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\begin{center}
\vskip 1cm{\LARGE\bf A Function Related to the \\
\vskip .1in
Rumor Sequence Conjecture}
\vskip 1cm
\large
Bruce Dearden, Joel Iiams, and Jerry Metzger\\
University of North Dakota\\
Department of Mathematics\\
Witmer Hall Room 313\\
101 Cornell Street Stop 8376\\
Grand Forks, ND 58202-8376\\
USA\\
\href{mailto:bruce.dearden@und.edu}{\tt bruce.dearden@und.edu}\\
\href{mailto:joel.iiams@und.edu}{\tt joel.iiams@und.edu}\\
\href{mailto:jerry.metzger@und.edu}{\tt jerry.metzger@und.edu}\\
\end{center}

\vskip .2in

\begin{abstract}
 For an integer $b\geq 2$ and for $x\in [0,1)$, define 
 $\rho_b(x) = \sum_{n=0}^{\infty} \frac{\{\mskip -5mu\{b^nx\}\mskip -5mu\}}{b^n}$,
 where $\{\mskip -5mu\{t\}\mskip -5mu\}$ denotes the fractional part of the real number $t$. 
A number of 
properties of $\rho_b$ are derived, and then a connection between $\rho_b$ and the rumor
conjecture is established.  To form a rumor sequence $\{z_n\}$, first select integers
$b\geq 2$ and $k\geq 1$. Then  
select an integer $z_0$, and for $n\geq 1$ let $z_n = bz_{n-1} \bmod{(n+k)}$, where the right 
side is the least non-negative residue of $bz_{n-1}$ modulo  $n+k$. The rumor sequence
conjecture asserts that all such rumor sequences are eventually $0$. A condition on
$\rho_b$ is shown to be equivalent to the rumor conjecture. 
\end{abstract}

                                                                                

\section{Introduction}

In this note, $b\geq 2$ is a fixed integer. For $x\in [0,1)$, define $\rho_b(x) =\sum_{n=0}^{\infty} \frac{\{\mskip -5mu\{b^nx\}\mskip -5mu\}}{b^n}$, where $\{\mskip -5mu\{t\}\mskip -5mu\}$
denotes the fractional part of the real number $t$. 

A {\it $b$-adic rational} is a rational which can be written as a quotient of an integer and a non-negative
power of $b$. When a $b${-}adic is written in the form
$\frac{a}{b^m}$, and $m&gt;0$,  it will be assumed $b$ does not divide $a$.

The $\rho_b$ function is similar to the well known Takagi function $\tau(x)$  defined by
$\tau(x) = \sum_{n=0}^\infty \frac{{&lt;\mskip-10mu&lt;}2^n x{&gt;\mskip -10mu&gt;}}{2^n}$, where 
${&lt;\mskip -10mu&lt;}t{&gt;\mskip -10mu&gt;}$ is the distance from $t$ to the nearest integer. 
Whereas the summand of the Takagi
function is a triangle wave, the summand of $\rho_b(x)$ is a sawtooth wave.
(See Figures \ref{taggraph}, \ref{rhograph}.)
The function $y={&lt;\mskip -10mu&lt;}2^n x{&gt;\mskip -10mu&gt;}$ is continuous and periodic with period $2^{-n}$, and it follows that
$\tau$ is continuous. It turns out $\tau$ (see Figure \ref{Takagi}) is also nowhere differentiable.

\begin{figure}[ht]
 \begin{minipage}[]{0.4\linewidth}
 \hspace{.2truein}
 \includegraphics[width=2truein, height=1.5truein, viewport= 0 0 240 250]{takagisummand.eps}
 \caption{$y={&lt;\mskip -10mu&lt;}2^2 x{&gt;\mskip -10mu&gt;}$}
 \label{taggraph}
\end{minipage}
\hspace{.5truein}
\begin{minipage}[]{0.5\linewidth}
 \hspace{.4truein}
 \includegraphics[width=2truein, height=1.5truein, viewport= 0 0 240 250]{rhosummand.eps}
 \caption{$y = \{\mskip -5mu\{2^2x\}\mskip -5mu\}$}
 \label{rhograph}
\end{minipage}
\end{figure}

\begin{figure}[ht]
\begin{minipage}[]{0.4\linewidth}
 \hspace{.2truein}
 \includegraphics[width=2truein, height=2.5truein, viewport= 0 0 240 250]{takagigraph.eps}
 \caption{$y=\tau(x)$}
 \label{Takagi}
\end{minipage}
\hspace{.5truein}
\begin{minipage}[]{0.5\linewidth}
 \hspace{.4truein}
 \includegraphics[width=2.5truein,height=2.5truein]{rhograph.eps} 
 \caption{$y = \rho_2(x)$}
 \label{syncgraph}
\end{minipage}
\end{figure}
 

The Takagi function has interesting analytic properties not shared by $\rho_b$. For example, while
the Takagi function is continuous, $\rho_b$ is easily seen to be continuous except 
at the $b${-}adics as suggested by Figure \ref{syncgraph} for the case $b=2$. In general, 
at each $b${-}adic,  $\frac{a}{b^m}$,  $\rho_b$ is right continuous with a jump 
discontinuity of $-\frac{1}{b^{m-1}(b-1)}$. 
On the other hand,  the $\rho_b$ function has interesting number theoretic 
features not shared by $\tau$ as we will show in the 
following section.

 
In the final section of this note, 
we show that the $\rho_b$ function is related to  the rumor conjecture 
described in Dearden and Metzger \cite{dm}, and finally 
a conjecture concerning $\rho_b$  that is equivalent to the rumor sequence conjecture is stated.

\section{Arithmetic Properties of the \texorpdfstring{$\rho_b$}{rho} Function}

An alternative expression for $\rho_b(x)$ can be given in terms of the base{-}$b$ expansion of $x$. In this note we will 
follow the usual convention that 
 base{-}$b$ expansions  of $b${-}adics terminate rather than end with an infinite string of $(b-1)$'s.


\begin{theorem}\label{thm:takagi-like}
If $\sum_{j=1}^{\infty}\frac{d_j}{b^j}$ is the base{-}$b$ expansion of $x\in [0,1)$, then  
$\rho_b(x) =\sum_{j=1}^{\infty}\frac{jd_j}{b^j}$.
\end{theorem}
\begin{proof}
 Let $x=\sum_{j=1}^\infty \frac{d_j}{b^j} \in [0,1)$.  Then 
\begin{align*}
 \rho_b(x) &amp; = \sum_{n=0}^\infty \frac{\{\mskip -5mu\{b^nx\}\mskip -5mu\}}{b^n}  
   = \sum_{n=0}^\infty \frac{\{\mskip -5mu\{b^n \sum_{j\geq1}d_j b^{-j}\}\mskip -5mu\}}{b^n}\\
    &amp;= \sum_{n=0}^\infty \frac{1}{b^n} \left\{\mskip -8mu\left\{\sum_{j=1}^n d_j b^{n-j} + \sum_{j&gt;n} d_jb^{n-j} \right\}\mskip -8mu\right\}
    = \sum_{n=0}^\infty \frac{1}{b^n} \sum_{j&gt;n} d_j b^{n-j}\\
    &amp;= \sum_{n=0}^\infty \sum_{j&gt;n} \frac{d_j}{b^j}
    = \sum_{j=1}^\infty \sum_{n=0}^{j-1} \frac{d_j}{b^j}
    = \sum_{j=1}^\infty \frac{j d_j}{b^j}.\\
\end{align*}
\end{proof}


\begin{theorem}\label{thm: rho-range}
 The range of $\rho_b$ is $[0, \frac{b}{b-1})$.
\end{theorem}

\begin{proof}
 
Let $y \in [0, \frac{b}{b-1})$ be given. Integers $d_j\in D=\{0,1,\dots,b-1\}$ are  selected recursively as follows.
First let $d_1$ be the largest integer in $D$ such that $\frac{d_1}{b}\leq  y$. Assuming $d_1,d_2,\dots,d_{i-1}$ have been selected, 
take $d_i$ to be the largest integer in $D$ such that $\frac{i d_i}{b^i} \leq y - \sum_{j=1}^{i-1}\frac{j d_j}{b^j}$. In this way,
the base{-}$b$ expansion of a number $x= \sum_{j=1}^{\infty} \frac{d_j}{b^j}$ is constructed.

We now show that this expansion of $x$
does not end in an infinite sequence of $(b-1)$'s, and consequently $x\in [0,1)$ and  
$\rho_b(x) = \sum_{j=1}^{\infty} \frac{j d_j}{b^j}$. To this end, by way of contradiction,  assume the expansion does
end with an infinite sequence of $(b-1)$'s. It can not be that all the digits, $d_j$, are $b-1$ since, if they were, we
would have
\[
\sum_{j=1}^{\infty} \frac{j d_j}{b^j} = \sum_{j=1}^{\infty} \frac{j(b-1)}{b^j} = (b-1)
\sum_{j=1}^{\infty}\frac{j}{b^j} = (b-1)\frac{b}{(b-1)^2}= \frac{b}{b-1},
\]
but $\sum_{j=1}^{\infty}\frac{j d_j}{b^j}\leq y &lt; \frac{b}{b-1}$.
 So there must be a last digit, $d_L$, that is less than $b-1$.
It follows that for all $m&gt;L$, 
\[
\frac{m(b-1)}{b^m} \leq y - \sum_{j=1}^{L-1} \frac{j d_j}{b^j} - \frac{L d_L} {b^L} - \sum_{j=L+1}^{m-1} \frac{j(b-1)}{b^j}.
\]
Hence, for all $m&gt;L$, we have
\[
 \frac{L d_L}{b^L} + \sum_{j=L+1}^{m} \frac{j(b-1)}{b^j}  \leq y - \sum_{j=1}^{L-1} \frac{j d_j}{b^j}.
\]
Consequently
\[
 \frac{L d_L}{b^L} + \sum_{j=L+1}^{\infty} \frac{j(b-1)}{b^j} \leq y - \sum_{j=1}^{L-1} \frac{j d_j}{b^j}.
\]
Now
\[
\sum_{j=L+1}^{\infty} \frac{j(b-1)}{b^j} = \frac{(L+1)b-L}{b^L (b-1)}&gt;\frac{L}{b^L}.
\]
Thus 
\[
 \frac{L(d_L+1)}{b^L}=\frac{L d_L}{b^L} + \frac{L}{b^L} \leq y - \sum_{j=1}^{L-1} \frac{j d_j}{b^j},
\]
contradicting the choice of $d_L$.

For any $i$ with $d_i&lt;b-1$, we have

\[
 \frac{i d_i}{b^i} \leq y - \sum_{j=1}^{i-1} \frac{j d_j}{b^j} &lt;  \frac{i (d_i + 1)}{b^i}.
\]
Since that holds for infinitely many $i$, and since $\sum_{j=1}^{\infty} \frac{j d_j}{b^j}$ is a positive
series, it follows that $\rho_b(x) = \sum_{j=0}^{\infty} \frac{j d_j}{b^j} = y$.

\end{proof}
The $x$ constructed in the proof above is the largest of the inverses of the given $y$ under $\rho_b$. Call the $x$ so
constructed the {\it greedy inverse image} of $y$.  
In order to construct a valid base-$b$ number $x$ as the greedy inverse of a given $y$,
 we explicitly required each $d_i$ to be an element of the set $D=\{0,1,\dots,b-1\}$, rather than using
a floor function, as in
\[
  \left\lfloor \frac{b^i\left( y - \sum_{j=1}^{i-1} \frac{j d_j}{b^j} \right)}{i} \right\rfloor.
\]
Since this integer may be larger than $b-1$, the restriction on $d_i$ was needed.
We now show that $d_i$ is eventually given by this floor function expression.


\begin{corollary}\label{cor:greedy=rumor}
 With the notation as in the proof of Theorem \ref{thm: rho-range}, for large enough $i$,
\[
 d_i=\left\lfloor \frac{b^i\left(y-\sum_{j=1}^{i-1} \frac{jd_j}{b^j}\right)}{i}\right\rfloor \text{ and }
     0 \leq y -\sum_{j=1}^i\frac{jd_j}{b^j} &lt; \frac{i}{b^i}.
\]
\end{corollary}

\begin{proof}
We show, for i large enough, that the quantity $z= b^i\left( y - \sum_{j=1}^{i-1} \frac{j d_j}{b^j} \right)/i$ is less than $b$,
and, thus, $\lfloor z \rfloor \in D$.
From the proof of Theorem \ref{thm: rho-range}, there is an integer $n$ with $d_n&lt;b-1$, where
$d_n$  is the largest integer in $D=\{0,1,\dots,b-1\}$ such that
\[
 \frac{n d_n}{b^n} \leq y - \sum_{j=1}^{n-1} \frac{j d_j}{b^j}.
\]
Thus, we see that
\begin{equation}\label{eq:d_{n+1}}
 \frac{n d_n}{b^n} \leq y - \sum_{j=1}^{n-1} \frac{j d_j}{b^j} &lt;  \frac{n (d_n + 1)}{b^n}.
\end{equation}
Equivalently, we have
\[
 d_n \leq \frac{b^n\left(y-\sum_{j=1}^{n-1} \frac{j d_j}{b^j}\right)}{n} &lt; d_n + 1.
\]
Now, since $d_n+1&lt;b$, we have that
\[
 \left\lfloor \frac{b^n\left(y-\sum_{j=1}^{n-1} \frac{j d_j}{b^j}\right)}{n}  \right\rfloor \in D.
\]
Therefore, $d_n$ may be expressed as
\[
 d_n =  \left\lfloor \frac{b^n\left(y-\sum_{j=1}^{n-1} \frac{j d_j}{b^j}\right)}{n}  \right\rfloor.
\]
And, rearranging \eqref{eq:d_{n+1}} we see that
\[ 
           0 \leq y -\sum_{j=1}^n\frac{jd_j}{b^j} &lt; \frac{n}{b^n}.
\]
Inductively, consider any $i\geq n$ where
\begin{equation}\label{eq:y_i}
 0 \leq y-\sum_{j=1}^{i} \frac{jd_j}{b^j} &lt; \frac{i}{b^i}.
\end{equation}
By definition, $d_{i+1}$ is the greatest integer in  $D$ such that
\begin{equation}\label{eq:d_{i+1}_def}
 d_{i+1} \leq \frac{b^{i+1}\left(y-\sum_{j=1}^{i} \frac{j d_j}{b^j}\right)}{i+1}.
\end{equation}
Moreover, using \eqref{eq:y_i}, we have
\[
  \frac{b^{i+1}\left(y-\sum_{j=1}^{i} \frac{j d_j}{b^j}\right)}{i+1}  &lt;\frac{i}{i+1}b &lt;b.
\]
Hence, as before, we have that
\[
 d_{i+1} =  \left\lfloor \frac{b^{i+1}\left(y-\sum_{j=1}^{i} \frac{j d_j}{b^j}\right)}{i+1}  \right\rfloor,
\]
since the value of the floor expression is an element of the set $D$.
Finally, from \eqref{eq:d_{i+1}_def}, we have
\[
 \frac{(i+1)d_{i+1}}{b^{i+1}} \leq y - \sum_{j=1}^{i} \frac{jd_j}{b^j} &lt; \frac{(i+1)(d_{i+1}+1)}{b^{i+1}}.
\]
From which we have
\[
 0 \leq y - \sum_{j=1}^{i+1} \frac{j d_j}{b^j} &lt; \frac{i+1}{b^{i+1}},
\]
completing the induction.

\end{proof}


There are several easily verified functional identities satisfied by $\rho_b$ stated  in the next theorem.

\begin{theorem}\label{thm: fun-equs} The following identities hold for $\rho_b$:

\begin{enumerate}

 \item[(a)] For the $b${-}adic  $x=\frac{a}{b^m} \in [0,1)$,  
  $\rho_b(x) + \rho_b(1-x) = \frac{b}{b-1}- \frac{1}{b^{m-1}(b-1)}.$
  
  \item[(b)] For any non-$b${-}adic  $x\in [0,1)$, $\rho_b(x) + \rho_b(1-x) = \frac{b}{b-1}$.
  
  \item[(c)] For $x\in [0,1)$ and integer $m\geq 1$, 
   $\rho_b\left(\frac{x}{b^m}\right) = \frac{m}{b^m}x + \frac{1}{b^m}\rho_b(x)$.
  
  \item[(d)] If $b^m x \in [0,1)$, then $\rho_b(b^mx) = b^m\rho_b(x) - mb^mx$.
  
 \end{enumerate}
 
\end{theorem}


\begin{theorem}\label{lem:rat-images}
 Suppose $\frac{s}{t}$ is a rational number in lowest terms with $\operatorname{gcd}(t,b)=1$. If $\rho_b(\frac{s}{t}) = \frac{u}{v}$,
a rational in lowest terms, then (1) there is a divisor $t^{\prime}&gt;1$ of $t$ such that $(t^{\prime})^2$ divides $v$, and (2) $b$ divides $u$.
 
\end{theorem}

\begin{proof}
Since $t$ is relatively prime to $b$, the base{-}$b$ expansion of $\frac{s}{t}$ is purely periodic. Let $r$ be the order 
of $b$ modulo $t$, so that $r$ is the period of that expansion. That means there is an integer $c$ so that $ct = b^r - 1$.  
Then
\[
 \frac{s}{t}=\frac{cs}{ct}=\frac{cs}{b^r-1}=\sum_{m\geq1}\frac{cs}{b^{mr}}=\sum_{m\geq1}\frac{\sum_{i=1}^r b^{r-i}d_i}{b^{mr}},
\]
where
\[
\frac{s}{t} = \sum_{j\geq1}\frac{d_j}{b^j}=\sum_{m\geq0}\sum_{i=1}^r \frac{d_i}{b^{mr+i}} \text{\quad has period $r$}.
\]
First, calculate $\rho_b(\frac{s}{t})$ as follows:
\begin{align*}
 \frac{u}{v} = \rho_b\left(\frac{s}{t}\right) &amp;= \sum_{j\geq1}\frac{jd_j}{b^j} = \sum_{m\geq0}\sum_{i=1}^r \frac{(mr+i)d_{mr+i}}{b^{mr+i}}
                                                                 = \sum_{m\geq0}\sum_{i=1}^r \frac{(mr+i)d_{i}}{b^{mr+i}}\\
       &amp;= \sum_{m\geq0}\frac{1}{b^{mr}}\left(mr\sum_{i=1}^r\frac{d_i}{b^i}   + \sum_{i=1}^r\frac{id_i}{b^i}\right)\\
       &amp;= \sum_{m\geq0}\frac{1}{(b^r)^{m+1}}\left(mr\sum_{i=1}^rb^{r-i}d_i + \sum_{i=1}^r ib^{r-i}d_i\right)\\
       &amp;= \sum_{m\geq0}\frac{1}{(b^r)^{m+1}}\left(mrcs + w\right),\text{where $w=\sum_{1\leq i \leq r}ib^{r-i}d_i$}\\
       &amp;= \frac{rcs}{(b^r-1)^2} + \frac{w}{b^r-1} = \frac{rcs + (b^r-1)w}{(b^r-1)^2} = \frac{rcs + ctw}{c^2t^2}\\
       &amp;= \frac{rs + t w}{ct^2}.
\end{align*}
Let $d=\gcd(t,r)$ and define $t'$ and $r'$ by $t=t'd$ and $r=r'd$. Then, we have
\[
\rho_b\left(\frac{s}{t}\right) = \frac{r's + t'w}{c t t'}=\frac{r's + t'w}{c d (t')^2}.
\]
Since $r$ divides $\varphi(t)$ we have 
\[
 r\leq\varphi(t) &lt; t, \text{ hence, } 1\leq r' &lt; t'.
\]
In particular, we have that $t'\not=1$.

Since $t'$ is relatively prime to both $s$ and $r'$, we have that 
$(t')^2$ does not cancel when the fraction is reduced to lowest terms. That completes the proof of (1).

For the proof of (2), calculate $\rho_b(\frac{s}{t})$ as

\begin{align*}
 \frac{u}{v} = \rho_b\left(\frac{s}{t}\right) &amp;= \sum_{m\geq0}\sum_{i=1}^r \frac{(mr+i)d_{i}}{b^{mr+i}} = 
                \sum_{i=1}^r\sum_{m\geq 0} \frac{(mr+i)d_{i}}{b^{mr+i}} \\
          &amp;=\frac{1}{(b^r - 1)^2} \sum_{i=1}^r d_i b^{r-i}(r-i+b^ri). \\
\end{align*}

Note that $b$ is a factor of each term in the sum, including the term when $i=r$. Since $b$ is relatively prime to $b^r-1$,
it follows that $b$ divides $u$.

\end{proof}

\begin{corollary}
 There are rationals in the range $[0,\frac{b}{b-1})$ of $\rho_b$ that are not images of any rationals in its domain.
\end{corollary}


\begin{example}
For $b \not= 3$, the rational $\frac{1}{3}$ cannot be the image of a rational under $\rho_b$.
\end{example} 
The conditions given in Theorem {\ref{lem:rat-images}} apparently do not completely characterize the rationals that are images
of rationals. In particular, for $b=2$ we suspect that among $\frac{2k}{9},\ k = 1,2,4,5,7,8$,
only $\frac{8}{9}= \rho_2(\frac{1}{3})$ and $\frac{10}{9}=\rho_2(\frac{2}{3})$ have  rational inverse images.


In Theorem \ref{thm:mad_like} we derive an  expression for $\rho_b(\frac{a}{b^r})$ analogous to one for the Takagi function 
given by Maddock \cite{zm}. 
If the base-$b$ expansion of the positive integer $a$ is given by $ a=\sum_{i=0}^{m-1} e_ib^i$, 
define $\sigma_b(a)$ by
\begin{equation*}\label{eq:sigma def}
 \sigma_b(a) = \sum_{i=0}^{m-1} ie_ib^i.
\end{equation*}

It is easy to check that $\sigma_b(a)$ can be written in a way  that does not
specifically involve the base-$b$ expansion:
\begin{equation*}
 \sigma_b(a) = \sum_{j\geq1} b^j\left\lfloor \frac{a}{b^j} \right\rfloor = \sum_{1\leq b^j \leq a} 
(a - (a\bmod b^j)). 
\end{equation*}
 The $\sigma_b$ functions are related to several sequences in Sloane's OEIS database.  Specifically,
sequence
\seqnum{A080277} is $a+\sigma_2(a)=\sum_{j\geq0} 2^j\lfloor\frac{a}{2^j}\rfloor$, 
while \seqnum{A080333} is
$a+\sigma_3(a)=\sum_{j\geq0}3^j\lfloor\frac{a}{3^j}\rfloor$. 
Also, the sums  $s_a= \sum_{1\leq b^j \leq a} (a\bmod b^j)$ 
appear  in OEIS for $b=2$ and $b=3$ as \seqnum{A049802} and \seqnum{A049803} respectively.

\begin{theorem}\label{thm:mad_like}
 For the $b$-adic rational $\frac{a}{b^r}$, where $0\leq a &lt; b^r$, we have
\begin{equation*}\label{eq:rho_sigma}
 \rho_b\left(\frac{a}{b^r}\right) = \frac{ra-\sigma_b(a)}{b^r}.
\end{equation*}
\end{theorem}
\begin{proof}
 Let the base-$b$ expansion of $a$ be  $a=\sum_{i=0}^{r-1} e_ib^i$. We then have
\begin{align*}
 \rho_b\left(\frac{a}{b^r}\right) &amp;= \rho_b\left( \sum_{i=0}^{r-1} \frac{e_i}{b^{r-i}} \right)
                       = \sum_{i=0}^{r-1} \frac{(r-i)e_i}{b^{r-i}},\\
   &amp;= \frac{1}{b^r}\left[ \sum_{i=0}^{r-1} re_ib^i -\sum_{i=0}^{r-1} ie_ib^i \right],\\
   &amp;= \frac{1}{b^r}\left[ra - \sigma_b(a)   \right].
\end{align*}
\end{proof}

\begin{theorem}\label{thm:sigma-exp} Consider the rational number $s/t$ in reduced form with $t$ relatively prime to $b$. 
Let $r={\textnormal{ord}}_t(b)$, $ct=b^r-1$, and $a=cs$.
Then,
\begin{equation*}
 \rho_b\left(\frac{s}{t}\right)=\rho_b\left(\frac{a}{b^r-1}\right)=\frac{rb^ra}{(b^r-1)^2} - \frac{\sigma_b(a)}{b^r-1}.
\end{equation*}
\end{theorem}
\begin{proof}
 Given the  base-$b$ expansion $a=\sum_{i=0}^{r-1} e_i b^i$, we have
\[
 \frac{a}{b^r-1}=\sum_{k\geq1} \sum_{i=0}^{r-1} \frac{e_i}{b^{rk-i}}.
\]
Hence, we calculate
\begin{align*}
 \rho_b\left(\frac{s}{t}\right) = \rho_b\left(\frac{a}{b^r-1}\right) &amp;= \sum_{k\geq1} \sum_{i=0}^{r-1} (rk-i)\frac{e_i}{b^{rk-i}},\\
   &amp;= \sum_{k\geq1} \frac{1}{b^{rk}}\left[\sum_{i=0}^{r-1} rk e_i b^i - \sum_{i=0}^{r-1} i e_i b^i  \right],\\
  &amp;= \sum_{k\geq1}\frac{1}{b^{rk}}\left[ kra - \sigma_b(a)  \right],\\
  &amp;= \frac{rb^ra}{(b^r-1)^2} - \frac{\sigma_b(a)}{b^r-1}.
\end{align*}
\end{proof}

Theorem \ref{thm:sigma-exp}  leads to a relation between two values of $\rho_b$. With $s,t,r,a$ as in the
proof of that  theorem, we see

\begin{align*}
 \rho_b\left(\frac{s}{t}\right) = \rho_b\left(\frac{a}{b^r-1}\right) 
     &amp;= \frac{rb^ra}{(b^r-1)^2} - \frac{\sigma_b(a)}{b^r-1}\\
     &amp;= \frac{rb^ra-(b^r-1)\sigma_b\left(a\right)}{(b^r-1)^2}\\
     &amp;= \frac{ra+(b^r-1)\left(ra-\sigma_b(a)\right)}{(b^r-1)^2}\\
     &amp;= \frac{ra}{(b^r-1)^2}+ \frac{b^r}{b^r-1}\rho_b\left(\frac{a}{b^r}\right).
\end{align*}



\section{The Connection Between \texorpdfstring{$\rho_b$}{rho} and Rumor Sequences}

In Dearden and Metzger \cite{dm}, rumor sequences 
({\bf\underbar{ru}}nning {\bf\underbar{mo}}dulus {\bf\underbar{r}}ecursive  sequences) 
were introduced as follows:

Let $b\geq 2$ and $k\geq 1$ be integers. To construct an {\it (integer) rumor sequence}
select an integer $z_0$, and for $n\geq 1$ let $z_n = bz_{n-1} \bmod{(n+k)}$, where the right 
side is the least non-negative residue of $bz_{n-1}$ modulo  $n+k$. The rumor sequence
conjecture asserts that all such integer rumor sequences are eventually $0$. Since the
conjecture concerns only the eventual behavior of such sequences and since $0\leq z_1&lt; k+1$,
nothing is lost by restricting $z_0$ to the interval $[0,k)$.

To establish a connection between the rumor sequence conjecture and the $\rho_b$ function, 
it is convenient to generalize the notion of integer rumor sequences to {\it real} rumor sequences. 

Let $b\geq 2$ and $k\geq 1$ be integers. To construct a {\it (real) rumor sequence},
select any real number $x_0$ and for $n\geq 1$ let $x_n = bx_{n-1} \bmod{(n+k)}$ where 
the right hand side is taken to be 
\begin{equation}\label{eq:real_rumor}
bx_{n-1} - (n+k)\left\lfloor \frac{bx_{n-1}}{n+k} \right\rfloor.
\end{equation}
As with integer rumors, there is no loss if $x_0$ is restricted to the interval $[0,k)$. The 
real and integer rumors are identical when $x_0 = z_0$ is an integer.

It will be shown that the rumor conjecture for integer rumor sequences is true if and only if
the greedy inverse image under $\rho_b$ of every $b${-}adic rational is a $b${-}adic rational.
It is worth noting that, in general, not all inverse images of a $b${-}adic under $\rho_b$ 
need be $b${-}adic.

\begin{example}
Consider the $3${-}adic rational $y = \frac{2}{3}$ in the range of $\rho_3$. 
With $b=3$, let the greedy $\rho_3$ inverse image of $\frac{5}{6}$ be $x$. 
Since $6$ is not divisible by a square greater than $1$, $x$ must be irrational.
It follows that  $1-x$ is irrational and, by Theorem \ref{thm: fun-equs}(b), we see
\[
\rho_3(1-x)= \frac{3}{2}-\frac{5}{6} = \frac{2}{3}.
\] 
\end{example}

\begin{theorem}\label{thm:sync=rho} For $b\geq 2$,  all integer rumor sequences are eventually $0$ if and only if the greedy 
inverse image under $\rho_b$ of every $b${-}adic is $b${-}adic.
\end{theorem}

\begin{proof}
Suppose that all integer rumor sequences are eventually zero, and let $y=a/b^m$ be a $b$-adic rational
in $\left[0,b/(b-1)\right)$.  By Corollary \ref{cor:greedy=rumor}, there is an integer $n$ so that for
$k\geq n$ we have
\[
 d_k=\left\lfloor \frac{b^k\left( y-\sum_{j=1}^{k-1} jd_j/b^j \right)}{k} \right\rfloor 
\text{ and }
 0 \leq y-\sum_{j=1}^{k} \frac{jd_j}{b^j} &lt; \frac{k}{b^k}.
\]
Now, consider the real rumor sequence with initial value $x_0\in[0,n)$ given by
\[
 x_0 = b^n\left( y-\sum_{j=1}^n \frac{jd_j}{b^j} \right).
\]
Applying the rumor recursion \eqref{eq:real_rumor}, we have
\begin{align*}
 x_1 &amp;= bx_0 -(n+1)\left\lfloor \frac{b x_0}{n+1} \right\rfloor \\
     &amp;= b^{n+1}\left( y -\sum_{j=1}^{n} \frac{j d_j}{b^j} \right) 
         - (n+1) \left\lfloor \frac{b^{n+1}\left( y - \sum_{j=1}^{n} j d_j /b^j\right)}{n+1} \right\rfloor\\
     &amp;= b^{n+1}\left( y - \sum_{j=1}^{n} \frac{j d_j}{b^j} \right) - (n+1)d_{n+1}, \text{ by Corollary \ref{cor:greedy=rumor}}\\
     &amp;= b^{n+1}\left( y - \sum_{j=1}^{n+1} \frac{j d_j}{b^j} \right).
\end{align*}
More generally, induction shows that, for all $i\geq0$, we have
\[
 x_i = b^{n+i}\left( y -\sum_{j=1}^{n+i} \frac{j d_j}{b^j}   \right)
        = b^{n+i}\left( \frac{a}{b^m} -\sum_{j=1}^{n+i} \frac{j d_j}{b^j}   \right).
\]
Now, for $i\geq m-n$, the sequence $x_i$ is obtained from an integer rumor recursion, and by our assumption
that integer rumor sequence is eventually zero, say from term $i_0$ on.  That means the 
greedy inverse image under  $\rho_b$ of 
the $b$-adic rational $a/b^m = \sum_{j=1}^{n+i_0} j d_j/b^j$ is
the $b$-adic rational
\[
 \upsilon = \sum_{j=1}^{n+i_0} \frac{d_j}{b^j} = \frac{\sum_{j=1}^{n+i_0} d_j b^{n+i_0-j}}{b^{n+i_0}}.
\]


Conversely, suppose that the greedy inverse image of  $b$-adic rationals in $\left[0,b/(b-1)\right)$ 
are $b$-adic rationals. Consider an integer rumor recursion with initial value $z_0$  in $[0,k)$.
By our assumption the greedy inverse of the $b$-adic rational $y=z_0/b^k$ is a $b$-adic rational
$\sum_{j=1}^n j/b^j$, where 
\[
 y=\sum_{j=1}^n \frac{j d_j}{b^j}, \text{ with } d_j\in\{0,1,\dots,b-1\}.
\]
Since $f(x) = x/b^x$ is a nondecreasing function on positive integers for all integers $b\geq 2$, we have
$z_0/b^k &lt; k/b^k \leq m/b^m$ for all $m=1,2,3,\dots,k$.  Therefore, it follows that
\[
 0 \leq \frac{z_0}{b^k} -\sum_{j=1}^{m-1} \frac{j d_j}{b^j}&lt; \frac{m}{b^m}, \text{ for $m=1,2,\dots,k$ }.
\]
Hence, 
\[
 d_j = 0 \text{ for $j=1,2,\dots,k$}.
\]
It follows that 
\begin{equation}\label{eq:z_0/b^k}
 \frac{z_0}{b^k}=y = \sum_{j=k+1}^{n-k} \frac{(k+i)d_{k+i}}{b^{k+i}}.
\end{equation}

Moreover, for all $m=1,2,\dots,n-k$, we have
\[
 \frac{(k+m) d_{k+m}}{b^{k+m}} \leq \frac{z_0}{b^k} - \sum_{i=1}^{m-1} \frac{(k+i) d_{k+i}}{b^{k+i}}
   &lt; \frac{(k+m)(d_{k+m} +1)}{b^{k+m}}.
\]
Multiplying through by $b^k$ gives
\[
 \frac{(k+m) d_{k+m}}{b^{m}} \leq z_0 - \sum_{i=1}^{m-1} \frac{(k+i) d_{k+i}}{b^{i}}
   &lt; \frac{(k+m)(d_{k+m} +1)}{b^{m}}.
\]
In particular, for $m=1$ we have
\[
 \frac{(k+1)d_{k+1}}{b} \leq z_0  &lt; \frac{(k+1)d_{k+1}}{b}
\]
or
\[
    d_{k+1} \leq \left\lfloor \frac{b z_0}{k+1} \right\rfloor &lt; d_{k+1}+1.
\]
It follows that
\[
 z_1 = b z_0 -(k+1)\left\lfloor \frac{b z_0}{k+1} \right\rfloor = b z_0 - (k+1)d_{k+1}.
\]
Hence,
\[
 \frac{z_1}{b} = z_0 -\frac{(k+1)d_{k+1}}{b}.
\]
In general, induction shows that, for all $m\geq 1$,
\[
 \frac{z_m}{b^m}= z_0 - \sum_{i=1}^{m} \frac{(k+i)d_{k+i}}{b^i}.
\]
Therefore, by equation \eqref{eq:z_0/b^k}, we have
\[
 \frac{z_{n-k}}{b^{n-k}} = z_0 - \sum_{i=1}^{n-k} \frac{(k+i)d_{k+i}}{b^i} = 0.
\]
Thus, any integer rumor sequence is eventually zero.

\end{proof} 


The following corollary follows immediately from the proof of Theorem \ref{thm:sync=rho}.

\begin{corollary}
Let $b\geq 2$ be an integer. The integer rumor sequence with initial term $z_0$, where $0\leq z_0 &lt; k$,
is eventually $0$ if and only if the greedy inverse image of $\frac{z_0}{b^k}$ under $\rho_b$ is $b${-}adic.
\end{corollary}
 

\begin{conjecture}
The greedy inverse image of every $b${-}adic under $\rho_b$ is $b${-}adic.
\end{conjecture} 


\section{Acknowledgments}
The authors would like to thank the referee for the careful reading and helpful suggestions which improved the paper.


\begin{thebibliography}{bib1}

\bibitem{dm} B.~Dearden and J.~Metzger, Running modulus recursions,  
{\em J. Integer Seq.} {\bf 13} (1) (2010), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL13/Dearden/dearden3.html}{Article 10.1.6}.


\bibitem{zm} Z. Maddock,
Level sets of the Takagi function: Hausdorff dimension,
{\it Monatshefte f\"ur Math.},
\textbf{160} (2010), 167--186.


\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B37; Secondary  11A50.

\noindent \emph{Keywords: } recurrence sequence, recurrence relation modulo $m$, 
running modulus recursion, Takagi function

\bigskip
\hrule
\bigskip

\noindent (Related to sequences
\seqnum{A049802},
\seqnum{A049803},
\seqnum{A080277},
\seqnum{A080333}, and
\seqnum{A177356}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 1 2010;
revised version received  February 1 2011;
Published in {\it Journal of Integer Sequences}, February 19 2011.

\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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