\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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 $m>0$, 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 $\tau(x) = \sum_{n=0}^\infty \frac{{<\mskip-10mu<}2^n x{>\mskip -10mu>}}{2^n}$, where ${<\mskip -10mu<}t{>\mskip -10mu>}$ 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 $y={<\mskip -10mu<}2^n x{>\mskip -10mu>}$ 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{$y={<\mskip -10mu<}2^2 x{>\mskip -10mu>}$} \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) & = \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}\\ &= \sum_{n=0}^\infty \frac{1}{b^n} \left\{\mskip -8mu\left\{\sum_{j=1}^n d_j b^{n-j} + \sum_{j>n} d_jb^{n-j} \right\}\mskip -8mu\right\} = \sum_{n=0}^\infty \frac{1}{b^n} \sum_{j>n} d_j b^{n-j}\\ &= \sum_{n=0}^\infty \sum_{j>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 $\sum_{j=1}^{\infty}\frac{j d_j}{b^j}\leq y < \frac{b}{b-1}$. So there must be a last digit, $d_L$, that is less than $b-1$. It follows that for all $m>L$, \[ \frac{m(b-1)}{b^m} \leq y - \sum_{j=1}^{L-1} \frac{j d_j}{b^j} - \frac{L d_L} {b^L} - \sum_{j=L+1}^{m-1} \frac{j(b-1)}{b^j}. \] Hence, for all $m>L$, 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 \[ \sum_{j=L+1}^{\infty} \frac{j(b-1)}{b^j} = \frac{(L+1)b-L}{b^L (b-1)}>\frac{L}{b^L}. \] 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 $d_i1$ 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) &= \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}}\\ &= \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)\\ &= \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)\\ &= \sum_{m\geq0}\frac{1}{(b^r)^{m+1}}\left(mrcs + w\right),\text{where $w=\sum_{1\leq i \leq r}ib^{r-i}d_i$}\\ &= \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}\\ &= \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 \[ r\leq\varphi(t) < t, \text{ hence, } 1\leq r' < t'. \] 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) &= \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}} \\ &=\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 $0\leq a < b^r$, 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) &= \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}},\\ &= \frac{1}{b^r}\left[ \sum_{i=0}^{r-1} re_ib^i -\sum_{i=0}^{r-1} ie_ib^i \right],\\ &= \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) &= \sum_{k\geq1} \sum_{i=0}^{r-1} (rk-i)\frac{e_i}{b^{rk-i}},\\ &= \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],\\ &= \sum_{k\geq1}\frac{1}{b^{rk}}\left[ kra - \sigma_b(a) \right],\\ &= \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) &= \frac{rb^ra}{(b^r-1)^2} - \frac{\sigma_b(a)}{b^r-1}\\ &= \frac{rb^ra-(b^r-1)\sigma_b\left(a\right)}{(b^r-1)^2}\\ &= \frac{ra+(b^r-1)\left(ra-\sigma_b(a)\right)}{(b^r-1)^2}\\ &= \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 $0\leq z_1< k+1$, 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 \[ d_k=\left\lfloor \frac{b^k\left( y-\sum_{j=1}^{k-1} jd_j/b^j \right)}{k} \right\rfloor \text{ and } 0 \leq y-\sum_{j=1}^{k} \frac{jd_j}{b^j} < \frac{k}{b^k}. \] 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 &= bx_0 -(n+1)\left\lfloor \frac{b x_0}{n+1} \right\rfloor \\ &= 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\\ &= 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}}\\ &= 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 $z_0/b^k < k/b^k \leq m/b^m$ for all $m=1,2,3,\dots,k$. Therefore, it follows that \[ 0 \leq \frac{z_0}{b^k} -\sum_{j=1}^{m-1} \frac{j d_j}{b^j}< \frac{m}{b^m}, \text{ for $m=1,2,\dots,k$ }. \] 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 \[ \frac{(k+m) d_{k+m}}{b^{k+m}} \leq \frac{z_0}{b^k} - \sum_{i=1}^{m-1} \frac{(k+i) d_{k+i}}{b^{k+i}} < \frac{(k+m)(d_{k+m} +1)}{b^{k+m}}. \] Multiplying through by $b^k$ gives \[ \frac{(k+m) d_{k+m}}{b^{m}} \leq z_0 - \sum_{i=1}^{m-1} \frac{(k+i) d_{k+i}}{b^{i}} < \frac{(k+m)(d_{k+m} +1)}{b^{m}}. \] In particular, for $m=1$ we have \[ \frac{(k+1)d_{k+1}}{b} \leq z_0 < \frac{(k+1)d_{k+1}}{b} \] or \[ d_{k+1} \leq \left\lfloor \frac{b z_0}{k+1} \right\rfloor < d_{k+1}+1. \] 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 $0\leq z_0 < k$, 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 \end{document}