Are there an infinite number of primes of the form $\lfloor \pi n \rfloor$?

Solution 1:

There is a conjecture on the least prime in an arithmetic progression that would imply the answer to your question is yes. For coprime positive integers $a$, $d$, write $p(a,d)$ for the smallest prime congruent to $a$ modulo $d$.

Conjecture 1: For every $\epsilon>0$, the bound $p(a,d) =O_\epsilon(d^{1+\epsilon})$ holds for all $a$, $d$.

The statement is known to hold if $1+\epsilon$ is replaced by $5$, and if we assume GRH it is know to hold with $1+\epsilon$ replaced by $2+\epsilon$. The conjecture and related results are discussed here.

Proof that Conjecture $1$ implies an affirmative answer to your question:. There are infinitely many pairs of coprime positive integers $p$, $q$ such that $$ 0<\frac{p}{q}-r< \frac{1}{q^2}. $$ For such $p$, $q$, the integers $$ \lfloor qr\rfloor,\lfloor 2qr\rfloor,\ldots,\lfloor q^2r\rfloor $$ form an arithmetic progression with initial term $p-1$ and common difference $p$. The largest term is about $p^2/r$, so once $p$ is sufficiently large, one of these terms must be prime.

Unconditionally, this argument shows that there are infinitely many primes of the form $\lfloor rn\rfloor$ if the irrationality measure $\mu(r)$ satisfies $\mu(r)> 5$, and on GRH it is sufficient that $\mu(r)>2$.

Solution 2:

Current mathematics can prove your result unconditionally: If a prime $p$ satisfies $\{\frac{1}{\pi} p \} > 1-\frac{1}{\pi}$, then it is of the form $\lfloor n \pi \rfloor$ (in fact on can check that this condition on the fractional part is equivalent to $p$ being of the form $\lfloor \pi n \rfloor$). Indeed, if $\{\frac{1}{\pi} p \} > 1-\frac{1}{\pi}$ set $n = \lceil \frac{p}{\pi} \rceil$.

Then $\pi n \ge \pi \frac{p}{\pi} = p$

While $\pi n < \pi (\frac{p}{\pi}+\frac{1}{\pi}) = p+1$

Hence $\lfloor \pi n \rfloor = p$, so $p$ is of this form.

However it is known that for any irrational $\alpha, \{ \alpha p \}$ is dense in $[0,1]$ (in fact equidistributed) as $p$ varies over the primes. See for instance this MO question.

So not only we can prove that there are infinitely many primes of the form $\lfloor \pi n \rfloor$, but (as expected!) a proportion $\frac{1}{\pi}$ of primes are of this form, since a proportion $\frac{1}{\pi}$ of the fractional parts of $\{\pi p\}$ will lie in the interval $(1-\frac{1}{\pi},1)$