1 Introduction
Let
$\mathbb {N}$
be the set of all nonnegative integers. For a set
$A\subseteq \mathbb {N}$
, define the additive representation function
and the ordered representation functions
For each
$i\in \{1,2,3\}$
, Sárközy asked whether there exist two sets
$A,B\subseteq \mathbb {N}$
with infinite symmetric difference, that is,
and
for all sufficiently large integers n. For
$i=1$
, Dombi [Reference Dombi3] observed that
$R_1(A, 2m)$
is odd if and only if
$m\in A$
. So
$R_1(A, 2m)\neq R_1(B, 2m)$
for all integers
$m\in (A\cup B)\setminus (A\cap B),$
and the answer is negative for
$i=1$
. In 2002, Dombi [Reference Dombi3] proved that the answer is positive for
$i=2$
. In 2003, Chen and Wang [Reference Chen and Wang2] proved that the answer is also positive for
$i=3$
. Later, many authors studied this problem by means of many different methods. In particular, Sándor [Reference Sándor5] and Tang [Reference Tang6] proved the following nice results in different ways.
Theorem 1.1 [Reference Sándor5, Reference Tang6].
Let N be a positive integer. The equality
$R_2(A, n) = R_2(\mathbb {N}\setminus A, n)$
holds for
$n \ge 2N-1$
if and only if
$|A \cap [0, 2N-1]| = N$
and
$2m \in A \Leftrightarrow m \in A, 2m + 1 \in A \Leftrightarrow m \notin A$
for
$m \ge N$
.
Theorem 1.2 [Reference Sándor5, Reference Tang6].
Let N be a positive integer. The equality
$R_3(A, n) = R_3(\mathbb {N}\setminus A, n)$
holds for
$n \ge 2N-1$
if and only if
$|A \cap [0, 2N-1]| = N$
and
$2m \in A \Leftrightarrow m \notin A, 2m + 1 \in A \Leftrightarrow m \in A$
for
$m \ge N$
.
From Dombi’s discussion, it can be seen that the reason why the answer for
$i=1$
is negative is because the parity of
$R_1(A,n)$
and
$R_1(\mathbb {N}\setminus A,n)$
is different when n is even. The authors of this paper and Cheng [Reference Yan, Li, Li and Cheng9] avoided the case where n is even, and they determined the structure of the set
$A\subseteq \mathbb {N}$
such that
$R_1(A,2k-1)=R_1(\mathbb {N}\setminus A,2k-1)$
for all integers
$k\ge K$
. In addition, they obtained the following results.
Theorem 1.3 [Reference Yan, Li, Li and Cheng9].
Let
$A\subseteq \mathbb {N}$
, K be a positive integer and let g be an even integer with
$-2K\le g\le 2K$
. The equations
hold for all integers
$k\ge K$
if and only if
where
$A_0\subseteq [0,2K-1]$
and
$|A_0|=K+g/2$
.
Theorem 1.4 [Reference Yan, Li, Li and Cheng9].
Let
$A\subseteq \mathbb {N}$
, let K be a nonnegative integer and let g be an odd integer with
$-2K-1\le g\le 2K+1$
. The equations
hold for all integers
$k\ge K$
if and only if
where
$A_0\subseteq [0,2K]$
and
$|A_0|=K+(g+1)/2$
.
There are many results about the ordered representation functions
$R_2(A,n)$
and
$R_3(A,n)$
. Cao-Labora et al. [Reference Cao-Labora, Rué and Spiegel1] showed that the ordered representation functions satisfy an Erdős–Fuchs-type result. Tang [Reference Tang7] considered Problem 6 of Erdős et al. [Reference Erdős, Sárközy and Sós4] about locally small and locally large values of the ordered representation functions. In this paper, we consider the following problem related to ordered representation functions.
Problem 1.5. Let
$g_1,g_2$
be nonnegative integers with
$g_1\neq g_2$
. For
$i\in \{2,3\}$
, do there exist infinite sets
$A\subseteq \mathbb {N}$
and
$T\subseteq \mathbb {N}$
such that
and
for all sufficiently large integers n?
If
$-N\le g_1=g_2\le N$
, the structure of a set
$A\subseteq \mathbb {N}$
such that
$R_i(A,n)-R_i(\mathbb {N}\setminus A,n)=g_1$
for all integers
$n\ge 2N-1$
was provided in [Reference Yan8], where N is a positive integer. In this paper, we prove that if
$g_1\neq g_2$
and
$T=2\mathbb {N}$
, then such a set A in Problem 1.5 does not exist.
Theorem 1.6. Let
$g_1,g_2$
be nonnegative integers with
$g_1\neq g_2$
. Then there does not exist any set
$A\subseteq \mathbb {N}$
such that
for all sufficiently large integers k.
Theorem 1.7. Let
$g_1,g_2$
be nonnegative integers with
$g_1\neq g_2$
. Then there does not exist any set
$A\subseteq \mathbb {N}$
such that
for all sufficiently large integers k.
2 Proofs
In this section, we denote the generating function associated with the set A by
$f_A(x)$
, that is,
where
$\chi _A(x)$
is the characteristic function of A, that is,
$\chi _A(x)=1$
if
$x\in A$
and
$\chi _A(x)=0$
if
$x\notin A$
. Clearly, if
$B=\mathbb {N}\setminus A$
, then
$\chi _A(i)+\chi _B(i)=1$
and
is the generating function associated with the set B.
Lemma 2.1. Let
$A\subseteq \mathbb {N}$
. Then
Proof. Since
and
it follows that
If
$n=2k$
is an even integer with
$k\ge 0$
, then
If
$n=2k-1$
is an odd integer with
$k\ge 1$
, then
$\chi _A({n}/{2})=0$
and
Hence,
This completes the proof of Lemma 2.1.
Lemma 2.2. Let
$A\subseteq \mathbb {N}$
. Then
Proof. Since
and
it follows that
If
$n=2k$
is an even integer with
$k\ge 0$
, then
If
$n=2k-1$
is an odd integer with
$k\ge 1$
, then
$\chi _A({n}/{2})=0$
and
Hence,
This completes the proof of Lemma 2.2.
It should be mentioned that Lemmas 2.1 and 2.2 can be obtained from the proof in [Reference Tang6]; we provide a different proof using generating functions.
Proof of Theorem 1.6.
Suppose that there exists a set
$A\subseteq \mathbb {N}$
such that
for all integers
$k\ge K$
. By Lemma 2.1,
and
for all integers
$k\ge K$
. Consequently,
Since
$-1\le 1-\chi _A(k)-\chi _A(2k+1)\le 1$
and
$g_1\neq g_2$
, it follows that
$g_1-g_2=\pm 1$
. If
$g_1-g_2=1$
, then, by (2.3),
$\chi _A(k)+\chi _A(2k+1)=0$
. So
$\chi _A(k)=0$
for all integers
$k\ge K$
. By (2.2), for all integers
$k\ge K$
,
which is impossible since
$g_2$
and K are fixed integers and k can be sufficiently large. If
$g_1-g_2=-1$
, then, by (2.3),
$\chi _A(k)+\chi _A(2k+1)=2$
. So
$\chi _A(k)=1$
for all integers
$k\ge K$
. By (2.2), for all integers
$k\ge K$
,
which is impossible since
$g_2$
and K are fixed integers and k can be sufficiently large. Hence, there does not exist any set
$A\subseteq \mathbb {N}$
such that (2.1) holds for all sufficiently large integers k.
Proof of Theorem 1.7.
Suppose that there exists a set
$A\subseteq \mathbb {N}$
such that
for all integers
$k\ge K$
. By Lemma 2.2,
and
for all integers
$k\ge K$
. Then
Since
$-1\le \chi _A(k)-\chi _A(2k+1) \le 1$
and
$g_1\neq g_2$
, it follows that
$g_1-g_2=\pm 1$
. If
$g_1-g_2=1$
, by (2.5), then
$\chi _A(k)-\chi _A(2k+1)=1$
. So
$\chi _A(k)=1$
and
$\chi _A(2k+1)=0$
for all integers
$k\ge K$
, which is impossible. If
$g_1-g_2=-1$
, by (2.5), then
$\chi _A(k)-\chi _A(2k+1)=-1$
. So
$\chi _A(k)=0$
and
$\chi _A(2k+1)=1$
for all integers
$k\ge K$
, which is impossible. Hence, there does not exist any set
$A\subseteq \mathbb {N}$
such that (2.4) for all sufficiently large integers k.















