Hostname: page-component-74656f9884-8r8nv Total loading time: 0 Render date: 2026-11-09T01:03:08.656Z Has data issue: false hasContentIssue false
  • English
  • Français

ON A PROBLEM OF NATHANSON ON NONMINIMAL ADDITIVE COMPLEMENTS

Part of: Sequences and sets

Published online by Cambridge University Press:  15 December 2025

SHI-QIANG CHEN Open the ORCID record for SHI-QIANG CHEN [Opens in a new window]  and
YUCHEN DING Open the ORCID record for YUCHEN DING [Opens in a new window]
SHI-QIANG CHEN
Affiliation:
School of Mathematics and Statistics, Anhui Normal University , PR China e-mail: csq20180327@163.com
YUCHEN DING*
Affiliation:
School of Mathematical Sciences, Yangzhou University , PR China HUN-REN Alfréd Rényi Institute of Mathematics , Budapest, Hungary
*
e-mail: ycding@yzu.edu.cn
Rights & Permissions [Opens in a new window]

Abstract

Let C and W be two sets of integers. If $C+W=\mathbb {Z}$, then C is called an additive complement to W. We further call C a minimal additive complement to W if no proper subset of C is an additive complement to W. Answering a problem of Nathanson in part, we give sufficient conditions to show that W has no minimal additive complements. Our result extends a result of Chen and Yang [‘On a problem of Nathanson related to minimal additive complements’, SIAM J. Discrete Math. 26 (2012), 1532–1536].

Keywords

integer setsminimal additive complement

MSC classification

Primary: 11B13: Additive bases, including sumsets

Information

Type
Research Article
Information
Bulletin of the Australian Mathematical Society , Volume 114 , Issue 1 , August 2026 , pp. 6 - 13
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited.
Copyright
© The Author(s), 2025. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc

1 Introduction

Let $\mathbb {Z}$ double struck upper Z be the set of all integers. For $C,W\subseteq \mathbb {Z}$ upper C comma upper W subset of or equal to double struck upper Z and $c\in \mathbb {Z}$ c element of double struck upper Z , define

$$ \begin{align*} C+W:= \{c+w: c\in C,~w\in W\} \quad \text{and} \quad cW:=\{cw: c\in C\}. \end{align*} $$ upper C plus upper W colon equals left brace c plus w colon c element of upper C comma w element of upper W right brace and c upper W colon equals StartSet c w colon c element of upper C EndSet period

For the set $W=\{w\}$ upper W equals StartSet w EndSet consisting of a single element, we just write $C+\{w\}$ upper C plus StartSet w EndSet as $C+w$ upper C plus w for simplicity. If $C+W=\mathbb {Z}$ upper C plus upper W equals double struck upper Z , then C is called an additive complement to W. Furthermore, if no proper subset of C is an additive complement to W, then C is called a minimal additive complement to W. Sometimes, for simplicity, a minimal additive complement is just called a minimal complement throughout this article.

In 2011, Nathanson [Reference Nathanson9, Theorem 8] proved the following interesting theorem.

Theorem 1.1 (Nathanson’s theorem).

Let W be a nonempty, finite set of integers. In  $\mathbb {Z}$ double struck upper Z , every complement to W contains a minimal complement to W.

Nathanson [Reference Nathanson9, Problem 11] also posed the following open problem.

Problem 1.2. Let W be an infinite set of integers. Does there exist a minimal complement to W? Does there exist a complement to W that does not contain a minimal complement?

In 2012, Chen and Yang [Reference Chen and Yang5, Theorem 1] considered Problem 1.2. For the situation $\inf W=-\infty $ inf upper W equals negative infinity and $\sup W=+\infty $ sup upper W equals plus infinity , they made the following progress.

Theorem 1.3 (Chen–Yang).

Let W be a set of integers with $\inf W=-\infty $ inf upper W equals negative infinity and $\sup W=+\infty $ sup upper W equals plus infinity . Then, there exists a minimal complement to W.

After Theorem 1.3, it therefore remains to deal with the case $\inf W>-\infty $ inf upper W greater than negative infinity or $\sup W<+\infty $ sup upper W less than plus infinity . Without loss of generality, we may assume that $\inf W>-\infty $ inf upper W greater than negative infinity since, otherwise, we could consider $-W$ negative upper W instead of W. If $\inf W=w_{0}>-\infty $ inf upper W equals w 0 greater than negative infinity , then we can consider the set

$$ \begin{align*} W-(w_{0}-1):=\{w-(w_{0}-1):w\in W\} \end{align*} $$ upper W minus left parenthesis w 0 minus 1 right parenthesis colon equals StartSet w minus left parenthesis w 0 minus 1 right parenthesis colon w element of upper W EndSet

instead of W. Thus, without loss generality, we may further assume that $\inf W=1$ inf upper W equals 1 . Based on these observations, Chen and Yang [Reference Chen and Yang5, Theorem 2] proved the following theorem.

Theorem 1.4 (Chen–Yang).

Let $W=\{1=w_1<w_2<\cdots \}$ upper W equals left brace 1 equals w 1 less than w 2 less than midline horizontal ellipsis right brace be a set of integers and

$$ \begin{align*}\overline{W}=\mathbb{Z}^{+}\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots\},\end{align*} $$ upper W overbar equals double struck upper Z Superscript plus Baseline minus upper W equals left brace w 1 overbar less than w 2 overbar less than midline horizontal ellipsis right brace comma

where $\mathbb {Z}^{+}$ double struck upper Z Superscript plus is the set of all positive integers.

  1. (a) If $\limsup _{i\rightarrow +\infty }(w_{i+1}-w_i)=+\infty $ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline minus w Subscript i Baseline right parenthesis equals plus infinity , then there exists a minimal complement to W.

  2. (b) If $\lim _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $ limit Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis equals plus infinity , then there does not exist a minimal complement to W.

In 2019, Kiss et al. [Reference Kiss, Sándor and Yang7] pointed out that if

$$ \begin{align*} W=\bigcup\limits_{k=0}^{+\infty} [10^k,2\times 10^k], \end{align*} $$ upper W equals union Underscript k equals 0 Overscript plus infinity Endscripts left bracket 10 Superscript k Baseline comma 2 times 10 Superscript k Baseline right bracket comma

then it is clear that

$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)=+\infty \quad \text{and} \quad \limsup_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_i})=+\infty. \end{align*} $$ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline minus w Subscript i Baseline right parenthesis equals plus infinity and limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis equals plus infinity period

Hence, the criterion $\lim _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $ limit Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis equals plus infinity in Theorem 1.4(b) cannot be simply improved to $\limsup _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis equals plus infinity since this will lead to a contradiction between Theorems 1.4(a) and 1.4(b) for this W. Therefore, it is meaningful to think of the following question asked by Yang (private communication).

Problem 1.5. Let $W=\{1=w_1<w_2<\cdots \}$ upper W equals left brace 1 equals w 1 less than w 2 less than midline horizontal ellipsis right brace be a set of integers and

$$ \begin{align*} \overline{W}=\mathbb{Z}^+\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots\}. \end{align*} $$ upper W overbar equals double struck upper Z Superscript plus Baseline minus upper W equals left brace w 1 overbar less than w 2 overbar less than midline horizontal ellipsis right brace period

Determine the structure of sets W with $\limsup _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis equals plus infinity which do not have a minimal complement.

Along the same lines, Kiss et al. [Reference Kiss, Sándor and Yang7] investigated certain sets W with

$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)<+\infty \end{align*} $$ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline minus w Subscript i Baseline right parenthesis less than plus infinity

which have a minimal complement. Precisely, let X be a set of integers. If there exists a positive integer T such that $x+T\in X$ x plus upper T element of upper X for all sufficiently large integers $x\in X$ x element of upper X , then X is said to be eventually periodic with period T. If W is eventually periodic with $|\mathbb {N}\setminus W|=+\infty $ StartAbsoluteValue double struck upper N minus upper W EndAbsoluteValue equals plus infinity , then there is some integer $T_W$ upper T Subscript upper W so that $w+T_W\in W$ w plus upper T Subscript upper W element of upper W for sufficiently large $w\in W$ w element of upper W , which means that

$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)\le T_W<+\infty. \end{align*} $$ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline minus w Subscript i Baseline right parenthesis less than or equals upper T Subscript upper W Baseline less than plus infinity period

Kiss, Sándor and Yang focused on eventually periodic sets and provided sufficient conditions and necessary conditions for eventually periodic sets to have a minimal additive complement.

For other related results about minimal additive complements, one may also refer to [Reference Alon, Kravitz and Larson1–Reference Chen and Fang4, Reference Davenport6–Reference Kwon8, Reference Ruzsa10–Reference Zhou12].

Extending Theorem 1.4(b), we answer Problem 1.5 by the following theorem which can be viewed as a partial solution to Problem 1.2.

Theorem 1.6. Let $W=\{1=w_1<w_2<\cdots \}$ upper W equals left brace 1 equals w 1 less than w 2 less than midline horizontal ellipsis right brace be a set of integers and

$$ \begin{align*} \overline{W}=\mathbb{Z}^+\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots\}. \end{align*} $$ upper W overbar equals double struck upper Z Superscript plus Baseline minus upper W equals left brace w 1 overbar less than w 2 overbar less than midline horizontal ellipsis right brace period

If there exists a sequence $i_1<i_2<\cdots $ i 1 less than i 2 less than midline horizontal ellipsis such that

$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \quad \text{and} \quad \limsup_{t\rightarrow+\infty}|\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|<+\infty, \end{align*} $$ limit Underscript t right arrow plus infinity Endscripts left parenthesis w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Subscript Baseline overbar right parenthesis equals plus infinity and limit sup Underscript t right arrow plus infinity Endscripts StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript t Subscript Baseline overbar comma w Subscript i Sub Subscript t plus 1 Subscript Baseline overbar right parenthesis EndAbsoluteValue less than plus infinity comma

then there does not exist a minimal complement to W.

The following example shows that our new theorem has some advantages compared with Theorem 1.4.

Example 1.7. Let

$$ \begin{align*} W=\{1\}\cup \bigcup\limits_{k=4}^{+\infty}[2^k+9,2^{k+1})=\{1=w_1<w_2<\cdots\}. \end{align*} $$ upper W equals StartSet 1 EndSet union union Underscript k equals 4 Overscript plus infinity Endscripts left bracket 2 Superscript k Baseline plus 9 comma 2 Superscript k plus 1 Baseline right parenthesis equals left brace 1 equals w 1 less than w 2 less than midline horizontal ellipsis right brace period

It is plain that

$$ \begin{align*} \overline{W}=\mathbb{Z}^+\setminus W=\bigcup\limits_{k=1}^{+\infty}[2^k,2^k+8]=\{\overline{w_1}<\overline{w_2}<\cdots\}. \end{align*} $$ upper W overbar equals double struck upper Z Superscript plus Baseline minus upper W equals union Underscript k equals 1 Overscript plus infinity Endscripts left bracket 2 Superscript k Baseline comma 2 Superscript k Baseline plus 8 right bracket equals left brace w 1 overbar less than w 2 overbar less than midline horizontal ellipsis right brace period

Clearly,

$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)\le 10 \quad \text{and} \quad \liminf_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_i})\le 1. \end{align*} $$ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline minus w Subscript i Baseline right parenthesis less than or equals 10 and limit inf Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis less than or equals 1 period

Thus, both Theorems 1.4(a) and 1.4(b) fail to determine whether such a W has a minimal additive complement. However, it is easy to see that

$$ \begin{align*} \lim_{k\rightarrow\infty}(2^{k+1}-(2^k+8))=+\infty \end{align*} $$ limit Underscript k right arrow infinity Endscripts left parenthesis 2 Superscript k plus 1 Baseline minus left parenthesis 2 Superscript k Baseline plus 8 right parenthesis right parenthesis equals plus infinity

and

$$ \begin{align*} \lim_{k\rightarrow\infty}|\overline{W}\cap(2^k+8,2^{k+1}+8)|=8. \end{align*} $$ limit Underscript k right arrow infinity Endscripts StartAbsoluteValue upper W overbar intersection left parenthesis 2 Superscript k Baseline plus 8 comma 2 Superscript k plus 1 Baseline plus 8 right parenthesis EndAbsoluteValue equals 8 period

Therefore, W does not have a minimal additive complement, thanks to the criterion from Theorem 1.6 with $\overline {w_{i_{k}}}=2^k+8$ w Subscript i Sub Subscript k Baseline overbar equals 2 Superscript k Baseline plus 8 and $\overline {w_{i_{k+1}}}=2^{k+1}+8$ w Subscript i Sub Subscript k plus 1 Baseline overbar equals 2 Superscript k plus 1 Baseline plus 8 .

Remark 1.8. We now give a simple explanation to show that the criterion given in Theorem 1.6 is essentially optimal. We first note that

$$ \begin{align*} \limsup_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_i})\ge \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \end{align*} $$ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline overbar minus w Subscript i Baseline overbar right parenthesis greater than or equals limit Underscript t right arrow plus infinity Endscripts left parenthesis w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Subscript Baseline overbar right parenthesis equals plus infinity

for those W assumed in Theorem 1.6 which are consistent with the requirement of Problem 1.5. We next show that the condition

$$ \begin{align*} \limsup_{t\rightarrow+\infty}|\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|<+\infty \end{align*} $$ limit sup Underscript t right arrow plus infinity Endscripts StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript t Subscript Baseline overbar comma w Subscript i Sub Subscript t plus 1 Subscript Baseline overbar right parenthesis EndAbsoluteValue less than plus infinity

cannot be neglected. To see this, let $W=\bigcup _{k=0}^{+\infty } [10^k,2\times 10^k]$ upper W equals union Underscript k equals 0 Overscript plus infinity Endscripts left bracket 10 Superscript k Baseline comma 2 times 10 Superscript k Baseline right bracket . It is clear that for this W, there exists a sequence $i_1<i_2<\cdots $ i 1 less than i 2 less than midline horizontal ellipsis with $\overline {w_{i_t}}=10^{t+1}-1$ w Subscript i Sub Subscript t Baseline overbar equals 10 Superscript t plus 1 Baseline minus 1 such that

$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \quad \text{and} \quad \limsup_{t\rightarrow+\infty} |\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|=+\infty. \end{align*} $$ limit Underscript t right arrow plus infinity Endscripts left parenthesis w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Subscript Baseline overbar right parenthesis equals plus infinity and limit sup Underscript t right arrow plus infinity Endscripts StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript t Subscript Baseline overbar comma w Subscript i Sub Subscript t plus 1 Subscript Baseline overbar right parenthesis EndAbsoluteValue equals plus infinity period

Then, there exists a minimal additive complement to W by Theorem 1.4(a) since $\limsup _{i\rightarrow +\infty }(w_{i+1}-w_i)=+\infty $ limit sup Underscript i right arrow plus infinity Endscripts left parenthesis w Subscript i plus 1 Baseline minus w Subscript i Baseline right parenthesis equals plus infinity .

2 Proof of Theorem 1.6

The proof of Theorem 1.6 follows from that of Theorem 1.4(b) with refinements.

Suppose to the contrary that C is a minimal additive complement to W. Then, we clearly have $\inf C=-\infty $ inf upper C equals negative infinity . Let

$$ \begin{align*}C\cap (-\infty,-\overline{w_{i_1+1}})=\{c_1>c_2>\cdots\}.\end{align*} $$ upper C intersection left parenthesis negative infinity comma minus w Subscript i 1 plus 1 Baseline overbar right parenthesis equals left brace c 1 greater than c 2 greater than midline horizontal ellipsis right brace period

For any $s\in \mathbb {Z}^+$ s element of double struck upper Z Superscript plus , we assume that

(2.1) $$ \begin{align} \overline{w_{i_{u_s-1}+1}}\leq -c_s<\overline{w_{i_{u_s}+1}}. \end{align} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar less than or equals minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar period

Since $\limsup _{t\rightarrow +\infty }|\overline {W}\cap (\overline {w_{i_{t}}},\overline {w_{i_{t+1}}})|<+\infty $ limit sup Underscript t right arrow plus infinity Endscripts StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript t Subscript Baseline overbar comma w Subscript i Sub Subscript t plus 1 Subscript Baseline overbar right parenthesis EndAbsoluteValue less than plus infinity , there is a constant K such that

$$ \begin{align*} \limsup_{t\rightarrow+\infty}|\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|=K, \end{align*} $$ limit sup Underscript t right arrow plus infinity Endscripts StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript t Subscript Baseline overbar comma w Subscript i Sub Subscript t plus 1 Subscript Baseline overbar right parenthesis EndAbsoluteValue equals upper K comma

which means that there is some $t_W$ t Subscript upper W such that for any $t\ge t_W$ t greater than or equals t Subscript upper W ,

(2.2) $$ \begin{align} |\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|\le K. \end{align} $$ StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript t Subscript Baseline overbar comma w Subscript i Sub Subscript t plus 1 Subscript Baseline overbar right parenthesis EndAbsoluteValue less than or equals upper K period

Our proof will be separated into three cases.

Case 1: $\liminf _{s\rightarrow +\infty } (-c_s-\overline {w_{i_{u_s-1}+1}})<+\infty $ limit inf Underscript s right arrow plus infinity Endscripts left parenthesis minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar right parenthesis less than plus infinity . Let

$$ \begin{align*} h:=\liminf_{s\rightarrow+\infty} (-c_s-\overline{w_{i_{u_s-1}+1}}). \end{align*} $$ h colon equals limit inf Underscript s right arrow plus infinity Endscripts left parenthesis minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar right parenthesis period

Then, $0\le h<+\infty $ 0 less than or equals h less than plus infinity . Define

$$ \begin{align*} H:=\{s:-c_s-\overline{w_{i_{u_s-1}+1}}=h\}. \end{align*} $$ upper H colon equals StartSet s colon minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar equals h EndSet period

It is clear that H is an infinite set since $-c_s-\overline {w_{i_{u_s-1}+1}}$ minus c Subscript s minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar are all integers. Let $s_0\in H$ s 0 element of upper H be any given integer. Then, there exist integers $n_{s_0}\in \mathbb {Z}$ n Subscript s 0 Baseline element of double struck upper Z and $w_{s_0}\in W$ w Subscript s 0 Baseline element of upper W such that

(2.3) $$ \begin{align} n_{s_0}=c_{s_0}+w_{s_0} \quad \text{but}\quad n_{s_0}\neq c+w \end{align} $$ n Subscript s 0 Baseline equals c Subscript s 0 Baseline plus w Subscript s 0 Baseline but n Subscript s 0 Baseline not equals c plus w

for any $w\in W$ w element of upper W and $c\in C$ c element of upper C with $c\neq c_{s_0}$ c not equals c Subscript s 0 , for otherwise, $C\setminus \{c_{s_0}\}$ upper C minus StartSet c Subscript s 0 Baseline EndSet will be an additive complement to W, which contradicts the minimal property of C. Recall that ${\inf C=-\infty }$ inf upper C equals negative infinity and

$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \end{align*} $$ limit Underscript t right arrow plus infinity Endscripts left parenthesis w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Subscript Baseline overbar right parenthesis equals plus infinity

from the condition of our theorem. There exists an integer $s_0^*> s_0$ s 0 Superscript asterisk Baseline greater than s 0 such that

(2.4) $$ \begin{align} n_{s_0}-c_{s}>0 \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline greater than 0

for all $s\in H$ s element of upper H with $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk and

(2.5) $$ \begin{align} \overline{w_{i_t+1}}-\overline{w_{i_t}}>|n_{s_0}|+h \end{align} $$ w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus h

for all $t\geq u_{s_0^*}-1$ t greater than or equals u Subscript s 0 Sub Superscript asterisk Baseline minus 1 . From (2.3) and (2.4),

(2.6) $$ \begin{align} n_{s_0}-c_{s}\in \overline{W} \end{align} $$ n Subscript s 0 Baseline minus c Subscript s element of upper W overbar

for $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk . From (2.5) with $t=u_s$ t equals u Subscript s ,

$$ \begin{align*} \overline{w_{i_{u_s}+1}}-\overline{w_{i_{u_s}}}>|n_{s_0}|+h=|n_{s_0}|-c_s-\overline{w_{i_{u_s-1}+1}}\geq|n_{s_0}|-c_s-\overline{w_{i_{u_s}}} \end{align*} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript u Sub Sub Subscript s Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus h equals StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar greater than or equals StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Baseline overbar

providing that $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk . In other words,

(2.7) $$ \begin{align} n_{s_0}-c_s\leq |n_{s_0}|-c_s<\overline{w_{i_{u_s}+1}}. \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline less than or equals StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar period

However, again from (2.5) with $t=u_s-1$ t equals u Subscript s Baseline minus 1 ,

(2.8) $$ \begin{align} \overline{w_{i_{u_s-1}+1}}-\overline{w_{i_{u_s-1}}}>|n_{s_0}|+h\geq -n_{s_0}. \end{align} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus h greater than or equals minus n Subscript s 0 Baseline period

So by (2.1) and (2.8),

(2.9) $$ \begin{align} n_{s_0}-c_s\ge n_{s_0}+\overline{w_{i_{u_{s}-1}+1}}>\overline{w_{i_{u_{s}-1}}}, \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline greater than or equals n Subscript s 0 Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar greater than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar comma

provided that $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk . We conclude from (2.6), (2.7) and (2.9) that

(2.10) $$ \begin{align} \overline{w_{i_{u_{s}-1}}}<n_{s_0}-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_0}-c_{s}\in \overline{W} \end{align} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar less than n Subscript s 0 Baseline minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar and n Subscript s 0 Baseline minus c Subscript s Baseline element of upper W overbar

for all $s_0,s\in H$ s 0 comma s element of upper H with $s\ge s_0^*> s_0$ s greater than or equals s 0 Superscript asterisk Baseline greater than s 0 .

Now, we take $K+2$ upper K plus 2 different integers $s_{0,1},s_{0,2},\ldots ,s_{0,K+2}\in H$ s Subscript 0 comma 1 Baseline comma s Subscript 0 comma 2 Baseline comma ellipsis comma s Subscript 0 comma upper K plus 2 Baseline element of upper H . Then, there are $K+2$ upper K plus 2 integers $s_{0,1}^*,s_{0,2}^*,\ldots ,s_{0,K+2}^*$ s Subscript 0 comma 1 Superscript asterisk Baseline comma s Subscript 0 comma 2 Superscript asterisk Baseline comma ellipsis comma s Subscript 0 comma upper K plus 2 Superscript asterisk such that for any j with $1\le j\le K+2$ 1 less than or equals j less than or equals upper K plus 2 ,

(2.11) $$ \begin{align} \overline{w_{i_{u_{s}-1}}}<n_{s_{0,j}}-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_{0,j}}-c_{s}\in \overline{W} \quad (\mbox{for all} ~s\ge s_{0,j}^*) \end{align} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar less than n Subscript s Sub Subscript 0 comma j Subscript Baseline minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar and n Subscript s Sub Subscript 0 comma j Subscript Baseline minus c Subscript s Baseline element of upper W overbar left parenthesis for all s greater than or equals s Subscript 0 comma j Superscript asterisk Baseline right parenthesis

from (2.10). Now, let s be a sufficiently large integer so that

$$ \begin{align*}s>\max\{s_{0,1}^*,s_{0,2}^*,\ldots,s_{0,K+2}^*\}\end{align*} $$ s greater than max left brace s Subscript 0 comma 1 Superscript asterisk Baseline comma s Subscript 0 comma 2 Superscript asterisk Baseline comma ellipsis comma s Subscript 0 comma upper K plus 2 Superscript asterisk Baseline right brace

and $u_s-1\ge t_W$ u Subscript s Baseline minus 1 greater than or equals t Subscript upper W . Then, for such s,

$$ \begin{align*} n_{s_{0,j}}-c_s\in \overline{W}\cap(\overline{w_{i_{u_s-1}}},\overline{w_{i_{u_s}+1}}) \quad (\mbox{for} ~1\le j\le K+2) \end{align*} $$ n Subscript s Sub Subscript 0 comma j Subscript Baseline minus c Subscript s Baseline element of upper W overbar intersection left parenthesis w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar comma w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar right parenthesis left parenthesis for 1 less than or equals j less than or equals upper K plus 2 right parenthesis

from (2.11). Moreover, from (2.2),

$$ \begin{align*} |\overline{W}\cap(\overline{w_{i_{u_s-1}}},\overline{w_{i_{u_s}+1}})| = |\overline{W}\cap(\overline{w_{i_{u_s-1}}},\overline{w_{i_{u_s}}})|+|\overline{W}\cap[\overline{w_{i_{u_s}}},\overline{w_{i_{u_{s}}+1}})| \le K+1, \end{align*} $$ StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar comma w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar right parenthesis EndAbsoluteValue equals StartAbsoluteValue upper W overbar intersection left parenthesis w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript Baseline overbar comma w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript Baseline overbar right parenthesis EndAbsoluteValue plus StartAbsoluteValue upper W overbar intersection left bracket w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript Baseline overbar comma w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar right parenthesis EndAbsoluteValue less than or equals upper K plus 1 comma

from which it follows that there are at least two integers $j_1, j_2$ j 1 comma j 2 with $1\le j_1<j_2\le K+2$ 1 less than or equals j 1 less than j 2 less than or equals upper K plus 2 such that

$$ \begin{align*} n_{s_{0,{j_1}}}-c_s=n_{s_{0,{j_2}}}-c_s \quad \text{that is}\quad n_{s_{0,{j_1}}}=n_{s_{0,{j_2}}}. \end{align*} $$ n Subscript s Sub Subscript 0 comma j 1 Subscript Baseline minus c Subscript s Baseline equals n Subscript s Sub Subscript 0 comma j 2 Subscript Baseline minus c Subscript s Baseline that is n Subscript s Sub Subscript 0 comma j 1 Subscript Baseline equals n Subscript s Sub Subscript 0 comma j 2 Subscript Baseline period

By (2.3), we clearly have

$$ \begin{align*} n_{s_{0,{j_1}}}=c_{s_{0,{j_1}}}+w_{s_{0,{j_1}}} \quad \text{and} \quad n_{s_{0,{j_2}}}=c_{s_{0,{j_2}}}+w_{s_{0,{j_2}}}, \end{align*} $$ n Subscript s Sub Subscript 0 comma j 1 Subscript Baseline equals c Subscript s Sub Subscript 0 comma j 1 Subscript Baseline plus w Subscript s Sub Subscript 0 comma j 1 Subscript Baseline and n Subscript s Sub Subscript 0 comma j 2 Subscript Baseline equals c Subscript s Sub Subscript 0 comma j 2 Subscript Baseline plus w Subscript s Sub Subscript 0 comma j 2 Subscript Baseline comma

which is a contradiction with (2.3) since $n_{s_{0,{j_1}}}=n_{s_{0,{j_2}}}$ n Subscript s Sub Subscript 0 comma j 1 Baseline equals n Subscript s Sub Subscript 0 comma j 2 whereas $c_{s_{0,{j_1}}}\neq c_{s_{0,{j_2}}}$ c Subscript s Sub Subscript 0 comma j 1 Baseline not equals c Subscript s Sub Subscript 0 comma j 2 .

Case 2: $\liminf _{s\rightarrow +\infty } (c_s+\overline {w_{i_{u_s}+1}})<+\infty $ limit inf Underscript s right arrow plus infinity Endscripts left parenthesis c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar right parenthesis less than plus infinity . The proof is similar to Case 1. Let

$$ \begin{align*} \ell:=\liminf_{s\rightarrow+\infty} (c_s+\overline{w_{i_{u_s}+1}}). \end{align*} $$ script l colon equals limit inf Underscript s right arrow plus infinity Endscripts left parenthesis c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar right parenthesis period

Then, $0\le \ell <+\infty $ 0 less than or equals script l less than plus infinity . Define

$$ \begin{align*} L:=\{s:c_s+\overline{w_{i_{u_s}+1}}=\ell\}. \end{align*} $$ upper L colon equals StartSet s colon c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar equals script l EndSet period

It is clear that L is an infinite set since $c_s+\overline {w_{i_{u_s}+1}}$ c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Subscript plus 1 Baseline overbar are all integers. Let $s_0\in H$ s 0 element of upper H be any given integer. Then, there exist integers $n_{s_0}\in \mathbb {Z}$ n Subscript s 0 Baseline element of double struck upper Z and $w_{s_0}\in W$ w Subscript s 0 Baseline element of upper W such that

(2.12) $$ \begin{align} n_{s_0}=c_{s_0}+w_{s_0} \quad \text{but}\quad n_{s_0}\neq c+w \end{align} $$ n Subscript s 0 Baseline equals c Subscript s 0 Baseline plus w Subscript s 0 Baseline but n Subscript s 0 Baseline not equals c plus w

for any $w\in W$ w element of upper W and $c\in C$ c element of upper C with $c\neq c_{s_0}$ c not equals c Subscript s 0 , for otherwise, $C\setminus \{c_{s_0}\}$ upper C minus StartSet c Subscript s 0 Baseline EndSet will be an additive complement to W, which contradicts with the minimal property of C. Since ${\inf C=-\infty }$ inf upper C equals negative infinity and

$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty, \end{align*} $$ limit Underscript t right arrow plus infinity Endscripts left parenthesis w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Subscript Baseline overbar right parenthesis equals plus infinity comma

from the condition of our theorem, there exists an integer $s_0^*> s_0$ s 0 Superscript asterisk Baseline greater than s 0 such that

(2.13) $$ \begin{align} n_{s_0}-c_{s}>0 \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline greater than 0

for all $s\in L$ s element of upper L with $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk and

(2.14) $$ \begin{align} \overline{w_{i_t+1}}-\overline{w_{i_t}}>|n_{s_0}|+\ell \end{align} $$ w Subscript i Sub Subscript t Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript t Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus script l

for all $t\geq u_{s_0^*}$ t greater than or equals u Subscript s 0 Sub Superscript asterisk . From (2.12) and (2.13),

(2.15) $$ \begin{align} n_{s_0}-c_{s}\in \overline{W} \end{align} $$ n Subscript s 0 Baseline minus c Subscript s element of upper W overbar

for $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk . By (2.14) with $t=u_s+1$ t equals u Subscript s Baseline plus 1 ,

(2.16) $$ \begin{align} \overline{w_{i_{u_s+1}+1}}-\overline{w_{i_{{u_s}+1}}}>|n_{s_0}|+\ell\geq n_{s_0}, \end{align} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript plus 1 Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript plus 1 Subscript Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus script l greater than or equals n Subscript s 0 Baseline comma

providing that $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk . From (2.1) and (2.16) for $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk ,

(2.17) $$ \begin{align} n_{s_0}-c_s<n_{s_0}+\overline{w_{i_{u_s}+1}}\leq n_{s_0}+\overline{w_{i_{u_s+1}}}<\overline{w_{i_{u_s+1}+1}}. \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline less than n Subscript s 0 Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar less than or equals n Subscript s 0 Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript plus 1 Subscript Baseline overbar less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript plus 1 Subscript plus 1 Baseline overbar period

However, using (2.16) again,

$$ \begin{align*} \overline{w_{i_{u_s}+1}}-\overline{w_{i_{u_s}}}>|n_{s_0}|+\ell= |n_{s_0}|+c_s+\overline{w_{i_{u_s}+1}}, \end{align*} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus script l equals StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue plus c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar comma

by the definition of $\ell $ script l , from which it follows that

(2.18) $$ \begin{align} n_{s_0}-c_s \geq -|n_{s_0}|-c_s>\overline{w_{i_{u_{s}}}}, \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline greater than or equals minus StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue minus c Subscript s Baseline greater than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript Baseline overbar comma

provided that $s\ge s_0^*$ s greater than or equals s 0 Superscript asterisk . We conclude from (2.15), (2.17) and (2.18) that

$$ \begin{align*} \overline{w_{i_{u_{s}}}}<n_{s_0}-c_s<\overline{w_{i_{u_s+1}+1}} \quad \text{and} \quad n_{s_0}-c_{s}\in \overline{W} \end{align*} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript Baseline overbar less than n Subscript s 0 Baseline minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript plus 1 Subscript plus 1 Baseline overbar and n Subscript s 0 Baseline minus c Subscript s Baseline element of upper W overbar

for all $s_0,s\in L$ s 0 comma s element of upper L with $s\ge s_0^*> s_0$ s greater than or equals s 0 Superscript asterisk Baseline greater than s 0 .

The remaining arguments are the same as in the last paragraph of Case 1, so we omit them.

Case 3: $ \liminf _{s\rightarrow +\infty } (-c_s-\overline {w_{i_{u_s-1}+1}})= \liminf _{s\rightarrow +\infty } (c_s+\overline {w_{i_{u_s}+1}})=+\infty $ limit inf Underscript s right arrow plus infinity Endscripts left parenthesis minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar right parenthesis equals limit inf Underscript s right arrow plus infinity Endscripts left parenthesis c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar right parenthesis equals plus infinity . Since C is a minimal complement to W, for any given integer $s_0\geq 1$ s 0 greater than or equals 1 , there exist integers $n_{s_0}\in \mathbb {Z}$ n Subscript s 0 Baseline element of double struck upper Z and $w_{s_0}\in W$ w Subscript s 0 Baseline element of upper W such that

(2.19) $$ \begin{align} n_{s_0}=c_{s_0}+w_{s_0} \quad \text{but}\quad n_{s_0}\neq c+w \end{align} $$ n Subscript s 0 Baseline equals c Subscript s 0 Baseline plus w Subscript s 0 Baseline but n Subscript s 0 Baseline not equals c plus w

for any $w\in W$ w element of upper W and $c\in C$ c element of upper C with $c\neq c_{s_0}$ c not equals c Subscript s 0 . By the conditions of Case 3, there exist infinitely many integers $s>s_0$ s greater than s 0 such that

$$ \begin{align*} -c_{s}-\overline{w_{i_{u_s-1}+1}}>|n_{s_0}| \quad \text{and} \quad c_s+\overline{w_{i_{u_s}+1}}>|n_{s_0}|, \end{align*} $$ minus c Subscript s Baseline minus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue and c Subscript s Baseline plus w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar greater than StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue comma

which implies that

(2.20) $$ \begin{align} n_{s_0}-c_s\leq |n_{s_0}|-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_0}-c_s\geq -|n_{s_0}|-c_s>\overline{w_{i_{u_s-1}+1}}. \end{align} $$ n Subscript s 0 Baseline minus c Subscript s Baseline less than or equals StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar and n Subscript s 0 Baseline minus c Subscript s Baseline greater than or equals minus StartAbsoluteValue n Subscript s 0 Baseline EndAbsoluteValue minus c Subscript s Baseline greater than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar period

We conclude from (2.19) and (2.20) that

$$ \begin{align*} \overline{w_{i_{u_{s}-1}+1}}<n_{s_0}-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_0}-c_{s}\in \overline{W} \end{align*} $$ w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript minus 1 Subscript plus 1 Baseline overbar less than n Subscript s 0 Baseline minus c Subscript s Baseline less than w Subscript i Sub Subscript u Sub Sub Subscript s Sub Subscript Subscript plus 1 Baseline overbar and n Subscript s 0 Baseline minus c Subscript s Baseline element of upper W overbar

for any $s>s_0$ s greater than s 0 .

Using similar arguments to those in Case 1, the validity of the last case clearly follows.

Footnotes

The first author is supported by National Natural Science Foundation of China (Grant No. 12301003), Anhui Provincial Natural Science Foundation (Grant No. 2308085QA02) and University Natural Science Research Project of Anhui Province (Grant No. 2022AH050171). The second author is supported by National Natural Science Foundation of China (Grant No. 12201544), Natural Science Foundation of Jiangsu Province, China (Grant No. BK20210784), China Postdoctoral Science Foundation (Grant No. 2022M710121).

References

Alon, N., Kravitz, N. and Larson, M., ‘Inverse problems for minimal complements and maximal supplements’, J. Number Theory 223 (2021), 307–324.CrossRefGoogle Scholar
Biswas, A. and Saha, J. P., ‘On minimal complements in groups’, Ramanujan J. 55 (2021), 823–847.CrossRefGoogle Scholar
Burcroff, A. and Luntzlara, N., ‘Sets arising as minimal additive complements in the integers’, J. Number Theory 247 (2023), 15–34.CrossRefGoogle Scholar
Chen, Y.-G. and Fang, J.-H., ‘On additive complements, II’, Proc. Amer. Math. Soc. 139 (2011), 881–883.CrossRefGoogle Scholar
Chen, Y.-G. and Yang, Q.-H., ‘On a problem of Nathanson related to minimal additive complements’, SIAM J. Discrete Math. 26 (2012), 1532–1536.CrossRefGoogle Scholar
Davenport, H., ‘On the addition of residue classes’, J. Lond. Math. Soc. 1 (1935), 30–32.CrossRefGoogle Scholar
Kiss, S. Z., Sándor, C. and Yang, Q.-H., ‘On minimal additive complements of integers’, J. Comb. Theory, Ser. A 162 (2019), 344–353.CrossRefGoogle Scholar
Kwon, A., ‘A note on minimal additive complements of integers’, Discrete Math. 342 (2019), 1912–1918.CrossRefGoogle Scholar
Nathanson, M. B., ‘Problems in additive number theory, IV: nets in groups and shortest length $g$ g -adic representations’, Int. J. Number Theory 7 (2011), 1999–2017.CrossRefGoogle Scholar
Ruzsa, I. Z., ‘Additive completion of lacunary sequences’, Combinatorica 21 (2001), 279–291.CrossRefGoogle Scholar
Weinstein, G., ‘Minimal complementary sets’, Trans. Amer. Math. Soc. 212 (1975), 131–137.CrossRefGoogle Scholar
Zhou, F., ‘On eventually periodic sets as minimal additive complements’, Electron. J. Combin. 30 (2023), Article no 4.34.CrossRefGoogle Scholar