An identity involving certain Hardy sums and Ramanujan’s sum

Abstract

The main purpose of this paper is using the properties of Gauss sums and the mean value theorem of the Dirichlet L-functions to study one kind of a hybrid mean value problem involving certain Hardy sums and Ramanujan’s sum and to give an exact computational formula for it.

MSC:11L05, 11L10.

1 Introduction

Let c be a natural number and d be an integer prime to c. The classical Dedekind sums

$$S(d,c)=\sum _{j=1}^c\left(\left(\frac{j}{c}\right)\right)\left(\left(\frac{dj}{c}\right)\right),$$

where

$$((x))=\{\begin{array}{ll}x-[x]-\frac{1}{2},& \text{if }x\text{ is not an integer;}\\ 0,& \text{if }x\text{ is an integer,}\end{array}$$

describe the behavior of the logarithm of the eta-function (see [1] and [2]) under modular transformations. Funakura [3] gave an analogous transformation formula for the logarithm of the classical theta-function

$$\theta (z)=\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}exp(\pi i{n}^{2}z),Im(z)>0.$$

That is, put $Vz=(az+b)(cz+d)$ with $a,b,c,d\in Z$, $c>0$, and $ad-bc=1$. Then we have

$$log\theta (Vz)=log\theta (z)+\frac{1}{2}log(cz+d)-\frac{1}{4}\pi i+\frac{1}{4}\pi i{S}_1(d,c),$$

(1.1)

where $S_1(d,c)$ is defined as

$$S_1(d,c)=\sum _{j=1}^{c-1}{(-1)}^{j+1+[\frac{dj}{c}]}.$$

The sums $S_1(d,c)$ (and certain related ones) are sometimes called Hardy sums. They are closely connected with Dedekind sums. Regarding the properties of $S_1(d,c)$ and related sums, some authors studied them and obtained many interesting results; see [4–7] and [8]. For example, Wenpeng Zhang [7] proved the following conclusion: Let p be an odd prime. Then, for any fixed positive integer m, we have the asymptotic formula

$$\sum _{h=1}^{p-1}|S_1(h,p){|}^{2m}=p^{2m}\cdot \frac{{\zeta }^2(2m)(1-\frac{1}{4^m})}{\zeta (4m)(1+\frac{1}{4^m})}+O(p^{2m-1}\cdot exp\left(\frac{6lnp}{lnlnp}\right)),$$

where $\zeta (s)$ is the Riemann zeta-function and $exp(y)=e^y$.

On the other hand, we define Ramanujan’s sum $R_q(c)$ as follows (see Theorem 8.6 of [9]):

$$R_q(c)=\underset{(k,q)=1}{\sum _{k=1}^q}{e}^{\frac{2\pi ikc}{q}}=\sum _{d|(c,q)}d\mu (q/d),$$

where $\mu (n)$ is the famous Möbius function. Some related properties of $R_q(c)$ can also be found in [9, 10] and [11].

The main purpose of this paper is using the properties of Gauss sums and the mean square value theorem of the Dirichlet L-functions to study a hybrid mean value problem involving certain Hardy sums and Ramanujan’s sum and to give an exact computational formula for it. That is, we shall prove the following.

Theorem Let $q>1$ be an odd square-full number. Then we have the identity

$$\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot S_1(2h,q)={\varphi }^2(q)\cdot \prod _{p|q}(1+\frac{1}{p}),$$

where $\varphi (q)$ denotes the Euler function, and ${\prod }_{p|q}$ denotes the product over all distinct prime divisors of q.

It is very interesting that the value is equal to zero in our theorem if we use $R_q(h+1)$ instead of $R_q(2h+1)$.

For a general odd number $q\ge 3$, whether there exits a computational formula for

$$\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot S_1(2h,q)$$

is an open problem. Interested readers are welcome to study it with us.

2 Several lemmas

In this section, we shall give several lemmas, which are necessary in the proof of our theorem. Hereinafter, we shall use many properties of Gauss sums and character sums, all of which can be found in [9], so they will not be repeated here. First we have the following lemma.

Lemma 1 Let $q>1$ be an odd number, and let χ be any non-principal character $\mathrm{mod}q$. Then, for any integer m with $(m,q)=1$, we have the identity

$$\sum _{c=1}^q\chi (c)\cdot R_q(mc+1)=\overline{\chi }(m)\cdot \tau (\chi )\cdot \tau (\overline{\chi }),$$

where $\tau (\chi )={\sum }_{a=1}^{q-1}\chi (a)e(\frac{a}{q})$ denotes the classical Gauss sums, and $e(y)=e^{2\pi iy}$.

Proof For any non-principal character $\chi modq$, from the definition of $R_q(c)$ and the properties of Gauss sums, we have

$$\begin{array}{rl}\sum _{c=1}^q\chi (c)\cdot R_q(mc+1)& =\sum _{c=1}^q\underset{(b,q)=1}{\sum _{b=1}^q}\chi (c)e\left(\frac{b(mc+1)}{q}\right)\\ =\underset{(b,q)=1}{\sum _{b=1}^q}e\left(\frac{b}{q}\right)\sum _{c=1}^q\chi (c)e\left(\frac{mbc}{q}\right)\\ =\overline{\chi }(m)\underset{(b,q)=1}{\sum _{b=1}^q}\overline{\chi }(b)e\left(\frac{b}{q}\right)\sum _{c=1}^q\chi (mbc)e\left(\frac{mbc}{q}\right)\\ =\overline{\chi }(m)\cdot \tau (\chi )\cdot \sum _{b=1}^q\overline{\chi }(b)e\left(\frac{b}{q}\right)=\overline{\chi }(m)\cdot \tau (\chi )\cdot \tau (\overline{\chi }).\end{array}$$

This proves Lemma 1. □

Lemma 2 Let $q>2$ be an integer. Then, for any integer a with $(a,q)=1$, we have the identity

$$S(a,q)=\frac{1}{{\pi }^{2}q}\sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (a){|L(1,\chi )|}^2,$$

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

Proof See Lemma 2 of [12]. □

Lemma 3 Let $q>0$ and $(h,q)=1$. Then we have the identity

$$S_1(h,q)=-8S(h+q,2q)+4S(h,q).$$

Proof This formula is an immediate consequence of (5.9) and (5.10) in [6]. □

Lemma 4 Let $q>1$ be an odd number and $0<h<q$ with $(h,q)=1$. Then we have the identity

$$S_1(2h,q)=-20\cdot S(2h,q)+8\cdot S(4h,q)+8\cdot S(h,q).$$

Proof From Lemma 2 and Lemma 3, we have

$$\begin{array}{rl}{S}_1(2h,q)=& -8\cdot S(2h+q,2q)+4\cdot S(2h,q)\\ =& -\frac{4}{{\pi }^{2}q}\sum _{d|2q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (2h+q){|L(1,\chi )|}^2\\ +\frac{4}{{\pi }^{2}q}\sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (2h){|L(1,\chi )|}^2\\ =& -\frac{4}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{{(2d)}^2}{\varphi (2d)}\underset{\chi (-1)=-1}{\sum _{\chi mod2d}}\chi (2h+d){|L(1,\chi )|}^2\\ =& -\frac{16}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (2h+q)\lambda (2h+q){|L(1,\chi \lambda )|}^2\\ =& -\frac{16}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (2h){|L(1,\chi \lambda )|}^2,\end{array}$$

(2.1)

where λ is the principal character $\mathrm{mod}2$.

From the Euler infinite product formula, we have

$$\begin{array}{rl}{|L(1,\chi \lambda )|}^2& =\prod _{p_1}|1-\frac{\chi (p_1)\lambda (p_1)}{p_1}{|}^{-2}=\prod _{p_1>2}|1-\frac{\chi (p_1)}{p_1}{|}^{-2}\\ =|1-\frac{\chi (2)}{2}{|}^2\cdot \prod _{p_1}|1-\frac{\chi (p_1)}{p_1}{|}^{-2}=(\frac{5}{4}-\frac{\chi (2)}{2}-\frac{\overline{\chi }(2)}{2})\cdot {|L(1,\chi )|}^2,\end{array}$$

(2.2)

where ${\prod }_{p_1}$ denotes the product over all primes $p_1$.

Now, combining (2.1), (2.2) and Lemma 2, we have the identity

$$\begin{array}{rl}{S}_1(2h,q)& =-\frac{16}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (2h){|L(1,\chi \lambda )|}^2\\ =-\frac{16}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\chi (2h)(\frac{5}{4}-\frac{\chi (2)}{2}-\frac{\overline{\chi }(2)}{2})\cdot {|L(1,\chi )|}^2\\ =-20\cdot S(2h,q)+8\cdot S(4h,q)+8\cdot S(h,q).\end{array}$$

This proves Lemma 4. □

Lemma 5 Let $q>1$ be a square-full number (i.e., $q\ge 4$ and a prime $p|q$ implies $p^2|q$). Then, for any non-primitive character $\chi modq$, we have the identity

$$\sum _{a=1}^q\chi (a)e\left(\frac{a}{q}\right)=0.$$

Proof It is clear that from the multiplicative properties of Gauss sums we know that we only need to prove $q=p^{\alpha }$, a power of prime, where $\alpha \ge 2$. Suppose that χ is a non-primitive character $\mathrm{mod}{p}^{\alpha }$, then χ must be a character $\mathrm{mod}{p}^{\alpha -1}$. So, from the definition of Gauss sums and the properties of a complete residue system $\mathrm{mod}{p}^{\alpha }$ and trigonometric sums, we have

$$\begin{array}{rl}\sum _{a=1}^{p^{\alpha }}\chi (a)e\left(\frac{a}{p^{\alpha }}\right)& =\sum _{r=0}^{p-1}\sum _{a=1}^{p^{\alpha -1}}\chi (r{p}^{\alpha -1}+a)e\left(\frac{r{p}^{\alpha -1}+a}{p^{\alpha }}\right)\\ =\sum _{a=1}^{p^{\alpha -1}}\chi (a)e\left(\frac{a}{p^{\alpha }}\right)\sum _{r=0}^{p-1}e\left(\frac{r}{p}\right)=0.\end{array}$$

This proves Lemma 5. □

Lemma 6 Let $q>3$ be an odd square-full number. Then we have the identities

$$\begin{array}{c}\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}{|L(1,\chi )|}^2=\frac{{\pi }^2}{12}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p});\hfill \\ \underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}\chi (2)\cdot {|L(1,\chi )|}^2=\frac{{\pi }^2}{24}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p}),\hfill \end{array}$$

where ${\sum }^{\ast }\underset{}{{}_{\chi modq}{}^{}\chi (-1)=-1}$ denotes the summation over all odd primitive characters $\chi modq$.

Proof From the definition of $S(a,q)$, we have the computational formula

$$S(1,q)=\sum _{a=1}^{q-1}{(\frac{a}{q}-\frac{1}{2})}^2=\frac{(q-1)(q-2)}{12q}.$$

(2.3)

From the reciprocity formula of $S(a,q)$, we know that for any positive integer a with $(a,q)=1$, we have the identity (see [4])

$$S(a,q)+S(q,a)=\frac{q^2+a^2+1}{12aq}-\frac{1}{4}.$$

Applying this formula, we have

$$S(2,q)=\frac{q^2+2^2+1}{24q}-\frac{1}{4}-S(q,2)=\frac{(q-1)(q-5)}{24q}.$$

(2.4)

From (2.3), Lemma 2 with $a=1$ and the Möbius inversion formula, we have

$$\begin{array}{rl}\frac{q^2}{\varphi (q)}\underset{\chi (-1)=-1}{\sum _{\chi modq}}{|L(1,\chi )|}^2& ={\pi }^2\cdot \sum _{d|q}\mu (d)\cdot \frac{q}{d}\cdot S(1,\frac{q}{d})={\pi }^2\cdot \sum _{d|q}\mu (d)\cdot \frac{(\frac{q}{d}-1)(\frac{q}{d}-2)}{12}\\ =\frac{{\pi }^2}{12}\cdot \varphi (q)\cdot [q\cdot \prod _{p|q}(1+\frac{1}{p})-3]\end{array}$$

or

$$\underset{\chi (-1)=-1}{\sum _{\chi modq}}{|L(1,\chi )|}^2=\frac{{\pi }^2}{12}\cdot \frac{{\varphi }^2(q)}{q^2}\cdot [q\cdot \prod _{p|q}(1+\frac{1}{p})-3].$$

(2.5)

Then, using formula (2.5) and the Möbius inversion formula, we have

$$\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}{|L(1,\chi )|}^2=\sum _{d|q}\mu (d)\left(\underset{\chi (-1)=-1}{\sum _{\chi modq/d}}{|L(1,\chi )|}^2\right)=\frac{{\pi }^2}{12}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p}).$$

(2.6)

From (2.4), Lemma 2 with $a=2$ and the Möbius inversion formula, we also have

$$\begin{array}{rl}\frac{q^2}{\varphi (q)}\underset{\chi (-1)=-1}{\sum _{\chi modq}}\chi (2)\cdot {|L(1,\chi )|}^2& ={\pi }^2\cdot \sum _{d|q}\mu (d)\cdot \frac{q}{d}\cdot S(2,\frac{q}{d})\\ ={\pi }^2\cdot \sum _{d|q}\mu (d)\cdot \frac{(\frac{q}{d}-1)(\frac{q}{d}-5)}{24}\\ =\frac{{\pi }^2}{24}\cdot \varphi (q)\cdot [q\cdot \prod _{p|q}(1+\frac{1}{p})-6]\end{array}$$

or

$$\underset{\chi (-1)=-1}{\sum _{\chi modq}}\chi (2)\cdot {|L(1,\chi )|}^2=\frac{{\pi }^2}{24}\cdot \frac{{\varphi }^2(q)}{q^2}\cdot [q\cdot \prod _{p|q}(1+\frac{1}{p})-6].$$

(2.7)

Then, using (2.7) and the Möbius inversion formula, we have

$$\begin{array}{rl}\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}\chi (2)\cdot {|L(1,\chi )|}^2& =\sum _{d|q}\mu (d)(\underset{\chi (-1)=-1}{\sum _{\chi modq/d}}\chi (2)\cdot {|L(1,\chi )|}^2)\\ =\frac{{\pi }^2}{24}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p}).\end{array}$$

(2.8)

Now Lemma 6 follows from (2.6) and (2.8). □

3 Proof of the theorems

In this section, we shall complete the proof of our theorem. Note that if χ is a primitive character $\mathrm{mod}q$, then $|\tau (\chi )|=\sqrt{q}$, and $\tau (\chi )\cdot \tau (\overline{\chi })=\overline{\chi }(-1)\cdot \tau (\chi )\cdot \overline{\tau (\chi )}=\overline{\chi }(-1)\cdot q$. From Lemma 1, Lemma 2, Lemma 4, Lemma 5 and Lemma 6, we have

$$\begin{array}{r}\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot S_1(2h,q)\\ =\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot (-20\cdot S(2h,q)+8\cdot S(4h,q)+8\cdot S(h,q))\\ =-\frac{20}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot \chi (h)\cdot \chi (2)\cdot {|L(1,\chi )|}^2\\ +\frac{8}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot \chi (h)\cdot \chi (4)\cdot {|L(1,\chi )|}^2\\ +\frac{8}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\underset{(h,q)=1}{\sum _{h=1}^q}{R}_q(2h+1)\cdot \chi (h)\cdot {|L(1,\chi )|}^2\\ =-\frac{20}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\tau (\chi )\cdot \tau (\overline{\chi })\cdot {|L(1,\chi )|}^2\\ +\frac{8}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\tau (\chi )\cdot \tau (\overline{\chi })\cdot \chi (2)\cdot {|L(1,\chi )|}^2\\ +\frac{8}{{\pi }^{2}q}\cdot \sum _{d|q}\frac{d^2}{\varphi (d)}\underset{\chi (-1)=-1}{\sum _{\chi modd}}\tau (\chi )\cdot \tau (\overline{\chi })\cdot \overline{\chi }(2)\cdot {|L(1,\chi )|}^2\\ =\frac{20}{{\pi }^2}\cdot \frac{q^2}{\varphi (q)}\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}{|L(1,\chi )|}^2-\frac{8}{{\pi }^2}\cdot \frac{q^2}{\varphi (q)}\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}\chi (2)\cdot {|L(1,\chi )|}^2\\ -\frac{8}{{\pi }^2}\cdot \frac{q^2}{\varphi (q)}\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}\overline{\chi }(2)\cdot {|L(1,\chi )|}^2\\ =\frac{20}{{\pi }^2}\cdot \frac{q^2}{\varphi (q)}\cdot \frac{{\pi }^2}{12}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p})-\frac{8}{{\pi }^2}\cdot \frac{q^2}{\varphi (q)}\cdot \frac{{\pi }^2}{24}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p})\\ -\frac{8}{{\pi }^2}\cdot \frac{q^2}{\varphi (q)}\cdot \frac{{\pi }^2}{24}\cdot \frac{{\varphi }^3(q)}{q^2}\prod _{p|q}(1+\frac{1}{p})\\ ={\varphi }^2(q)\cdot \prod _{p|q}(1+\frac{1}{p}),\end{array}$$

where we have used the identity

$$\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}\chi (2)\cdot {|L(1,\chi )|}^2=\underset{\chi (-1)=-1}{\underset{\chi modq}{{\sum }^{\ast }}}\overline{\chi }(2)\cdot {|L(1,\chi )|}^2.$$

This completes the proof of our theorem.