Infinitely many primes of the form $\lfloor \sqrt {3} \cdot n \rfloor $? How to prove or disprove following statement :

There are infinitely many primes of the form : $\lfloor \sqrt {3} \cdot n \rfloor $ 

Note: This is a problem I made myself.
There is a theorem that states :
$\lfloor nx \rfloor =
\begin{cases}
 n\lfloor x \rfloor, & \text{if } 0 \leq \{x\} < \frac{1}{n} \\
 n\lfloor x \rfloor +1, & \text{if } \frac{1}{n} \leq \{x\} < \frac{2}{n} \\
 n\lfloor x \rfloor +2, & \text{if } \frac{2}{n} \leq \{x\} < \frac{3}{n} \\
 \vdots \\
 n\lfloor x \rfloor +n-1, & \text{if } \frac{n-1}{n} \leq \{x\} < 1 \\
\end{cases}$
where $\{x\}$ is a non-integer part of $x$ .
Hence :
$\lfloor \sqrt{3}\cdot n \rfloor =
\begin{cases}
 n, & \text{if } 0 \leq \sqrt{3}-1 < \frac{1}{n} \\
 n +1, & \text{if } \frac{1}{n} \leq \sqrt{3}-1 < \frac{2}{n} \\
 n +2, & \text{if } \frac{2}{n} \leq \sqrt{3}-1 < \frac{3}{n} \\
 \vdots \\
 2n-1, & \text{if } \frac{n-1}{n} \leq \sqrt{3}-1 < 1 \\
\end{cases}$
How can I proceed from here ?
 A: 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 $\lfloor\alpha\,n\rfloor$.
