# The lower bound of weighted representation function

Shi-Qiang Chen111 E-mail<EMAIL_ADDRESS>(S.-Q. Chen).
School of Mathematics and Statistics,
Anhui Normal University, Wuhu 241002, P. R. China

Abstract. For any given set $A$ of nonnegative integers and for any given two
positive integers $k_{1},k_{2}$, $R_{k_{1},k_{2}}(A,n)$ is defined as the
number of solutions of the equation $n=k_{1}a_{1}+k_{2}a_{2}$ with
$a_{1},a_{2}\in A$. In this paper, we prove that if integer $k\geq 2$ and set
$A\subseteq\mathbb{N}$ such that $R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus
A,n)$ holds for all integers $n\geq n_{0}$, then $R_{1,k}(A,n)\gg\log n$. 2020
Mathematics Subject Classification: 11B13

Keywords: Partition; weighted representation function

##  1 Introduction

Let $\mathbb{N}$ be the set of all nonnegative integers. For a given set
$A\subseteq\mathbb{N}$, $n\in\mathbb{N}$, representation functions
$R_{1}(A,n)$, $R_{2}(A,n)$ and $R_{3}(A,n)$ are defined as

$R_{1}(A,n)=\mid\\{(a,a^{\prime}):n=a+a^{\prime},~{}a,a^{\prime}\in A\\}\mid,$
$R_{2}(A,n)=\mid\\{(a,a^{\prime}):n=a+a^{\prime},~{}a<a^{\prime},~{}a,a^{\prime}\in
A\\}\mid,$ $R_{3}(A,n)=\mid\\{(a,a^{\prime}):n=a+a^{\prime},~{}a\leq
a^{\prime},~{}a,a^{\prime}\in A\\}\mid,$

respectively. Sárközy once asked the following question$:$ for
$i\in\\{1,2,3\\}$, are there two sets of nonnegative integers $A$ and $B$ such
that

$|(A\cup B)\setminus(A\cap B)|=+\infty,$

$R_{i}(A,n)=R_{i}(B,n)$ for all sufficiently large integers $n$? This problem
of Sárközy has been solved completely. Recently, many researchers have
obtained many profound results around this problem of Sárközy. For related
research, please refer to [1]-[5], [7]-[10] .

For any given two positive integers $k_{1},k_{2}$ and set
$A\subseteq\mathbb{N}$, weighted representation function
$R_{k_{1},k_{2}}(A,n)$ is defined as the number of solutions of the equation
$n=k_{1}a_{1}+k_{2}a_{2}$ with $a_{1},a_{2}\in A$.

In 2012, Yang and Chen [11] studied weighted representation function. They
proved that if $k_{1}$ and $k_{2}$ are two integers with $k_{2}>k_{1}\geq 2$
and $(k_{1},k_{2})=1$, then there does not exist set $A\subseteq\mathbb{N}$
such that $R_{k_{1},k_{2}}(A,n)=R_{k_{1},k_{2}}(\mathbb{N}\setminus A,n)$ for
all sufficiently large integers $n$, and if $k$ is an integer with $k\geq 2$,
then exists a set $A\subseteq\mathbb{N}$ such that
$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)$ for all integers $n\geq 1$.
They also asked the following question.

######  Problem 1.

Let $k$ be an integer with $k\geq 2$ and $A\subseteq\mathbb{N}$ such that
$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)$ for all integers $n\geq
n_{0}$. Is it true that $R_{1,k}(A,n)\geq 1$ for all sufficiently larger
integers $n$? Is it true that $R_{1,k}(A,n)\rightarrow\infty$ as
$n\rightarrow\infty$?

In 2016, Qu [6] solved this problem affirmatively and proved that the
following result.

######  Theorem A.

(See [6, Theorem 1].) Let $k$ be an integer with $k>1$ and
$A\subseteq\mathbb{N}$ such that $R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus
A,n)$ for all integers $n\geq n_{0}$. Then $R_{1,k}(A,n)\rightarrow\infty$ as
$n\rightarrow\infty$.

In this paper, we continue to focus on Problem 1 and give the lower bound of
weighted representation function.

######  Theorem 1.1.

Let $k$ be an integer with $k\geq 2$ and $A\subseteq\mathbb{N}$ such that
$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)$ holds for all integers $n\geq
n_{0}$. Then $R_{1,k}(A,n)\gg\log n$.

Throughout this paper, the characteristic function of the set
$A\subseteq\mathbb{N}$ is denoted by

$\displaystyle\chi(t)=\begin{cases}0&\mbox{ $t\not\in A$},\\\ 1&\mbox{ $t\in
A$}.\\\ \end{cases}$

Let $C(x)$ be the set of nonnegative integers in $C$ which are less than or
equal to $x$. For positive integer $k$ and sets $A,B\subseteq\mathbb{N}$,
define $kA=\\{ka:a\in A\\}$ and $A+B=\\{a+b:a\in A,~{}b\in B\\}$.

##  2 Lemmas

######  Lemma 2.1.

(See [11, Lemma 2].) Let $k\geq 2$ be an integer and $A\subseteq\mathbb{N}$.
Then $R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)$ holds for all integers
$n\geq n_{0}$ if and only if the following two conditions hold:

(a) for all $n_{0}\leq n<k+n_{0}$, we have

$\underset{a_{1}+ka_{2}=n}{\underset{a_{1}\geq 0,a_{2}\geq
0}{\sum}}1=\underset{a_{1}+ka_{2}=n}{\underset{a_{1}\geq 0,a_{2}\geq
0}{\sum}}\chi(a_{1})+\underset{a_{1}+ka_{2}=n}{\underset{a_{1}\geq 0,a_{2}\geq
0}{\sum}}\chi(a_{2});$ (2.1)

(b) for all $n\geq k+n_{0}$, we have

$\chi(n)+\chi\left(\left\lfloor\frac{n}{k}\right\rfloor\right)=1.$ (2.2)

######  Lemma 2.2.

Let $k\geq 2$ be an integer and $A\subseteq\mathbb{N}$. Then
$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)$ holds for all integers $n\geq
n_{0}$, then for any $n\geq\lfloor\frac{n_{0}+k}{k}\rfloor+1$, we have

$\displaystyle\chi(n)+\chi(k^{i}n+j)=1,~{}~{}~{}j=0,\ldots,k^{i}-1,~{}~{}~{}\text{if
$i$ is odd};$
$\displaystyle\chi(n)=\chi(k^{i}n+j),~{}~{}~{}~{}~{}~{}j=0,\ldots,k^{i}-1,~{}~{}~{}~{}~{}\text{if
$i$ is even}.$ (2.3)

###### Proof.

We now use induction on $i$ to prove that (2.2) is true. By (2.2), we have

$\chi(n)+\chi(kn+j)=1,~{}~{}~{}j=0,\ldots,k-1.$ (2.4)

Therefore, (2.2) is true for $i=1$.

Next, we assume that (2.2) is true for $i=s$, we are going to prove the truth
of (2.2) for $i=s+1$. If $s+1$ is even, then by the induction hypothesis on
$i=s$, we have

$\chi(n)+\chi(k^{s}n+j)=1,~{}~{}~{}j=0,\ldots,k^{s}-1.$ (2.5)

By (2.2), we have

$\chi(k^{s}n+j)+\chi(k(k^{s}n+j)+u)=1,~{}~{}~{}j=0,\ldots,k^{s}-1;u=0,\ldots,k-1.$

It follows from (2.5) that

$\chi(n)=\chi(k(k^{s}n+j)+u),~{}~{}~{}j=0,\ldots,k^{s}-1;u=0,\ldots,k-1,$

that is

$\chi(n)=\chi(k^{s+1}n+j),~{}~{}~{}j=0,\ldots,k^{s+1}-1.$ (2.6)

If $s+1$ is odd, then by the induction hypothesis on $i=s$, we have

$\chi(n)=\chi(k^{s}n+j),~{}~{}~{}j=0,\ldots,k^{s}-1.$ (2.7)

By (2.2), we have

$\chi(k^{s}n+j)+\chi(k(k^{s}n+j)+u)=1,~{}~{}~{}j=0,\ldots,k^{s}-1;u=0,\ldots,k-1,$

It follows from (2.7) that

$\chi(n)+\chi(k(k^{s}n+j)+u)=1,~{}~{}~{}j=0,\ldots,k^{s}-1;u=0,\ldots,k-1,$

that is

$\chi(n)+\chi(k^{s+1}n+j)=1,~{}~{}~{}j=0,\ldots,k^{s+1}-1.$ (2.8)

Up to now, (2.2) has been proved.

This completes the proof of Lemma 2.2. ∎

##  3 Proof of Theorem 1.1

Let $T=\lfloor\frac{n_{0}+k}{k}\rfloor+1$. Given an odd
$j\in[0,\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{2}\rfloor]$, for any
sufficiently larger integer $n$, there exist an integer $i$ such that

$k^{i}(k^{j}+1)T\leq n<k^{i+1}(k^{j}+1)T.$ (3.1)

Now we are going to prove $i+j=\lfloor\log_{k}{\frac{n}{T}}\rfloor$ or
$\lfloor\log_{k}{\frac{n}{T}}\rfloor-1$. In deed, if
$i+j\geq\lfloor\log_{k}{\frac{n}{T}}\rfloor+1$, then

$\frac{n}{T}=k^{\log_{k}{\frac{n}{T}}}<k^{\lfloor\log_{k}{\frac{n}{T}}\rfloor+1}\leq
k^{i+j}<k^{i+j}+k^{i}\leq\frac{n}{T},$

a contradiction. If $i+j\leq\lfloor\log_{k}{\frac{n}{T}}\rfloor-2$, then

$\frac{n}{T}<k^{i+j+1}+k^{i+1}\leq 2k^{i+j+1}\leq
2k^{\lfloor\log_{k}{\frac{n}{T}}\rfloor-1}\leq
2k^{\log_{k}{\frac{n}{T}}-1}\leq k^{\log_{k}{\frac{n}{T}}}=\frac{n}{T},$

a contradiction. By (3.1), there exist $T\leq t\leq kT-1$ and $0\leq
r<k^{i}(k^{j}+1)$ such that

$n=k^{i}(k^{j}+1)t+r.$

According to the value of $r$, we divide into the following two cases for
discussion: Case1. $0\leq r\leq k^{i+j}+k^{i}-k-1$. Noting that $j$ is an odd,
by (2.2), we have

$[k^{i+j-1}t,k^{i+j-1}t+k^{i+j-1}-1]\cup[k^{i}t,k^{i}t+k^{i}-1]\subseteq
A~{}\text{or}~{}\mathbb{N}\setminus A.$

Then

$[k^{i}(k^{j}+1)t,k^{i}(k^{j}+1)t+(k^{i+j}+k^{i}-k-1)]\subseteq
A+kA~{}\text{or}~{}(\mathbb{N}\setminus A)+k(\mathbb{N}\setminus A),$

it follows that $n\in A+kA~{}\text{or}~{}(\mathbb{N}\setminus
A)+k(\mathbb{N}\setminus A)$, which implies that

$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)\geq 1.$

Up to now, we proved that for a given odd
$j\in[0,\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{2}\rfloor]$, we have

$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)\geq 1.$ (3.2)

It is clear that for any two different odds $j_{1},j_{2}$ such that
$j_{1},j_{2}\in[0,\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{2}\rfloor]$
and integers $i_{1},i_{2}$ such that

$i_{1}+j_{1}=K_{1},~{}~{}~{}~{}i_{2}+j_{2}=K_{2},$

where
$K_{1},K_{2}\in\\{\lfloor\log_{k}{\frac{n}{T}}\rfloor,\lfloor\log_{k}{\frac{n}{T}}\rfloor-1\\},$
we have

$i_{1}\neq i_{2}.$ (3.3)

In deed, assume that $j_{1}<j_{2}$, since

$1=-1+2\leq K_{1}-K_{2}-j_{1}+j_{2}=i_{1}-i_{2},$

it follows that $i_{1}\neq i_{2}$. By (3.3), we have

$[k^{i_{1}}t,k^{i_{1}}t+k^{i_{1}}-1]\cap[k^{i_{2}}t,k^{i_{2}}t+k^{i_{2}}-1]=\emptyset.$
(3.4)

Therefore, by (3.2) and (3.4), we have

$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus
A,n)\geq\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{4}\rfloor\gg\log n.$

Case 2. $(k^{i+j}+k^{i}-k-1)+1\leq r\leq k^{i}(k^{j}+1)-1$. Since

$|A\cap\\{T,kT\\}|=1,$

it follows that

$|A(kT)|\geq 1,~{}|(\mathbb{N}\setminus A)(kT)|\geq 1.$ (3.5)

Let $r=k^{i+j}+k^{i}-k-1+s,~{}s\in[1,k]$. Then

$n=k^{i}((k+1)t)+k^{i+j}+k^{i}-k-1+s=k^{i}((k+1)t+k^{j})+k^{i}-k-1+s.$ (3.6)

By (2.2), we have

$[k^{i}((k+1)t+k^{j}),k^{i}((k+1)t+k^{j})+k^{i}-1]\subseteq
A~{}\text{or}~{}\mathbb{N}\setminus A.$

By (3.5), we can choose $a\in[0,kT]$ such that

$\\{a\\}\cup[k^{i}((k+1)t+k^{j}),k^{i}((k+1)t+k^{j})+k^{i}-1]\subseteq
A~{}\text{or}~{}\mathbb{N}\setminus A.$ (3.7)

Since $j\in[0,\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{2}\rfloor]$,
it follows from

$i+j=\lfloor\log_{k}{\frac{n}{T}}\rfloor~{}~{}\text{or}~{}~{}\lfloor\log_{k}{\frac{n}{T}}\rfloor-1$

that

$k^{i}-k-1\geq
k^{\lfloor\frac{\lfloor\log_{k}\frac{n}{T}\rfloor}{2}\rfloor-1}-k-1\geq
k^{2}T\geq ka$

for any sufficiently larger $n$. It follows from (3.6) and (3.7) that

$k^{i}((k+1)t+k^{j})+s\leq n-ka\leq k^{i}((k+1)t+k^{j})+k^{i}-k-1+s,$

which implies that $n\in A+kA~{}\text{or}~{}(\mathbb{N}\setminus
A)+k(\mathbb{N}\setminus A)$, and so

$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)\geq 1.$

Up to now, we proved that for any given odd
$j\in[0,\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{2}\rfloor]$, we have

$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n)\geq 1.$ (3.8)

By (3.3), we have

$[k^{i_{1}}((k+1)t+k),k^{i_{1}}((k+1)t+k)+k^{i_{1}}-1]\cap[k^{i_{2}}((k+1)t+k),k^{i_{2}}((k+1)t+k)+k^{i_{2}}-1]=\emptyset.$
(3.9)

Therefore, by (3.8) and (3.9), we have

$R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus
A,n)\geq\lfloor\frac{\lfloor\log_{k}{\frac{n}{T}}\rfloor}{4}\rfloor\gg\log n.$

This completes the proof of Theorem 1.1.

## References

  * [1] Y.G. Chen and V.F. Lev, Integer sets with identical representation functions, Integers 16(2016), A36.
  * [2] G. Dombi, Additive properties of certain sets, Acta Arith. 103(2002), 137-146.
  * [3] V.F. Lev, Reconstructing integer sets from their representation functions, Electron. J. Combin. 11(2004), R78.
  * [4] S.Z. Kiss and C. Sándor, Partitions of the set of nonnegative integers with the same representation functions, Discrete Math. 340(2017), 1154-1161.
  * [5] Z.H. Qu, A remark on weighted representation functions, Taiwanese J. Math. 18(2014), 1713-1719.
  * [6] Z.H. Qu, A note on representation functions with different weights, Collq. Math. 143(2016), 105-112.
  * [7] E. Rozgonyi and C. Sándor, An extension of Nathanson’s theorem on representation functions, Combinatorica 37(2017), 521-537.
  * [8] C. Sándor, Partitions of natural numbers and their representation functions, Integers 4(2004), A18.
  * [9] M. Tang, Partitions of the set of natural numbers and their representation functions, Discrete Math. 308(2008), 2614-2616.
  * [10] M. Tang, Partitions of natural numbers and their representation functions, Chinese Ann. Math. Ser A 37(2016), 41-46. For English version, see Chinese J. Contemp. Math. 37(2016), 39-44.
  * [11] Q.H. Yang and Y.G. Chen, Partitions of natural numbers with the same weightes representation functions, J. Number Theory 132(2012), 3047-3055.