%%%% THIS FILE IS AUTOMATICALLY GENERATED %%%% DO NOT EDIT THIS FILE DIRECTLY, %%%% ONLY EDIT THE SOURCE, tom-39/document.tex. %%%% %% Standard package list \documentclass[letterpaper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[english]{babel} \usepackage[top=3cm, bottom=3cm, left=3.5cm, right=3.5cm]{geometry} \usepackage[onehalfspacing]{setspace} \usepackage{amsmath,amssymb,amsthm,wasysym} \usepackage{nicefrac,booktabs} \usepackage{mathptmx} \usepackage{cite} \usepackage[colorlinks=true]{hyperref} %% Various helpers for Tom's papers \newcommand{\gs}{\textnormal{gs}} \newcommand{\ord}{\textnormal{ord}} \newcommand{\Exp}{\textnormal{Exp}} \newcommand{\Log}{\textnormal{Log}} \newcommand{\lcm}{\textnormal{lcm}} \newcommand{\range}{\textnormal{range}} \newcommand{\NR}{\textnormal{NR}} \newcommand{\Mod}[1]{\left(\textnormal{mod}~#1\right)} \newcommand{\ap}[2]{\left\langle #1;#2 \right\rangle} \newcommand{\summ}[1]{\sum_{k=1}^m{#1}} \newcommand{\bt}[1]{{{#1}\mathbb{N}}} \newcommand{\fp}[1]{{\left\lbrace{#1}\right\rbrace}} \newcommand{\intv}[1]{{\left[1,{#1}\right]}} %% Lifted from http://stackoverflow.com/questions/2767389/referencing-a-theorem-like-environment-by-its-name %% This lets me do things like "Theorem A" and have the references work properly. \makeatletter \let\@old@begintheorem=\@begintheorem \def\@begintheorem#1#2[#3]{% \gdef\@thm@name{#3}% \@old@begintheorem{#1}{#2}[#3]% } \def\namedthmlabel#1{\begingroup \edef\@currentlabel{\@thm@name}% \label{#1}\endgroup } \makeatother % end lift \newtheoremstyle{namedthrm} {}{}{}{}{}{}{ } % This last space needs to be there {\bf\thmname{#1} \thmnote{#3}.} %% End reference hack %% Document start \date{} \begin{document} %% Content start \newtheorem{cor}{Corollary} \newtheorem{lemma}{Lemma} \newtheorem{thm}{Theorem} \newtheorem{fact}{Fact} \newtheorem{q}{Problem} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem*{remark}{Remark} \newtheorem*{ack}{Acknowledgements} \newtheorem*{conj}{Conjecture} \title{Arithmetic Progressions in Lacunary Sets} \author{T. C. Brown and A. R. Freedman} \date{} \maketitle \begin{center}{\small {\bf Citation data:} T.C. Brown and A.R. Freedman, \emph{Arithmetic progressions in lacunary sets}, Rocky Mountain J. Math. \textbf{17} (1987), no.~3, 587--596.}\bigskip\end{center} \begin{abstract} We make some observations concerning the conjecture of Erd\H{o}s that if the sum of the reciprocals of a set $A$ of positive integers diverges, then $A$ contains arbitrarily long arithmetic progressions. We show, for example, that one can assume without loss of generality that $A$ is lacunary. We also show that several special cases of the conjecture are true. \end{abstract} \section{Introduction} \label{sec: 1} The now famous theorem of Szemer\'{e}di~\cite{szemeredi1975} is often stated: \emph{(a) If the density of a set $A$ of natural numbers is positive, then $A$ contains arbitrarily long arithmetic progressions.} \vspace{12pt} Let us call a set $A$ of natural numbers $k$-good if $A$ contains a $k$-term arithmetic progression. Call $A$ $\omega$-good if $A$ is $k$-good for all $k \geq 1$. We define four density functions as follows: For a set $A$ and natural numbers $m,n$, let $A[m,n]$ be the cardinality of the set $A \cap \{m, m+1, m + 2, \dots, n\}$. Then define \begin{align*} \underline{\delta}(A) &= \liminf_n \frac{A[1,n]}{n}, \\ \overline{\delta}(A) &= \limsup_n \frac{A[1,n]}{n}, \\ \underline{u}(A) &= \lim_n \min_{m \geq 0} \frac{A[m + 1, m + n]}{n} \: \textup{and} \\ \overline{u}(A) &= \lim_n \max_{m \geq 0} \frac{A[m + 1, m + n]}{n}. \end{align*} It can be seen that the limits in the definitions of $\underline{u}$ and $\overline{u}$ always exist. These four ``asymptotic" set functions are called the lower and upper ``ordinary" and the lower and upper ``uniform" density of the set $A$ respectively. They are related by \[ \underline{u}(A) \leq \underline{\delta}(A) \leq \overline{\delta}(A) \leq \overline{u}(A) \] for any set $A$. Szemer\'{e}di actually proved: \emph{(b) If $\overline{u}(A) > 0$, then $A$ is $\omega$-good.} Hence we also have \emph{(c) If $\overline{\delta}(A) > 0$, then $A$ is $\omega$-good.} In fact, Szemer\'{e}di proved the following ``finite" result (which we state in a general form to be used later): \emph{(d) Let $\varepsilon > 0$ and $k \in N = \{1,2,3,\dots\}$. Then there exists an $n_0 \in N$ such that if $P$ is any arithmetic progression of length $|P| \geq n_0$ and $A \subset P$ with $|A| \geq \varepsilon |P|$, then $A$ is $k$-good.} \vspace{12pt} It is not hard to prove (without assuming the truth of any of the statements) that (b), (c), and (d) are equivalent. Erd\H{o}s has conjectured that the following stronger statement holds: \emph{(e) If $A \subseteq N$ and $\sum_{A} \frac{1}{a} = \infty$, then $A$ is $\omega$-good.} \vspace{12pt} By $\sum_A (1/a)$ we mean of course $\sum_{a \in A} (1/a)$. The proof (or disproof) of (e) is, at present, out of sight. In fact, it has not even been proved that $\sum_A (1/a) = \infty$ implies that $A$ is $3$-good (compare Roth~\cite{roth1953}). That (e) $\Rightarrow$ (c) can be seen as follows: If $\overline{\delta}(A) = \varepsilon > 0$, then there exists a sequence of natural numbers $0 = n_0 < n_1 < n_2 < \cdots$, such that, for each $i$, \[ \frac{A[1,n_i]}{n_i} > \frac{\varepsilon}{2} \quad \textup{and} \quad \frac{n_{i-1}}{n_i} < \frac{\varepsilon}{4}. \] Then \begin{align*} \sum_A \frac{1}{a} &\geq \sum_{\substack{a \in A \\ a \leq n_k}} \frac{1}{a} \geq \sum_{i=1}^k \frac{A[n_{i-1} + 1, n_i]}{n_i} \geq \sum_{i=1}^k \frac{A[1,n_i] - n_{i-1}}{n_i} \\ &\geq k(\frac{\varepsilon}{2} - \frac{\varepsilon}{4}) = \frac{k\varepsilon}{4} \to \infty \quad (k \to \infty) \end{align*} and so $\sum_A(1/a) = \infty$. Assuming (e), it follows that $A$ is $\omega$-good. \vspace{12pt} Hence Erd\H{o}s' conjecture is indeed stronger than Szemer\'{e}di's theorem. Note also that Erd\H{o}s' conjecture, if true, would immediately answer in the affirmative the long-standing question of whether or not the primes are $\omega$-good. \vspace{12pt} In the next section we make some observations regarding this conjecture, and we show that several special cases of the conjecture are true. Other observations can be found in Gerver~\cite{gerver1977,gerver+ramsey1979} and Wagstaff~\cite{wagstaff1979}. \section{Main results} \label{sec: 2} (2.1). First we consider the ``finite form" of Erd\H{o}s' conjecture. \begin{thm} \label{thm: 1} Fix $k$, and assume that for all sets $A \subseteq N$, if $\sum_A(1/a) = \infty$ then $A$ is $k$-good. Under this assumption, there exists $T$ such that if $\sum_A(1/a) > T$, then $A$ is $k$-good. (Gerver~\cite{gerver1977} has this result under the stronger hypothesis that if $\sum_A(1/a) = \infty$ then $A$ is $(k + 1)$-good.) \end{thm} \begin{proof}We may assume $k \geq 3$. Suppose the theorem is false. We will construct a set $A$ such that $\sum_A(1/a) = \infty$ and $A$ is not $k$-good. Choose a finite set $A_0$ such that $A_0$ is not $k$-good and $\sum_A(1/a) > 1$. Let $p_1$ be prime, $p_1 > 2\max A_0$, and choose a finite subset $A_1$ of $\{tp_1: t \geq 1\}$ such that $A_1$ is not $k$-good and $\sum_{A_1} (1/a) > 1$. Let $p_2$ be prime, $p_2 > 2\max A_1$, and choose a finite subset $A_2$ of $\{tp_2: t \geq 1\}$ such that $A_2$ is not $k$-good and $\sum_{A_2} (1/a) > 1$. Continuing in this way, we obtain finite sets $A_0,A_1,\dots$ such that for each $i \geq 0$, $A_i$ is not $k$-good, $\min A_{i + 1} \geq p_{i + 1} > 2 \max A_i$, each elements of $A_{i + 1}$ is a multiple of $p_{i + 1}$, and $\sum_{A_i} (1/a) > 1$. Let $A = \bigcup A_i$. It is clear that $\sum_A (1/a) = \infty$. To show that $A$ is not $k$-good, it suffices to show that every 3-term arithmetic progression contained in $A$ must be contained in a single set $A_i$. To this end, suppose that $x < y < z$, with $x,y,z \in A$ and $z - y = y - x$. Let $y \in A_i$. Then $z \in A_i$ also, since otherwise $z - y \geq \min A_{i + 1} - \max A_i > \max A_i > y - x$. Thus $y,z \in A_i \subset \{tp_i: t \geq 1\}$. Hence $x$ is divisible by $p_i$, so $x \geq p_i > \max A_{i-1}$, and $x \in A_i$. This finishes the proof of Theorem~\ref{thm: 1}. \end{proof} \begin{cor} \label{cor: 1} The following statement is equivalent to statement (e): (f) For each $k \in N$, there exists $T \in N$ such that if $\sum_A(1/a) > T$, then $A$ is $k$-good. \end{cor} We state next a lemma which will be useful later. \begin{lemma} \label{lemma: 1} Let $F_1,F_2,\dots$ be a sequence of finite subsets of $N$ such that for each $i$, $F_i$ is not $k$-good and $\min F_{i + 1} \geq 2 \max F_i$. Then $F = \bigcup F_i$ is not $( k + 1)$-good. \end{lemma} (The proof of Lemma~\ref{lemma: 1} is contained in the proof of Theorem~\ref{thm: 1} above). \vspace{12pt} (2.2). Now we define an increasing sequence, $a_1 < a_2 < a_3< \cdots$, of natural numbers to be lacunary if $d_n = a_{n + 1} - a_n \to \infty$ as $n \to \infty$ and to be $M$-lacunary if, furthermore, $d_n \leq d_{n + 1}$ for all $n$. We shall think of such a sequence simultaneously as a sequence and as a subset of $N$. Any lacunary sequence $A$ has $\overline{u}(A) = 0$ (see~\cite{freedman+sember1981}), so that Szemer\'{e}di's theorem does not apply. A subsequence of a lacunary sequence is lacunary, but the corresponding statement, unfortunately, does not hold for $M$-lacunary sequences. It is known that if the real series $\sum t_i$ is not absolutely convergent, then there exists a lacunary sequence $B$ such that $\sum_{i \in B} t_i $ diverges (see Freedman and Sember~\cite{freedman+sember1981}). It follows that if $A \subseteq N$ and $\sum_A(1/a) = \infty$, then there exists a lacunary sequence $B \subseteq A$ such that $\sum_B (1/b) = \infty$. Thus we have the following. \begin{thm} \label{thm: 2} The following statement is equivalent to statement (e). (g) If $A$ is a lacunary sequence and $\sum_A(1/a) = \infty$, then $A$ is $\omega$-good. \end{thm} Hence we need only investigate lacunary sequences when contemplating the Erd\H{o}s conjecture. It can also be shown that if $\sum t_i = \infty$ and $t_i \geq 0$ for all $i$, then there exists an $M$-lacunary sequence $B$ such that $\sum_{i \in B} t_i = \infty$. (We omit the rather cumbersome proof of this statement.) But notice that this does not imply that statement (h) below is equivalent to statement (e)! This is too bad--because we now prove (h). \begin{thm} \label{thm: 3} The following statement is true: (h) If $A$ is $M$-lacunary and $\sum_A (1/a) = \infty$, then $A$ is $\omega$-good. \end{thm} \begin{proof} Let $A = \{a_1 < a_2 < a_3 < \cdots\}$ be an $M$-lacunary sequence with infinite reciprocal sum. Assume there is a $k$ such that $d_i < d_{i + k}$ for each $i$, where $d_n = a_{n + 1} - a_n$, $n \geq 1$. We show that $a_{i + jk} \geq j^2/2$ for all $i \geq 1$, $j \geq 0$. Indeed, \begin{align*} a_{i + jk} &= a_i + d_i + d_{i + 1} + \cdots + d_{i + jk - 1} \\ &\geq d_i + d_{i + k} + d_{i + 2k} + \cdots + d_{i + (j-1)k} \\ &> 1 + 2 + 3+ \cdots + j > j^2/2. \end{align*} (Note that to obtain the first inequality we have merely omitted some terms from the sum.) But then \begin{align*} \sum_{i=1}^\infty \frac{1}{a_i} &= \sum_{j=0}^\infty \frac{1}{a_{1 + jk}} + \sum_{j=0}^\infty \frac{1}{a_{2 + jk}} + \cdots + \sum_{j=0}^\infty \frac{1}{a_{k + jk}} \\ &\leq k(1 + \sum_{j=1}^\infty \frac{2}{j^2}) < \infty, \quad \textup{a contradiction.} \end{align*} Hence, for each $k$, there is an $i$ such that $d_i = d_{i + k}$, whence $a_i, a_{i + 1}, \dots,a_{i + k + 1}$ are in arithmetic progression and $A$ is $\omega$-good. \end{proof} \vspace{12pt} The following is an immediate corollary. \begin{cor} \label{cor: 2} If $A$ is a finite union of $M$-lacunary sets and $\sum_A (1/a) = \infty$, then $A$ is $\omega$-good. \end{cor} (2.3). We now use some slightly expanded arguments to show that statement (g) holds for some special sequences which are not $M$-lacunary (but are nearly so). \begin{thm} \label{thm: 4} Let $A = \{a_1 < a_2 < a_3 < \cdots \}$ be any set. Suppose there are intervals $I_n = [s_n,t_n]$ with $t_n < s_{n + 1}$ such that \[ \sum_{n=1}^\infty \frac{1}{\sqrt{a_{s_n}}} < \infty, \quad \sum_{k \in \bigcup I_n} \frac{1}{a_k} = \infty. \] Suppose further that for each $n$, $d_k \leq d_{k + 1}$ if $s_n \leq k < t_n$. Then $A$ is $\omega$-good. \end{thm} \begin{proof} We will arrive at a contradiction if we assume that there is a $K \in N$, such that $d_i < d_{i + K}$ whenever $i, i + K$ belong to the same interval $I_j$. Then, for any $K$, we have that there exists an $i$ such that $d_i = d_{i + 1} = \cdots = d_{i + K}$ so that $a_i,a_{i + 1}, \dots,a_{i + K + 1}$ are in arithmetic progression. To get the required contradiction we proceed as follows: If $n, n + K, n + 2K, \dots, n + cK \in I_i$, then \begin{align*} \frac{1}{a_n} &+ \frac{1}{a_{n + K}} + \frac{1}{a_{n + 2K}} + \cdots + \frac{1}{a_{n + cK}} \\ &\leq \frac{1}{a_n} + \frac{1}{a_n + d_n} + \frac{1}{a_n + d_n + d_{n + K}} + \cdots \\ &+ \frac{1}{a_n + d_n + d_{n + K} + \cdots + d_{n + (c-1)K}} \\ &< \sum_{j=0}^\infty \frac{1}{a_{n} + (j^2/2)} < \frac{b}{\sqrt{a_n}} \leq \frac{b}{\sqrt{a_{s_i}}} \quad \textup{($b$ constant)}. \end{align*} Hence, \[ \sum_{K \in I_i} \frac{1}{a_k} < \frac{Kb}{\sqrt{a_{s_i}}} \quad \textup{and} \quad \sum_{k \in \bigcup I_i} \frac{1}{a_k} < Kb \sum_{i=1}^\infty \frac{1}{\sqrt{a_{s_i}}} < \infty, \] contrary to assumption. \end{proof} Using a similar technique we can prove the following theorem. \begin{thm} \label{thm: 5} Let $A = \{a_1 0$. \begin{thm} \label{thm: 6} There exists a set $A$ such that $\overline{\lambda}(A) > 0$ and $A$ is not $\omega$-good. \end{thm} \begin{proof} Let $B_i = \{1!, 2!, \dots,i!\}$. $B_i$ is not $3$-good. Let $(H_i)$ be the sequence of sets \[ (B_1, B_1,B_2,B_1,B_2,B_3,B_1,B_2,B_3,B_4,B_1, \dots). \] Let $f_i$ be an increasing sequence of integers such that $f_1 = 0$ and \[ \min(f_{i + 1} + H_{i + 1}) \geq 2 \max(f_i + H_i) \] and define $A = \bigcup_i (f_i + H_i)$. By Lemma~\ref{lemma: 1}, $A$ is not $4$-good. (By choosing $f_i$ sufficiently quickly increasing one can even make $A$ not 3-good.) Finally, $\overline{\lambda}(A) = 1$ since otherwise $A = L_1 \cup L_2 \cup \cdots \cup L_k$ where each $L_j$ is a lacunary sequence. Whenever $H_i = B_{k + 1}$ we have $|f_i + H_i| > k$ and so some $L_j$ has at least two members in $f_i + H_i$. Hence we may find a fixed $j$ such that \[ |L_j \cap (f_i + B_{k + 1})| \geq 2 \] for infinitely many $i$. Then $L_j$ has infinitely many differences $d_t < (k + 1)!$, and so $L_j$ is not lacunary. \end{proof} (2.5). Let us consider ``relative density", that is, ``the density of $A$ relative to $B$" where $A \subseteq B$. The definitions are \begin{align*} \underline{\delta}(A|B) &= \liminf_{i \to \infty} \frac{A[1,b_i]}{i} \quad \textup{and} \\ \underline{u}(A| B) &= \lim_{n \to \infty} \min_{m \geq 0} \frac{A[b_{m + 1}, b_{m + n}]}{n}. \end{align*} $\overline{\delta}(A|B)$ and $\overline{u}(A|B)$ are obtained by replacing ``inf" with ``sup" and ``min" with ``max" respectively. One can show, as before, for any $A, B, A \subseteq B$, that \[ \underline{u}(A|B) \leq \underline{\delta}(A|B) \leq \overline{\delta}(A|B) \leq \overline{u}(A|B). \] Let $B$ be $M$-lacunary and $\sum_B 1/b = \infty$. Then, by Theorem~\ref{thm: 3}, $B$ is $\omega$-good. We ask whether $A \subseteq B$ and the relative density of $A$ positive imply that $A$ is also $\omega$-good. The answer is ``yes" if $\underline{u}(A|B) > 0$ (Theorem~\ref{thm: 7}), ``no" if $\overline{\delta}(A|B) > 0$ (Theorem~\ref{thm: 8}) and the question is open for $\underline{\delta}(A|B) > 0$. \begin{thm} \label{thm: 7} If $B$ is $M$-lacunary, $\sum_B 1/b = \infty$, $A \subseteq B$ and $\underline{u}(A|B) > 0$ then $A$ is $\omega$-good. \end{thm} \begin{proof} By (the proof of) Theorem~\ref{thm: 3} there are arbitrarily large $n,m$ such that \[ P = \{b_{m + 1}, b_{m + 2}, \dots, b_{m + n}\} \] is an arithmetic progression. By the definition of $\underline{u}(A|B)$ we have $|A \cap P| \geq \varepsilon P$ where $\varepsilon = (1/2)\underline{u}(A|B)$ and $|P|$ is arbitrarily large. Thus, by Szemer\'{e}di's theorem (d) we have, for any $k$, that $|A \cap P|$ is $k$-good if $|P|$ is sufficiently large. Hence $A$ is $\omega$-good. \end{proof} \begin{thm} \label{thm: 8} There exists an $M$-lacunary sequence $B$ with $\sum_B = 1/b = \infty$ and an $A \subseteq B$ with $\overline{\delta}(A|B) > 0$ ($=1$ in fact) such that $A$ is not 3-good. \end{thm} \begin{proof} (leaving most of the details to the reader). Let $F = \{1!,2!,3!,\dots\}$, $b_1 = 1$ and define $b_{n + 1} = b_n + d_n$ where the $d_n$'s have the following properties: For all $i,d_i \in F$ and $d_i \leq d_{i + 1}$. Furthermore, the set of natural numbers $N$ can be partitioned into consecutive pairwise disjoint intervals $J_1,J_2,J_3,\dots$ such that if $r$ is odd, then for $i \in J_r$, $d_i = d_{i + 1}$ and $\sum_{i \in J_r} 1/b_i \geq 1$, and, if $r$ is even, then, for $i \in J_r$, $d_i < d_{i + 1}$, $b_i > 2b_{i-1}$ and $|J_r| > (\max J_{r-1})^2$. Clearly $B = \{b_1,b_2,\dots\}$ is $M$-lacunary and $\sum_B 1/b = \infty$. Let $A = \{b_k: k \in \bigcup_r J_{2r}\}$. Then \[ \overline{\delta}(A|B) \geq \lim_r \frac{|J_{2r}|}{|J_{2r}| + \max J_{2r - 1}} = 1. \] One can also see that $A$ is not 3-good since $a_i > 2a_{i-1}$ holds. \end{proof} (2.6). Theorems~\ref{thm: 4},\ref{thm: 5}, and~\ref{thm: 7} notwithstanding, it seems to be difficult to generalize the notion of $M$-lacunary even slightly and still prove the corresponding case of the Erd\H{o}s conjecture. In this connection let us define a lacunary sequence $A$ to be $M_k$-lacunary (where $k \geq 0$) if, for all $i,j,i \leq j$, we have $d_i \leq d_j + k$. Clearly, the $M_0$-lacunary sequences are just the $M$-lacunary sequences. For no $k \ne 0$ are we able to prove that $M_k$-lacunary and $\sum_A (1/a) = \infty$ imply $\omega$-good. We can show if $A$ is $M_1$-or $M_2$-lacunary with $\sum_A(1/a) = \infty$ then $A$ is 3-good. We prove first a lemma which may have independent interest: \begin{lemma} \label{lemma: 2} If $A = \{a_1 0$, there exists an $i$ such that $d_{i + j} \leq d_i$ for $j = 0,1,\dots,t$. (Of course, $d_n = a_{n + 1} - a_n$.) \end{lemma} \begin{proof} The method is familiar by now: Suppose there is a $t$ such that, for each $i$, there exists $j \in[1,t]$ with $d_i < d_{i + j}$. Then we can find a sequence $(j_n)$ such that \[ d_1 < d_{1 + j_1} < d_{1 + j_1 + j_2} < \cdots (j_n \in [1,t]). \] It follows that \[ \sum_A \frac{1}{a} \leq t \sum_{s = 0}^\infty \frac{1}{a_{1 + (j_1 + \cdots + j_s)}} \leq t\sum_{s=0}^\infty \frac{1}{a_1 + (1 + 2 + \cdots + s)} < \infty. \qedhere \] \end{proof} \begin{thm} \label{thm: 9} Let $A$ be $M_1$-or $M_2$-lacunary and $\sum_A(1/a) = \infty$. Then $A$ is 3-good. \end{thm} \begin{proof} By the definition of $M_k$-lacunary and Lemma~\ref{lemma: 2} we have: for any $t > 0$ there is an $i$ such that \[ d_i - e \leq d_{i + j} \leq d_i, \quad j = 0,1,\dots,t, \] where $e = 1$ or 2. Hence, in the sequence $(d_i)$, we have arbitrarily long blocks where the $d_i$ take on only two (in case $e = 1$) or three (in case $e = 2$) values. Such long blocks must contain two consecutive subblocks with identical composition (see Pleasants~\cite{pleasants1970}). These two subblocks will determine three terms of the sequence $A$ in arithmetic progression. \end{proof} This last result suggests a conjecture which is related to van der Waerden's theorem on arithmetic progressions and would immediately imply that $M_k$-lacunary with $\sum_A (1/a) = \infty$ implies that $A$ is $3$-good. \begin{conj} Let $x_i$ be a sequence of positive integers with $1 \leq x_i \leq K$. Then there are two consecutive intervals $I,J$ of the same length, with $\sum_{i \in I} x_i = \sum_{j \in J} x_j$. Equivalently, if $a_1 < a_2 < \cdots$ satisfy $a_{n+1} - a_n \leq K$, all $n$, then there exist $x< y < z$ such that $x + z = 2y$ and $a_x +a_z = 2a_y$. \end{conj} \nocite{erdos1975,gerver+ramsey1979,wroblewski1984} \bibliographystyle{amsplain} \bibliography{tom-all} \end{document}