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\begin{document}
\vspace*{-40pt} 
\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 6 
(2006), \#A14} 
\vskip 40pt

\begin{center}
{\bf ON A LINEAR DIOPHANTINE PROBLEM OF FROBENIUS}
\vskip 20pt
{\bf Amitabha Tripathi}\\
{\smallit Department of Mathematics, Indian Institute of Technology, Hauz Khas, New Delhi --
          110016, India}\\
{\tt atripath@maths.iitd.ac.in}\\
\end{center}
\vskip 30pt
\centerline{\smallit Received: 3/17/06, Revised: 4/17/06,
 Accepted: 4/24/06,
Published: 5/3/06}
\vskip 30pt

\centerline{\bf Abstract}

\noindent Let $a_1,a_2,\ldots,a_k$ be positive and pairwise coprime
integers with product $P$. For each $i$, $1 \le i \le k$, set
$A_i=P/a_i$. We find closed form expressions for the functions
$g(A_1,A_2,\ldots,A_k)$ and $n(A_1,A_2,\ldots,A_k)$ that denote the
{\em largest\/} (respectively, the {\em number\/} of) $N$ such that the
equation
$A_1x_1+A_2x_2+\cdots+A_kx_k = N $
has no solution in nonnegative integers $x_i$. This is a special
case of the well-known {\em Coin Exchange Problem\/} of Frobenius.

\pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 6 (2006),
\#A14\hfill}

\thispagestyle{empty}
\baselineskip=15pt
\vskip 30pt

\section*{\normalsize 1. Introduction}

Given positive integers $a_1,a_2,\ldots,a_k$, relatively prime, it
is well-known that for all sufficiently large $N$ the equation
\begin{equation}
a_1x_1+a_2x_2+\cdots+a_kx_k = N
\end{equation}
has a solution with nonnegative integers $x_i$. If we denote by
$g(a_1,a_2,\ldots,a_k)$ the {\em largest\/} integer $N$ such that
$(1)$ has no solution in nonnegative integers, then it is a
well-known result of Sylvester that $g(a_1,a_2)=a_1a_2-a_1-a_2$. The
related functions $n(a_1,a_2,\ldots,a_k)$ and
$s(a_1,a_2,\ldots,a_k)$ denote the number of positive integers $N$
for which $(1)$ has no solution and the sum of such integers,
respectively. While it is well-known that
$n(a_1,a_2)=(a_1-1)(a_2-1)/2$, the corresponding result
$s(a_1,a_2)=(a_1-1)(a_2-1)(2a_1a_2-a_1-a_2-1)/12$ is more recent and
less known \cite{BS93a}. Except when the $a_i$'s are in arithmetic
progression \cite{Bat58,Gra73,Rob56,Tri94} or in certain other
particular cases with three or more variables
\cite{Bra42,BS62,NW72,Rob57,Rod78,Rod79,Sel77,SB78}, there is no
closed form expression for either $g$ or $n$. More information on
this problem may be found in the recently published monograph \cite{Ram06}. 


The purpose of this note is to obtain a formula for the functions
$g$ and $n$ in a special case. More specifically, we shall
henceforth assume that the $a_i$'s are {\em pairwise\/} coprime with
product $P$, and set $A_i=P/a_i$ for $1 \le i \le k$. We determine
$g(A_1,A_2,\ldots,A_k)$ and $n(A_1,A_2,\ldots,A_k)$ by two methods.
The first method uses a reduction formula while the second method is
direct. We note that $g(A_1,A_2)=g(a_1,a_2)$ and
$n(A_1,A_2)=n(a_1,a_2)$. 

We close by showing that the set ${\S}^{\star}$ introduced in
\cite{Tri03} has exactly one element in the special case we are
dealing with. Since it is known (and easy to see from the definition
of ${\S}^{\star}$) that $g \in {\S}^{\star}$, we have further
confirmation of the result for $g$ in the special case. 
\vskip 30pt

\section*{\normalsize 2. Main Results}

For the sake of completeness, we prove two well-known results that help in evaluating
the functions $g$ and $n$ in the general case. 

\noindent {\bf Lemma 1 \cite{{BS62,Sel77}}.} Let
$\gcd(a_1,a_2,\ldots,a_k)=1$, and for $1 \le j \le a_1-1$, let $m_j$
denote the {\em least\/} positive integer $N$ congruent to $j$ mod
$a_1$ such that $(1)$ has a solution in nonnegative integers. Then
\begin{itemize}
\item[{\rm (a)}] 
$g(a_1,a_2,\dots,a_k) = \ds\max_{1 \le j \le a_1-1}\:m_j - a_1$;
\item[{\rm (b)}]
$n(a_1,a_2,\ldots,a_k) = \ds\frac{1}{a_1}\:\ds\sum_{j=1}^{a_1-1}\:(m_j-j) = \ds\frac{1}{a_1}\:
\ds\sum_{j=1}^{a_1-1}\:m_j - \ds\frac{a_1-1}{2}$.
\end{itemize}


\noindent{\it Proof.}
\begin{itemize}
\item[{\rm (a)}]
From the definition of $m_i$ it follows that $m_i-a_1$ is not
representable by $a_1,\ldots,a_k$ in nonnegative integers for each
$i$, $1 \le i \le a_1$. On the other hand, any $N$ greater than each
$m_i-a_1$ and congruent to $j$ mod $a_1$ must be at least $m_j$, and
hence representable by $a_1,\ldots,a_k$ in nonnegative integers.
\item[{\rm (b)}]
Since the numbers congruent to $j$ mod $a_1$ and not representable
by $a_1,\ldots,a_k$ in nonnegative integers form an arithmetic
progression with first term $j$, last term $m_j-a_1$ and common
difference $a_1$, their number is given by $(m_j-j)/a_1$. The second
part of the lemma now easily follows. \hfill $\Box$
\end{itemize}


\noindent {\bf Lemma 2 \cite{{Joh60,Rod78}}.}
Let $a_1,a_2,\ldots,a_k$ be positive integers. If $\gcd(a_2,\ldots,a_k)=d$ and $a_j=da_j^{\:\pr}$ for each $j>1$, then
\begin{itemize}
\item[{\rm (a)}]
$g(a_1,a_2,\ldots,a_k) = d\,g(a_1,a_2^{\:\pr},\ldots,a_k^{\:\pr})+a_1(d-1)$;
\item[{\rm (b)}]
$n(a_1,a_2,\ldots,a_k) = d\,n(a_1,a_2^{\:\pr},\ldots,a_k^{\:\pr})+\frac{1}{2}(a_1-1)(d-1)$;
\end{itemize}


\noindent{\it Proof.} As in Lemma 1, for each $j$, $1 \le j \le
a_1-1$, let $m_j$ and $m_j^{\:\pr}$ denote the {\em least\/}
positive integer congruent to $j$ mod $a_1$ representable as a
nonnegative linear combination of $a_1,a_2,\ldots,a_k$ and
$a_1,a_2^{\:\pr},\ldots,a_k^{\:\pr}$, respectively. Since each such
$m_j$ and $m_j^{\:\pr}$ must also be representable as a nonnegative
linear combination of $a_2,\ldots,a_k$ and of
$a_2^{\:\pr},\ldots,a_k^{\:\pr}$, respectively, it follows that
$\{m_j: 1 \le j \le a_1-1\}=\{dm_j^{\:\pr}:1\le j \le a_1-1\}$. We
now apply Lemma 1.

\noindent
For part (a) we have
\[ \begin{array}{lll}
  g(a_1,a_2,\ldots,a_k) & = & \ds\max_{1 \le j \le a_1-1}\:m_j - a_1 \\[10pt]
                        & = & d \left( \ds\max_{1 \le j \le a_1-1}\:
                              m_j^{\:\pr} - a_1\right) + a_1(d-1) \\[15pt]
                        & = & d\,g(a_1,a_2^{\:\pr},\ldots,a_k^{\:\pr}) +
                              a_1(d-1).
   \end{array} \]
For part (b) we have
\[ \begin{array}{lll}
   n(a_1,a_2,\ldots,a_k) & = & \ds\frac{1}{a_1}\:\sum_{j=1}^{a_1-1}\:m_j -
                               \ds\frac{1}{2}(a_1-1) \\[15pt]
                         & = & d \left( \ds\frac{1}{a_1}\:\sum_{j=1}^{a_1-1}
                               \:m_j^{\:\pr} - \ds\frac{1}{2}(a_1-1) \right)
                               + \ds\frac{1}{2}(a_1-1)(d-1) \\[20pt]
                         & = & d\,n(a_1,a_2^{\:\pr},\ldots,a_k^{\:\pr}) +
                               \frac{1}{2}(a_1-1)(d-1).
   \end{array} \]
\hfill$\Box$


\noindent{\bf Theorem 1.}
Let $a_1,a_2,\ldots,a_k$ be pairwise coprime, positive integers with product $P$. Let
$A_i=P/a_i$ for $1 \le i \le k$. Let ${\s}_r$ denote the sum of the products of the $a_i$'s
taken $r$ at a time, so that ${\s}_k=P$ and ${\s}_{k-1}=A_1+A_2+\cdots+A_k$. Then
\begin{itemize}
\item[{\rm (a)}]
$g(A_1,A_2,\ldots,A_k) = (k-1){\s}_k-{\s}_{k-1}$;
\item[{\rm (b)}]
$n(A_1,A_2,\ldots,A_k) = \ds\frac{1}{2}\{(k-1){\s}_k-{\s}_{k-1}+1\}$.
\end{itemize}
\vspace*{0.1in}

\noindent{\it Proof.} This is a direct consequence of Lemma 2. We
induct on $k$. If $k=2$, these are just the well-known results
mentioned in the Introduction. We observe that $A_k$ is a multiple
of $A_j/a_k=A_j^{\:\pr}$ for each $j \ne k$ since
$A_j|a_kA_k={\s}_k$ and $a_k|A_j$ if $j \ne k$.

\noindent For part (a), by the induction hypothesis, we have
\[ \begin{array}{lll}
   g(A_1,A_2,\ldots,A_k) & = &
   a_k\,g\left( \ds\frac{A_1}{a_k},\frac{A_2}{a_k},\ldots,\frac{A_{k-1}}
   {a_k},A_k \right) + A_k(a_k-1) \\[10pt]
   & = &
   a_k\,g\left( \ds\frac{A_1}{a_k},\frac{A_2}{a_k},\ldots,\frac{A_{k-1}}
   {a_k} \right) + {\s}_k - A_k \\[10pt]
   & = &
   a_k\,g(A_1^{\:\pr},A_2^{\:\pr},\ldots,A_{k-1}^{\:\pr})+ {\s}_k - A_k
   \\[5pt]
   & = &
   (k-2){\s}_k - ({\s}_{k-1}-A_k) + {\s}_k - A_k  \\[5pt]
   & = &
   (k-1){\s}_k - {\s}_{k-1}.
   \end{array}
\]
For part (b), by the induction hypothesis, we have
\[ \begin{array}{lll}
   n(A_1,A_2,\ldots,A_k) & = &
   a_k\,n\left( \ds\frac{A_1}{a_k},\frac{A_2}{a_k},\ldots,\frac{A_{k-1}}
   {a_k},A_k \right) + \ds\frac{1}{2}(a_k-1)(A_k-1) \\[10pt]
   & = &
   a_k\,n\left( \ds\frac{A_1}{a_k},\frac{A_2}{a_k},\ldots,\frac{A_{k-1}}
   {a_k} \right) + \ds\frac{1}{2}{\s}_k - \frac{1}{2}a_k - \frac{1}{2}A_k +
   \frac{1}{2} \\[10pt]
   & = &
   a_k\,n(A_1^{\:\pr},A_2^{\:\pr},\ldots,A_{k-1}^{\:\pr}) + \ds\frac{1}{2}
   {\s}_k - \frac{1}{2}a_k - \frac{1}{2}A_k + \frac{1}{2} \\[10pt]
   & = &
   \ds\frac{1}{2}\{(k-2){\s}_k - ({\s}_{k-1}-A_k) + a_k + {\s}_k-a_k-A_k+1\}
   \\[8pt]
   & = &
   \ds\frac{1}{2}\{(k-1){\s}_k - {\s}_{k-1} + 1\}.
   \end{array}
\]
 \hfill $\Box$


The proof of Theorem 1 given above is based on Lemma 2. It is indeed possible to give an independent proof. Using the notation of Theorem 1, we give a \\[10pt]
{\bf Second proof of Theorem 1.} Let $a_1,a_2,\ldots,a_k$ be pairwise coprime, positive integers. Let ${\s}_r$ denote the sum of the products of the $a_i$'s taken $r$ at a time, and let $A_j={\s}_k/a_j$ for $1 \le j \le k$.
 Then
$ g(A_1,A_2,\ldots,A_k) = (k-1){\s}_k-{\s}_{k-1}. $


\noindent{\it Proof.} If each $x_j \ge 0$ and
\begin{equation}
A_1x_1+A_2x_2+\cdots+A_kx_k = (k-1){\s}_k-{\s}_{k-1},
\end{equation}
$A_jx_j \equiv -A_j\mod{a_j}$, so that $x_j \ge a_j-1$ since $\gcd(a_j,A_j)=1$. But then
\[ \sum_{j=1}^k\:A_jx_j \ge \sum_{j=1}^k\:A_j(a_j-1) \ge k{\s}_k-{\s}_{k-1}, \]
and $(2)$ has no solution in nonnegative integers. \\[5pt]
Since the $A_ix_i+A_jx_j=A_i(x_i+a_i)+A_j(x_j-a_j)$, and since $\gcd(A_1,A_2,\ldots,A_k)=1$, we can always write {\em any\/} $N$ in the form $A_1x_1+A_2x_2+\cdots+A_kx_k$ with $0 \le x_j \le a_j-1$ for $1 \le j \le k-1$. Now, if $N>(k-1){\s}_k-{\s}_{k-1}$ and we choose $x_j$ as above, then
\[ x_k = \frac{N-\sum_{j=1}^{k-1}\,A_jx_j}{A_k} > \frac{\sum_{j=1}^{k-1}\:A_j
   (a_j-x_j-1)}{A_k} -1 \ge -1. \]
Thus $x_k \ge 0$, and every $N$ greater than $(k-1){\s}_k-{\s}_{k-1}$ is expressible as a nonnegative linear combination of the $A_j$'s. \hfill$\Box$


\noindent {\bf Lemma 3.} Let $a_1,a_2,\ldots,a_k$ be pairwise
coprime, positive integers, and let $A_j={\s}_k/a_j$ for $1 \le j
\le k$. If $p,q$ are integers such that $p+q=g(A_1,A_2,\ldots,A_k)$,
then exactly one of the equations
$A_1x_1+A_2x_2+\cdots+A_kx_k=p \;{\rm and}\;
A_1x_1+A_2x_2+\cdots+A_kx_k=q $ is solvable in nonnegative integers
$x_j$.


\noindent{\it Proof.} If both the equations had a solution, so would
$g(A_1,A_2,\ldots,A_k)$, contradicting its definition. Suppose
$A_1x_1+A_2x_2+\cdots+A_kx_k=p$ has no solution in nonnegative
integers. Choose $x_j$ such that $0 \le x_j \le a_j-1$ for $1 \le j
\le k-1$. But then $x_k<0$, and
\[ q = (k-1){\s}_k-{\s}_{k-1}-p = \sum_{j=1}^{k-1}\:A_j(a_j-x_j-1) + A_k(-x_k) \]
is expressible in the given form, proving the lemma.
\hfill$\Box$


\noindent{\bf Corollary 1.}
Let $a_1,a_2,\ldots,a_k$ be pairwise coprime, positive integers. Let ${\s}_r$ denote the sum
of the products of the $a_i$'s taken $r$ at a time, and let $A_j={\s}_k/a_j$ for $1 \le j \le k$.
Then
$n(A_1,A_2,\ldots,A_k) = \frac{1}{2}\{ (k-1){\s}_k-{\s}_{k-1}+1 \}.$


\noindent{\it Proof.} If we pair $p$ with $q$ whenever
$p+q=g(A_1,A_2,\ldots,A_k)$ and $p,q \ge 0$, by Lemma 1,
\[ n(A_1,A_2,\ldots,A_k) = \frac{1}{2}\left\{ 1+g(A_1,A_2,\ldots,A_k)
   \right\}. \]
The corollary now follows from Theorem 1. \hfill$\Box$


The evaluation of $g$ given in Theorem 1 can also be derived by explicitly determining the set
${\S}^{\star}$, introduced in \cite{Tri03}, since $g(a_1,a_2,\ldots,a_k)$ is the largest element
in ${\S}^{\star}(a_1,a_2,\ldots,a_k)$. For positive and coprime integers $a_1,a_2,\ldots,a_k$,
let ${\G}^{\star}$ denote the positive integers in the set $\{a_1x_1+a_2x_2+\cdots+a_kx_k:x_j
\ge 0\}$. Then
\[ {\S}^{\star}(a_1,a_2,\ldots,a_k):= \{n \notin {\G}^{\star}:n+{\G}^{\star} \subset
   {\G}^{\star} \} \subseteq \{m_j-a_1:1 \le j \le a_1-1\}. \]
Moreover,
\begin{equation}
m_j-a_1 \in {\S}^{\star}(a_1,a_2,\ldots,a_k) \Liff m_j+m_i>m_{j+i} \;\;{\rm for}\;\;1 \le i \le a_1-1.
\end{equation}
We refer to \cite{Tri03} for the more notations and results.
 With the notations above, we show that
${\S}^{\star}(A_1,A_2,\ldots,A_k)=\{(k-1){\s}_k-{\s}_{k-1}\}$ for each $k
\ge 2$. Since $g(a_1,a_2,\ldots,a_k) \in
{\S}^{\star}(a_1,a_2,\ldots,a_k)$, this further verifies the first result
of Theorem 1. 

\noindent{\bf Theorem 2.}
Let $a_1,a_2,\ldots,a_k$ be pairwise coprime, positive integers. Let ${\s}_r$ denote the sum of the products of the $a_i$'s taken $r$ at a time, and let $A_j={\s}_k/a_j$ for $1 \le j \le k$. Then
$ {\S}^{\star}(A_1,A_2,\ldots,A_k) = \{(k-1){\s}_k-{\s}_{k-1}\} $
for $k \ge 2$. 

\noindent{\it Proof.} We prove the result by inducting on $k$. The
case $k=2$ is a special case of the main result in \cite{Tri03}.
Given pairwise coprime, positive integers $a_1,a_2,\ldots,a_k$,
define integers $A_1,A_2,\ldots,A_k$ as above. As in the proof of
Lemma 2, for each $j$, $1 \le j \le A_k-1$, let $M_j$ and
$M_j^{\:\pr}$ denote the {\em least\/} positive integer congruent to
$j$ mod $A_k$ representable as a nonnegative linear combination of
$A_1,A_2,\ldots,A_k$ and
$A_1^{\:\pr},A_2^{\:\pr},\ldots,A_{k-1}^{\:\pr},A_k$, respectively,
where $A_j^{\:\pr}=A_j/a_k$ for $1 \le j \le k-1$. Then $\{M_j:1 \le
j \le A_k-1\}=\{a_kM_j^{\:\pr}:1 \le j \le A_k-1\}$. Observe that
each $A_i^{\:\pr}$ divides $A_k$, and that
$\{A_1^{\:\pr},A_2^{\:\pr},\ldots,A_{k-1}^{\:\pr}\}$ is just the set
of $A_i$'s corresponding to $a_1,a_2,\ldots,a_{k-1}$. From $(3)$,
$M_j-A_k \in {\S}^{\star}(A_1,A_2,\ldots,A_k)$ if and only if
$M_j+M_i>M_{j+i}$ for $1 \le i \le A_k-1$, which holds precisely
when $M_j^{\:\pr}+M_i^{\:\pr}>M_{j+i}^{\:\pr}$ for $1 \le i \le
A_k-1$. Thus $M_j-A_k \in {\S}^{\star}(A_1,A_2,\ldots,A_k)$ if and
only if $M_j^{\:\pr}-A_k \in
{\S}^{\star}(A_1^{\:\pr},A_2^{\:\pr},\ldots,A_{k-1}^{\:\pr},A_k)={\S}^{\star}(A_1^{\:\pr},A_2^{\:\pr},\ldots,A_{k-1}^{\:\pr})$,
which is the set $\{(k-2)a_1a_2\cdots
a_{k-1}-(A_1^{\:\pr}+\cdots+A_{k-1}^{\:\pr})\}$, by the induction
hypothesis. It now follows that
${\S}^{\star}(A_1,A_2,\ldots,A_k)=\{a_kM_j^{\:\pr}-A_k\}=\{(k-2)a_1a_2\cdots
a_k-a_k(A_1^{\:\pr}+\cdots+A_{k-1}^{\:\pr})+a_kA_k-A_k\}=\{(k-1)a_1a_2\cdots
a_k-(A_1+A_2+\cdots+A_k)\}$, as desired. \hfill$\Box$
\vskip 20pt

\noindent{\bf Acknowledgment.}
The author wishes to thank the referee for some valuable comments and for
pointing out the eighth reference. 


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\end{document}