Are there an infinite number of primes of the form $\lfloor \pi n \rfloor$? Are there an infinite number of primes 
of the form
$\lfloor \pi n \rfloor$?
This is sort of
a clickbait title.
I would really like to show that,
for any real irrational
$r > 1$,
there are an infinite number of primes
of the form
$\lfloor r n \rfloor$
for positive integer $n$.
I can show that
there are an
infinite number of primes
of the form
$\lfloor r n \rfloor$
for
$r < 1+1/g$
for a fixed $g > 1$,
but the best known value
of $g$
is $246$ unconditionally
and
$6$ assuming
an unproved conjecture.
To prove this,
I use the idea
of Beatty sequences
(https://en.wikipedia.org/wiki/Beatty_sequence).
For a real $r > 1$,
let
$B(r)
=\{\lfloor nr \rfloor
\mid n \in \mathbb{N}^+\}$.
($\mathbb{N}^+$
is the set of positive integers.)
Then
Beatty's theorem states that
$B(r)$ and
$B(r/(r-1))$
make up a
disjoint partition
of $\mathbb{N}^+$.
If there are
only a finite number of primes
in $B(r)$,
then all the primes
above a certain value
are in
$B(r/(r-1))$.
We have
$\lfloor (n+1)r \rfloor
=\lfloor nr+r \rfloor
\ge \lfloor nr \rfloor +\lfloor r \rfloor
$.
Therefore,
if $r > 3$,
$B(r)$
can not contain
any twin primes.
Therefore,
if there are an
infinite number of twin primes,
$B(r/(r-1))$
must contain
an infinite number
of primes
for all 
$r > 3$,
or
$r/(r-1)
\lt 3/2
$.
The use of
an unproved conjecture
can be removed
at the cost of
a weaker conclusion
by using the recent results
on prime gaps
(https://en.wikipedia.org/wiki/Prime_gap).
It has been shown that
there is a constant $g$
such that
there are an
infinite number of
consecutive primes
that differ by
at most $g$.
It has been shown that
$g \le 246$
unconditionally
and that,
assuming the
Elliott–Halberstam conjecture
(https://en.wikipedia.org/wiki/Elliott%E2%80%93Halberstam_conjecture),
$g \le 12$
($g \le 6$
assuming a generalized form
of the conjecture.).
Arguing as before,
if $r > g+1$,
then
$B(r)$
can not contain
all the primes
above any finite value.
For,
if it did,
there are an
infinite numer of
consecutive primes
$p$ and $q$
such that
$q-p \le g$,
and $p$ and $q$
can not both be in
$B(r)$.
Therefore
$B(r/(r-1))$
must contain an
infinite number of primes
for $r > g+1$.
Restating,
$B(r)$
must contain an
infinite number of primes
for $r < (g+1)/g
=1+1/g$.
I don't know how to go
beyond this.
 A: 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 $\mu(r)>2$.
A: 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 $(1-\frac{1}{\pi},1)$
