63j+13}{42}}}\hfill \\ \hfill & \ll x^{1+\epsilon }\left(x^{{\displaystyle \frac{5}{7}}+\epsilon }{T}^{{\displaystyle \frac{42j^2+63j+13}{147}}-1+\epsilon }+x^{1+\epsilon }{T}^{-1}\right)\hfill \\ \hfill & \ll x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{2j^2+3j}{7}}-{\displaystyle \frac{134}{147}}+\epsilon }+x^{2+2\epsilon }{T}^{-1}.\hfill \end{array}$$

By using Hölder’s inequality and Lemmas 2, 4, 6, we can get the contribution of left vertical line integral $U_1$ in absolute value is

$$\begin{array}{cc}\hfill \ll & {\int }_{{\displaystyle \frac{12}{7}}-iT+\epsilon }^{{\displaystyle \frac{12}{7}}+iT+\epsilon }{\displaystyle \frac{\left|\zeta (\frac{5}{7}+it+\epsilon )L(\frac{5}{7}+it+\epsilon ,{\mathrm{sym}}^{4}f\otimes {\tilde{\chi }}_0)\dots L(\frac{5}{7}+it+\epsilon ,{\mathrm{sym}}^{4j}f\otimes {\tilde{\chi }}_0)\right|}{|\frac{12}{7}+it+\epsilon |}}{x}^{{\displaystyle \frac{12}{7}}+\epsilon }dt\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}+\epsilon }+{\displaystyle \frac{x^{\frac{12}{7}+\epsilon }}{T}}{\left({\int }_{10\le \left|t\right|\le T}{\left|\zeta \left({\displaystyle \frac{5}{7}}+it+\epsilon \right)\right|}^{12}\right)}^{{\displaystyle \frac{1}{12}}}{\left({\int }_{10\le \left|t\right|\le T}{\left|L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{4}f\otimes {\tilde{\chi }}_0\right)\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \times {\left({\int }_{10\le \left|t\right|\le T}{\left|L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{8}f\otimes {\tilde{\chi }}_0\right)\right|}^2\right)}^{{\displaystyle \frac{5}{12}}}\underset{10\le \left|t\right|\le T}{max}\left|L{\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{8}f\otimes {\tilde{\chi }}_0\right)}^{{\displaystyle \frac{1}{6}}}\right|\hfill \\ \hfill & \times \underset{10\le \left|t\right|\le T}{max}\left|L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{12}f\otimes {\tilde{\chi }}_0\right)\dots L\left({\displaystyle \frac{5}{7}}+it+\epsilon ,{\mathrm{sym}}^{4j}f\otimes {\tilde{\chi }}_0\right)\right|\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}+\epsilon }+{\displaystyle \frac{x^{\frac{12}{7}+\epsilon }}{T}}\left(T^{{\displaystyle \frac{1}{12}}+\epsilon }{T}^{{\displaystyle \frac{5}{7}}+\epsilon }{T}^{{\displaystyle \frac{15}{14}}+\epsilon }{T}^{{\displaystyle \frac{3}{14}}+\epsilon }{T}^{{\displaystyle \frac{13+17+\dots +(4j+1)}{7}}+\epsilon }\right)\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}+\epsilon }+x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{2j^2+3j}{7}}-{\displaystyle \frac{11}{12}}+\epsilon }\hfill \\ \hfill \ll & x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{2j^2+3j}{7}}-{\displaystyle \frac{11}{12}}+\epsilon }.\hfill \end{array}$$

Therefore, we can get

$$\begin{array}{c}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^3\left(n\right)r_4\left(n\right)=c_1{x}^2+O\left(x^{{\displaystyle \frac{12}{7}}}{T}^{{\displaystyle \frac{2j^2+3j}{7}}-{\displaystyle \frac{134}{147}}+\epsilon }+x^{2+2\epsilon }{T}^{-1}\right)\end{array}$$

Taking $T=x^{{\displaystyle \frac{42}{42j^2+63j+13}}}$, then we have

$$\begin{array}{c}\hfill \sum _{n\le x}{\lambda }_{{\mathrm{sym}}^{j}f}^4\left(n\right)r_4\left(n\right)=c_2{x}^2+O\left(x^{2-{\displaystyle \frac{42}{42j^2+63j+13}}+2\epsilon }\right).\end{array}$$

Thus we complete the proof of Theorem 1.