# 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.

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 $\mu(r)>2$.

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 $(1-\frac{1}{\pi},1)$