Mills' Theorem

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


 * $\floor x$ denotes the floor function of $x$
 * $\N$ denotes the set of all natural numbers.

Proof
We define $\map f x$ as a prime-representing function :
 * $\forall x \in \N: \map f x \in \Bbb P$

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
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:
 * $\forall n \in \N_{>0}: {P_n}^3 < P_{n + 1} < \paren {P_n + 1}^3 - 1$

Let us define two mappings $u, v: \N \to \Bbb P$ as:
 * $\forall n \in \N: \map u n = {P_n}^{3^{-n} }$
 * $\forall n \in \N: \map v n = \paren {P_n + 1}^{3^{-n} }$

It is trivial that $\map v n > \map u n$.

Lemma 3
It follows trivially that $\map u n$ is bounded and strictly monotone.

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

From Lemma 2 and Lemma 3, we have:
 * $\map u n < A < \map v n$

The result follows.