On some properties of representation functions related to the Erd\H{o}s-Tur\'an conjecture

TL;DR

This paper investigates properties of representation functions related to the Erdős-Turán conjecture, providing bounds on the number of elements with zero representation and bounds for certain representation counts in modular arithmetic.

Contribution

It establishes new lower bounds on the number of elements with zero representation when the representation function is bounded by 5, improving previous results.

Findings

If R_A(n) ≤ 5, then at least a quarter of the elements have zero representation.

Provides upper bounds for the number of elements with representation counts 2 and 4.

Improves bounds related to the Erdős-Turán conjecture in modular settings.

Abstract

For a set $A\subseteq \mathbb{N}$ and $n\in \mathbb{N}$, let $R_A(n)$ denote the number of ordered pairs $(a,a')\in A\times A$ such that $a+a'=n$. The celebrated Erd\H{o}s-Tur\'an conjecture says that, if $R_A(n)\ge 1$ for all sufficiently large integers $n$, then the representation function $R_A(n)$ cannot be bounded. For any positive integer $m$, Ruzsa's number $R_m$ is defined to be the least positive integer $r$ such that there exists a set $A\subseteq \mathbb{Z}_m$ with $1\le R_A(n)\le r$ for all $n\in \mathbb{Z}_m$. In 2008, Chen proved that $R_{m}\le 288$ for all positive integers $m$. Recently the authors proved that $R_m\ge 6$ for all integers $m\ge 36$. In this paper, we prove that if $A\subseteq \mathbb{Z}_m$ satisfies $R_A(n)\le 5$ for all $n\in \mathbb{Z}_m$, then $|\{g:g\in \mathbb{Z}_m, R_A(g)=0\}|\ge \frac{1}{4}m-\sqrt{5m}$. This improves a recent result of Li and Chen. We also give upper bounds of $|\{g:g\in \mathbb{Z}_m, R_A(g)=i\}|$ for $i=2,4$.

Equations

$$R_{A,B}(g)=♯\{(a,b):a\in A,b\in B,a+b=g\}.$$

$$|\{g:g\in G,R_A(g)=0\}|\ge \frac{7}{32}m-\frac{1}{2}\sqrt{10m}-1.$$

$$\sum _{g\in G}(R_A(g)-k{)}^2\ge km-(2k-1)|A|+k^2-k.$$

$$\sum _{g\in G}(R_A(g)-k{)}^2$$

$$\sum _{g\in G}(R_A(g)-k{)}^2\ge k(m-1)-(2k-1)|A|+k^2=km-(2k-1)|A|+k^2-k,$$

$$\sum _{g\in G\setminus \{0\}}{R}_{A,-A}(g{)}^2\ge \underset{\begin{array}{c}{k}_1,k_2,\dots ,k_{m-1}\in \mathbb{N}\\ {\sum }_{i=1}^{m-1}{k}_i=|A{|}^2-|A|\end{array}}{\min }\sum _{i=1}^{m-1}{k}_i^2.$$

$$\sum _{g\in G\setminus \{0\}}{R}_{A,-A}(g{)}^2$$

$$\sum _{g\in G}(R_A(g)-k{)}^2$$

$$\sum _{g\in G}(R_A(g)-3{)}^2$$

$$|S_0|+|S_1|+|S_2|+|S_3|+|S_4|+|S_5|=|\{g:g\in G,0\le R_A(g)\le 5\}|=m,$$

$$|S_1|+|S_3|+|S_5|=|\{g:g\in G,2\nmid R_A(g)\}|=|\{2a:a\in A\}|\le |A|.$$

$$\sum _{g\in G}(R_A(g)-3{)}^2\le 8|S_0|+3|A|+m.$$

$$\sum _{g\in G}(R_A(g)-3{)}^2\ge 3m-5|A|+6.$$

$$|S_0|\ge \frac{1}{4}m-|A|+\frac{3}{4}\ge \frac{1}{4}m-\sqrt{5m}.$$

$$A_l=2B\cup (2B+2l+1)\text{mod}m,\text{where}2B=\{2b:b\in B\}.$$

$$I(P)=\left\{\begin{array}[]{l l}1,&\text{if the statement}~{}P~{}\text{is true};\\ 0,&\text{if the statement}~{}P~{}\text{is false}.\end{array}\right.$$

$$X_{\text{odd}}^l=\{2k+1:2k+1\in {\mathbb{Z}}_m\text{and}{R}_{A_l}(2k+1)=0\},$$

$$X_{\text{even}}^l=\{2k:2k\in {\mathbb{Z}}_m\text{and}{R}_{A_l}(2k)=0\}.$$

$$|X_{odd}^l|$$

$$\sum _{l=0}^{p^2+p}|X_{\text{even}}^l|$$

$$|X_{even}^l|\le \frac{1}{4}\cdot \frac{(p^2-p{)}^2}{p^2+p+1}<\frac{1}{4}(p^2+p+1)=\frac{1}{8}m.$$

$$|S_0|=|X_{odd}^l|+|X_{even}^l|<\frac{1}{4}m+\frac{1}{8}m=\frac{3}{8}m.$$

$$\sum _{g\in G}(R_A(g)-2{)}^2\ge 2m-3|A|+2.$$

$$\sum _{g\in G}(R_A(g)-2{)}^2$$

$$|S_0|+|S_1|+|S_2|+|S_3|+|S_4|+|S_5|=m,$$

$$|S_2|\le \frac{1}{2}m+3\sqrt{5m}.$$

$$|S_2|=\left(\genfrac{}{}{0ex}{}{|A|}{2}\right)=\frac{1}{2}(p^2+p+1)-\frac{1}{2}=\frac{1}{2}m-\frac{1}{2}.$$

$$\sum _{g\in G}(R_A(g)-4{)}^2\ge 4m-7|A|+12.$$

$$\sum _{g\in G}(R_A(g)-4{)}^2$$

$$m\ge |S_0|+|S_4|\ge \frac{|S_4|-4|A|-3}{3}+|S_4|=\frac{4}{3}|S_4|-\frac{4}{3}|A|-1.$$

Full text

2010 Mathematics Subject Classification: Primary 11B34,11B13.

Keywords and phrases: Representation function, Ruzsa’s number, Erdős-Turán conjecture

1. Institute of Mathematics, Budapest University of Technology and Economics, H-1529 B.O. Box, Hungary

2. School of Mathematics and Statistics, Nanjing University of Information

Science and Technology, Nanjing 210044, China

1 Introduction

Let $G$ be an abelian group. For any set $A,B\subseteq G$, let

[TABLE]

Let $R_{A}(g)=R_{A,A}(g)$. If $A\subseteq\mathbb{N}$ and $R_{A}(n)\geq 1$ for all sufficiently large integers $n$, then we say that $A$ is a basis of $\mathbb{N}$. The celebrated Erdős-Turán conjecture [7] states that if $A$ is a basis of $\mathbb{N}$, then $R_{A}(n)$ cannot be bounded. Erdős [6] proved that there exists a basis $A$ and two constants $c_{1},c_{2}>0$ such that $c_{1}\log n\leq R_{A}(n)\leq c_{2}\log n$ for all sufficiently large integers $n$. Recently, Dubickas [5] gave the explicit values of $c_{1}$ and $c_{2}$. In 2003, Nathanson [15] proved that the Erdős-Turán conjecture does not hold on $\mathbb{Z}$. In fact, he proved that there exists a set $A\subseteq\mathbb{Z}$ such that $1\leq R_{A}(n)\leq 2$ for all integers $n$. In the same year, Grekos et al. [8] proved that if $R_{A}(n)\geq 1$ for all $n$, then $\limsup_{n\rightarrow\infty}R_{A}(n)\geq 6.$ Later, Borwein et al. [1] improved 6 to 8. In 2013, Konstantoulas [11] proved that if the upper density $\overline{d}(\mathbb{N}\setminus(A+A))$ of the set of numbers not represented as sums of two numbers of $A$ is less than $1/10$, then $R_{A}(n)>5$ for infinitely many natural numbers $n$. Chen [3] proved that there exists a basis $A$ of $\mathbb{N}$ such that the set of $n$ with $R_{A}(n)=2$ has density one. Later, the second author [20] and Tang [19] generalized Chen’s result. For the analogue of Erdős-Turán conjecture in groups, one can refer to [9], [10] and [12].

For a positive integer $m$, let $\mathbb{Z}_{m}$ be the set of residue classes mod $m$. If $R_{A}(n)\geq 1$ for all $n\in\mathbb{Z}_{m}$, then $A$ is called an additive basis of $\mathbb{Z}_{m}$.

In 1990, Ruzsa [16] found a basis $A$ of $\mathbb{N}$ for which $R_{A}(n)$ is bounded in the square mean. Ruzsa’s method implies that there exists a constant $C$ such that for any positive integer $m$, there exists an additive basis $A$ of $\mathbb{Z}_{m}$ with $R_{A}(n)\leq C$ for all $n\in\mathbb{Z}_{m}$. For each positive integer $m$, Chen [2] defined Ruzsa’s number $R_{m}$ to be the least positive integer $r$ such that there exists an additive basis $A$ of $\mathbb{Z}_{m}$ with $R_{A}(n)\leq r$ for all $n\in\mathbb{Z}_{m}$. In the same paper, Chen proved that $R_{m}\leq 288$ for all positive integer $m$ and $R_{2p^{2}}\leq 48$ for all primes $p$.

In 2016, the authors [17] proved that if $m\geq 36$, then $R_{m}\geq 6$. That is, if $m\geq 36$ and $A\subseteq\mathbb{Z}_{m}$ satisfies $R_{A}(n)\leq 5$ for all integers $n$, then there exists a $n_{0}\in\mathbb{Z}_{m}$ such that $R_{A}(n_{0})=0$. Recently, Li and Chen (see [14, Corollary 1.3]) gave a quantitative version of this result.

Li & Chen’s Theorem. Let $G$ be a finite abelian group with $|G|=m$ and $A\subseteq G$. If $R_{A}(g)\leq 5$ for all $g\in G$, then

[TABLE]

In this paper, we improve Li and Chen’s theorem and also give an example on the other hand. For convenience, for a fixed nonnegative integer $i$, we denote the set $\{g:g\in G,R_{A}(g)=i\}$ by $S_{i}$.

Theorem 1**.**

(a) Let $G$ be a finite abelian group with $|G|=m$ and $A\subseteq G$. If $R_{A}(g)\leq 5$ for all $g\in G$, then $|S_{0}|\geq\frac{1}{4}m-\sqrt{5m}$.

(b) Let $p$ be a prime and $m=2(p^{2}+p+1)$. Then there exists a subset $A\subseteq\mathbb{Z}_{m}$ such that $R_{A}(n)\leq 5$ for all $n\in\mathbb{Z}_{m}$ and $|S_{0}|<\frac{3}{8}m$.

If $R_{A}(g)\leq 5$ for all $g\in G$, then by $|S_{0}|+|S_{2}|+|S_{4}|\leq m$ and Theorem 1 (a), we see that $|S_{2}|+|S_{4}|\leq\frac{3}{4}m+\sqrt{5m}$. In the next two theorems, we give upper bounds for $|S_{2}|$ and $|S_{4}|$ respectively.

Theorem 2**.**

(a) Let $A\subseteq G$ satisfy $R_{A}(g)\leq 5$ for all $g\in G$. Then $|S_{2}|\leq\frac{1}{2}m+3\sqrt{5m}$.

(b) Let $p$ be a prime and $m=p^{2}+p+1$. Then there exists a subset $A\subseteq\mathbb{Z}_{m}$ such that $R_{A}(n)\leq 2$ for all $n\in\mathbb{Z}_{m}$ and $|S_{2}|=\frac{1}{2}m-\frac{1}{2}$.

Remark 1**.**

The example in Theorem 2 (b) shows that Theorem 2 (a) is nearly best possible.

If $R_{A}(g)\leq 5$ for all $g\in G$, by the statement before Theorem 2, we have $|S_{2}|+|S_{4}|\leq\frac{3}{4}m+\sqrt{5m}$, and so $|S_{4}|\leq\frac{3}{4}m+O(\sqrt{m})$. It seems difficult to improve this upper bound. In the following, we will prove this result by a weak condition $R_{A}(g)\leq 7$ for all $g\in G$.

Theorem 3**.**

(a) Let $A\subseteq G$ satisfy $R_{A}(g)\leq 7$ for all $g\in G$. Then $|S_{4}|\leq\frac{3}{4}m+O(\sqrt{m})$.

(b) Let $p$ be a prime and $m=2(p^{2}+p+1)$. Then there exists a subset $A\subseteq\mathbb{Z}_{m}$ such that $R_{A}(n)\leq 4$ for all $n\in\mathbb{Z}_{m}$ and $|S_{4}|=\frac{1}{2}m-1$.

2 Preliminary Lemmas

Lemma 1**.**

(See [17, Lemma 3].) Let $A\subseteq G$ and $c$ be a positive integer. If $R_{A}(g)\leq c$ for all $g\in G$, then $|A|\leq\sqrt{cm}$.

Lemma 2**.**

(See [18, Singer’s Theorem].) If $l$ is a prime power, then there exists $A\subseteq\mathbb{Z}_{l^{2}+l+1}$ such that $R_{A,-A}(n)=1$ for all $n\in\mathbb{Z}_{l^{2}+l+1},~{}n\not=\overline{0}$.

Lemma 3**.**

If $A$ is a subset of $G$, then for any positive integer $k$ we have

[TABLE]

Proof.

We use Lev and Sárközy’s argument (see [13]) in the following.

[TABLE]

If $\frac{|A|^{2}-|A|}{m-1}\geq k$ or $\frac{|A|^{2}-|A|}{m-1}\leq k-1$, then we have

[TABLE]

and the result is true.

If $k-1<\frac{|A|^{2}-|A|}{m-1}<k$, then

[TABLE]

It is known that if $\sum_{i=1}^{m-1}k_{i}$ is fixed, where $k_{i}\in\mathbb{N}$, then $\sum_{i=1}^{m-1}k_{i}^{2}$ gets the minimal value when $|k_{i}-k_{j}|\leq 1$ for all $1\leq i,j\leq m-1$. Let $|A|^{2}-|A|=q(m-1)+r$, where $q,r$ are nonnegative integers and $0\leq r<m-1$. Then $q=\left\lfloor\frac{|A|^{2}-|A|}{m-1}\right\rfloor$ and $r=\left\{\frac{|A|^{2}-|A|}{m-1}\right\}(m-1)$. Hence

[TABLE]

Therefore,

[TABLE]

∎

3 Proofs

Proof of Theorem 1.

Let $A$ be a given subset of $G$ such that $R_{A}(g)\leq 5$ for all $g\in G$. Then

[TABLE]

It is clear that

[TABLE]

[TABLE]

Hence we have

[TABLE]

On the other hand, by Lemma 3, taking $k=3$, we have

[TABLE]

Therefore, by (1),(2) and Lemma 1, it follows that

[TABLE]

Now we prove part (b). Let $p$ be a prime number and $m=2(p^{2}+p+1)$. By Lemma 2, there is a set $B\subseteq\mathbb{Z}_{p^{2}+p+1}$ such that $R_{B,-B}(n)=1$ for all $n\in\mathbb{Z}_{p^{2}+p+1}$ and $n\not=\overline{0}$. Then for any integer $l$ with $0\leq l\leq p^{2}+p$, we define

[TABLE]

Now we first prove that $R_{A_{l}}(n)\leq 4$ for all $n\in\mathbb{Z}_{m}$.

If $2\mid n$, then $n=a_{1}+a_{2}$ with $a_{1},a_{2}\in 2B$ or $a_{1},a_{2}\in 2B+2l+1$. Hence $R_{A_{l}}(n)=R_{B}(\frac{n}{2})+R_{B}(\frac{n}{2}-(2l+1))\leq 2+2=4$.

If $2\nmid n$, then $n=a_{1}+a_{2}$ with $a_{1}\in 2B,a_{2}\in 2B+2l+1$ or $a_{1}\in 2B+2l+1,a_{2}\in 2B$. Hence, $R_{A_{l}}(n)=R_{B}(\frac{n-2l-1}{2})\times 2\leq 4$.

Therefore, $R_{A_{l}}(n)\leq 4$ for all $n\in\mathbb{Z}_{m}$.

Let $P$ be a statement and we define

[TABLE]

Let

[TABLE]

[TABLE]

Then $S_{0}=X_{\text{odd}}^{l}\cup X_{\text{even}}^{l}$. It is clear that $R_{A_{l}}(2n+1)=0$ if and only if $R_{B}(n-l)=0$. Then

[TABLE]

and

[TABLE]

Hence there is an integer $l$ such that

[TABLE]

Therefore, for this integer $l$,

[TABLE]

∎

Proof of Theorem 2.

By Lemma 3, taking $k=2$, we have

[TABLE]

On the other hand,

[TABLE]

Hence, by (3) and (3), we have $|S_{0}|+|S_{4}|\geq\frac{1}{2}m-3|A|+\frac{1}{2}\geq\frac{1}{2}m-3\sqrt{5m}$. Since

[TABLE]

it follows that

[TABLE]

Now we prove the part (b). By Lemma 2, there exists a subset $A\subseteq\mathbb{Z}_{m}$ such that $R_{A,-A}(n)=1$ for all $n\in\mathbb{Z}_{m}$, $n\not=\overline{0}$. It is easy to see that $|A|=p+1$ and $R_{A}(n)\leq 2$ for all $n\in\mathbb{Z}_{m}$. Hence

[TABLE]

∎

Proof of Theorem 3.

By Lemma 3, taking $k=4$, we have

[TABLE]

On the other hand, by $|S_{1}|+|S_{3}|+|S_{5}|+|S_{7}|\leq|A|$, we have

[TABLE]

Hence $|S_{4}|\leq 4|A|+3|S_{0}|+3$. Since $\sum_{i=0}^{7}|S_{i}|=m$, it follows that

[TABLE]

By Lemma 1, we have

[TABLE]

Now we prove the part (b). Let $p$ be a prime and $m=2(p^{2}+p+1)$. By Lemma 2, there exists a subset $A_{p}\subseteq\mathbb{Z}_{p^{2}+p+1}$ such that $R_{A_{p},-A_{p}}(n)=1$ for all $n\not=\overline{0}$. Let $A=2A_{p}\cup(p^{2}+p+1+2A_{p})\subseteq\mathbb{Z}_{m}$.

If $2\mid n$, then $R_{A}(n)=R_{A_{p}}(\frac{n}{2})+R_{A_{p}}(\frac{n}{2})\leq 2+2=4$. If $2\nmid n$, then $R_{A}(n)=2R_{A_{p}}(\frac{n-(p^{2}+p+1)}{2})\leq 4$. Hence $R_{A}(n)\leq 4$ for all $n\in\mathbb{Z}_{m}$.

[TABLE]

∎

4 Acknowledgement

This work was done during the second author visiting to Budapest University of Technology and Economics. He would like to thank Dr. Sándor Kiss and Dr. Csaba Sándor for their warm hospitality.