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\vskip 1cm{\LARGE\bf 
On a Generalization of the Frobenius Number
}
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\large
Alexander Brown, Eleanor Dannenberg, Jennifer Fox, Joshua Hanna,
Katherine Keck,  Alexander Moore, Zachary Robbins,  Brandon Samples, and
James Stankewicz \\
Department of Mathematics \\
University of Georgia \\
Athens, GA 30602 \\
USA \\
\href{mailto:stankewicz@gmail.com}{\tt stankewicz@gmail.com} \\
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\begin{abstract}
We consider a generalization of the Frobenius problem, where the object
of interest is the greatest integer having exactly $j$
representations by a collection of positive relatively prime integers.
We prove an analogue of a theorem of Brauer and Shockley and show how
it can be used for computation.
\end{abstract}

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\section{Introduction}

The \emph{linear diophantine problem of Frobenius} has long been a
celebrated problem in number theory. Most simply put, the problem is to
find the \emph{Frobenius number} of $k$ positive relatively prime
integers $(a_1, \ldots, a_k)$, i.e., the greatest integer $M$ for which
there is no way to express $M$ as the non-negative integral linear
combination of the given $a_i$.

A generalization, which has drawn interest both from classical study of
the Frobenius problem (\cite[Problem A.2.6]{Alf}) and from the
perspective of partition functions and integer points in polytopes (as
in Beck and Robins \cite{1}), is to ask for the greatest integer $M$
that can be expressed in exactly $j$ different ways. We make this
precise with the following definitions:

A \emph{representation} of $M$ by a $k$-tuple $(a_1, \ldots, a_k)$ of non-negative, relatively prime integers is a solution 
$(x_1, \ldots, x_k)\in \mathbb{N}^k$ to the equation $M = \sum_{i=1}^k a_ix_i$.

We define the $j$-\emph{Frobenius number} of a $k$-tuple $(a_1, \ldots,
a_k)$ of relatively prime positive integers to be the greatest integer
$M$ with exactly $j$ representations of $M$ by $(a_1, \ldots, a_k)$ if
such a positive integer exists and zero otherwise. We refer to this
quantity as $g_j(a_1, \ldots, a_k)$.

Finally, we define $f_j(a_1, \ldots, a_k)$ exactly as we defined
$g_j(a_1, \ldots, a_k)$, except that we consider only \emph{positive
representations} $(x_1, \ldots, x_k)\in \mathbb{Z}^k_{&gt; 0}$.

Note that the $0$-Frobenius number of $(a_1, \ldots, a_k)$ is just the
classical Frobenius number. The purpose of this paper is to prove a
generalization of a result of Brauer and Shockley \cite{2} on
the classical Frobenius number.

\section{The Main Results}

Our main result is the following:

\begin{theorem}\label{bigresult} If $d=\gcd(a_2, \ldots, a_k)$ and $j\ge 0$, then either
$$g_j(a_1, a_2, \ldots,a_k) = d\cdot g_j(a_1, \frac{a_2}{d}, \ldots, \frac{a_k}{d}) + (d-1)a_1$$
or $g_j(a_1,a_2, \ldots,a_k) = g_j(a_1, \frac{a_2}{d}, \ldots, \frac{a_k}{d}) = 0.$\end{theorem}



\begin{lemma}\label{lemma}

If $f_j(a_1, \ldots, a_k)$ is nonzero, there exist integers $x_2, \ldots, x_k &gt; 0$ such that $$f_j(a_1, \ldots, a_k) = \sum_{i=2}^k a_i x_i.$$

\end{lemma}

\begin{proof}Let $f_j := f_j(a_1, \ldots, a_k)$. By the definition of $f_j$, we can write $f_j = \displaystyle\sum_{i=1}^k a_i x_{i,\ell}$ with $x_{i,\ell} &gt; 0$ for $1 \leq \ell \leq j$.  Since
$$f_j + a_1 = \displaystyle \sum_{i=1}^k a_i x_{i,\ell} + a_1= a_1(x_{1,\ell} + 1) + \displaystyle \sum_{i=2}^k a_i x_{i,\ell},$$
we obtain at least $j$ positive representations of $f_j + a_1$.  As $f_j$ is the largest number with exactly $j$ positive representations,  there must be at least $j+1$ distinct ways to represent $f_j + a_1$.  Specifically, we have $f_j + a_1 = \displaystyle \sum_{i=1}^k a_i x'_{i,\ell}$ with $x'_{i,\ell} &gt; 0$ for all $1 \leq \ell \leq j+1$.  Subtract $a_1$ from both sides of these $j+1$ equations to obtain $f_j = (x'_{1,{\ell}} - 1)a_1 + \displaystyle \sum_{i=2}^k a_i x'_{i,\ell}$.  Evidently, there exists some $\ell_0 \in [1, j+1]$ for which $x'_{1,{\ell_0}} - 1 = 0$ because $f_j$ cannot have $j+1$ positive representations.  Therefore, $f_j(a_1, \ldots, a_k) = \displaystyle \sum_{i=2}^k a_i x'_{i,{\ell_0}}.$\end{proof}

\begin{theorem}\label{theorem}

If $\mathrm{gcd}(a_2,\ldots,a_k)=d$, then $$f_j(a_1,a_2,\ldots,a_k)=d\cdot f_j(a_1,\dfrac{a_2}{d},\ldots,\dfrac{a_k}{d}).$$ 

\end{theorem}

\begin{proof}Let $a_i=da'_i$ for $i=2,\ldots,k$ and $N=f_j(a_1,\ldots,a_k)$.\\
Assuming $N&gt;0$, we know by Lemma \ref{lemma} that $$N=\sum_{i=2}^k a_i x_i=d\sum_{i=2}^k a'_i x_i$$ with $x_i&gt;0$. Let $N'=\sum_{i=2}^k a'_i x_i$. We want to show that $N'=f_j(a_1,a'_2,\ldots,a'_k)$ and will do this in three steps.\\ \\
\textbf{Step 1:}  First, we know that $N'$ does not have $j+1$ or more positive representations by   $a_1,a'_2,\ldots,a'_k$.  If $N'$ could be so represented, then for $1\leq l\leq j+1$ we would have
$$N'=a_1y_{{1,\ell}} + \sum_{i=2}^k a'_i y_{{i,\ell}}.$$
Multiplying this equation by $d$ immediately produces too many representations of $N$ and thus a contradiction.\\ \\
\textbf{Step 2:} Next, we know that
$$f_j(a_1,\ldots, a_k) =N=a_1x_{{1,\ell}} + \sum_{i=2}^k a_i x_{{i,\ell}}$$ 
for $1\leq l\leq j$ and $x_i&gt;0$, so
$$\dfrac{N}{d}=\dfrac{a_1x_{{1,\ell}}}{d} + \sum_{i=2}^k \dfrac{a_i x_{{i,\ell}}}{d}.$$
Since $d|N$ and $d|a_i$ for $i\ge 2$, we must have $d|a_1x_{1,\ell}$ for $1\le \ell \le j$. In addition, $\mathrm{gcd}(a_1,d)=1$ so we must have $d|x_{1,\ell}$ for $1\le \ell \le j$. So
$$ N'=a_1 \dfrac{x_{1,\ell}}{d} + \sum_{i=2}^k a'_i x_{{i,\ell}},$$
hence $N'$ has at least $j$ distinct positive representations.  But we have already shown that $N'$ cannot have $j+1$ or more positive representations, thus $N'$ has exactly $j$ positive representations.\\ \\
\textbf{Step 3:}  Finally we will show that $N'$ is the largest number with exactly $j$ positive representations by $a_1, a_2', \dots, a_k'$.  Consider any $n&gt;N'$.  Since $dn&gt;dN'=N$, we know that $dn$ can be represented as a linear combination of $a_1,\ldots,a_k$ in exactly $X$ ways with $X\not=j$.\\
Thus, for $1\leq l\leq X$ and $X\not=j$ we have
$$dn=a_1x_{{1,\ell}} + \sum_{i=2}^k a_i x_{{i,\ell}}$$
and as in Step 2,
$$n=a_1(\dfrac{x_{{1,\ell}}}{d}) + \sum_{i=2}^k a'_i x_{{i,\ell}}.$$
If $X&gt;j$ then we certainly do not have exactly $j$ representations, so assume $X&lt;j$. Assume now that we can write $n=a_1y_1 + \sum_{i=2}^k a'_i y_i$
where $y_i\not=x_{i,\ell}$ for any such $\ell$.  By multiplying by $d$ we get a new representation for $dn$, which is a contradiction because $dn$ is represented in exactly $X\not=j$ ways.  
\\\\
Therefore $N'$ is the greatest number with exactly $j$ positive representations and so 
$$N'=f_j(a_1,a'_2,\ldots,a'_k).$$
Thus
$$f_j(a_1,a_2,\ldots,a_k)=d\cdot f_j(a_1,\dfrac{a_2}{d},\ldots,\dfrac{a_k}{d}).$$\end{proof}

Having established our results about $f_j(a_1, \ldots, a_k)$, we show that we can translate these results to results about the $j$-Frobenius numbers.

\begin{lemma}\label{lemma2}

Either $f_j(a_1, \ldots, a_k) = g_j(a_1, \ldots, a_k)=0$ or, $$f_j(a_1, \ldots, a_k) = g_j(a_1, \ldots, a_k) + \displaystyle \sum_{i=1}^k a_i.$$

\end{lemma}

\begin{proof}For ease, write $f_j$ for $f_j(a_1, \ldots, a_k)$, $g_j$ for $g_j(a_1, \ldots, a_k)$, and $K=\sum_{i=1}^k a_i$.\\ \\
Any representation $(y_1, \ldots, y_k)$ of $M$ gives a representation $(y_1+1, \ldots, y_k +1)$ of $M+K$. Moreover, adding or subtracting $K$ preserves the distinctness of representations because it adjusts every coefficient $y_i$ by $1$. Therefore if $M$ has $j$ representations, $M+K$ has at least $j$ positive representations. Likewise, every positive representation of $M+K$ gives a representation of $M$. Thus $f_j =0$ if and only if $g_j=0$. Assume now that $f_j$ and $g_j$ are both nonzero.\\ \\
Suppose that $f_j &lt; g_j+ K$.  By definition, we can find exactly $j$ representations $(y_1, \ldots, y_k)$ for $g_j$ and $g_j$ has exactly $j$ representations if and only if $g_j + K$ has exactly $j$ \emph{positive} representations $(x_1, \ldots, x_k)$. However, by assumption $g_j + K &gt;f_j$ and $g_j +K$ has exactly $j$ positive representations.  This contradicts the definition of $f_j$, hence $f_j \geq g_j+ K$.\\ \\
Suppose that $f_j &gt; g_j+ K$.  By definition, we can find exactly $j$ positive representations $(x_1, \ldots, x_k)$ for $f_j$.  The same argument as above shows that $f_j - K$ has exactly $j$ representations in contradiction to the definition of $g_j$.  Thus $f_j \leq g_j+ K$.\end{proof}

\textbf{Proof of Theorem \ref{bigresult}:} Combine Theorem \ref{theorem} with Lemma \ref{lemma2}.

\begin{corollary}Let $a_1, a_2$ be coprime positive integers and let $m$ be a positive integer. Suppose that $g_j = g_j(a_1,a_2,ma_1a_2)\ne 0$. Then \begin{itemize}
\item $g_{j} = (j+1)a_1a_2 - a_1-a_2$ for $j&lt; m+1$

\item $g_{m+1} = 0$ and

\item $g_{m+2} = (m+2)a_1a_2-a_1-a_2$.\end{itemize}\end{corollary}

\begin{proof} Theorem \ref{bigresult} tells us that if $g_j(1,1,m) \ne 0$ then \begin{eqnarray*}
g_{j}(a_1, a_2, ma_1a_2) &amp;= &amp; a_2(g_{j}(a_1,1,ma_1)) + (a_2-1)a_1\\
&amp; = &amp; a_2 \left( a_1g_{j}(1,1,m) + (a_1 -1)1\right) + (a_2 -1)a_1\\
&amp; = &amp; a_1a_2 (g_{j}(1,1,m) + 2) -a_1 - a_2 .\\ 
\end{eqnarray*}

Following Beck and Robins in their proof of \cite[Proposition 1]{1}, we can use the values of the restricted partition function $p_{1,1,m}(k)$ to determine $g_{j}(1,1,m)$. Furthermore we can determine the relevant values with the Taylor series $\frac{1}{(1-t)^2(1-t^m)} = \sum_{k=0}^\infty p_{1,1,m}(k)t^k$. Now recall that for $k&lt;m$, $p_{1,1,m}(k) = p_{1,1}(k) = k+1$ but $p_{1,1,m}(m) = m+2$ and for all $k&gt;m$, $p_{1,1,m}(k)&gt;m+2$.  Note that no number is represented $m+1$ times. Thus $g_{m+1}(1,1,m) = 0$, $g_j(1,1,m) = j-1$ for $j&lt;m$ and $g_{m+2}(1,1,m) = m$.\end{proof}

\begin{remark}
It is a consequence of the asymptotics in Nathanson
\cite{3} that for a given tuple, there may be many $j$ for which $g_j
=0$, so the ordering $g_0&lt;g_1&lt; \cdots$ may not hold. In the process of
discovering the theorems of this paper, we noted the somewhat stranger
occurrence of tuples where $0&lt; g_{j+1} &lt; g_j$.

Take, for instance, the 3-tuple (3, 5, 8). The order $g_0&lt;g_1&lt;\cdots$
holds until $g_{14}=52$ and $g_{15}=51$. As should also be the case,
the 3-tuple increased by a factor of $d=2$ creates the new
``dependent'' 3-tuple (3, 10, 16), which fails to hold order in the
same position with $g_{14}=107$ and $g_{15}=105$. A few independent
examples are as follows:\\

$g_{17}(2, 5, 7) = 43$ and $g_{18}(2, 5, 7) = 42$,

$g_{38}(2, 5, 17)=103$ and $g_{39}(2, 5, 17)=102$,

$g_{35}(4, 7, 19)=181$ and $g_{36}(4, 7, 19)=180$, and

$g_{38}(9, 11, 20)=376$ and $g_{39}(9, 11, 20)=369$.\\

We do not as of yet know a lower bound on $j$ for the above to occur.
Indeed, in every case we have computed, if $g_0,g_1&gt;0$ then $g_1&gt;g_0$,
but to date neither a proof or a counterexample has presented itself.
\end{remark}

\section{Acknowledgments}

The research done in this note was completed as part of the
undergraduate number theory VIGRE research seminar directed by
Professor Dino Lorenzini and assisted by graduate students Brandon
Samples and James Stankewicz at the University of Georgia in Fall 2008.
Brandon Samples and James Stankewicz were partially supported by NSF
VIGRE grant DMS-0738586. Support for the seminar was provided by the
Mathematics Department's NSF VIGRE grant.


\begin{thebibliography}{4}

\bibitem{Alf} J. L. Ram\'{\i}rez Alfons\'{\i}n, \emph{The Diophantine Frobenius Problem,} Oxford University Press, 2005.

\bibitem{1} M. Beck and S. Robins, 
A formula related to the Frobenius Problem in two dimensions, in {\it Number Theory (New York 2003)}, Springer, New York, 2004, pp.\ 17--23.

\bibitem{2} A. Brauer and J. E. Shockley, On a problem of Frobenius,  \emph{J. Reine Angew. Math.} {\bf 211} (1962) 215--220.

\bibitem{3} M. Nathanson, Partitions with parts in a finite set,
\emph{Proc. Amer. Math. Soc.} {\bf 128} (2000) 1269--1273.

\end{thebibliography}

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2000 {\it Mathematics Subject Classification}:
Primary 11D45, Secondary 45A05.

\noindent \emph{Keywords: } 
Frobenius problem, counting solutions of Diophantine equations,
linear integral equations.


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\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11D45; Secondary 45A05.

\noindent \emph{Keywords: }
Frobenius problem,
counting solutions of Diophantine equations, linear integral equations.

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\vspace*{+.1in}
\noindent
Received  April 29 2009;
revised version received January 7 2010.  
Published in {\it Journal of Integer Sequences}, January 8 2010.

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