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A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\newcommand{\dist}{\text{dist}}\dist(A) = \{b-a:a < b \in A\}$. ${}{}{}$

Is there a Golomb set $A\subseteq \N$ with $\dist(A) = \N\setminus\{0\}$? If not, is there a Golomb set $A$ with $$\liminf_{n\to\infty}\frac{|\dist(A)\cap\{1,\ldots,n\}|}{n} = 1?$$

A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\newcommand{\dist}{\text{dist}}\dist(A) = \{b-a:a < b \in A\}$.

Is there a Golomb set $A\subseteq \N$ with $\dist(A) = \N\setminus\{0\}$? If not, is there a Golomb set $A$ with $$\liminf_{n\to\infty}\frac{|\dist(A)\cap\{1,\ldots,n\}|}{n} = 1?$$

A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\newcommand{\dist}{\text{dist}}\dist(A) = \{b-a:a < b \in A\}$.

Is there a Golomb set $A\subseteq \N$ with $\dist(A) = \N\setminus\{0\}$? If not, is there a Golomb set $A$ with $$\lim\inf_{n\to\infty}\frac{|\dist(A)\cap\{1,\ldots,n+1\}|}{n+1} = 1?$$

A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\newcommand{\dist}{\text{dist}}\dist(A) = \{b-a:a < b \in A\}$. ${}{}{}$

Is there a Golomb set $A\subseteq \N$ with $\dist(A) = \N\setminus\{0\}$? If not, is there a Golomb set $A$ with $$\liminf_{n\to\infty}\frac{|\dist(A)\cap\{1,\ldots,n\}|}{n} = 1?$$

A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\newcommand{\dist}{\text{dist}}\dist(A) = \{b-a:a < b \in A\}$.

Is there a Golomb set $A\subseteq \N$ with $\dist(A) = \N\setminus\{0\}$? If not, is there a Golomb set $A$ with $$\lim\inf_{n\to\infty}\frac{|\dist(A)\cap\{1,\ldots,n+1\}|}{n+1} = 1?$$

A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\newcommand{\dist}{\text{dist}}\dist(A) = \{b-a:a < b \in A\}$.

Is there a Golomb set $A\subseteq \N$ with $\dist(A) = \N\setminus\{0\}$? If not, is there a Golomb set $A$ with $$\lim\inf_{n\to\infty}\frac{|\dist(A)\cap\{1,\ldots,n+1\}|}{n+1} = 1?$$