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%Sam Speed   email: samspeed@bellsouth.net









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\title{{\protect\large An application of formal power series}}

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\maketitle

\begin{center}
  {\Large \textbf{Problem}}
\end{center}



Show that each nonnegative integer has a unique representation in (exactly) one of the forms
 $1/3( 4^k(6n+1)-1)$  or $1/3(2\cdot4^k(6n+5) -1)$ for $n\ge 0, k\ge 0$.

%maybe $1/3(4^k(12n+10) -1)$














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\hrule

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\begin{center}
  {\Large \textbf{Solution to Problem}}
\end{center}



First prove the generalized proposition:

\noindent\textbf{Proposition.} Let $2\leq u\in \mathbb{N} := \{0,1,2,3,\dots\}$, the nonnegative integers,

and $w\in \{1,2,\dots, u-1\}$, then

\[ \sum_{k=0}^\infty  \Bigl(\frac{x^{w\cdot\frac{u^k-1}{u-1}}}{1-x^{u^{k+1}}}  \sum_{\substack{i=0\\ i\neq w}}^{u-1} x^{i\cdot u^k}\Bigr)
   \;=\; \frac{1}{1-x}.\]


where the sums are \emph{formal power series},
for which the best introductions is the first chapter of the first volume of
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\emph{Applied and Computational Complex Analysis} (Wiley-Interscience, 1974) by  Peter  Henrici.

\medskip
\noindent\textbf{Proof of Generalized Proposition.} Assume the initial conditions on $u$ and $w$ and show that the sum telescopes.

Let $F_k :=   \frac{x^{w\cdot\frac{u^k-1}{u-1}}}{1-x^{u^{k+1}}}  \sum\limits_{i=0}^{u-1} x^{i\cdot u^k} = \frac{x^{w\cdot\frac{u^k-1}{u-1}}}{1-x^{u^{k}}}$
\;and\; $f_k := x^{w\cdot u^k}\frac{x^{w\cdot\frac{u^k-1}{u-1}}}{1-x^{u^{k+1}}}$, for all $k$.

Then $\sum\limits_{k=0}^\infty  \Bigl(\frac{x^{w\cdot\frac{u^k-1}{u-1}}}{1-x^{u^{k+1}}}
\sum\limits_{\substack{i=0\\ i\neq w}}^{u-1} x^{i\cdot u^k}\Bigr)= \sum\limits_{k=0}^\infty (F_k-f_k) = F_0= \frac{1}{1-x}$,

since $F_{k+1}= \frac{x^{w\cdot\frac{u^{k+1}-1}{u-1}}}{1-x^{u^{k+1}}} = x^{w\cdot u^k}\frac{x^{w\cdot\frac{u^{k}-1}{u-1}}}{1-x^{u^{k+1}}}
= f_k$.\qed

\bigskip


So we have some partitions of the nonnegative integers.

\medskip
Now the solution to the problem follows by letting $u=4$ and $w=1$ and observing that

$\frac{1}{1-x} = \sum\limits_{m=0}^\infty x^m =
\sum\limits_{k=0}^\infty   \frac{x^{\frac{4^k-1}{3}}}{1-x^{4\cdot4^{k}}}
( 1+x^{2\cdot4^{k}} +x^{3\cdot4^{k}} )\\
 = \sum\limits_{k=0}^\infty \bigl(\frac{x^{\frac{4^k-1}{3}}}{1-x^{4\cdot4^{k}}} ( 1+x^{2\cdot4^{k}} )
+ x^{\frac{4^k-1}{3}+3\cdot4^k}\sum\limits_{n=0}^\infty x^{4\cdot4^kn}\bigr)\\
 = \sum\limits_{k=0}^\infty \bigl( \frac{x^{\frac{4^k-1}{3}}}{1-x^{2\cdot4^{k}}}
 +  \sum\limits_{n=0}^\infty x^{\frac{10\cdot4^k-1+12\cdot4^kn}{3}} \bigr)\\
= \sum\limits_{k=0}^\infty \bigl( \sum\limits_{n=0}^\infty x^{\frac{4^k(6n+1)-1}{3}}
+ \sum\limits_{n=0}^\infty x^{\frac{2\cdot4^k(6n+5)-1}{3}} \bigr)$. \qed




\bigskip\bigskip

\noindent\textbf{Remarks.}

The preceding shows that the famous function $f$ on $\mathbb{N}$ has a range of
$6\mathbb{N}+1 \;\cup\; 6\mathbb{N}+5$, where $f(x) := 3x+1$ with all positive powers of 2 removed.

%\medskip
%More information on this function is at
%\texttt{https://umdrive.memphis.edu/samspeed/}.





\bigskip\bigskip

%\hfill\texttt{Sam E Speed}   %\; University of Memphis

%\hfill\texttt{email: samspeed@bellsouth.net}

%\hfill\texttt{file: mathtex1/hailr/ammprob-f4,1,2.tex}



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