Union of Two Arithmetic Progressions with the Same Common Difference Is Not Sum-dominant

TL;DR

This paper proves that the union of two arithmetic progressions with the same common difference is not sum-dominant, improves bounds on the size of sum-dominant subsets, and investigates partitioning properties related to sum-dominance.

Contribution

It establishes that such unions are not sum-dominant, refines the minimal size of sum-dominant sets, and bounds the threshold for partitioning sets into sum-dominant subsets.

Findings

Union of two arithmetic progressions with same difference is not sum-dominant.

Sum-dominant sets in an arithmetic progression have size constraints: at least 7 elements.

Threshold R for partitioning into sum-dominant subsets is between 24 and 145.

Abstract

Given a finite set $A\subseteq \mathbb{N}$, define the sum set $$A+A = \{a_i+a_j\mid a_i,a_j\in A\}$$ and the difference set $$A-A = \{a_i-a_j\mid a_i,a_j\in A\}.$$ The set $A$ is said to be sum-dominant if $|A+A|>|A-A|$. We prove the following results. 1) The union of two arithmetic progressions (with the same common difference) is not sum-dominant. This result partially proves a conjecture proposed by the author in a previous paper; that is, the union of any two arbitrary arithmetic progressions is not sum-dominant. 2) Hegarty proved that a sum-dominant set must have at least $8$ elements with computers' help. The author of the current paper provided a human-verifiable proof that a sum-dominant set must have at least $7$ elements. A natural question is about the largest cardinality of sum-dominant subsets of an arithmetic progression. Fix $n\ge 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$ that contain(s) $0$ and $n-1$. Then $n-7\le N\le n-4$; that is, from an arithmetic progression of length $n\ge 16$, we need to discard at least $4$ and at most $7$ elements (in a clever way) to have the largest sum-dominant set(s). 3) Let $R\in \mathbb{N}$ have the property that for all $r\ge R$, $\{1,2,\ldots,r\}$ can be partitioned into $3$ sum-dominant subsets, while $\{1,2,\ldots,R-1\}$ cannot. Then $24\le R\le 145$. This result answers a question by the author et al. in another paper on whether we can find a stricter upper bound for $R$.

Equations

$$A+A=\{a_i+a_j\mid a_i,a_j\in A\}$$

$$A-A=\{a_i-a_j\mid a_i,a_j\in A\}.$$

$$(\max P_1+1+P_2)\backslash (\max P_1+P_2)=\{\max P_1+\max P_2+1\}.$$

$$k<k+1<\cdots <k+|P_1|+|P_2|-2.$$

$$(|P_1|+1)+3(|P_2|-1)=|P_1|+3|P_2|-2$$

$$2(|P_1|+|P_2|-1)\ge |P_1|+3|P_2|-2,$$

$$m_2-\max P_1<m_2-\max P_1+1<\cdots <m_2$$

$$2k+2(|P_2|-1)-(k+1+3(|P_2|-1))=k-|P_2|\ge 1.$$

$$m_2-n<m_2-(n-1)<\cdots <m_2-0,$$

$$n+1-m_2<n+2-m_2<\cdots <\max P_1-m_2.$$

$$|P_1|+1+3(|P_2|-1)=|P_1|+3|P_2|-2,$$

$$2|P_1|+2(|P_2|-1)=2|P_1|+2|P_2|-2.$$

$$|P_1|+3|P_2|-2\le 2|P_1|+2|P_2|-2.$$

$$K$$

$$=\{0,1,2,4\}\cup \{7,8,\dots ,m\}\cup \{m+4,m+6,m+7\}$$

$$L_1$$

$$R_1$$

$$L_2$$

$$R_2$$

$$R_1^{\prime }$$

$$R_2^{\prime }$$

$$O_{11}$$

$$O_{12}$$

$$O_{21}$$

$$O_{22}$$

$$\cup \{102+m,103+m,104+m\}.$$

$$A_1^{\prime }$$

$$A_2^{\prime }$$

$$M_1$$

$$M_2$$

$$A_1^{\prime }$$

$$=\{1,2,3,4,8,9,11,13,14,15,20,24\}\cup \{25,27,29,\dots ,61\}$$

$$\cup \{62,71,72,84\}\cup \{85,87,\dots ,121\}$$

$$\cup \{122,126,131,132,133,136,138,142,143,144,145\}$$

$$\text{mbox}with|A_1^{\prime }+A_1^{\prime }|-|A_1^{\prime }-A_1^{\prime }|=2,$$

$$A_2^{\prime }$$

$$=\{5,6,7,10,12,16,17,18,19,21,22,23\}\cup \{26,28,30,\dots ,60\}$$

$$\cup \{63,64,65,67,74,75,76,79,81,82,83\}\cup \{86,88,\dots ,120\}\cup \{123,124,125\}$$

$$\cup \{127,128,129,130,134,135,137,139,140,141\}$$

$$\text{mbox}with|A_2^{\prime }+A_2^{\prime }|-|A_2^{\prime }-A_2^{\prime }|=2,$$

$$S$$

Full text

**UNION OF TWO ARITHMETIC PROGRESSIONS WITH THE SAME COMMON DIFFERENCE IS NOT SUM-DOMINANT

Hùng Việt Chu

**

Received: , Revised: , Accepted: , Published:

Abstract

Given a finite set $A\subseteq\mathbb{N}$, define the sum set

[TABLE]

and the difference set

[TABLE]

The set $A$ is said to be sum-dominant if $|A+A|>|A-A|$. We prove the following results

The union of two arithmetic progressions (with the same common difference) is not sum-dominant. This result partially proves a conjecture proposed by the author in a previous paper; that is, the union of any two arbitrary arithmetic progressions is not sum-dominant. 2. 2.

Hegarty proved that a sum-dominant set must have at least $8$ elements with computers’ help. The author of the current paper provided a human-verifiable proof that a sum-dominant set must have at least $7$ elements. A natural question is about the largest cardinality of sum-dominant subsets of an arithmetic progression. Fix $n\geq 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$ that contain(s) [math] and $n-1$. Then $n-7\leq N\leq n-4$; that is, from an arithmetic progression of length $n\geq 16$, we need to discard at least $4$ and at most $7$ elements (in a clever way) to have the largest sum-dominant set(s). 3. 3.

Let $R\in\mathbb{N}$ have the property that for all $r\geq R$, $\{1,2,\ldots,r\}$ can be partitioned into $3$ sum-dominant subsets, while $\{1,2,\ldots,R-1\}$ cannot. Then $24\leq R\leq 145$. This result answers a question by the author et al. in another paper on whether we can find a stricter upper bound for $R$.

2010 Mathematics Subject Classification: Primary 11P99.

*Keywords: *sum-dominant set, MSTD set, arithmetic progression.

1 Introduction

1.1 Background and main results

Given a finite set $A\subseteq\mathbb{N}$, define $A+A=\{a_{i}+a_{j}\,|\,a_{i},a_{j}\in A\}$ and $A-A=\{a_{i}-a_{j}\,|\,a_{i},a_{j}\in A\}$. The set $A$ is said to be

• •

sum-dominant, if $|A+A|>|A-A|$;

• •

balanced, if $|A+A|=|A-A|$; and

• •

difference-dominant, if $|A+A|<|A-A|$.

Because addition is commutative, while subtraction is not, sum-dominant sets are very rare. However, it was first proved by Martin and O’Bryant [13] that as $n\rightarrow\infty$, the proportion of sum-dominant subsets of $\{0,1,2,\ldots,n-1\}$ is bounded below by a positive constant (about $2\cdot 10^{-7}$), which was later improved by Zhao [25] to about $4\cdot 10^{-4}$. However, these works used the probabilistic method and did not give explicit constructions of sum-dominant sets. Later, Miller et al. [15] constructed a family of density $\Theta(1/n^{4})$111A more refined analysis improves the bound to $\Theta(1/n^{2})$ [11]. and Zhao [24] gave a family of density $\Theta(1/n)$. The last few years have seen an explosion of papers exploring properties of sum-dominant sets: see [7, 10, 12, 19, 20, 21, 22] for history and overview, [8, 14, 15, 19, 24] for explicit constructions, [5, 9, 13, 25] for positive lower bounds for the percentage of sum-dominant sets, [11, 16] for generalized sum-dominant sets, and [1, 4, 6, 17, 25] for extensions to other settings.

We know that numbers from an arithmetic progression do not form a sum-dominant set. (We prove in the next section.) It is natural to ask whether numbers from the union of several arithmetic progressions produce a sum-dominant set. Our first result is that the union of two arithmetic progressions with the same common difference is not sum-dominant.

Theorem 1.1**.**

The union of two arithmetic progressions $P_{1}$ and $P_{2}$ (with the same common difference) is not sum-dominant.

This result partially proves the conjecture by the author of the current paper [2] that the union of any two arbitrary arithmetic progressions is not sum-dominant. (The author is motivated by the anonymous referee’s comment that the conjecture was marvelous and tantalizing.) Note that $\{0,2\}\cup\{3,7,11,\ldots,4k-1\}\cup\{4k,4k+2\}$ for $k\geq 5$ is sum-dominant [19], and the set is the union of three arithmetic progressions. Hence, [2, Conjecture 17] is the most we can do.

Our next result concerns the cardinality of a sum-dominant set. Hegarty [8] proved that a sum-dominant set must have at least $8$ elements with the help of computers. The author of the current paper provided a human-understandable proof that a sum-dominant set must have at least $7$ elements [2, 3]. Another natural question is about the largest cardinality of a sum-dominant set. It is well-known that a sum-dominant set can be arbitrarily large, so we put a restriction on the size of the set to have the following result

Theorem 1.2**.**

Fix $n\geq 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$ that contain(s) [math] and $n-1$. Then $n-7\leq N\leq n-4$.

The theorem implies that from an arithmetic progression of length at least $16$, we need to discard at least $4$ elements and not more than $7$ elements (in a clever way) to have the largest sum-dominant set(s). A corollary is that if we want to search for all sum-dominant subsets of $\{0,1,\ldots,n-1\}$, we only need to look for subsets of size between $8$ and $n-4$.

Conjecture 1.3**.**

Fix $n\geq 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$ that contain(s) [math] and $n-1$. Then $N=n-7$.

We run a computer program to find that the conjecture holds for all $16\leq n\leq 34$. For $n\leq 14$, $N$ does not exist. For $n=15$, $N=9$, corresponding to the set $\{0,1,2,4,5,9,12,13,14\}$; that is, we discard $6$ elements.

Our final result is related to the partition of an arithmetic progression into sum-dominant subsets. Asada et al. proved that as $r\rightarrow\infty$, the proportion of $2$-decompositions of $\{1,2,\ldots,r\}$ into sum-dominant subsets is bounded below by a positive constant [1]. Continuing the work, the author of the current paper with Luntzlara, Miller, and Shao proved that it is possible to partition $\{1,2\ldots,r\}$ (for $n$ sufficiently large) into $k\geq 3$ sum-dominant subsets. By defining $R$ to be the smallest integer such that for all $r\geq R$, $\{1,2,\ldots,r\}$ can be $k$-decomposed into MSTD subsets, while $\{1,2,\ldots,R-1\}$ cannot, the authors established rough lower and upper bounds for $R$. However, the upper bound when $k=3$ is very loose because it depends on a sum-dominant subset $A$ with $|A+A|-|A-A|\geq 10|A|$.

Theorem 1.4**.**

Let $R\in\mathbb{N}$ have the property that for $r\geq R$, $\{1,2,\ldots,r\}$ can be partitioned into $3$ sum-dominant subsets, while $\{1,2,\ldots,R-1\}$ cannot. Then $24\leq R\leq 145$.

This theorem answers a question raised by the author of the current paper et al. about whether we can find a more efficient way to decompose $\{1,2,\ldots,r\}$ into $3$ sum-dominant sets. We find a smaller upper bound by a new way of partitioning $\{1,2,\ldots,n\}$ into $3$ sum-dominant subsets. Our construction is similar to that of Miller et al. [15] and utilizes the fact that their construction allows a long run of missing elements. The long run of missing elements is where we can insert a fixed sum-dominant set in.

1.2 Notation

We introduce some notation. Let $A$ and $B$ be sets. We write $A\rightarrow B$ to mean the introduction of elements in $A$ to $B$. We also use a different notation to write a set, which was first introduced by Spohn [23]. Given a set $S=\{m_{1},m_{2},\ldots,m_{n}\}$, we arrange its elements in increasing order and find the differences between two consecutive numbers to form a sequence. Suppose that $m_{1}<m_{2}<\cdots<m_{n}$, then our sequence is $m_{2}-m_{1},m_{3}-m_{2},m_{4}-m_{3},\ldots,m_{n}-m_{n-1}$, and we represent $S=(m_{1}\,|\,m_{2}-m_{1},m_{3}-m_{2},m_{4}-m_{3},\ldots,m_{n}-m_{n-1})=(m_{1}\,|\,a_{1},\ldots,a_{n-1})$, where $a_{i}=m_{i+1}-m_{i}$. Any difference in $S-S$ must be equal to at least a sum $a_{i}+\cdots+a_{j}$ for some $1\leq i\leq j\leq n-1$. Take $S=\{3,2,15,10,9\}$, for example. We arrange the elements in increasing order to have $2$, $3$, $9$, $10$, $15$, form a sequence by looking at the difference between two consecutive numbers: $1$, $6$, $1$, $5$, and write $S=(2\,|\,1,6,1,5)$. All information about a set is preserved in this notation.

An arithmetic progression is a sequence of the form $(a,a+d,a+2d,a+3d,\ldots,a+kd)$ for any arbitrary numbers $a$, $k$, and the common difference $d$. Because sum-dominance is preserved under affine transformations, we can safely assume that our arithmetic progressions contain nonnegative numbers with $1$ being the common difference. To emphasize, all arithmetic progressions we consider will have nonnegative numbers and have the same common difference, which is $1$.

2 Important Results

We use the definition of a symmetric set given by Nathanson [18]: a set $A$ is symmetric if there exists a number $a$ such that $a-A=A$. If so, we say that the set $A$ is symmetric about $a$. The following proposition was proved by Nathanson [18].

Proposition 2.1**.**

A symmetric set is balanced.

Proof.

Let $A$ be a symmetric set about $a$. We have $|A+A|=|A+(a-A)|=|a+(A-A)|=|A-A|$. Hence, $A$ is balanced. ∎

Though symmetric sets are not sum-dominant, adding a few numbers into these sets (in a clever way) can produce sum-dominant sets. Examples of such a technique were provided by Hegarty [8] and Nathanson [19].

Corollary 2.2**.**

A set of numbers from an arithmetic progression is not sum-dominant.

Note that a set of numbers from an arithmetic progression is symmetric about the sum of the maximum and the minimum of the arithmetic progression. For example, the set $E=\{3,5,7,9,11\}$ is symmetric about $14$. The following lemma is proved by Macdonald and Street [14].

Lemma 2.3**.**

Given a finite set $A=(0\,|\,a_{1},a_{2},\ldots,a_{n})$, the following claims hold.

• (1)

If $a_{i}\leq 2$ for all $i$, then $A$ is not sum-dominant.

• (2)

If $a_{i}\in\{1,m\}$ and the first and last times that $1$ occurs as a difference, it occurs in a block of at least $m-1$ consecutive differences, then $A$ is not sum-dominant.

The following lemma is trivial but very useful in our proof of Theorem 1.1.

Lemma 2.4**.**

The following claims hold.

• (1)

Given an arithmetic progression $P_{1}$, $\{\max P_{1}+1\}\rightarrow P_{1}$ gives $2$ new sums.

• (2)

Given arithmetic progressions $P_{1}$ and $P_{2}$, $\{\max P_{1}+1\}\rightarrow(P_{1}\cup P_{2})$ gives at most $3$ new sums.

Proof.

We first prove item 1. Without loss of generality, assume $P_{1}=\{0,1,\ldots,n\}$ for some $n\geq 0$. Denote $Q_{1}=P_{1}\cup\{n+1\}$. Then $P_{1}+P_{1}=\{0,1,\ldots,2n\}$ and $Q_{1}+Q_{1}=\{0,1,\ldots,2n+2\}$. Clearly, $|Q_{1}+Q_{1}|-|P_{1}+P_{1}|=2$.

We proceed to prove item 2. New sums come from the interactions of $\max P_{1}+1$ with $P_{1}$, with $P_{2}$, and with itself. By item 1, the interactions of $\max P_{1}+1$ with $P_{1}$ and itself give at most $2$ new sums. We consider the interactions of $\max P_{1}+1$ with $P_{2}$. We have

[TABLE]

Therefore, the interactions of $\{\max P_{1}+1\}$ with $P_{2}$ gives at most $1$ new sum. In total, we have at most $3$ new sums, as desired. ∎

3 Proof of Theorem 1.1

Because sum-dominance is preserved under affine transformations, without loss of generality, assume that $0=\min P_{1}\leq\min P_{2}$ and $|P_{1}|\geq|P_{2}|$. Let $m_{i}$ and $M_{i}$ denote $\min P_{i}$ and $\max P_{i}$, respectively. Finally, we only consider $P_{1}\cap P_{2}=\emptyset$ because if $P_{1}\cap P_{2}\neq\emptyset$, $P_{1}\cup P_{2}$ is an arithmetic progression222Recall that $P_{1}$ and $P_{2}$ have the same common difference., which does not form a sum-dominant set by Corollary 2.2. Our proof considers $P_{1}$ as the original set and sees how $P_{2}\rightarrow P_{1}$ changes the number of sums and differences.

3.1 Part I. $\mathbf{\max P_{1}<\min P_{2}}$

Let $k=\min P_{2}-\max P_{1}$. If $k=1$, $P_{1}\cup P_{2}$ is an arithmetic progression, not a sum-dominant set. We consider two cases corresponding to $k<1$ and $k>1$.

Case I.1: $k<1$. We consider $P_{2}\rightarrow P_{1}$. The set of new positive and distinct differences includes

[TABLE]

Hence, the number of new differences is at least $2(|P_{1}|+|P_{2}|-1)$. Now, we count the number of new sums. Consider $m_{2}\rightarrow P_{1}$. We have at most $|P_{1}|+1$ new sums. Due to Lemma 2.4, $m_{2}+j\rightarrow P_{1}\cup\{m_{2},\ldots,m_{2}+j-1\}$ gives at most $3$ new sums for all $j\geq 1$. Therefore, $P_{2}\rightarrow P_{1}$ gives at most

[TABLE]

new sums.

Because $|P_{1}|\geq|P_{2}|$, we have

[TABLE]

and so, we do not have a sum-dominant set.

Case I.2: $k>1$. If $k$ is not a multiple of $1$, then with the same reasoning as Case I.1, we are done. If $k$ is a multiple of $1$, we consider two following subcases.

Subcase I.2.1: $k>\max P_{1}$. Then $m_{2}\rightarrow P_{1}$ gives $|P_{1}|$ new positive differences

[TABLE]

while at most $|P_{1}|+1$ new sums. Due to Lemma 2.4, $m_{2}+j\rightarrow P_{1}\cup\{m_{2},\ldots,m_{2}+j-1\}$ gives at most $3$ new sums and at least $2$ new differences $\pm(m_{2}+j)$ for all $j\geq 1$. Therefore, $P_{2}\rightarrow P_{1}$ gives at most $|P_{1}|+1+3(|P_{2}|-1)$ new sums while at least $2|P_{1}|+2(|P_{2}|-1)$ new differences. Because $|P_{1}|\geq|P_{2}|$, the number of new differences is not smaller than the number of new sums, and so, $P_{1}\cup P_{2}$ is not sum-dominant.

Subcase I.2.2: $k\leq\max P_{1}$. If $|P_{2}|\geq k$, we are done due to item 2 Lemma 2.3. So, we consider $|P_{2}|\leq k-1$. Consider $m_{2}\rightarrow P_{1}$. By [2, Proposition 7], $m_{2}\rightarrow P_{1}$ gives $2k$ new differences and $k+1$ new sums. Due to Lemma 2.4, $m_{2}+j\rightarrow P_{1}\cup\{m_{2},\ldots,m_{2}+j-1\}$ gives at most $3$ new sums and $2$ new differences $\pm(m_{2}+j)$ for all $j\geq 1$. The total number of new sums is at most $k+1+3(|P_{2}|-1)$, while the number of new differences is at least $2k+2(|P_{2}|-1)$. We have

[TABLE]

Hence, $P_{1}\cup P_{2}$ is not sum-dominant.

3.2 Part II. $\mathbf{\max P_{1}>\min P_{2}}$

If $m_{2}-1/2\in\mathbb{Z}$, we consider $2(P_{1}\cup P_{2})$. Because the difference between any two consecutive numbers in increasing order is either $1$ or $2$, by item 1 Lemma 2.3, we do not have a sum-dominant set. Hence, we assume that $m_{2}-1/2\notin\mathbb{Z}$. Suppose that $n<m_{2}<n+1$ for some $n\in P_{1}$. The following are new and pairwise distinct positive differences from $m_{2}\rightarrow P_{1}$

[TABLE]

Hence, we have at least $2|P_{1}|$ new differences. On the other hand, $m_{2}\rightarrow P_{1}$ gives at most $|P_{1}|+1$ new sums. Due to Lemma 2.4, $m_{2}+j\rightarrow P_{1}\cup\{m_{2},\ldots,m_{2}+j-1\}$ gives at most $3$ new sums and at least $2$ new differences $\pm(m_{2}+j)$ for all $j\geq 1$. Hence, the total number of new sums as a result of $P_{2}\rightarrow P_{1}$ is at most

[TABLE]

while the number of new differences is at least

[TABLE]

Because $|P_{1}|\geq|P_{2}|$, we have

[TABLE]

Therefore, $P_{1}\cup P_{2}$ is not sum-dominant.

We finish our proof.

4 Proof of Theorem 1.2

Lemma 4.1**.**

For $m\geq 9$, the set

[TABLE]

is sum-dominant. Note that $K$ is a sum-dominant subset of $\{0,1,\ldots,m+7\}$ after we discard $7$ numbers from the arithmetic progression.

Proof.

Observe that $K-K=\{\pm 0,\pm 1,\ldots,\pm(m+7)\}\backslash\{\pm(m+1)\}$, while $K+K=\{0,1,\ldots,2m+14\}\backslash\{2m+9\}$. Hence, $|K+K|-|K-K|=1$. ∎

We now prove Theorem 1.2. Fix $n\geq 16$. Let $N$ be the cardinality of the largest sum-dominant subset(s) of $\{0,1,\ldots,n-1\}$. Lemma 4.1 proves the lower bound for $N$ in Theorem 1.2; that is, $N\geq n-7$. We proceed to show that $N\leq n-4$.

If $N=n$, we have the arithmetic progression $\{0,1,\ldots,n-1\}$, which is not sum-dominant.

If $N=n-1$, we do not have a sum-dominant set due to item 1 Lemma 2.3.

If $N=n-2$, we have two cases. If the two missing numbers are not next to each other, we do not have a sum-dominant set due to item 1 Lemma 2.3. If the two missing numbers are next to each other, we do not have a sum-dominant set due to Theorem 1.1.

If $N=n-3$, we have three cases.

Case 4.1: If the three missing numbers are consecutive, then we do not have a sum-dominant set due to Theorem 1.1. 2. 2.

Case 4.2: If no two numbers are next to each other, then we do not have a sum-dominant set due to item 1 Lemma 2.3. 3. 3.

Case 4.3: Two numbers are next to each other, while the other is not next to any of these numbers. Let the two numbers that are next to each other be $k$ and $k+1$ for some $k\geq 0$. Without loss of generality, assume that the third number is $k+p$ such that $k+p>k+2$.

1. (a)

If $k=0$, we have the set $\{2,3,\ldots,n-1\}\backslash\{k+p\}$, which is not sum-dominant due to item 1 Lemma 2.3. 2. (b)

If $k+p=n-1$, we are back to the case $N=n-2$. 3. (c)

Suppose that $k>0$ and $k+p<n-1$. We have all differences in $\{0,1,\ldots,n-1\}\backslash\{k,k+1,k+p\}$ by looking at $\{0,1,\ldots,n-1\}\backslash\{k,k+1,k+p\}-0$. If we do not have any missing differences, then we are done.

If $k=1$, because $n\geq 16$ and we miss only $3$ numbers, it must be that we have three consecutive numbers in our set. So, we have differences of $1$ and $2$, and so, $k$ and $k+1$ are in the difference set. Hence, we miss at most $2$ differences, which are $\pm(k+p)$. However, we also miss at least $2$ sums, which are $1$ and $2$. Therefore, we do not have a sum-dominant set.

If $k=2$, then $1$ is in our set. We have $k+p$ by looking at $(k+p+1)-1$ and $k+1$ by looking at $(k+2)-1$. Because $n\geq 16$ and we miss only $3$ numbers, it must be that we have three consecutive numbers in our set. So, we have a difference of $2$, and so, $k$ is in the difference set. We are done.

If $k>2$, then $1$ and $2$ are in our set. We have $k+p$ by looking at $(k+p+1)-1$, $k+1$ by looking at $(k+2)-1$, and $k$ by looking at $(k+2)-2$. We are done.

5 Proof of Theorem 1.4

We will use the construction discussed in [4, Theorem 1.1] to partition $\{1,2,\ldots,r\}$ into $3$ sum-dominant subsets. Following the construction, we fix $n=k=20$ and set

[TABLE]

Note that in [4, Theorem 1.1], $A_{1}=L_{1}\cup R_{1}$ and $A_{2}=L_{2}\cup R_{2}$. Pick $m\geq 21$. Set

[TABLE]

Let $M_{1}\subseteq\{66,67,\ldots,59+m\}\backslash\{66,68,69,70,73,77,78,80\}$ such that within $M_{1}$, there exists a sequence of pairs of consecutive elements, where consecutive pairs are not more than $39$ apart and the sequence starts with a pair in $\{66,67,\ldots,101\}$ and ends with a pair in $\{24+m,25+m,\ldots,59+m\}$. Let $M_{2}\subseteq\{66,67,\ldots,59+m\}$ such that within $M_{2}$, there exists a sequence of triplets of consecutive elements, where consecutive triplets are not more than $40$ apart and the sequence starts with a triplet in $\{66,67,\ldots,105\}$ and ends with a triplet in $\{20+m,21+m,\ldots,59+m\}$. Also, $M_{1}\cap M_{2}=\emptyset$ and $M_{1}\cup M_{2}=\{66,67,\ldots,59+m\}\backslash\{66,68,69,70,73,77,78,80\}$. Then

[TABLE]

are both sum-dominant and along with $S=\{66,68,69,70,73,77,78,80\}$ partition $\{1,124+m\}$.

Example 5.1**.**

Let $m=21$. Set

[TABLE]

We partition $\{1,145\}$ into three following sum-dominant sets

[TABLE]

Example 5.1 proves the upper bound of $145$ for $R$ in our Theorem 1.4.

6 Future Research

We end with a list of questions for future research.

• •

Is Conjecture 1.3 correct?

• •

Is Conjecture [2, Conjecture 17] correct?

• •

Is it true that for every fixed $k\geq 2$, as $r\rightarrow\infty$, the proportion of $k$-decompositions of $\{1,2,\ldots,r\}$ into sum-dominant subsets is bounded below by a positive constant?