Are there infinitely many primes of the form $\lfloor e x\rfloor$ for $x\in\mathbb{Z}^+$?

Question

Construct a function $f(x)=\lfloor e x\rfloor$. For each positive integer $x$, $f(x)$ will be a positive integer. Among these integers $f(x)$, are there an infinite number of primes?

Answer 1

Something is known in considerably greater generality. Let $\lVert x\rVert=\min_{n\in\mathbb{Z}}|x-n|$. For $\gamma\in\mathbb{R}-\mathbb{Q}$, define

$$\tau(\gamma) = \sup\Big\{\rho\in\mathbb{R}\colon \liminf_{n\to\infty} n^{\rho}\lVert\gamma n\rVert=0\Big\},$$

where the limit inferior is as $n\to\infty$ along the integers. Dirichlet's approximation theorem implies that for all irrational $\gamma$, we have $\tau(\gamma)\geq 1$. Khinchin and Roth proved that $\tau(\gamma)=1$ for almost all real numbers $\gamma$ (in the sense of Lebesgue measure), and $\tau(\gamma)=1$ for all (real) irrational algebraic integers $\gamma$.

If $\gamma>1$ and $\tau(\gamma)<\infty$, then there is an asymptotic prime number theorem that counts the primes of the form $p=\lfloor \gamma n+\beta\rfloor$, where $n\geq 1$ is an integer and $\beta\in\mathbb{R}$ is fixed. A standard source for this is the paper Prime numbers with Beatty sequences by Banks and Shparlinski.