Infinitely many primes of the form $\lfloor \sqrt {3} \cdot n \rfloor $?

Solution 1:

This is too long to be a comment, so I will write it as an answer.

Given $n\in\mathbb{N}$, define the interval $$ I_n=\Bigl[\frac{n}{\sqrt3},\frac{n+1}{\sqrt3}\Bigr]. $$ If $I_n$ contains an integer, then $$ n=\Bigl\lfloor\sqrt3\,\Bigl\lceil\frac{n}{\sqrt3}\Bigr\rceil\Bigr\rfloor. $$ If we could prove that there are infinitely many primes $p$ such that $I_p\cap\mathbb{N}\ne\emptyset$, then we could answer the question in the affirmative. I wil give a probabilistic argument, assuming that the fractional parts of $p/\sqrt3$, where $p$ runs over all primes, are uniformly distributed in $[0,1]$. In that case, the probability that $I_p\cap\mathbb{N}\ne\emptyset$ would be the width of the interval $I_p$, which is $1/\sqrt3=0.577\dots$ Thus, about $57\%$ of all primes should satisfy $I_p\cap\mathbb{N}\ne\emptyset$. Computation shows that $576874$ of the first $10^6$ primes verify the condition.

Edit

According to Aryabhata's comment, $\bigl\{\,\{p\,\sqrt3\,\bigr\}: p \text{ is prime}\}$ is uniformly distributed in $[0,1]$. Then there are an infinite number of primes $p$ such that $I_p\cap\mathbb{N}\ne\emptyset$ and $$ p=\Bigl\lfloor\sqrt3\,\Bigl\lceil\frac{p}{\sqrt3}\Bigr\rceil\Bigr\rfloor. $$

The same argument shows that given an irrational $\alpha>0$ there are infinite primes of the form $\lfloor\alpha\,n\rfloor$.