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\begin{center}
\vskip 1cm{\LARGE\bf Families of Rational Numbers with \\
\vskip .1in
Predictable Engel Product Expansions}
\vskip 1cm
\large
R.~S.~Melham\\
Department of Mathematical Sciences\\
University of Technology, Sydney\\
PO Box 123, Broadway, NSW 2007\\
Australia\\
\href{mailto:ray.melham@uts.edu.au}{\tt ray.melham@uts.edu.au} \\
\end{center}

\vskip .2 in

\begin{abstract}
In 1987 Knopfmacher and Knopfmacher published new infinite product
expansions for real numbers $0&lt;A&lt;1$ and $A&gt;1$. They called these
Engel product expansions. At that time they had
difficulty finding rational $0&lt;A&lt;1$ for which the Engel product
expansion is predictable. Later, in 1993, Arnold Knopfmacher
presented many such families of rationals. In this paper we add to
Arnold Knopfmacher's list of such families.
\end{abstract}


\section{Introduction}\label{sec1}


Knopfmacher and Knopfmacher \cite{knop1} proved that every positive
real number $0&lt;A&lt;1$ has an expansion of the form


\begin{equation}\label{expan}
    A=\prod_{n=1}^{\infty}\left(1-\frac{1}{a_{1}a_{2} \cdots a_{n}}\right),
\end{equation}

\noindent where $a_{n}$ is a positive integer for $n\geq1$,
$a_{1}\geq2$, $a_{n+1}\geq a_{n}-1$ for $n\geq1$, and $a_{n}\geq2$
infinitely often.

These representations were called Engel product expansions. To
abbreviate (\ref{expan}), Knopfmacher and Knopfmacher wrote

 \[
  A=((a_{1}, a_{2}, a_{3}, \ldots)).
 \]

\noindent They remarked that there do not appear to be simple cases
where the digits $\{a_{i}\}$ could be found with explicit formulae.
Later, Arnold Knopfmacher \cite{knop2} gave several families of
rationals where, from some point onwards, the digits $\{a_{i}\}$ in
the expansion (\ref{expan}) satisfy the recurrence

\begin{equation}\label{rec}
     a_{n+1}=\left(a_{n}+1\right)a_{n-1}-1.
\end{equation}

\noindent To illustrate, one of Knopfmacher's main results is the
following:

\begin{theorem}\label{knop}
Let $a$ and $m$ be positive integers with $a\geq3$ if $m=1$, and
$a\geq2$ otherwise. Then

 \[
 \frac{\left(a-1\right)m-1}{am-1}=\left(\left(a_{1}, a_{2}, a_{3}, \ldots\right)\right),
 \]

\noindent where $a_{1}=a$, $a_{2}=(a-1)m$, $a_{3}=am-1$,
$a_{4}=a_{1}a_{2}-1$, and (\ref{rec})
 applies for $n\geq4$.
\end{theorem}

Knopfmacher and Knopfmacher \cite{knop1} showed that the expansion (\ref{expan}) is
unique if it has non-decreasing digits. Therefore, in Theorem
\ref{knop} a unique expansion is guaranteed for every $a$ provided
$m\geq2$.


Knopfmacher \cite[page~426]{knop2} stated that it would be
interesting to characterize in some way which rational numbers have
expansions where the digits $\{a_{i}\}$ ultimately satisfy the
nonlinear recurrence (\ref{rec}). Based on our investigations it
seems difficult to resolve this completely. We have, however,
managed to add to Knopfmacher's collection of such rationals. We
present our results as families of rationals that require two or
more parameters to describe them. Indeed, we have found many
one-parameter families, but, to conserve space, we do not present
them here.

In Section \ref{sec2} we state and prove one of our main results. In
Sections  \ref{sec3} and  \ref{sec4} we state further results.


\section{A Result With Sample Proof}\label{sec2}

Before proceeding we give the greedy-type algorithm from
\cite{knop1} which was used to derive the expansion (\ref{expan}):

\noindent Given any $0&lt;A&lt;1$, let $A_{1}=A$,
$a_{1}=1+\left\lfloor\frac{1}{1-A_{1}}\right\rfloor$, where $\lfloor
x\rfloor$ denotes the floor of the real number $x$. Then recursively
define, for $n\geq2$,

\[
 a_{n} = 1+\left\lfloor\frac{1}{\left(1-A_{n}\right)a_{1}a_{2}\cdots a_{n-1}}\right\rfloor,
\]


\noindent where

 \[
   A_{n+1}=\frac{a_{1}a_{2}\cdots a_{n}}{a_{1}a_{2} \cdots a_{n}-1}A_{n}.
 \]

The theorem that follows is similar in spirit to Theorem \ref{knop}
above. The proof we present proceeds along the same lines as
Knopfmacher's beautiful proof of Theorem \ref{knop}.

\begin{theorem}\label{mel1}
Let $a$ and $m$ be positive integers with $a\geq2$ if $m=1$, and
$a\geq1$ otherwise. Then

 \[
 \frac{\left(2a-1\right)m-1}{4am-2}=\left(\left(a_{1}, a_{2}, a_{3}, \ldots\right)\right),
 \]

\noindent where $a_{1}=2$, $a_{2}=a$, $a_{3}=(2a-1)m$, $a_{4}=2am-1$
, $a_{5}=a_{1}a_{2}a_{3}-1$, and (\ref{rec}) applies for $n\geq5$.
\end{theorem}


\begin{proof}
With the use of the algorithm above we write down $A_{1}$, $a_{1}$,
$A_{2}$, $a_{2}$, $A_{3}$, $a_{3}$, $A_{4}$, and $a_{4}$. We present
each $A_{i}$ in a manner that makes it easy to find $a_{i}$.


 \[
   A_{1}=\frac{(2a-1)m-1}{4am-2}=1-\frac{(2a+1)m-1}{2(2am-1)},
 \]

 \[
   a_{1}=1+\left\lfloor\frac{1}{1-A_{1}}\right\rfloor=1+\left\lfloor\frac{2(2am-1)}{(2a+1)m-1}\right\rfloor=1+\left\lfloor2-\frac{2m}{(2a+1)m-1}\right\rfloor=2,
 \]

 \[
   A_{2}=\frac{a_{1}}{a_{1}-1}A_{1}=2\left(1-\frac{(2a+1)m-1}{2(2am-1)}\right)=1-\frac{m}{2am-1},
 \]

 \[
   a_{2}=1+\left\lfloor\frac{1}{\left(1-A_{2}\right)a_{1}}\right\rfloor=1+\left\lfloor\frac{2am-1}{2m}\right\rfloor=1+\left\lfloor a-\frac{1}{2m}\right\rfloor=a,
 \]

 \[
   A_{3}=\frac{a_{1}a_{2}}{a_{1}a_{2}-1}A_{2}=\frac{2a}{2a-1}\left(1-\frac{m}{2am-1}\right)=1-\frac{1}{(2a-1)(2am-1)},
 \]


\begin{eqnarray*}
  a_{3} &amp;=&amp; 1+\left\lfloor\frac{1}{\left(1-A_{3}\right)a_{1}a_{2}}\right\rfloor =1+\left\lfloor\frac{(2a-1)(2am-1)}{2a}\right\rfloor\\
        &amp;=&amp; 1+\left\lfloor(2a-1)m-\frac{2a-1}{2a}\right\rfloor=(2a-1)m,
\end{eqnarray*}

\begin{eqnarray*}
  A_{4} &amp;=&amp; \frac{a_{1}a_{2}a_{3}}{a_{1}a_{2}a_{3}-1}A_{3}=\frac{2a(2a-1)m}{2a(2a-1)m-1}\left(1-\frac{1}{(2a-1)(2am-1)}\right) \\
        &amp;=&amp; 1-\frac{1}{(2am-1)(2a(2a-1)m-1)},
\end{eqnarray*}

\begin{eqnarray*}
  a_{4} &amp;=&amp; 1+\left\lfloor\frac{1}{\left(1-A_{4}\right)a_{1}a_{2}a_{3}}\right\rfloor=1+\left\lfloor\frac{(2am-1)(2a(2a-1)m-1)}{2a(2a-1)m}\right\rfloor  \\
        &amp;=&amp; 1+\left\lfloor2am-1-\frac{2am-1}{2a(2a-1)m}\right\rfloor=2am-1.
\end{eqnarray*}


We define a sequence $\left\{b_{n}\right\}$ as follows:

 \[
   \left(b_{1}, b_{2}, b_{3}, b_{4}\right)=\left(2, a, (2a-1)m, 2am-1\right),\;\; b_{n+2}=b_{1}\cdots b_{n}-1,\;\; n\geq3.
 \]

\noindent We prove the following assertions for $n\geq4$ with the
use of induction.

\bigskip
\noindent (i)\; $\displaystyle A_{n}=1-\frac{1}{b_{n}b_{n+1}}$\;
(ii)\; $a_{n}=b_{n}$\; (iii)\; $
b_{n+2}=\left(b_{n+1}+1\right)b_{n}-1$.

\bigskip
Firstly, for $n=4$

\bigskip
\noindent (i)\; $\displaystyle A_{4}=\displaystyle
1-\frac{1}{a_{4}(a_{1}a_{2}a_{3}-1)}=\displaystyle
1-\frac{1}{b_{4}(b_{1}b_{2}b_{3}-1)}=\displaystyle
\frac{1}{b_{4}b_{5}}$;

\bigskip
\noindent (ii)\; $a_{4}=b_{4}$ by the definition of $b_{4}$;

\bigskip
\noindent (iii)\; $b_{6}=(b_{1}b_{2}b_{3})b_{4}-1=(b_{5}+1)b_{4}-1$.

\bigskip
Next we show that the validity of each assertion for $n\geq4$
implies its validity for $n+1$.

\begin{eqnarray*}
  A_{n+1} &amp;=&amp; \frac{a_{1}\cdots a_{n}}{a_{1}\cdots a_{n}-1} A_{n}=\left(1+\frac{1}{b_{n+2}}\right)\left(1-\frac{1}{b_{n}b_{n+1}}\right)\\
          &amp;=&amp; 1-\frac{b_{n+2}-b_{n}b_{n+1}+1}{b_{n}b_{n+1}b_{n+2}}=1-\frac{1}{b_{n+1}b_{n+2}},
\end{eqnarray*}

\noindent where we have used (iii) in the last step. This proves
(i).

\begin{eqnarray*}
  a_{n+1} &amp;=&amp; 1 +\left\lfloor\frac{1}{\left(1-A_{n+1}\right)a_{1}\cdots a_{n}}\right\rfloor=1+\left\lfloor\frac{b_{n+1}b_{n+2}}{b_{n+2}+1}\right\rfloor\\
   &amp;=&amp; 1+\left\lfloor b_{n+1}-\frac{b_{n+1}}{b_{n+2}+1}\right\rfloor=b_{n+1},
\end{eqnarray*}

\noindent since $b_{n}\geq0$ for all $n$. This proves (ii).

\begin{eqnarray*}
  b_{n+3} &amp;=&amp; (b_{1}\cdots b_{n})b_{n+1}-1=(b_{n+2}+1)b_{n+1}-1.
\end{eqnarray*}

\noindent This establishes (iii), and the proof is complete.

\end{proof}


Note that, according to the discussion in the paragraph that follows
the statement of Theorem \ref{knop}, a unique expansion is
guaranteed in Theorem \ref{mel1} provided $a\geq2$.


\section{Miscellaneous Results}\label{sec3}

In this section, and the next, we state our results without proof
since all proofs are similar to that shown above. Knopfmacher
\cite{knop2} took $1^{a}$ to denote $a$ consecutive occurrences of
the digit 1, and we do the same.

Knopfmacher gave the following theorem with $m=1$.

\begin{theorem}\label{mel2}
Let $m$ be a positive integer, and let $a\geq1$ and $b\geq0$ be
integers. Then

 \[
 \frac{m}{2^{b}\left(2^{a}m+1\right)}=\left(\left(2, 1^{a+b-1}, 2^{a-1}m+1, 2^{a}m+1, 2^{a}m+1, a_{3}, a_{4}, a_{5}, \ldots\right)\right),
 \]

\noindent where $a_{1}=a_{2}=2^{a}m+1$, and (\ref{rec}) applies for
$n\geq2$.
\end{theorem}

Due to the presence of $a^{2}$, the following theorem differs in
nature from those above.

\begin{theorem}\label{mel3}
Let $a$ and $m$ be positive integers with $a\geq2$ and $m\geq1$.
Then

 \[
 \frac{(a-1)am-1}{a^{2}m}=\left(\left(a_{1}, a_{2}, a_{3}, \ldots\right)\right),
 \]

\noindent where $a_{1}=a$, $a_{2}=(a-1)m+1$, $a_{3}=(a-1)m$,
$a_{4}=a_{1}a_{2}-1$, and (\ref{rec})
 applies for $n\geq4$.
\end{theorem}


In the next three theorems $a_{1}$ and $a_{2}$ immediately precede
$a_{3}$, and (\ref{rec}) applies for $n\geq2$.

\begin{theorem}\label{mel4}
Let $a\geq1$ and $k$ be integers. Then the expansion for
$\displaystyle \frac{m-1}{2^{a}m}$ is

 \[
 \left(\left(2, 1^{a-1}, k+1, k, 2k+1, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=2k,\; k\geq1;
 \]

 \[
 \left(\left(2, 1^{a-1}, k+1, 2k+1, 2k+1, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=2k+1,\; k\geq1;
 \]
\end{theorem}

\begin{theorem}\label{mel5}
Let $a\geq3$ and $k$ be integers. Then the expansion for
$\displaystyle \frac{3m-1}{2^{a}m}$ is

 \[
 \left(\left(2, 1^{a-3}, 2, 3k+1, 3k, 12k+3, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k,\; k\geq1;
 \]

 \[
 \left(\left(2, 1^{a-3}, 2, 3k+1, 12k+3, 12k+3, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k+1,\; k\geq0;
 \]

  \[
 \left(\left(2, 1^{a-3}, 2, 3k+2, 6k+3, 12k+7, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k+2,\; k\geq0;
 \]

  \[
 \left(\left(2, 1^{a-3}, 2, 3k+3, 4k+3, 12k+11, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k+3,\; k\geq0;
 \]
\end{theorem}

\begin{theorem}\label{mel6}
Let $a\geq5$ and $k$ be integers. Then the expansion for
$\displaystyle \frac{9m-1}{2^{a}m}$ is

 \[
 \left(\left(2, 1^{a-5}, 2, 1, 9k+1, 9k, 36k+3, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k,\; k\geq1;
 \]

 \[
 \left(\left(2, 1^{a-5}, 2, 1, 9k+3, 12k+3, 36k+11, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k+1,\; k\geq0;
 \]

  \[
 \left(\left(2, 1^{a-5}, 2, 1, 9k+5, 18k+9, 36k+19, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k+2,\; k\geq0;
 \]

  \[
 \left(\left(2, 1^{a-5}, 2, 1, 9k+7, 36k+27, 36k+27, a_{3}, a_{4}, a_{5}, \ldots\right)\right), m=4k+3,\; k\geq0;
 \]
\end{theorem}


\section{A Host of Results of a Particular Type}\label{sec4}


Knopfmacher \cite{knop2}, in Theorem 3(iii), gave an
expansion for $\displaystyle\frac{1}{2^{a}\left(2^{n}+1\right)}$ when $a\geq0$ and $n\geq1$.
Inspired by this, we decided to search for triples of positive
integers $\left(b, c, d\right)$ such that the expansion for
$\displaystyle\frac{1}{2^{a}\left(2^{bn+c}+2d+1\right)}$ is
predictable for all integers $a\geq0$ and $n\geq0$. We restricted
our search to $1\leq d\leq 50$, first fixing $d$ and then searching
for appropriate pairs $\left(b, c\right)$. Some $d$ are associated
with many pairs $\left(b, c\right)$. For example, associated with
$d=48$ are the following eight pairs $\left(b, c\right)$: $(48,
24)$, $(48, 27)$, $(240, 121)$, $(336, 172)$, $(336, 173)$, $(528,
270)$, $(528, 303)$, and $(1200, 602)$. On the other hand, for some
$d$ we could find no pair $\left(b, c\right)$.


In total we found one-hundred and five triples as described in the
previous paragraph. For seventy-nine of these triples $\left(b, c,
d\right)$ we found that the expansion has the form

\begin{equation}\label{4a}
     \frac{1}{2^{a}\left(2^{bn+c}+2d+1\right)}=\left(\left(2, 1^{a+bn+c-1}, a_{1}, a_{2}, a_{3}, \ldots\right)\right),
\end{equation}

\noindent in which\\

\begin{eqnarray*}
  a_{1} &amp;=&amp; \left(2^{bn+c-1}+2d+1+e\right)/\left(2d+1\right), \\
  a_{2} &amp;=&amp; \left(2^{bn+c}+2d+1\right)/\left(2d+1+2e\right), \\
  a_{3} &amp;=&amp; 2a_{1}-1,
\end{eqnarray*}

\noindent and (\ref{rec}) applies for $n\geq3$.  Here $e$ is a
relatively small positive integer that depends upon the triple
$\left(b, c, d\right)$, and can be found by observation.

In twenty-four of our triples, $b=2c$. In these cases it is
convenient to write the triples as $\left(2b, b, d\right)$. In all
such cases we found that

\begin{equation}\label{4b}
     \frac{1}{2^{a}\left(2^{2bn+b}+2d+1\right)}=\left(\left(2, 1^{a+2bn+b-1}, a_{1}, a_{2}, a_{3}, a_{4} , \ldots\right)\right),
\end{equation}

\noindent in which\\


\begin{eqnarray*}
  a_{1} &amp;=&amp; \left(2^{2bn+b-1}+d+1\right)/\left(2d+1\right), \\
  a_{2} &amp;=&amp; 2^{2bn+b}+1, \\
  a_{3} &amp;=&amp; a_{2}+2d, \\
  a_{4} &amp;=&amp; 2a_{1}a_{2}-1,
\end{eqnarray*}

\noindent and (\ref{rec}) applies for $n\geq4$.


The reader can now write down the expansion for
$\displaystyle\frac{1}{2^{a}\left(2^{2bn+b}+2d+1\right)}$
corresponding to the first of the eight triples mentioned above. For
this triple (\ref{4b}) applies. For the remaining seven triples
(\ref{4a}) applies, and the values of $e$ are, respectively, 4, 1,
8, 16, 32, 9, and 2. The reader can now use (\ref{4a}) to write down
the associated expansions.

Interestingly, we found two triples not of the form $\left(2b, b,
d\right)$, namely $(16, 4, 8)$ and $(16, 12, 8)$, but where the
associated expansions are found with the use of (\ref{4b}). For
these two triples we use (\ref{4b}) and replace each occurrence of
$2b$ by 16.


Some further examples of triples $\left(2b, b, d\right)$ where the
associated expansions are found with the use of (\ref{4b}) are $(2,
1, 1)$, $(4, 2, 2)$, $(6, 3, 4)$, $(10, 5, 5)$, $(12, 6, 6)$, and
$(18, 9, 9)$. Further examples of triples $\left(b, c, d\right)$
where the associated expansions are found with the use of (\ref{4a})
are $(12, 7, 2)$, $(12, 8, 2)$, $(12, 9, 6)$, $(24, 14, 8)$, $(40,
23, 8)$, and $(40, 24, 8)$. Here, for these six triples, the
corresponding values of $e$ are, respectively, 1, 2, 4, 2, 4, and 8.

\begin{thebibliography}{10}


\bibitem{knop1} A.~Knopfmacher and J.~Knopfmacher,
A new infinite product representation for real numbers,
 {\it Monatshefte Math.} {\bf 104} (1987), 29--44.

\bibitem{knop2} A.~Knopfmacher,
Rational numbers with predictable Engel product expansions,
in G.~E.~Bergum, A.~N.~Philippou, and A.~F.~Horadam, eds., {\it Applications of Fibonacci Numbers}
{\bf 5}, Kluwer Academic Publishers, 1993, pp. 421--427.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A67.

\noindent \emph{Keywords: }
Engel product expansion, Engel expansion.


\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received July 9 2009;
revised version received August 5 2009.
Published in {\it Journal of Integer Sequences} August 30 2009.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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