On the distribution of the Cantor-integers

Abstract

For any positive integer $p\ge 3$, let $A$ be a proper subset of $\{0,1,\dots ,p-1\}$ with $♯A=s\ge 2$. Suppose $h:\{0,1,\dots ,s-1\}\to A$ is a one-to-one map which is strictly increasing with $A=\{h(0),h(1),\dots ,h(s-1)\}$. We focus on so-called Cantor-integers $\{a_n{\}}_{n\ge 1}$, which consist of these positive integers $n$ such that all the digits in the $p$-ary expansion of $n$ belong to $A$. Let $\mathfrak{C}=\left\{\sum _{n\ge 1}\frac{{\epsilon }_n}{p^n}:{\epsilon }_n\in A\text{ for any positive integer }n\right\}$ be the appropriate Cantor set, and denote the classic self-similar measure supported on $\mathfrak{C}$ by ${\mu }_{\mathfrak{C}}$. Now that $n^{{\log }_{s}p}$ is the growth order of $a_n$ and ${\left\{\frac{a_n}{n^{{\log }_{s}p}}:n\ge 1\right\}}^{\prime }$ is precisely the set $\left\{\frac{x}{({\mu }_{\mathfrak{C}}([0,x]){)}^{{\log }_{s}p}}:x\in \mathfrak{C}\cap [\frac{h(1)}{p},1]\right\}$, where $E^{\prime }$ is the set of limit points of $E$, we show that ${\left\{\frac{a_n}{n^{{\log }_{s}p}}:n\ge 1\right\}}^{\prime }$ is just an interval $[m,M]$ with $m:=\inf \left\{\frac{a_n}{n^{{\log }_{s}p}}:n\ge 1\right\}$ and $M:=\sup \left\{\frac{a_n}{n^{{\log }_{s}p}}:n\ge 1\right\}$. In particular, $\left\{\frac{x}{({\mu }_{\mathfrak{C}}([0,x]){)}^{{\log }_{s}p}}:x\in \mathfrak{C}\backslash \{0\}\right\}=[m,M]$ if $0\in A$, and $m=\frac{q(s-1)+r}{p-1},M=\frac{q(p-1)+pr}{p-1}$ if the set $A$ consists of all the integers in $\{0,1,\dots ,p-1\}$ which have the same remainder $r\in \{0,1,\dots ,q-1\}$ modulus $q$ for some positive integer $q\ge 2$ (i.e. $h(x)=qx+r$). We further show that the sequence ${\left\{\frac{a_n}{n^{{\log }_{s}p}}\right\}}_{n\ge 1}$ is not uniformly distributed modulo 1, and it does not have the cumulative distribution function, but has the logarithmic distribution function (give by a specific Lebesgue integral).