\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage{mathdots} \usepackage{mathtools} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}} \newcommand{\G}{\mbox{$\mathcal{G}$}} \newcommand{\Z}{\ensuremath{\mathbb{Z}}} \newcommand{\N}{\mathbb{N}} \newcommand{\ph}{g} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \providecommand{\floor}[1]{\lfloor #1 \rfloor} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf When Sets Can and Cannot Have\\ \vskip .1in Sum-Dominant Subsets} \vskip 0.8cm %\large H\`ung Vi\d{\^e}t Chu\\ Department of Mathematics\\ Washington and Lee University \\ Lexington, VA 24450\\ USA \\ \href{mailto:chuh19@mail.wlu.edu}{\tt chuh19@mail.wlu.edu} \\ \ \\ Nathan McNew\\ Department of Mathematics\\ Towson University\\ Towson, MD 21252\\ USA \\ \href{mailto:nmcnew@towson.edu}{\tt nmcnew@towson.edu}\\ \ \\ Steven J. Miller\\ Department of Mathematics\\ Williams College \\ Williamstown, MA 01267\\ USA \\ \href{mailto:sjm1@williams.edu}{\tt sjm1@williams.edu} \\ \href{Steven.Miller.MC.96@aya.yale.edu}{\tt Steven.Miller.MC.96@aya.yale.edu}\\ \ \\ Victor Xu and Sean Zhang\\ Department of Mathematics\\ Carnegie Mellon University \\ Pittsburgh, PA 15213\\ USA \\ \href{mailto:vzx@andrew.cmu.edu}{\tt vzx@andrew.cmu.edu}\\ \href{mailto:xiaoronz@andrew.cmu.edu}{\tt xiaoronz@andrew.cmu.edu}\\ \end{center} \newpage \begin{abstract} A finite set of integers $A$ is a sum-dominant (also called a More Sums Than Differences or MSTD) set if $|A+A| > |A-A|$. While almost all subsets of $\{0, \dots, n\}$ are not sum-dominant, interestingly a small positive percentage are. We explore sufficient conditions on infinite sets of positive integers such that there are either no sum-dominant subsets, at most finitely many sum-dominant subsets, or infinitely many sum-dominant subsets. In particular, we prove no subset of the Fibonacci numbers is a sum-dominant set, establish conditions such that solutions to a recurrence relation have only finitely many sum-dominant subsets, and show there are infinitely many sum-dominant subsets of the primes. \end{abstract} \section{Introduction} For any finite set of natural numbers $A \subset \N$, we define the sumset \be A+A \ \coloneqq \ \{a + a' : a, a' \in A\}\ee and the difference set \be A-A \ \coloneqq \ \{a - a' : a, a' \in A\};\ee $A$ is sum-dominant (also called a More Sums Than Differences or MSTD set) if $|A+A| > |A-A|$ (if the two cardinalities are equal it is called balanced, and otherwise difference-dominant). As addition is commutative and subtraction is not, it was natural to conjecture that sum-dominant sets are rare. Conway gave the first example of such a set, $\{0, 2, 3, 4, 7, 11, 12, 14\}$, and this is the smallest such set. Later authors constructed infinite families, culminating in the work of Martin and O'Bryant, which proved a small positive proportion of subsets of $\{0, \dots, n\}$ are sum-dominant as $n\to\infty$, and Zhao, who estimated this percentage at around $4.5 \cdot 10^{-4}$. See \cite[pp.~172--174]{FP} and \cite{He, HM, Ma, MO, Na1, Na2, Na3, Ru1, Ru2, Zh3} for general overviews, examples, constructions, bounds on percentages and some generalizations, \cite{MOS, MPR, MS, Zh1} for some explicit constructions of infinite families of sum-dominant sets, and \cite{DKMMW, DKMMWW, MV, Zh2} for some extensions to other settings. Much of the above work looks at finite subsets of the natural numbers, or equivalently subsets of $\{0, 1, \dots, n\}$ as $n\to\infty$. We investigate the effect of restricting the initial set on the existence of sum-dominant subsets. In particular, given an infinite set $A = \{a_k\}_{=1}^\infty$, when does $A$ have no sum-dominant subsets, only finitely many sum-dominant subsets, or infinitely many sum-dominant subsets? \emph{We assume throughout the rest of the paper that every such sequence $A$ is strictly increasing and non-negative.} Our first result shows that if the sequence grows sufficiently rapidly and there are no `small' subsets which are sum-dominant, then there are no sum-dominant subsets. \begin{theorem}\label{thm:gen} Let $A = \{a_k\}_{k=1}^\infty$ be a strictly increasing sequence of non-negative numbers. If there exists a positive integer $r$ such that \begin{enumerate} \item $a_k > a_{k-1} + a_{k-r}$ for all $k\ge r+1$, and \item $A$ does not contain any sum-dominant set $S$ with $|S| \le 2r-1$, \end{enumerate} then $A$ contains no sum-dominant set. \end{theorem} We prove this in \S\ref{sec:noMSTD}. As the smallest sum-dominant set has 8 elements (see \cite{He}), the second condition is trivially true if $r \le 4$. In particular, we immediately obtain the following interesting result. \begin{corollary} No subset of the Fibonacci numbers $\{0, 1, 2, 3, 5, 8, \dots\}$ is a sum-dominant set. \end{corollary} The proof is trivial, and follows by taking $r=3$ and noting \be F_k \ = \ F_{k-1} + F_{k-2} \ > \ F_{k-1} + F_{k-3} \ee for $k \ge 4$. After defining a class of subsets we present a partial result on when there are at most finitely many sum-dominant subsets. \begin{definition}[Special Sum-Dominant Set] For a sum-dominant set $S$, we call $S$ a special sum-dominant set if $|S+S| - |S-S| \ge |S|$. \end{definition} We prove sum-dominant sets exist in \S\ref{sec:specialsumdom}. Note if $S$ is a special sum-dominant set then if $S' = S \cup \{x\}$ for any sufficiently large $x$ then $S'$ is also a sum-dominant set. We have the following result about a sequence having at most finitely many sum-dominant sets (see \S\ref{sec:finitelymany} for the proof). \begin{theorem}\label{thm:finite} Let $A = \{a_k\}_{k=1}^\infty$ be a strictly increasing sequence of non-negative numbers. If there exists a positive integer $s$ such that the sequence $\{a_k\}$ satisfies \begin{enumerate} \item $a_k > a_{k-1} + a_{k-3}$ for all $k\ge s$, and \item $\{a_1, \dots, a_{4s+6}\}$ has no special sum-dominant subsets, \end{enumerate} then $A$ contains at most finitely many sum-dominant sets. \end{theorem} The above results concern situations where there are not many sum-dominant sets; we end with an example of the opposite behavior. \begin{theorem}\label{thm:prime} There are infinitely many sum-dominant subsets of the primes. \end{theorem} We will see later that this result follows immediately from the Green-Tao Theorem \cite{GT}, which asserts that the primes contain arbitrarily long progressions. We also give a conditional proof in \S\ref{sec:primes}. There we assume the Hardy-Littlewood conjecture (see Conjecture \ref{conj:HL}) holds. The advantage of such an approach is that we have an explicit formula for the number of the needed prime tuples up to $x$, which gives a sense of how many such solutions exist in a given window. \section{Subsets with no sum-dominant sets}\label{sec:noMSTD} We prove Theorem \ref{thm:gen}, establishing a sufficient condition to ensure the non-existence of sum-dominant subsets. \begin{proof}[Proof of Theorem \ref{thm:gen}] Let $S=\{s_1,s_2,\dots, s_k\} = \{a_{g(1)}, a_{g(2)}, \dots, a_{g(k)}\}$ be a finite subset of $A$, where $g : \Z^+ \to \Z^+$ is an increasing function. We show that $S$ is not a sum-dominant set by strong induction on $g(k)$. We proceed by induction. We show that if $A$ has no sum-dominant subsets of size $k$, then it has no sum-dominant subsets of size $k+1$; as any sum-dominant set has only finitely many elements, this completes the proof. For the Basis Step, we know (see \cite{He}) that all sum-dominant sets have at least 8 elements, so any subset $S$ of $A$ with exactly $k$ elements is not a sum-dominant set if $k \le 7$; in particular, $S$ is not a sum-dominant set if $g(k) \le 7$. Thus we may assume for $g(k) \ge 8$ that all $S'$ of the form $\{s_1,\dots, s_{k-1}\}$ with $s_{k-1} < a_{g(k)}$ are not sum-dominant sets. The proof is completed by showing \begin{equation} S \ = \ S' \cup \{a_{g(k)}\} \ =\ \{s_{1},\dots, s_{k-1}, a_{g(k)}\}\end{equation} is not sum-dominant sets for any $a_{g(k)}$. We now turn to the inductive step. We know that $S'$ is not a sum-dominant set by the inductive assumption. Also, if $k \le 2r-1$ then $|S| \le 2r-1$ and $S$ is not a sum-dominant set by the second assumption of the theorem. If $k \ge 2r$, consider the number of new sums and differences obtained by adding $a_{g(k)}$. As we have at most $k$ new sums, the proof is completed by showing there are at least $k$ new differences. Since $k \ge 2r$, we have $k - \floor{\frac{k+1}{2}} \ge r$. Let $t = \floor{\frac{k+1}{2}}$. Then $t \le k - r$, which implies $s_{t} \le s_{k-r}$. The largest difference in absolute value between elements in $S$ is $s_{k-1}-s_1$; we now show that we have added at least $k+1$ distinct differences greater than $s_{k-1}-s_1$ in absolute value, which will complete the proof. We have \begin{align} a_{g(k)} - s_{t} &\ \ge \ a_{g(k)} - s_{k-r} \ = \ a_{g(k)} - a_{g(k-r)} \nonumber\\ &\ \ge \ a_{g(k)} - a_{g(k)-r} \nonumber\\ &\ > \ a_{g(k)-1}-a_1 & \text{(by the first assumption on $\{a_n\}$)} \nonumber\\ &\ \ge \ s_{k-1} - a_{1} \ \ge \ s_{k-1} - s_{1}. \end{align} Since $a_{g(k)} - s_{t} \ge s_{k-1}-s_1$, we know that $$a_{g(k)} - s_{t},\ \dots,\ a_{g(k)} - s_{2},\ a_{g(k)} - s_{1}$$ are $t$ differences greater than the greatest difference in $S'$. As we could subtract in the opposite order, $S$ contains at least \begin{equation} 2t\ =\ 2\left\lfloor\frac{k+1}{2}\right\rfloor\ \ge \ k\end{equation} new differences. Thus $S+S$ has at most $k$ more sums than $S'+S'$ but $S-S$ has at least $k$ more differences compared to $S'-S'$. Since $S'$ is not a sum-dominant set, we see that $S$ is not a sum-dominant set. \end{proof} \begin{remark} We thank the referee for the following alternative formulation of our proof. Given any infinite increasing sequence $\{a_{g(i)}\}$ that is a subset of a set $A$ satisfying $a_k > a_{k_1} + a_{k-r}$ for all $k > r$, let $S_k = \{a_{g(1)}, \dots, a_{g(k)}\}$ and $\Delta_k = |S_k - S_k| - |S_k + S_k|$. Similar arguments as above show that $\{\Delta_k\}$ is increasing for $k \ge 2r$. \end{remark} We immediately obtain the following. \begin{corollary}\label{cor:abs} Let $A = \{a_k\}_{k=1}^\infty$ be a strictly increasing sequence of non-negative numbers. If $a_k > a_{k-1} + a_{k-4}$ for all $k\ge 5$, then $A$ contains no sum-dominant subsets. \end{corollary} \begin{proof} From \cite{He} we know that all sum-dominant sets have at least 8 elements. When $r = 4$ the second condition of Theorem \ref{thm:gen} holds, completing the proof. \end{proof} For another example, we consider shifted geometric progressions. \begin{corollary}\label{thm:geo} Let $A = \{a_k\}_{k=1}^\infty$ with $a_k = c \rho^k + d$ for all $k\ge 1$, where $0 \neq c \in \N$, $d\in\N$, and $1 < \rho \in \N$. Then $A$ contains no sum-dominant subsets. \end{corollary} \begin{proof} Without loss of generality we may shift and assume $d=0$ and $c=1$; the result now follows immediately from simple algebra. \end{proof} \begin{remark} Note that if $\rho$ is an integer greater than the positive root of $x^4 - x^3 - 1$ (the characteristic polynomial associated to $a_k = a_{k-1} + a_{k-4}$ from Theorem \ref{thm:finite}, which is approximately 1.3803) then the above corollary holds for $\{c \rho^k + d\}$. \end{remark} \section{Subsets with finitely many sum-dominant sets}\label{sec:finitelymany} We start with some properties of special sum-dominant sets, and then prove Theorem \ref{thm:finite}. The arguments are similar to those used in proving Theorem \ref{thm:gen}. \emph{In this section, in particular in all the statements of the lemmas, we assume the conditions of Theorem \ref{thm:finite} hold.} Thus $A = \{a_k\}_{k=1}^\infty$ and there is an integer $s$ such that the sequence $\{a_k\}$ satisfies \begin{enumerate} \item $a_k > a_{k-1} + a_{k-3}$ for all $k\ge s$, and \item $\{a_1, \dots, a_{4s+6}\}$ has no special sum-dominant subsets. \end{enumerate} \subsection{Special sum-dominant sets}\label{sec:specialsumdom} Recall a sum-dominant set $S$ is special if $|S+S| - |S-S| \ge |S|$. For any $x \ge \sum_{a\in S} a$, adding $x$ creates $|S|+1$ new sums and $2|S|$ new differences. Let $S^* = S\cup \{x\}$. Then \begin{equation} |S^* + S^*| - |S^* - S^*|\ \ge\ |S| + (|S|+1) - 2|S|\ =\ 1,\end{equation} and $S^*$ is also a sum-dominant set. Hence, from one special sum-dominant set $S \subset \{a_n\}_{n=1}^\infty =: A$, we can generate infinitely many sum-dominant sets by adding any large integer in $A$. We immediately obtain the following converse. \begin{lemma}\label{lem:appendingelementtonotspecial} If a set $S$ is not a special sum-dominant set, then $|S+S| - |S-S| < |S|$, and by adding any large $x \ge \sum_{a\in S} a$, $S\cup \{x\}$ has at least as many differences as sums. Thus only finitely many sum-dominant sets can be generated by appending one integer from $A$ to a non-special sum-dominant set $S$. \end{lemma} Note that special sum-dominant sets exist. We use the base expansion method (see \cite{He}), which states that given a set $A$, for all $m$ sufficiently large if \be A_t \ = \ \left\{\sum_{i=1}^t a_i m^{i-1}: a_i \in A\right\} \ee then \be |A_t \pm A_t| \ = \ |A \pm A|^t; \ee the reason is that for $m$ large the various elements are clustered with different pairs of clusters yielding well-separated sums. To construct the desired special sum-dominant set, consider the smallest sum-dominant set $S = \{0, 2, 3, 4$, $7$, $11$, $12, 14\}$. Using the method of base expansion, taking $m = 10^{2017}$ we obtain $S_3$ containing $|S_3| = 8^3 = 512$ elements such that $|S_3 + S_3| = |S + S|^3 = 26^3 = 17576$ and $|S_3 - S_3| = |S - S|^3 = 25^3 = 15625$. Then $|S_3 + S_3| - |S_3 - S_3| > |S_3|$. \subsection{Finitely many sum-dominant sets on a sequence} If a sequence $A = \{a_n\}_{n=1}^\infty$ contains a special sum-dominant set $S$, then we can get infinitely many sum-dominant subsets on the sequence just by adding sufficiently large elements of $A$ to $S$. Therefore for a sequence $A$ to have at most finitely many sum-dominant subsets, it is necessary that it has no special sum-dominant sets. Using the result from the previous subsection, we can prove Theorem \ref{thm:finite}. We establish some notation before turning to the proof in the next subsection. We can write $A$ as the union of $A_1 = \{a_1, \dots, a_{s-1}\}$ and $A_2 = \{a_{s}, a_{s+1}, \dots \}$. We assume this is done with an $s \ge 5$ so that we can use Corollary \ref{cor:abs}, which implies that $A_2$ contains no sum-dominant sets. Thus any sum-dominant set must contain some elements from $A_1$. We prove a lemma about $A_2$. \begin{lemma}\label{lem:add} Let $S' = \{s_1, \dots, s_{k-1}\}$ be a subset of $A$ containing at least 3 elements $a_{r_1}, a_{r_2}, a_{r_3}$ in $A_2$, with $r_3 > r_2 > r_1$. Consider the index $\ph(k) > r_3$, and let $S = S' \cup \{a_{\ph(k)}\}$. Then either $S$ is not a sum-dominant set, or $S$ satisfies $|S-S| - |S+S| > |S' - S'| - |S' + S'|$. Thus the excess of sums to differences from $S$ is \emph{less than} the excess from $S'$. \end{lemma} \begin{proof} We follow a similar argument as in Theorem \ref{thm:gen}. If $k \le 7$, then $S$ is not a sum-dominant set. If $k \ge 8$, then $k - \floor{\frac{k+3}{2}} \ge 3$. Let $t = \floor{\frac{k+2}{2}}$. Then $t \le k - 3$, and $s_{t} \le s_{k-3}$, and \begin{align} a_{\ph(k)} - s_{t} &\ \ge\ a_{\ph(k)} - s_{k-3} = a_{\ph(k)} - a_{\ph(k-3)} \nonumber\\ &\ \ge\ a_{\ph(k)} - a_{\ph(k)-3} \nonumber\\ & \ >\ a_{\ph(k)-1} = a_{\ph(k)-1} - a_{1} & \text{(by assumption on $a$)} \nonumber\\ &\ \ge\ s_{k-1} - a_{1} \ge s_{k-1} - s_{1}. \end{align} In the set $S'$, the greatest difference is $s_{k-1}-s_1$. Since $a_{\ph(k)} - s_{t} \ge s_{k-1}-s_1$, we know that $a_{\ph(k)} - s_{t}, \dots, a_{\ph(k)} - s_{2}, a_{\ph(k)} - s_{1}$ are all differences greater than the greatest difference in $S'$. By a similar argument, $s_{t} - a_{\ph(k)}, \dots, s_{2} - a_{\ph(k)}, s_{1} - a_{\ph(k)}$ are all differences smaller than the smallest difference in $S'$. So $S$ contains at least $2t = 2\floor{\frac{k+3}{2}} > 2\cdot \frac{k+1}{2} = k+1$ new differences compared to $S'$, and $S$ satisfies \be |S-S| - |S+S| \ >\ |S' - S'| - |S' + S'|, \ee completing the proof. \end{proof} \subsection{Proof of Theorem \ref{thm:finite}} Recall that we write $A = A_1 \cup A_2$ with $A_1 = \{a_1$, $\dots$, $a_{s-1}\}$, $A_2 = \{a_{s}$, $a_{s+1}$, $\dots \}$, and by Corollary \ref{cor:abs} $A_2$ contains no sum-dominant sets (thus any sum-dominant set must contain some elements from $A_1$). We first prove a series of useful results which imply the main theorem. Our first result classifies the possible sum-dominant subsets of $A$. Since any such set must have at least one element of $A_1$ in it but not necessarily any elements of $A_2$, we use the subscript $n$ below to indicate how many elements of $A_2$ are in our sum-dominant set. \begin{lemma}[Classification of Sum-Dominant Subsets of $A$]\label{lem:classificationA} Notation as above, let $K_n$ be a sum-dominant subset of $A = A_1 \cup A_2$ with $n$ elements in $A_2$. Thus we may write $$K_n\ =\ S \cup \{a_{r_1}, \dots, a_{r_n}\}$$ for some $$S \ \subset \ A_1 \ = \ \{a_1, \dots, a_{s-1}\}, \ \ \ s \le r_1 < r_2 < \dots < r_n.$$ Set $$d\ =\ \max_{K_3} (|K_3 + K_3| - |K_3 - K_3|, 1).$$ Then $n \le d+3$. In other words, a sum-dominant subset of $A$ can have at most $d+3$ elements of $A_2$. \end{lemma} \begin{proof} Let $S_m$ be any subset of $A$ with $m$ elements of $A_2$. Lemma \ref{lem:add} tells us that for any $S_m$ with $m \ge 3$, when we add any new element $a_{r_{m+1}}$ to get $S_{m+1}$, either $S_{m+1}$ is not a sum-dominant set, or $$|S_{m+1}-S_{m+1}| - |S_{m+1}+S_{m+1}| \ \ge\ |S_m - S_m| - |S_m + S_m| + 1.$$ For an $n > d+3$, assume there exists a sum-dominant set; if so, denote it by $K_n$. For $3 \le k \le n$, define $S_k$ as the set obtained by deleting the $(n-k)$ largest elements from $K_n$ (equivalently, keeping only the $k$ smallest elements from $K_n$ which are in $A_2$). We prove that each $S_k$ is sum-dominant, and then show that this forces $S_n$ not to be sum-dominant; this contradiction proves the theorem as $K_n = S_n$. If $S_k$ is not a sum-dominant set for any $k \ge 3$, by Lemma \ref{lem:add} either $S_{k+1}$ is not a sum-dominant set, or $$|S_{k+1} - S_{k+1}| - |S_{k+1} + S_{k+1}|\ \ge \ |S_k - S_k| - |S_k + S_k| + 1\ \ge \ 0,$$ in which case $S_{k+1}$ is also not a sum-dominant set (because $S_k$ is not sum-dominant, the set $S_{k+1}$ generates at least as many differences as sums). As we are assuming $K_n$ (which is just $S_n$) is a sum-dominant set, we find $S_{n-1}$ is sum-dominant. Repeating the argument, we find that $S_{n-2}$ down to $S_3$ must also all be sum-dominant sets, and we have \be |S_n - S_n| - |S_n + S_n|\ \ge \ |S_3 - S_3| - |S_3 + S_3| + (n-3). \ee Since $S_3$ is one of the $K_3$'s (i.e., it is a sum-dominant subset of $A$ with exactly three elements of $A_2$), by the definition of $d$ the right hand side above is at least $n-3-d$. As we are assuming $n > d+3$ we see it is positive, and hence $S_n$ is not sum-dominant. As $S_n = K_n$ we see that $K_n$ is not a sum-dominant set, contradicting our assumption that there is a sum-dominant set $K_n$ with $n > d+3$, proving the theorem. \end{proof} \begin{lemma}\label{lem:precursortothm} For $n \ge 0$, let $k_n$ denote the number of subsets $K_n \subset A$ which are sum-dominant and contain exactly $n$ elements from $A_2$. We write \begin{equation}\label{eq:writingKn} K_n \ = \ S \cup \{a_{r_1}, \dots, a_{r_n}\} \ \ \ {\rm with}\ S \subset A_1.\end{equation} Then \begin{enumerate} \item $k_n$ is finite for all $n \ge 0$, and \item every $K_n$ is not a special sum-dominant set. \end{enumerate} \end{lemma} \begin{proof} We prove each part by induction. It is easier to do both claims simultaneously as we induct on $n$. We break the analysis into $n \in \{0, 1, 2, 3\}$ and $n \ge 4$. The proof for $n=0$ is immediate, while $n \in \{1,2,3\}$ follow by obtaining bounds on the indices permissible in a $K_n$, and then $n \ge 4$ follows by induction. We thus must check (1) and (2) for $n \le 3$. While the arguments for $n \le 3$ are all similar, it is convenient to handle each case differently so we can control the indices and use earlier results, in particular removing the largest element in $A_2$ yields a set which is not a special sum-dominant set. \\ \ \noindent \emph{Case $n=0$:} As $A_1$ is finite, it has finitely many subsets and thus $k_0$, which is the number of sum-dominant subsets of $A_1$, is finite (it is at most $2^{|A_1|}$). Further any $K_0$ is a subset of $$ A_1\ =\ \{a_1, \dots, a_{s-1}\},$$ which is a subset of \begin{equation}\label{eq:Aprime} A'\ = \ \{a_1, \dots, a_{4s+6}\}.\end{equation} As we have assumed $A'$ has no special sum-dominant set, no $K_0$ can be a special sum-dominant set. \\ \ \noindent \emph{Case $n=1$:} We start by obtaining upper bounds on $r_1$, the index of the smallest (and only) element in our set coming from $A_2$. Consider the index $4s$. We claim that \begin{equation} \label{eq:s-1} a_{4s} \ >\ \sum_{a \in A_1} a. \end{equation} This is because $|A_1| < s$ and $a_k > a_{k-1} + a_{k-3}$ for all $k\ge s$, and hence \begin{align*} \label{eq:s-1} \sum_{a \in A_1} a & \ <\ s\cdot a_{s} \nonumber\\ & \ <\ \frac{s}{2} \left(a_{s} + a_{s+2} \right) \ <\ \frac{s}{2} \cdot a_{s+3} \nonumber\\ & \ <\ \frac{s}{4} \left(a_{s+3} + a_{s+5} \right) \ <\ \frac{s}{4} \cdot a_{s+6} \dots \nonumber\\ & \ <\ \frac{s}{2^{\lceil \log_2 s\rceil}} a_{s+3\lceil\log_2(s)\rceil} \nonumber\\ & \ <\ a_{s+3s} \ =\ a_{4s} \end{align*} (by doing the above $\lceil\log_2 s\rceil$ times we ensure that $s/2^{\lceil \log_2 s\rceil} < 1$, and since $s \ge 1$ we have $3s \ge 3 \lceil\log_2(s)\rceil$). Therefore for all $r_{1}$ sufficiently large, \begin{equation} \label{eq:t-1} a_{r_{1}} \ >\ a_{4s} \ > \ \sum_{a \in A_1} a. \end{equation} Clearly there are only finitely many sum-dominant subsets $K_1$ with $r_1 \le 4s$; the analysis is completed by showing there are no sum-dominant sets with $r_1 > 4s$. Imagine there was a sum-dominant $K_1$ with $a_{r_1} > a_{4s}$. Then $K_1$ is the union of a set of elements $S=\{s_1, \dots, s_m\}$ in $A_1$ and $a_{r_1}$ in $A_2$. As $\sum_{s\in S} s < a_{r_1}$, by Lemma \ref{lem:appendingelementtonotspecial} we find $K_1$ is not a sum-dominant set. All that remains is to show none of the $K_1$ are special sum-dominant sets. This is immediate, as each sum-dominant $K_1$ is a subset of $\{a_1, \dots, a_{4s}\}$, which is a subset of $A'$ (defined in \eqref{eq:Aprime}). As we have assumed $A'$ has no special sum-dominant set, no $K_1$ can be a special sum-dominant set. \\ \ \noindent \emph{Case $n=2$:} Consider the index $4s+3$. If $K_2$ is a sum-dominant set then it has two elements, $a_{r_1} < a_{r_2}$, that are in $A_2$. We show that if $r_2 \ge 4s+3$ then there can be no sum-dominant sets, and thus there are only finitely many $K_2$. For all $r_{2} \ge 4s+3$, \begin{equation} \label{eq:t-2} a_{r_2} - a_{r_2 - 1} \ >\ a_{r_2-3} \ \ge\ a_{4s} \ >\ \sum_{a \in A_1} a. \end{equation} Assume there is a sum-dominant $K_2$ with $r_2 \ge 4s+3$. It contains some elements $S=\{s_1, \dots, s_m\}$ in $A_1$ and $a_{r_1}, a_{r_2}$ in $A_2$. We have $$a_{r_2} - a_{r_1}\ \ge\ a_{r_2} - a_{r_2-1} \ >\ \sum_{a\in S} a.$$ Therefore $a_{r_2} > \left( \sum_{a\in S} a\right) + a_{r_1}$, and $S\cup\{a_{r_1}\}$ is not a special sum-dominant set by the $n=1$ case\footnote{If $S' = S \cup \{a_{r_1}\}$ is sum-dominant then it is not special, while if it is not sum-dominant then clearly it is not a special sum-dominant set.}. Hence, by Lemma \ref{lem:appendingelementtonotspecial} we find $K_2 = (S\cup \{a_{r_1}\}) \cup \{a_{r_2}\}$ is not a sum-dominant set. Finally, as $K_2$ is a subset of $\{a_1, \dots, a_{4s+1}\}$, which is a subset of $A'$, by assumption $K_2$ is not a special sum-dominant set. \\ \ \noindent \emph{Case $n=3$:} Let $K_3$ be a sum-dominant set with three elements from $A_2$. We show that if $r_3 \ge 4s+6$ then there are no such $K_3$; as there are only finitely many sum-dominant sets with $r_3 < 4s+6$, this completes the counting proof in this case. Consider the index $4s+6$. For all $r_{3} \ge 4s+6$, \begin{equation} \label{eq:t-3} a_{r_3-3} - a_{r_3-4} \ >\ a_{r_3-6} \ \ge\ a_{4s} \ >\ \sum_{a \in A_1} a. \end{equation} Consider any $K_3$ with $r_3 \ge 4s+6$. We write $K_3$ as $S \cup \{a_{r_1}, a_{r_2}, a_{r_3}\}$ and $S \subset A_1$. If $|S| < 5$, we know that $|K_3| < 8$, and $K_3$ is not a sum-dominant set as such a set has at least 8 elements. We can therefore assume that $|S| \ge 5$. We have two cases.\\ \ \noindent \emph{Subcase 1: $r_2 \le r_3-3$:} Thus $$a_{r_3} - a_{r_2} - a_{r_1}\ \ge\ a_{r_3} - a_{r_3-3} - a_{r_3-4}\ \ge\ a_{r_3-1} - a_{r_3-4}\ \ge\ a_{r_3 -2} > a_{r_3 -6} \ >\ \sum_{a \in S} a.$$ As $S\cup\{a_{r_1}, a_{r_2}\}$ is not a special sum-dominant set by the $n=2$ case\footnote{As before, if it is sum-dominant it is not special, while if it is not sum-dominant it cannot be sum-dominant special; thus we have the needed inequalities concerning the sizes of the sets.}, adding $a_{r_3}$ with $$a_{r_3}\ >\ \left( \sum_{s\in S} s\right) + a_{r_1} + a_{r_2}$$ creates a non-sum-dominant set by Lemma \ref{lem:appendingelementtonotspecial}. \\ \ \noindent \emph{Subcase 2: $r_2 > r_3 - 3$:} Using \eqref{eq:t-3} we find $$a_{r_3} - a_{r_2}\ \ge\ a_{r_3} - a_{r_3-1}\ >\ \sum_{a\in S} a$$ and $$a_{r_2} - a_{r_1} \ > \ a_{r_3-2} - a_{r_3-3}\ >\ \sum_{a\in S} a.$$ Therefore the differences between $a_{r_1}$, $a_{r_2}$, $a_{r_3}$ are large relative to the sum of the elements in $S$, and our new sums and new differences are well-separated from the old sums and differences. Explicitly, $K_3 + K_3$ consists of $S + S$, $a_{r_1} + S$, $a_{r_2} + S$, $a_{r_3} + S$, plus at most 6 more elements (from the sums of the $a_r$'s), while $K_3 - K_3$ consists of $S-S$, $\pm(a_{r_1} - S)$, $\pm(a_{r_2} - S)$, $\pm(a_{r_3} - S)$, plus possibly some differences from the differences of the $a_r$'s. As $S$ is not a special sum-dominant set, we know $|S+S| - |S-S| < |S|$ (if $S$ is not sum-dominant the claim holds trivially, while if it is sum-dominant it holds because $S$ is not special). Thus for $K_3$ to be sum-dominant, we must have \begin{eqnarray} 0 & \ < \ & |K_3 + K_3| - |K_3 - K_3| \nonumber\\ & \le & \left(|S+S| + 3|S| + 6\right) - \left(|S-S| + 6|S|\right) \nonumber\\ & < & 6 -2|S|; \nonumber \end{eqnarray} as $|S| \ge 5$ this is impossible, and thus $K_3$ cannot be sum-dominant. Finally, as again $K_3$ is a subset of $A' = \{a_1, \dots, a_{4s+6}\}$, no $K_3$ is a special sum-dominant set.\\ \ \noindent \emph{Case $n\ge 4$ (inductive step):} We proceed by induction. We may assume that $k_n$ is finite for some $n\ge 3$, and must show that $k_{n+1}$ is finite. By the earlier cases we know there is an integer $t_n$ such that if $K_n$ is a sum-dominant subset of $A$ with exactly $n$ elements of $A_2$, then the largest index $r_n$ of an $a_i \in K_n$ is less than $t_n$. We claim that if $K_{n+1}$ is a sum-dominant subset of $A$ then each index is less than $t_{n+1}$, where $t_{n+1}$ is the smallest index such that if $r_{n+1} \ge t_{n+1}$ then \begin{equation} \label{eq:t} a_{r_{n+1}} \ >\ \sum_{i < r_{n}} a_i. \end{equation} We write $$K_{n+1} \ = \ S \cup \{a_{r_1}, \dots, a_{r_n}, a_{r_{n+1}}\}, \ \ S \subset A_1, \ \ \{a_{r_1}, \dots, a_{r_n}\} \subset A_2.$$ We show that if $r_{n+1} \ge t_{n+1}$ then $K_{n+1}$ is not sum-dominant. Let $S_n = K_{n+1} \setminus \{a_{r_{n+1}}\}$. We have two cases. \begin{itemize} \item If $r_n < t_{n}$, then by the inductive hypothesis $S_n$ is not a special sum-dominant set. So adding $a_{r_{n+1}} > \sum_{x\in S_{n}} x$ to $S_n$ gives a non-sum-dominant set by Lemma \ref{lem:appendingelementtonotspecial}. \item If $r_n \ge t_{n}$, then by the inductive hypothesis $S_n$ is not a sum-dominant set. So $|S_n - S_n| - |S_n + S_n| \ge 0$. Since $n \ge 3$, we can apply Lemma \ref{lem:add}, and either $K_{n+1} = S_n \cup \{a_{r_{n+1}}\}$ is not a sum-dominant set, or $$|K_{n+1} - K_{n+1}| - |K_{n+1} + K_{n+1}|\ >\ |K_n - K_n| - |K_n + K_n| > 0,$$ in which case $S_{n+1}$ is still not a sum-dominant set. \end{itemize} We conclude that for all sum-dominant sets $S_{n+1}$, we must have $r_{n+1} < t_{n+1}$. So $k_{n+1}$ is finite. Consider any sum-dominant set $K_{n+1} = S_n \cup \{a_{r_{n+1}}\}$. Applying lemma \ref{lem:add} again, we have $|K_{n+1} - K_{n+1}| - |K_{n+1} + K_{n+1}| > |S_n - S_n| - |S_n + S_n|$. We know, from inductive hypothesis, that $S_n$ is not a special sum-dominant set. Therefore all possible $K_{n+1}$ are not special sum-dominant sets. By induction, $k_n$ is finite for all $n\ge 0$, and all $K_n$ are not special sum-dominant sets. \end{proof} \begin{proof}[Proof of Theorem \ref{thm:finite}] By Lemma \ref{lem:classificationA} every sum-dominant subset of $A$ is of the form $K_0$, $K_1$, $K_2$, $\dots$, $K_{d+3}$ where the $K_n$ are as in \eqref{eq:writingKn}. By Lemma \ref{lem:precursortothm} there are only finitely many sets of the form $K_n$ for $n \le d+3$, and thus there are only finitely many sum-dominant subsets of $A$. \end{proof} \section{Sum-dominant subsets of the prime numbers}\label{sec:primes} We now investigate sum-dominant subsets of the primes. While Theorem \ref{thm:prime} follows immediately from the Green-Tao theorem, we first conditionally prove there are infinitely many sum-dominant subsets of the primes as this argument gives a better sense of what the `truth' should be (i.e., how far we must go before we find sum-dominant subsets). \subsection{Admissible prime tuples and prime constellations} We first consider the idea of prime $m$-tuples. A prime $m$-tuple $(b_1, b_2, \dots, b_m)$ represents a pattern of differences between prime numbers. An integer $n$ matches this pattern if $(b_1 + n, b_2 + n, \dots, b_m + n)$ are all primes. A prime $m$-tuple $(b_1, b_2, \dots, b_m)$ is called admissible if for all integers $k \ge 2$, $\{b_1, b_2, \dots, b_m\}$ does not cover all values modulo $k$. If a prime $m$-tuple is not admissible, whenever $n > k$ then at least one of $b_1 + n, b_2 + n, \dots, b_m + n$ is divisible by $k$ and greater than $k$, so this cannot be an $m$-tuple of prime numbers (in this case the only $n$ which can lead to an $m$-tuple of primes are $n \le k$, and there are only finitely many of these). It is conjectured in \cite{HL} that all admissible $m$-tuples are matched by infinitely many integers. \begin{conjecture}[Hardy-Littlewood \cite{HL}]\label{conj:HL} Let $b_1, b_2, \dots, b_m$ be $m$ distinct integers, $v_p(b) = v(p; b_1, b_2, \dots , b_m)$ the number of distinct residues of $b_1, b_2, \dots b_m$ to the modulus $p$, and $P(x$; $b_1$, $b_2$, $\dots$, $b_m)$ the number of integers $1 \le n \le x$ such that every element in $\{n + b_1, n + b_2, \dots , n + b_m\}$ is prime. Assume $(b_1, b_2, \dots, b_m)$ is admissible (thus $v_p(b) \neq p$ for all $p$). Then \begin{equation}\label{eq:HLforPx} P(x) \ \sim \ \mathfrak{S}(b_1, b_2, \dots, b_m) \int_2^x \frac{du}{(\log u)^m} \end{equation} when $x \to \infty$, where $$ \mathfrak{S}(b_1, b_2, \dots, b_m)\ =\ \prod_{p \ge 2} \left(\left( \frac{p}{p-1}\right)^{m-1} \frac{p-v_p(b)}{p-1} \right) \ \neq \ 0. $$ \end{conjecture} As $(b_1, b_2, \cdots, b_m)$ is an admissible $m$-tuple, $v(p; b_1, b_2, \dots, b_m)$ is never equal to $p$ and equals $m$ for $p> \max\{|b_i-b_j|\}$. The product $\mathfrak{S}(b_1, b_2, \dots, b_m)$ thus converges to a positive number as each factor is non-zero and is $1 + O_m(1/p^2)$. Therefore this conjecture implies that every admissible $m$-tuple is matched by infinitely many integers. \subsection{Infinitude of sum-dominant subsets of the primes} We now show the Hardy-Littlewood conjecture implies there are infinitely many subsets of the primes which are sum-dominant sets. \begin{theorem}\label{thm:HLprimes} If the Hardy-Littlewood conjecture holds for all admissible $m$-tuples then the primes have infinitely many sum-dominant subsets. \end{theorem} \begin{proof} Consider the smallest sum-dominant set $S=\{0, 2, 3, 4, 7, 11, 12, 14\}$. We know that $\{p, p+2s, p+3s, p+4s, p+7s, p+11s, p+12s, p+14s\}$ is a sum-dominant set for all positive integers $p, s$. Set $s=30$ and let $T = (0, 60, 90, 120, 210, 330, 360, 420)$. We deduce that if there are infinitely many $n$ such that $n+T = (n, n+60, n+90, n+120, n+210, n+330, n+360, n+420)$ is an 8-tuple of prime numbers, then there are infinitely many sum-dominant sets of prime numbers. We check that $T$ is an admissible prime 8-tuple. When $m>8$, the eight numbers in $T$ clearly don't cover all values modulo $m$. When $m \le 8$, one sees by straightforward computation that $T$ does not cover all values modulo $m$. By Conjecture \ref{conj:HL}, there are infinitely many integers $p$ such that every element of $\{p, p+60, p+90, p+120, p+210, p+330, p+360, p+420\}$ is prime. These are all sum-dominant sets, so there are infinitely many sum-dominant sets on primes. \end{proof} Of course, all we need is that the Hardy-Littlewood conjecture holds for one admissible $m$-tuple which has a sum-dominant subset. We may take $p=19$, which gives an explicit sum-dominant subset of the primes: $\{19, 79, 109, 139, 229, 349, 379, 439\}$ (a natural question is which sum-dominant subset of the primes has the smallest diameter). If one wishes, one can use the conjecture to get some lower bounds on the number of sum-dominant subsets of the primes at most $x$. The proof of Theorem \ref{thm:prime} follows similarly. \begin{proof}[Proof of Theorem \ref{thm:prime}] By the Green-Tao theorem, the primes contain arbitrarily long arithmetic progressions. Thus for each $N \ge 14$ there are infinitely many pairs $(p,d)$ such that \begin{equation} \{p, p + d, p + 2d, \dots, p + Nd\} \end{equation} are all prime. We can then take subsets as in the proof of Theorem \ref{thm:HLprimes}. \end{proof} \section{Future work} We list some natural topics for further research. \begin{itemize} \item Can the conditions in Theorem \ref{thm:gen} or \ref{thm:finite} be weakened? \item What is the smallest special sum-dominant set by diameter, and by cardinality? \item What is the smallest, in terms of its largest element, set of primes that is sum-dominant? \end{itemize} \section{Acknowledgments} The first named author was supported by Washington \& Lee University, and the third named author was partially supported by NSF grants DMS1265673 and DMS1561945. 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