Generalized lattice-point visibility in $\mathbb{N}^k$

Abstract

A lattice point $\left(r,s\right)\in {\mathbb{N}}^2$ is said to be visible from the origin if no other integer lattice point lies on the line segment joining the origin and $\left(r,s\right)$. It is a well-known result that the proportion of lattice points visible from the origin is given by $1∕\zeta \left(2\right)$, where $\zeta \left(s\right)={\sum }_{n=1}^{\infty }1∕n^s$ denotes the Riemann zeta function. Goins, Harris, Kubik and Mbirika generalized the notion of lattice-point visibility by saying that for a fixed $b\in \mathbb{N}$ a lattice point $\left(r,s\right)\in {\mathbb{N}}^2$ is $b$-visible from the origin if no other lattice point lies on the graph of a function $f\left(x\right)=m{x}^b$, for some $m\in \mathbb{Q}$, between the origin and $\left(r,s\right)$. In their analysis they establish that for a fixed $b\in \mathbb{N}$ the proportion of $b$-visible lattice points is $1∕\zeta \left(b+1\right)$, which generalizes the result in the classical lattice-point visibility setting. In this paper we give an $n$-dimensional notion of $b$-visibility that recovers the one presented by Goins et. al. in two dimensions, and the classical notion in $n$ dimensions. We prove that for a fixed $b=\left(b_1,b_2,\dots ,b_n\right)\in {\mathbb{N}}^n$ the proportion of $b$-visible lattice points is given by $1∕\zeta \left({\sum }_{i=1}^n{b}_i\right)$.

Moreover, we give a new notion of $b$-visibility for vectors

$$b=\left(b_1∕a_1,b_2∕a_2,\dots ,b_n∕a_n\right)\in {\left(\mathbb{Q}\setminus \left\{0\right\}\right)}^n,$$

with nonzero rational entries. In this case, our main result establishes that the proportion of $b$-visible points is $1∕\zeta \left({\sum }_{i\in J}\left|b_i\right|\right)$, where $J$ is the set of the indices $1\le i\le n$ for which $b_i∕a_i<0$. This result recovers a main theorem of Harris and Omar for $b\in \mathbb{Q}\setminus \left\{0\right\}$ in two dimensions, while showing that the proportion of $b$-visible points (in such cases) only depends on the negative entries of $b$.