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\begin{center}
\vskip 1cm{\LARGE\bf 
A Note on a Problem of Motzkin Regarding \\
\vskip .03in
Density of Integral Sets with Missing \\
\vskip .1in
Differences 
}
\vskip 1cm
\large
Ram Krishna Pandey\thanks{This work was done when at Department of Mathematics, Indian Institute of Technology, Delhi} \\
Department of Mathematics \\
Indian Institute of Technology \\
Patliputra Colony, Patna -- 800013 \\
India\\
\href{mailto:ram@iitp.ac.in}{\tt ram@iitp.ac.in}\\
\ \\
Amitabha Tripathi\thanks{\em Corresponding author} \\
Department of Mathematics \\
Indian Institute of Technology \\
Hauz Khas, New Delhi -- 110016 \\
India \\
\href{mailto:atripath@maths.iitd.ac.in}{\tt atripath@maths.iitd.ac.in}
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\vskip .2 in
\begin{abstract}
For a given set $M$ of positive integers, a problem of Motzkin asks to
determine the maximal density ${\mu}(M)$ among sets of nonnegative
integers in which no two elements differ by an element of $M$. The
problem is completely settled when $|M| \le 2$, and some partial
results are known for several families of $M$ when $|M| \ge 3$. In 1985
Rabinowitz \&amp; Proulx provided a lower bound for ${\mu}(\{a,b,a+b\})$
and conjectured that their bound was sharp. Liu \&amp; Zhu proved this
conjecture in 2004. For each $n \ge 1$, we determine
${\k}(\{a,b,n(a+b)\})$, which is a lower bound for
$\mu(\{a,b,n(a+b)\})$, and conjecture this to be the exact value of
${\mu}(\{a,b,n(a+b)\})$.
\end{abstract}




\section{Introduction}

For $x \in {\R}$ and a set $S$ of nonnegative integers, let $S(x)$ denote the number of elements $n \in S$ such that $n \le x$. The upper and lower densities of $S$, denoted by $\o{\d}(S)$ and $\ul{\d}(S)$ respectively, are given by
\[ \o{\d}(S):= \limsup_{x \imp \infty} \frac{S(x)}{x}, \quad \ul{\d}(S):= \liminf_{x \imp \infty} \frac{S(x)}{x}. \]
If $\o{\d}(S)=\ul{\d}(S)$, we denote the common value by $\d(S)$, and say that $S$ has density $\d(S)$. Given a set of positive integers $M$, $S$ is said to be an $M$-set if $a \in S$, $b \in S$ imply $a-b \notin M$. Motzkin \cite{Mot} asked to determine $\mu(M)$ given by
\[ \mu(M):= \sup_S \o{\d}(S) \]
where $S$ varies over all $M$-sets. Cantor \&amp; Gordon \cite{CG73} showed the existence of $\mu(M)$ for any $M$, determined ${\mu}(M)$ when $|M| \le 2$, and gave the following lower bound for $\mu(M)$:
\begin{equation}\label{1}
\mu(M) \ge {\k}(M):= \sup_{\gcd(c,m)=1} \frac{1}{m} \min_i |cm_i|_m,
\end{equation}
where $m_i$ are the elements of $M$, and $|x|_m$ denotes the absolute value of the absolutely least remainder of $x$ modulo $m$. We use the following equivalent form for the lower bound for ${\mu}(M)$, due to Haralambis \cite{Har77}:
\begin{equation}\label{2}
{\k}(M) = \max_{\stackrel{\tiny m=m_i+m_j}{\tiny 1 \le k \le \frac{m}{2}}} \frac{1}{m} \min_{m_i \in M} |km_i|_m,
\end{equation}
where $m_i,m_j$ represent distinct elements of $M$. The following useful upper bound for ${\mu}(M)$ is due to Haralambis \cite{Har77}:
\begin{equation}\label{3}
{\mu}(M) \le \a
\end{equation}
provided there exists a positive integer $k$ such that $S(k) \le (k+1)\a$ for every $M$-set $S$ with $0 \in S$ and for some $\a \in [0,1]$.
The problem of Motzkin has a rich and diverse history but little progress towards the general problem has been made so far. Exact results for ${\mu}(M)$ have been few, and computation of ${\mu}(M)$ has only been completely possible when $|M| \le 2$. Cantor \&amp; Gordon \cite{CG73} showed that
\[ {\mu}(\{m\}) = \frac{1}{2}, \quad {\mu}(\{m_1,m_2\}) = \frac{\lf (m_1+m_2)/2 \rf}{m_1+m_2}. \]
There have, however, been a number of results that give the exact value or bounds for ${\mu}(M)$ in other cases; see \cite{LZ04} for an exhaustive bibliography.

Connections with coloring problems in graph theory have been found
useful in solving the Motzkin problem. One such connection, introduced
by Hale \cite{Hal80} and shown to be equivalent to the Motzkin problem
by Griggs \&amp; Liu \cite{GL94}, is the {\it $T$-coloring problem}.
Another connection with colorings of graphs involves the {\it
fractional chromatic number} of {\it distance graphs}.

The lower bound for ${\mu}(M)$, denoted by ${\k}(M)$ in \eqref{1}, is itself at the heart of a longstanding conjecture. The {\it Lonely Runner Conjecture} (LRC) stated independently by Wills \cite{Wil67} in the context of diophantine approximations and by Cusick \cite{Cus74} while studying view obstructions problems in $n$-dimensional geometry, was actually given this apt name by Bienia {\it et al} \cite{BGGST98}. Chen \cite{Che91} characterized $3$-sets $M$ for which $1/{\k}(M)$ is an integer and also obtained general bounds.

We consider the problem of determining ${\mu}(M)$ for the family $M=\{a,b,n(a+b)\}$, $n \ge 1$. By a result of Cantor \&amp; Gordon \cite{CG73}, we know that ${\mu}(kM)={\mu}(M)$. Thus, it is no loss of generality to assume that $\gcd(a,b)=1$, and that $a&lt;b$. We determine the value of ${\k}(M)$, which is the lower bound for $\mu(M)$ and conjecturally equal to it. This extends a result of Liu \&amp; Zhu [\cite{LZ04}, Theorem 5.1], wherein they determined ${\mu}(M)$ in the case $M=\{a,b,a+b\}$. Rabinowitz \&amp; Proulx \cite{RP85} provided a lower bound for ${\mu}(M)$ in this case and conjectured that their bound was sharp. An extensive list of work related to the Motzkin problem may be found in \cite{LZ04}.

\section{Main Result}

For the set $M=\{a,b,a+b\}$, Rabinowitz \&amp; Proulx \cite{RP85} conjectured the exact value of ${\mu}(M)$ in 1985, and Liu \&amp; Zhu \cite{LZ04} proved this conjecture in 2004. We determine ${\k}(M)$ for $M=\{a,b,n(a+b)\}$ where $n$ is a fixed positive integer. If the conjecture of Haralambis \cite{Har77} is true, then ${\k}(M)={\mu}(M)$ in this case, thus providing a generalization of the result of Liu \&amp; Zhu.

\begin{theorem}\label{T}
Let $M=\{a,b,n(a+b)\}$, where $a&lt;b$, $\gcd(a,b)=1$ and $n \ge 1$. Then
\begin{equation*}
{\k}(M) =
\begin{cases}
\frac{n(a+b-\l)}{2\big(a+n(a+b)\big)}, &amp; \text{if $\l \equiv a+b\pmod{2}$}; \\
\frac{n(a+b+\l-1)}{2\big(b+n(a+b)\big)}, &amp; \text{if $\l \not\equiv a+b\pmod{2}$},
\end{cases}
\end{equation*}
where $\l=1+\big\lf\frac{b-a}{2n+1}\big\rf$.
\end{theorem}

\begin{proof}
We use \eqref{2} to compute ${\k}(M)$. The choice $m=a+b$ trivially yields $\min\big\{|ax|_m,|bx|_m,|n(a+b)x|_m\big\}=0$ for each $x$. There are two choices remaining for $m$, and we determine ${\k}(M)$ by comparing the two rational numbers corresponding to these cases. In each of the two cases we need to compute ${\k}(M)$ with $m=a+n(a+b)$ and $m=b+n(a+b)$, and compare the two. Note that the definition of $\l$ implies
\[ b-a &lt; (2n+1)\l \le b-a+2n+1. \]
{\sc Case I:} {\bf \big($\l \equiv a+b$ (mod $2$)\big)} \\[10pt]
{\bf \em Subcase\/} (i): \big($m=a+n(a+b)$\big) Choose $x$ such that
\[ (a+b)x \equiv \ts\frac{a+b-\l}{2}\pmod{m}. \]
Then
\[ -ax \equiv n(a+b)x \equiv \ts n\frac{a+b-\l}{2} \equiv \ts\frac{m-(a+n\l)}{2}\pmod{m}, \]
and
\[ bx = (a+b)x-ax \equiv \ts\frac{m+\{b-(n+1)\l\}}{2}\pmod{m}. \]
Therefore
\begin{equation}\label{4}
\min \big\{|ax|_m, |bx|_m, |n(a+b)x|_m\big\} = \ts\frac{m-(a+n\l)}{2}
\end{equation}
since $b-(n+1)\l&lt;a+n\l$ if and only if $(2n+1)\l&gt;b-a$. \\[10pt]
Write $\ell:=a+n\l$. We show that
\[ \min \big\{|ay|_m, |by|_m, |n(a+b)y|_m\big\} \le \ts\frac{m-(a+n\l)}{2} = \ts\frac{m}{2}-\frac{\ell}{2} \]
for each $y$, $1 \le y \le \frac{m}{2}$. Let ${\I}:=\left(\frac{m}{2}-\frac{\ell}{2},\frac{m}{2}+\frac{\ell}{2}\right)$. We show that $-ay\bmod{m} \in \I$ and $by\bmod{m} \in \I$ is simultaneously impossible for $1 \le y \le \frac{m}{2}$. Suppose
\[ (a+b)y \equiv \ts\frac{a+b-\l}{2}+i\pmod{m}, \]
with $1 \le i \le m-1$. Then
\begin{eqnarray}\label{5}
-ay \equiv n(a+b)y \equiv \ts\frac{m}{2}-\ts\frac{\ell}{2}+ni\pmod{m}, \\ \nonumber
by \equiv (a+b)y-ay \equiv \ts\frac{m}{2}-\ts\frac{\ell}{2}+\ts\frac{a+b-\l}{2}+(n+1)i\pmod{m}
\end{eqnarray}
Thus $-ay\bmod{m} \in \I$ if and only if
\[ km+\ts\frac{m}{2}-\ts\frac{\ell}{2} &lt; \ts\frac{m}{2}-\ts\frac{\ell}{2}+ni &lt; km+\ts\frac{m}{2}+\ts\frac{\ell}{2} \]
for some integer $k$, with $0 \le k \le n-1$. This is equivalent to
\[ k\ts\frac{m}{n} &lt; i &lt; k\ts\frac{m}{n}+\ts\frac{\ell}{n}, \]
so that
\begin{equation}\label{6}
k(a+b)+1 \le i \le k(a+b)+a+\l-1.
\end{equation}

For $k(a+b)+1 \le i \le k(a+b)+a+\l-1$, we show that
\begin{eqnarray*}
km+\ts\frac{m}{2}+\ts\frac{\ell}{2} &lt; \ts\frac{m}{2}-\ts\frac{\ell}{2}+\ts\frac{a+b-\l}{2}+(n+1)\big\{k(a+b)+1\big\}
\end{eqnarray*}
and
\begin{eqnarray*}
\ts\frac{m}{2}-\ts\frac{\ell}{2}+\ts\frac{a+b-\l}{2}+(n+1)\big\{k(a+b)+a+\l-1\big\} &lt; (k+1)m+\ts\frac{m}{2}-\ts\frac{\ell}{2}.
\end{eqnarray*}
This will prove that
\[ km+\ts\frac{m}{2}+\ts\frac{\ell}{2} &lt; by &lt; (k+1)m+\ts\frac{m}{2}-\ts\frac{\ell}{2}, \]
so that $by\bmod{m} \notin \I$, as claimed. Each of the above two inequalities is easy to prove. Using the fact that $(n+1)(a+b)=m+b$, each inequality can be shown to hold if $(2n+1)\l&lt;(b-a)+2(n+1)$, which is true by the definition of $\l$. This completes the subcase when $m=a+n(a+b)$. \\[10pt]
{\bf \em Subcase\/} (ii): ($m=b+n(a+b)$) The argument in this subcase is similar to the one in subcase (i). We omit the calculation and state only the significant parts. Choose $x$ such that
\[ (a+b)x \equiv \ts\frac{a+b+\l}{2}\pmod{m}. \]
Then
\[ -bx \equiv n(a+b)x \equiv \ts\frac{m-(b-n\l)}{2}\pmod{m}, \]
and
\[ ax = (a+b)x-bx \equiv -\ts\frac{m-\{a+(n+1)\l\}}{2}\pmod{m}. \]
Therefore
\begin{equation}\label{7}
\min \big\{|ax|_m, |bx|_m, |n(a+b)x|_m\big\} = \ts\frac{m-\{(a+(n+1)\l\}}{2}
\end{equation}
since $b-n\l&lt;a+(n+1)\l$ if and only if $(2n+1)\l&gt;b-a$. \\[10pt]
\noindent As in subcase (i), we may show that
\[ \min \big\{|ay|_m, |by|_m, |n(a+b)y|_m\big\} \le \ts\frac{m-\{a+(n+1)\l\}}{2} \]
for each $y$, $1 \le y \le \frac{m}{2}$. The argument is similar and we omit the proof. This completes subcase (ii). \\[5pt]
To compute ${\k}(M)$ in Case I, we need to compare the expressions in \eqref{4} and \eqref{7}. If we let $m_1=a+n(a+b)$ and $m_2=b+n(a+b)$, then a lengthy but easy computation shows that
\[ \ts\frac{m_2-\{(a+(n+1)\l\}}{2m_2} = \ts\frac{1}{2}-\ts\frac{1}{2}\ts\frac{a+(n+1)\l}{b+n(a+b)} &lt;
   \ts\frac{1}{2}-\ts\frac{1}{2}\ts\frac{a+n\l}{a+n(a+b)} = \ts\frac{m_1-(a+n\l)}{2m_1}
\]
if and only if $(2n+1)\l&gt;b-a$. This completes the proof of Case I. \\[10pt]
{\sc Case II:} {\bf \big(${\l} \not\equiv a+b$ (mod $2$)\big)} \\[10pt]
{\bf \em Subcase\/} (i): ($m=a+n(a+b)$) Choose $x$ such that
\[ (a+b)x \equiv \ts\frac{a+b-\l+1}{2}\pmod{m}. \]
Then
\[ -ax \equiv n(a+b)x \equiv \ts\frac{m-\{a+n(\l-1)\}}{2}\pmod{m}, \]
and
\[ bx = (a+b)x-ax \equiv \ts\frac{m+\{b-(n+1)(\l-1)\}}{2}\pmod{m}. \]
Therefore
\begin{equation}\label{8}
\min \big\{|ax|_m, |bx|_m, |n(a+b)x|_m\big\} = \ts\frac{m-\{b-(n+1)(\l-1)\}}{2}
\end{equation}
since $a+n(\l-1) \le b-(n+1)(\l-1)$ if and only if $(2n+1)(\l-1) \le b-a$. \\[10pt]
As in subcase (i) of Case I, we may show that
\[ \min \big\{|ay|_m, |by|_m, |n(a+b)y|_m\big\} \le \ts\frac{m-\{b-(n+1)(\l-1)\}}{2} \]
for each $y$, $1 \le y \le \frac{m}{2}$. This completes subcase (i). \\[5pt]
{\bf \em Subcase\/} (ii): ($m=b+n(a+b)$) Choose $x$ such that
\[ (a+b)x \equiv \ts\frac{a+b+\l-1}{2}\pmod{m}. \]
Then
\[ -bx \equiv n(a+b)x \equiv \ts\frac{m-\{b-n(\l-1)\}}{2}\pmod{m}, \]
and
\[ ax = (a+b)x-bx \equiv \ts\frac{m+\{a+(n+1)(\l-1)\}}{2}\pmod{m}. \]
Therefore
\begin{equation}\label{9}
\min \big\{|ax|_m, |bx|_m, |n(a+b)x|_m\big\} = \ts\frac{m-\{b-n(\l-1)\}}{2}
\end{equation}
since $a+(n+1)(\l-1) \le b-n(\l-1)$ if and only if $(2n+1)(\l-1) \le b-a$. \\[10pt]
As in subcase (i) of Case I, we may show that
\[ \min \big\{|ay|_m, |by|_m, |n(a+b)y|_m\big\} \le \ts\frac{m-\{b-n(\l-1)\}}{2} \]
for each $y$, $1 \le y \le \frac{m}{2}$. This completes subcase (ii). \\[5pt]
To compute ${\k}(M)$ in Case II, we need to compare the expressions in \eqref{8} and \eqref{9}. If we let $m_1=a+n(a+b)$ and $m_2=b+n(a+b)$, then a lengthy but easy computation shows that
\[ \ts\frac{m_1-\{b-(n+1)(\l-1)\}}{2m_1} = \ts\frac{1}{2}-\ts\frac{1}{2}\ts\frac{b-(n+1)(\l-1)}{a+n(a+b)} \le
   \ts\frac{1}{2}-\ts\frac{1}{2}\ts\frac{b-n(\l-1)}{b+n(a+b)} = \ts\frac{m_2-\{b-n(\l-1)\}}{2m_2}
\]
if and only if $(2n+1)(\l-1) \le b-a$. This completes the proof of Case II, and of the theorem.
\end{proof}

\begin{corollary}\label{C} {\bf (Liu \&amp; Zhu \cite{LZ04})}  \\[5pt]
Let $M=\{a,b,a+b\}$, where $a&lt;b$ and $\gcd(a,b)=1$. Then
\begin{equation*}
{\k}(M) =
\begin{cases}
\frac{1}{3}, &amp; \text{if $b \equiv a\pmod{3}$}; \\
\frac{2a+b-1}{3(2a+b)}, &amp; \text{if $b \equiv a+1\pmod{3}$}; \\
\frac{a+2b-1}{3(a+2b)}, &amp; \text{if $b \equiv a+2\pmod{3}$}.
\end{cases}
\end{equation*}
\end{corollary}

\begin{proof}
This is a direct consequence of Theorem~\ref{T}. Set $b-a=3k+r$, where $0 \le r \le 2$. Then $\l=k+1$, so that we are in Case I when $b \equiv a+1$ (mod $3$) and Case II otherwise. The calculation is routine, and the details are omitted.
\end{proof}

\begin{remark}\label{R}
Haralambis \cite{Har77} conjectured that ${\mu}(M)={\k}(M)$ when $|M|=3$. If this is true, Theorem~\ref{T} actually determines ${\mu}(\{a,b,n(a+b)\})$. Observe that, if $a,b$ are of opposite parity and if $n \ge (b-a)/2$, Theorem~\ref{T} reduces to
\[ {\k}(\{a,b,n(a+b)\}) = \frac{a+b-1}{2(a+b)+(2a/n)}, \]
which is asymptotic to ${\mu}(\{a,b\})$. This may be an indication that the conjecture of Haralambis may hold, at least for the special case $M=\{a,b,n(a + b)\}$ when $n$ is large enough, and perhaps even for $M=\{a,b,c\}$ when $c$ is sufficiently large even if not of the form $n(a+b)$.
\end{remark}

\section{Acknowledgement}

The authors wish to thank the referee for his comments.

\bibliographystyle{amsplain}
\begin{thebibliography}{10}

\bibitem{BGGST98}
W. Bienia, L. Goddyn, P. Gvozdjak, A. Seb\"{o}, and M. Tarsi, Flows, view obstructions and the lonely runner, {\it J. Combin. Theory Ser. B} {\bf 72} (1998), 1--9.

\bibitem{BS08}
J. Barajas and O. Serra, The lonely runner with seven runners, {\it Electron. J. Combin.} {\bf 15} (2008), Research paper 48, 18 pages.

\bibitem{CG73}
D. G. Cantor and B. Gordon, Sequences of integers with missing differences, {\it J. Combin. Theory Ser. A} {\bf 14} (1973), 281--287.

\bibitem{CHZ98}
G. J. Chang, L. Huang, and X. Zhu, The circular chromatic numbers and the fractional chromatic numbers of distance graphs, {\it European J. Combin.} {\bf 19} (1998), 423--431.

\bibitem{CLZ99}
G. Chang, D. Liu, and X. Zhu, Distance graphs and $T$-colorings, {\it J. Combin. Theory Ser. B} {\bf 75} (1999), 159--169.

\bibitem{Che91}
Y. G. Chen, On a conjecture about diophantine approximations III, {\it J. Number Theory} {\bf 37} (1991), 181--198.

\bibitem{Cus74}
T. W. Cusick, View-obstruction problems in $n$-dimensional geometry, {\it J. Combin. Theory Ser. A} {\bf 16} (1974), 1--11.

\bibitem{EES85}
R. B. Eggleton, P. Erd\H{o}s, and D. K. Skilton, Coloring the real line, {\it J. Combin. Theory Ser. B} {\bf 39} (1985), 86--100.

\bibitem{GL94}
J. R. Griggs and D.-D. F. Liu, The channel assignment problem for mutually adjacent sites, {\it J. Combin. Theory Ser. A} {\bf 68} (1994), 169--183.

\bibitem{Hal80}
W. K. Hale, Frequency assignment: theory and applications, {\it Proceedings of the IEEE} {\bf 68} (1980), 310--320.

\bibitem{Har77}
N. M. Haralambis, Sets of integers with missing differences, {\it J. Combin. Theory Ser. A} {\bf 23} (1977), 22--33.

\bibitem{KK98}
A. Kemnitz and H. Kolberg, Coloring of integer distance graphs, {\it Discrete Math.} {\bf 191} (1998), 113--123.

\bibitem{LZ04}
D.-D. F. Liu and X. Zhu, Fractional chromatic number for distance graphs with large clique size, {\it J. Graph Theory} {\bf 47} (2004), 129--146.

\bibitem{LZ08}
D. D.-F. Liu and X. Zhu, Fractional chromatic number of distance graphs generated by two-interval sets, {\it European J. Combin.} {\bf 29} (7) (2008), 1733--1742.

\bibitem{Mot}
T. S. Motzkin, Unpublished problem collection.

\bibitem {RP85}
J. H. Rabinowitz and V. K. Proulx, An asymptotic approach to the channel assignment problem, {\it SIAM J. Alg. Disc. Methods} {\bf 6} (1985), 507--518.

\bibitem{Wil67}
J. M. Wills, Zwei S\"{a}tze uber inhomogene diophantische Approximation
von Irrationalzahlen, {\it Monatsh. Math.} {\bf 71} (1967), 263--269.

\end{thebibliography}



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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B05.

\noindent \emph{Keywords: } density, coloring.

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\vspace*{+.1in}
\noindent
Received January 6 2011;
revised version received May 18 2011.  
Published in {\it Journal of Integer Sequences}, June 1 2011.

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\noindent
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