On Optimal M-Sets Related to Motzkin’s Problem

Abstract

Let be a set of positive integers. A set of nonnegative integers is called an set if , then . If is an set with the maximal cardinality, then is called a maximal set of . If is a maximal set of for all integers , then we call an optimal set. In this paper, we study the existence of an optimal set.

1. Introduction

Let be the set of all nonnegative integers. For a positive real number and , we denote by the number of elements such that . The upper and lower densities of denoted by and , respectively, are given byIf , let , then we say that has density .

Given a set of positive integers, we call an set if then . Motzkin [1] presented the problem of determining the quantitywhere the supremum is taken over all sets . Cantor and Gordon [2] proved that if , and that if , then . Later, Haralambis [3] determined for many members of the families and . In 1999, Gupta and Tripathi [4] completely determined when is a finite arithmetic progression. In 2011, Pandey and Tripathi [5] investigated this quantity when is related to an arithmetic progression. After that, Pandey and Tripathi [6] and the first author and Tang [7] determined when is a geometric progression. For other results, one may refer to [8–11].

In 2010, Chen and Yang [12] and Khovanova and Konyagin [13] studied the upper density among sets of nonnegative integers in which no two elements have quotient belonging to .

Suppose is a graph and is a set of nonnegative integers. A coloring of is an assignment of a positive integer to each vertex of so that if and are joined by an edge of , then is not in . colorings were introduced by Hale [14] in connection with the channel assignment problem in communications. Motzkin’s problem has connections with the colouring problem. For more results on this topic, one can refer to [15–19].

Motzkin’s problem also has connections with some other problems, such as problems related to the fractional chromatic number of distance graphs and the Lonely Runner Conjecture. One can refer to [9, 20–29].

In this paper, for any integers and , we define . For an set , if for any set , then is said to be a maximal set of . If is a maximal set of for all integers , then we call an optimal set. If is finite, then an optimal set is infinite. For a finite set , let be the infinite sequence recursively by choosing to be the least integer exceeding such that is also an set. It is easy to see that, for a given set , if there exists an optimal set, then this set must be .

In this paper, we determine all the optimal set when .

Theorem 1. For any set with , there exists an optimal set.

Theorem 2. Let , where , and let . Then, there exists an optimal set if and only if or with an odd .

Remark 1. By Theorem 2, we know that there exists an optimal set when .

2. Proofs of Theorem 1

Before presenting the proof of Theorem 1, we first give some lemmas here. Lemma 1 is a small generalization of Corollary 1, which is used by Cantor and Gordon in [2].

Lemma 1. Let and , where is a positive integer. For any sets ,is an set. Conversely, for any set and ,is an set.

Proof. Suppose that there exist two elements such that for some positive integer . Let and with and . Then, by , we obtain . Hence, and . This contradicts that is an set.
Conversely, suppose that there exists an integer with such that is not an set. Then, there exist such that for some positive integer . Let , where . Then, . This contradicts that is an set.

Corollary 1. Let and , where is a positive integer. For any sets , the set is an set.
Corollary 1 follows from Lemma 1 immediately by taking .

For two sets and and an integer , let ,

Lemma 2. Let and , where is a positive integer. Then, is an optimal set if and only if is an optimal set.

Proof. We first prove the relationship between and , i.e.,Suppose that equality (5) does not hold. Let be the smallest integer such that . For any integer , let , where . Then, is equivalent to for all and . If , then must hold. Hence, we only need to consider the case . So, is equivalent to for all integers with and , that is, . Hence, is equivalent to , a contradiction.
Now, we shall prove is an optimal set if and only if is an optimal set.
Suppose that is an optimal set. If is not an optimal set, then there exists an integer and an set such that . Let . Then, . By (5), we have . By Lemma 1, is an set. Thus, is not an optimal set.
Now, suppose that is an optimal set. If is not an optimal set, then there exists an integer and an set such that .
Suppose that . Let be the maximal integer such that and . Then, by (5), we have . Thus,Suppose that . Let be the minimal integer such that and . Then, by (5), we have . Thus,Hence, without loss of generality, we may assume . By (5), we haveFor , let . Then, . Since , by (8), there exists an integer with such that . By Lemma 1, is an set. Thus, is not an optimal set.
Therefore, is an optimal set if and only if is an optimal set.
Proof of Theorem 1: it suffices to prove that is an optimal set when . By Lemma 2, we may assume that . Clearly, is the set of all even numbers. Since any set contains no consecutive integers, it follows that is an optimal set.

3. Proofs of Theorem 2

In order to prove Theorem 2, we first prepare some lemmas. Let denote the absolute value of the absolutely least remainder of (mod). The proof of Lemma 3 follows proofs of Theorem 1 and Theorem 4 (ii) in [2]. For the convenience of the reader, we present it here.

Lemma 3. (see [2], Theorem 1 and Theorem 4 (ii)). Let , where and . Then, there exists an set such that .

Proof. By and , we have , and then there exists such that . Thus, . Noting that , we obtain and . Hence, , and then,LetThen, and . Next, we shall prove that is an set. That is, if , then . We obtain , where . Then, if , we haveIt follows that or . However, by and (9), we havea contradiction.
In the next lemma, we study the period of .

Lemma 4. Let . Then, is periodic with period .

Proof. We shall prove that if and only if for all .
If , then . By the definition of , we have .
If , then we shall prove . Suppose that . Then, . If , by and the definition of , we have , a contradiction. Thus, we have , and then . Similarly, we also have and . Since and , by the definition of , it follows that , a contradiction. Hence, .
Therefore, is a periodic set with period .
In 2012, the authors of [30] study the difference sets containing integer powers. Lemma 5 is the main theorem in [30], which is used to prove Lemma 6.

Lemma 5. (see [30], Theorem 1). Let be positive integers and , where . Letwhere denotes the cardinality of even numbers in the interval . Then, we have the following:(i)If is a subset of with , then contains either 1 or (ii)There exists a subset of with such that contains no power of

Lemma 6. Let . Then, is an optimal set.

Proof. It is easy to see that, if is odd, then is the set of all even integers; if is even, thenHence, . By Lemma 5 (i), there is no set such that and is also an set. Hence, is an optimal set.
Proof of Theorem 2: it suffices for us to determine such that is an optimal set. By Lemma 2, we may assume . If , by Lemma 6, then is an optimal set. Now, we suppose that . Noting that , we write , where . If is odd, thenand so . If is even, thenIt follows that .

Case 1. . Let be the set of all even numbers contained in . Clearly, is an set and . Since and , we have . Hence, is not an optimal set.

Case 2. . By Lemma 3, there exists an set such that . It is easy to check that except for the following two cases: (a) is odd and and (b) is even and . Hence, is not an optimal set except for these two cases.
Now, we shall prove that is an optimal set in cases (a) and (b). By Lemma 4, is periodic with period . Hence, it suffices for us to prove that is a maximal set of for any integer . Subcase 1: is odd and . Then, . Suppose that is an set. Noting that , we have for all integers with and . Since It follows that is a maximal set of for all . Now, we consider the case . If , then we have and for all integers with and . Hence, Similarly, we have . On the other hand, by and , we have , a contradiction. Thus, is an optimal set in this case. Subcase 2: is even and . Then, . Suppose that is an set. By , we havefor all integers with and . However,It follows that is a maximal set of for all .
Now, we consider the case . If is a maximal set of , by , then is a maximal set of . Hence, it is enough for us to prove is a maximal set of . If , then we havefor all integers with and . In particular, , since . Hence,Similarly, we obtain . It is a contradiction with . Hence, is also an optimal set in this case.
Therefore, is an optimal set if and only if or , where is odd.
At the end of this paper, we pose a problem for further research.

Problem 1. Given an integer , determine all sets with such that there exists an optimal set.

Remark 2. If is optimal, then gives the value of . If is not optimal, we cannot find any information about the value of . Thus, Problem 1 and Motzkin’s problem are incomparable.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This study was supported by the National Natural Science Foundation for Youth of China (Grant no. 11501299), Natural Science Foundation of Jiangsu Province (Grant nos. BK20150889 and 15KJB110014), and the Startup Foundation for Introducing Talent of NUIST (Grant no. 2014r029).