Mills' Theorem

Theorem
There exists a real number $A$ such that $\left\lfloor{A^{3^n} }\right\rfloor$ is a prime number for all $n \in \N_{>0}$, where:


 * $\left\lfloor{x}\right\rfloor$ denotes the floor function of $x$
 * $\N$ denotes the set of all natural numbers.

Proof
We define $f \left({x}\right)$ as a prime-representing function :
 * $\forall x \in \N: f \left({x}\right) \in \Bbb P$

where:
 * $\N$ denotes the set of all natural numbers
 * $\Bbb P$ denotes the set of all prime numbers.

Let $p_n$ be the $n$th prime number.

From Difference between Consecutive Primes:
 * $p_{n+1} - p_n < K p_n^{5/8}$

where $K$ is an unknown but fixed positive integer.

Lemma 1

 * $\forall N > K^8 \in \Z: \exists p \in \Bbb P: N^3 < p < \left({N + 1}\right)^3 - 1$

Proof
Let $p_n$ be the greatest prime less than $N^3$.

Therefore:
 * $N^3 < p_{n+1} < \left({N + 1}\right)^3 - 1$

Let $P_0 > K^8$ be a prime number.

By Lemma 1, there exists an infinite sequence of primes:
 * $P_0, P_1, P_2, \ldots$

such that:
 * $\forall n \in \N_{>0}: {P_n}^3 < P_{n+1} < \left({P_n + 1}\right)^3 - 1$

Then we define two functions $u, v: \N \to \Bbb P$:
 * $\forall n \in \N: u \left({n}\right) = {P_n}^{3^{-n} }$
 * $\forall n \in \N: v \left({n}\right) = \left({P_n + 1}\right)^{3^{-n} }$

It is trivial that $v \left({n}\right) > u \left({n}\right)$.

Lemma 2

 * $\forall n \in \N_{>0}: u \left({n + 1}\right) > u \left({n}\right)$

Lemma 3

 * $\forall n \in \N_{>0}: v \left({n + 1}\right) < v \left({n}\right)$

Proof
It follows trivially that $u \left({n}\right)$ is bounded and strictly monotone.

Therefore, there exists a number $A$ which is defined as:
 * $A := \lim_{n \mathop \to \infty} u \left({n}\right)$

From Lemma 2 and Lemma 3, we have:
 * $u \left({n}\right) < A < v \left({n}\right)$

The result follows.