PARTITIONS OF NATURAL NUMBERS AND THEIR WEIGHTED REPRESENTATION FUNCTIONS

Abstract

For any positive integers $k_1,k_2$ and any set $A\subseteq \mathbb {N}$, let $R_{k_1,k_2}(A,n)$ be the number of solutions of the equation $n=k_1a_1+k_2a_2$ with $a_1,a_2\in A$. Let g be a fixed integer. We prove that if $k_1$ and $k_2$ are two integers with $2\le k_1<k_2$ and $(k_1,k_2)=1$, then there does not exist any set $A\subseteq \mathbb {N}$ such that $R_{k_1,k_2}(A,n)-R_{k_1,k_2}(\mathbb {N}\setminus A,n)=g$ for all sufficiently large integers n, and if $1=k_1<k_2$, then there exists a set A such that $R_{k_1,k_2}(A,n)-R_{k_1,k_2}(\mathbb {N}\setminus A,n)=1$ for all positive integers n.

1 Introduction

Let $\mathbb {N}$ be the set of all nonnegative integers. For a set $A\subseteq \mathbb {N}$ , let $R_{1}(A,n)$ , $R_{2}(A,n)$ and $R_{3}(A,n)$ denote the number of solutions of $a_1+a_2=n, a_1,a_2\in A$ ; $a_1+a_2=n, a_1, a_2\in A, a_1<a_2$ and $a_1+a_2=n, a_1,a_2\in A, a_1\leq a_2$ , respectively. For $i=1,2,3$ , Sárközy asked whether there exist two sets A and B with $|(A\cup B)\setminus (A\cap B)|=+\infty $ such that $R_{i}(A,n)=R_{i}(B,n)$ for all sufficiently large integers n. We call this problem the Sárközy problem. In 2002, Dombi [2] proved that the answer is negative for $i=1$ and positive for $i=2$ . For $i=3$ , Chen and Wang [1] proved that the answer is also positive. In 2004, Lev [3] provided a new proof by using generating functions. Later, Sándor [5] determined the partitions of $\mathbb {N}$ into two sets with the same representation functions by using generating functions. In 2008, Tang [6] provided a simple proof by using the characteristic function.

In 2012, Yang and Chen [7] first considered the Sárközy problem with weighted representation functions. For any positive integers $k_1,\ldots ,k_t$ and any set $A\subseteq \mathbb {N}$ , let $R_{k_1,\ldots ,k_t}(A,n)$ be the number of solutions of the equation $n=k_1a_1+\cdots +k_ta_t$ with $a_1,\ldots ,a_t\in A$ . They posed the following question.

Problem 1.1 [7, Problem 1]

Does there exist a set $A\subseteq \mathbb {N}$ such that $R_{k_1,\ldots ,k_t}(A,n)=R_{k_1,\ldots ,k_t}(\mathbb {N}\setminus A,n)$ for all $n\geq n_0$ ?

They answered this question for $t=2$ and proved the following results.

Theorem 1.2 [7, Theorem 1].

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 any 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.

Theorem 1.3 [7, Theorem 2].

If k is an integer with $k>1$ , then there exists a set $A\subseteq \mathbb {N}$ such that

(1.1) $$ \begin{align} R_{1,k}(A,n)=R_{1,k}(\mathbb{N}\setminus A,n) \end{align} $$

for all integers $n\geq 1$ .

Furthermore, if $0\in A$ , then (1.1) holds for all integers $n\geq 1$ if and only if

$$ \begin{align*}A=\{0\}\bigcup\bigg(\bigcup_{i=0}^{\infty}[(k+1)k^{2i},(k+1)k^{2i+1}-1]\bigg),\end{align*} $$

where $[x,y]=\{n:n\in \mathbb {Z},x\leq n\leq y\}$ .

Later, Li and Ma [4] proved the same results by using generating functions.

Let g be a fixed integer. In this paper, we consider whether there exists a set $A\subseteq \mathbb {N}$ such that $R_{k_1,k_2}(A,n)-R_{k_1,k_2}(\mathbb {N}\setminus A,n)=g$ for all $n\geq n_0$ . First, we answer this problem in the negative if $k_1$ and $k_2$ are two integers with $2\le k_1<k_2$ and $(k_1,k_2)=1$ .

Theorem 1.4. Let g be a fixed integer. If $k_1$ and $k_2$ are two integers with $2\le k_1<k_2$ and $(k_1,k_2)=1$ , then there does not exist any set $A\subseteq \mathbb {N}$ such that

$$ \begin{align*}R_{k_1,k_2}(A,n)-R_{k_1,k_2}(\mathbb{N}\setminus A,n)=g\end{align*} $$

for all sufficiently large integers n.

Similar to Theorem 1.3, we seek a set $A\subseteq \mathbb {N}$ such that $R_{1,k}(A,n)-R_{1,k}(\mathbb {N}\setminus A,n)=g$ for all integers $n\geq 1$ . In fact, if $|g|>1$ , then such a set A does not exist by the simple observation that $0\le R_{1,k}(A,n)\le 1$ and $0\le R_{1,k}(\mathbb {N}\setminus A,n)\le 1$ for all positive integers $n<k$ . So we only need to consider the case $g=1$ .

Theorem 1.5. If k is an integer with $k>1$ , then there exists a set $A\subseteq \mathbb {N}$ such that

(1.2) $$ \begin{align} R_{1,k}(A,n)-R_{1,k}(\mathbb{N}\setminus A,n)=1 \end{align} $$

for all integers $n\ge 1$ .

Furthermore, (1.2) holds for all integers $n\geq 1$ if and only if

$$ \begin{align*}A=\{0\}\bigcup\bigg(\bigcup_{i=0}^{\infty}[k^{2i},k^{2i+1}-1]\bigg).\end{align*} $$

2 Proofs

Lemma 2.1. Let $k_1<k_2$ be two positive integers, $\{a(n)\}_{n=-\infty }^{+\infty }$ be a sequence of integers with $a(n)=0$ for $n<0$ and $A\subseteq \mathbb {N}$ . Then the equality

(2.1) $$ \begin{align} R_{k_1,k_2}(A,n)-R_{k_1,k_2}(\mathbb{N}\setminus A,n)=a(n)\end{align} $$

holds for all nonnegative integers n if and only if

$$ \begin{align*}\chi_{A}\bigg(\bigg[\frac{n}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{n}{k_2}\bigg]\bigg)=1+\sum_{j=0}^{k_1-1}(a(n-j)-a(n-k_2-j))\end{align*} $$

holds for all nonnegative integers n, where $\chi _A(i)$ is the characteristic function of A, that is, $\chi _A(i) = 1$ if $i\in A$ and $\chi _A(i) = 0$ if $i\notin A$ .

Proof. Let $f(x)$ be the generating function associated with A, that is,

$$ \begin{align*}f(x)=\sum_{a\in A}x^{a}=\sum_{i=0}^{\infty}\chi_A(i)x^i.\end{align*} $$

Then,

$$ \begin{align*} &\sum_{n=0}^\infty (R_{k_1,k_2}(A,n) -R_{k_1,k_2}(\mathbb{N}\setminus A,n))x^n \\ &\quad=f(x^{k_1})f(x^{k_2})-\bigg(\frac{1}{1-x^{k_1}}-f(x^{k_1})\bigg)\bigg(\frac{1}{1-x^{k_2}}-f(x^{k_2})\bigg)\\ &\quad=\frac{f(x^{k_1})}{1-x^{k_2}}+\frac{f(x^{k_2})}{1-x^{k_1}}-\frac{1}{(1-x^{k_1})(1-x^{k_2})}. \end{align*} $$

Let

$$ \begin{align*}p(x)=\sum_{n=0}^{\infty}a(n)x^n.\end{align*} $$

It follows that (2.1) holds for all nonnegative integers n if and only if

$$ \begin{align*}\frac{f(x^{k_1})}{1-x^{k_2}}+\frac{f(x^{k_2})}{1-x^{k_1}}-\frac{1}{(1-x^{k_1})(1-x^{k_2})}=p(x),\end{align*} $$

that is,

(2.2) $$ \begin{align} f(x^{k_1})\frac{1-x^{k_1}}{1-x}+f(x^{k_2})\frac{1-x^{k_2}}{1-x}=\frac{1}{1-x}+(1-x^{k_2})\frac{1-x^{k_1}}{1-x}p(x). \end{align} $$

Note that

$$ \begin{align*} f(x^{k_1})\dfrac{1-x^{k_1}}{1-x}=(1+x+\cdots+x^{k_1-1})\displaystyle\sum_{n=0}^{\infty}\chi_A(n)x^{k_1n}=\displaystyle\sum_{n=0}^{\infty}\chi_{A}\bigg(\bigg[\dfrac{n}{k_1}\bigg]\bigg)\kern0.8pt x^n, \end{align*} $$

$$ \begin{align*} f(x^{k_2})\dfrac{1-x^{k_2}}{1-x}=(1+x+\cdots+x^{k_2-1})\displaystyle\sum_{n=0}^{\infty}\chi_A(n)x^{k_2n}=\displaystyle\sum_{n=0}^{\infty}\chi_{A}\bigg(\bigg[\dfrac{n}{k_2}\bigg]\bigg)\kern0.8pt x^n, \end{align*} $$

$$ \begin{align*} \dfrac{1}{1-x}=\displaystyle\sum_{n=0}^{\infty}x^n \end{align*} $$

and

$$ \begin{align*} (1-x^{k_2})\frac{1-x^{k_1}}{1-x}p(x) &=(1-x^{k_2})(1+x+\cdots+x^{k_1-1})\sum_{n=0}^{\infty}a(n)x^n\\ &=\sum_{n=0}^{\infty}\bigg(\sum_{j=0}^{k_1-1}(a(n-j)-a(n-k_2-j))\bigg)\kern0.8pt x^n. \end{align*} $$

It follows from (2.2) that for all nonnegative integers n,

$$ \begin{align*}\chi_{A}\bigg(\bigg[\frac{n}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{n}{k_2}\bigg]\bigg)=1+\sum_{j=0}^{k_1-1}(a(n-j)-a(n-k_2-j)).\end{align*} $$

This completes the proof of Lemma 2.1.

Lemma 2.2. Let $n_0$ be a positive integer and $k_1<k_2$ be two positive integers with $(k_1,k_2)=1$ and $A\subseteq \mathbb {N}$ be a set with

(2.3) $$ \begin{align} \chi_{A}\bigg(\bigg[\frac{i}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{i}{k_2}\bigg]\bigg)=1 \quad \text{for~all}~i\ge k_1+k_2+n_0. \end{align} $$

If $n\ge k_1+k_2+n_0$ and $\chi _{A}(n)+\chi _{A}(n+1)=1$ , then $k_2\mid n+1$ .

Proof. Since $\chi _{A}(n)+\chi _{A}(n+1)=1$ , it follows that

(2.4) $$ \begin{align} \chi_{A}\bigg(\bigg[\frac{(n+1)k_1-1}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{(n+1)k_1}{k_1}\bigg]\bigg)=\chi_{A}(n)+\chi_{A}(n+1)=1. \end{align} $$

By (2.3),

$$ \begin{align*}\chi_{A}\bigg(\bigg[\frac{(n+1)k_1-1}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{(n+1)k_1-1}{k_2}\bigg]\bigg)=1\end{align*} $$

and

$$ \begin{align*}\chi_{A}\bigg(\bigg[\frac{(n+1)k_1}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{(n+1)k_1}{k_2}\bigg]\bigg)=1.\end{align*} $$

It follows from (2.4) that

$$ \begin{align*}\chi_{A}\bigg(\bigg[\frac{(n+1)k_1-1}{k_2}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{(n+1)k_1}{k_2}\bigg]\bigg)=1.\end{align*} $$

Let t and r be integers with

$$ \begin{align*}(n+1)k_1=tk_2+r, \quad 0\le r\le k_2-1.\end{align*} $$

If $r\ge 1$ , then

$$ \begin{align*}1=\chi_{A}\bigg(\bigg[\frac{(n+1)k_1-1}{k_2}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{(n+1)k_1}{k_2}\bigg]\bigg)=2\chi_{A}(t),\end{align*} $$

which is a contradiction. Hence, $r=0$ and $(n+1)k_1=tk_2$ . Noting that $(k_1,k_2)=1$ , we have $k_2\mid n+1$ . This completes the proof of Lemma 2.2.

Proof of Theorem 1.4.

Let g be an integer and let $k_1,k_2$ be integers with $2\le k_1<k_2$ and $(k_1,k_2)=1$ . Suppose that

(2.5) $$ \begin{align} R_{k_1,k_2}(A,n)-R_{k_1,k_2}(\mathbb{N}\setminus A,n)=g \end{align} $$

for all integers $n\ge n_0$ . Let $\{a(n)\}_{n=-\infty }^{+\infty }$ be a sequence of integers with $a(n)=0$ for $n<0$ and $a(n)=g$ for all integers $n\ge n_0$ . It follows from Lemma 2.1 that for all integers $i\ge k_1+k_2+n_0$ ,

(2.6) $$ \begin{align} \chi_{A}\bigg(\bigg[\frac{i}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{i}{k_2}\bigg]\bigg)=1. \end{align} $$

If A is a finite set, then $R_{k_1,k_2}(A,n)=0$ for all sufficiently large integers n, and $R_{k_1,k_2} (\mathbb {N}\setminus A,n)$ cannot be a fixed constant as $n\rightarrow +\infty $ , which implies that (2.5) cannot hold. So A is an infinite set. Similarly, $\mathbb {N}\setminus A$ is also an infinite set.

Since $2\le k_1<k_2$ , it follows that there exists an integer $t>1$ such that $k_2< k_1^{t}$ . Note that both A and $\mathbb {N}\setminus A$ are infinite sets. So there exists an integer $n=k_1^{\alpha }k_2^{\beta }h-1>(k_1+k_2+n_0)^{t+1}$ such that $n\in A$ and $n+1\notin A$ , where $\alpha $ and $\beta $ are nonnegative integers and h is a positive integer with $(h,k_1k_2)=1$ . It follows from (2.6) and Lemma 2.2 that $k_2\mid n+1$ and $\beta \ge 1$ . Since

$$ \begin{align*}(k_1+k_2+n_0)^{t+1}<n<k_1^{\alpha}k_2^{\beta}h<k_1^{t(\alpha+\beta)}h,\end{align*} $$

it follows that $k_1^{\alpha +\beta }>k_1+k_2+n_0$ or $h>k_1+k_2+n_0$ . Hence, for any $0\le i\le \beta $ ,

(2.7) $$ \begin{align} k_1^{\alpha+i}k_2^{\beta-i}h\ge k_1^{\alpha+\beta}h>k_1+k_2+n_0. \end{align} $$

By (2.6),

(2.8) $$ \begin{align} \chi_{A}\bigg(\bigg[\frac{k_1^{\alpha+1}k_2^{\beta}h}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{k_1^{\alpha+1}k_2^{\beta}h}{k_2}\bigg]\bigg)=1 \end{align} $$

and

(2.9) $$ \begin{align} \chi_{A}\bigg(\bigg[\frac{k_1^{\alpha+1}k_2^{\beta}h-k_1}{k_1}\bigg]\bigg)+\chi_{A}\bigg(\bigg[\frac{k_1^{\alpha+1}k_2^{\beta}h-k_1}{k_2}\bigg]\bigg)=1. \end{align} $$

Since $k_1^{\alpha }k_2^{\beta }h=n+1\notin A$ and $k_1^{\alpha }k_2^{\beta }h-1=n\in A$ , it follows from (2.8) and (2.9) that

$$ \begin{align*}\chi_{A}(k_1^{\alpha+1}k_2^{\beta-1}h-1)+\chi_{A}(k_1^{\alpha+1}k_2^{\beta-1}h)=1. \end{align*} $$

By Lemma 2.2, $k_2\mid k_1^{\alpha +1}k_2^{\beta -1}h$ and so $\beta \ge 2$ . Continuing this procedure yields

$$ \begin{align*}\chi_{A}(k_1^{\alpha+\beta}h-1)+\chi_{A}(k_1^{\alpha+\beta}h)=1.\end{align*} $$

By (2.7) and Lemma 2.2, we also have $k_2\mid k_1^{\alpha +\beta }h$ , which is impossible. Hence, there does not exist any set $A\subseteq \mathbb {N}$ such that (2.5) holds for all sufficiently large integers n. This completes the proof of Theorem 1.4.

Proof of Theorem 1.5.

Suppose that there is a set A such that

(2.10) $$ \begin{align} R_{1,k}(A,n)-R_{1,k}(\mathbb{N}\setminus A,n)=1 \end{align} $$

for all integers $n\ge 1$ . Then $0 \in A$ and (2.10) holds for all integers $n\ge 0$ . Let $\{a(n)\}_{n=-\infty }^{+\infty }$ be a sequence of integers with $a(n)=0$ for $n<0$ and $a(n)=1$ for $n\ge 0$ . By Lemma 2.1,

$$ \begin{align*} R_{1,k}(A,n)-R_{1,k}(\mathbb{N}\setminus A,n)=a(n)\end{align*} $$

for all nonnegative integers n if and only if

$$ \begin{align*}\chi_{A}(n)+\chi_{A}\bigg(\bigg[\frac{n}{k}\bigg]\bigg)=1+a(n)-a(n-k)\end{align*} $$

for all nonnegative integers n, that is,

$$ \begin{align*} \chi_{A}(n)+\chi_{A}(0) & =2 \quad\text{for}~0\le n\le k-1, \\ \chi_{A}(n)+\chi_{A}\bigg(\bigg[\frac{n}{k}\bigg]\bigg) & =1 \quad\text{for}~n\ge k. \end{align*} $$

Thus,

$$ \begin{align*} A=\{0\}\bigcup\bigg(\bigcup_{i=0}^{\infty}[k^{2i},k^{2i+1}-1]\bigg).\\[-45pt] \end{align*} $$