User:Kc kennylau/sandbox

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


 * $\lfloor {x} \rfloor$ denotes the floor function of $x$; and
 * $\N$ denotes the set of all natural numbers

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

where $\N$ denotes the set of all natural numbers and $\Bbb{P}$ denotes the set of all prime numbers.

Let $p_n$ be the $n$th prime number. From Difference between Consecutive Primes, we have:
 * $p_{n+1} - p_n < K p_n^{5/8}$

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 the lemma, there exists an infinite sequence of primes, $P_0$, $P_1$, $P_2$, $\cdots$ such that:
 * $\forall n \in \N_{>0} : P_n^3 < P_{n+1} < \left({ P_n + 1 }\right)^3 - 1$

where $\N$ denotes the set of all natural numbers.

Then we define two functions:
 * $u_n := P_n^{3^{-n}}$
 * $v_n := \left({ P_n + 1 }\right)^{3^{-n}}$

It is trivial that $v_n > u_n$.

Lemma 2

 * $\forall n \in \N_{>0} : u_{n+1} > u_n$

where $\N$ denotes the set of all natural numbers.

Lemma 3

 * $\forall n \in \N_{>0} : v_{n+1} < v_n$

where $\N$ denotes the set of all natural numbers.