Mills' Theorem

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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.


We define $\map f x$ as a prime-representing function if and only if:

$\forall x \in \N: \map f x \in \Bbb P$


$\N$ 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

$\forall N > K^8 \in \Z: \exists p \in \Bbb P: N^3 < p < \paren {N + 1}^3 - 1$


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:

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

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 2

$\forall n \in \N_{>0}: \map u {n + 1} > \map u n$


Lemma 3

$\forall n \in \N_{>0}: \map v {n + 1} < \map v n$


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:

$\map u n < A < \map v n$
\(\ds \map u n\) \(<\) \(\ds A\) \(\ds < \paren n\)
\(\ds \leadsto \ \ \) \(\ds {P_n}^{3^{-n} }\) \(<\) \(\ds A\) \(\ds < \paren {P_n + 1}^{3^{-n} }\)
\(\ds \leadsto \ \ \) \(\ds P_n\) \(<\) \(\ds A^{3^n}\) \(\ds < P_n + 1\)

The result follows.


Source of Name

This entry was named for William Harold Mills.