W. H. MILLS
A function $f(x)$ is said to be a prime-representing function if $ƒ(x)$ is a prime number for all positive integral values of $x$. It will be shown that there exists a real number $A$ such that $A^{3^x}$ is a prime-representing function, where $\left [ R \right ]$ denotes the greatest integer less than or equal to $R$.
Let $p_n$ denote the $n-$th prime number.
A. E. Ingham has shown that
If $N$ is an integer greater than $K^8$ there exists a prime $p$ such that $N^3 \lt p \lt (N+1)^3-1$ .
Let $p_n$ be the greatest prime less than $N^3$. Then
$N^3 \lt p_{n+1} \lt p_n + K p_n^{\frac{5}{8}} \lt N^3 + KN^{\frac{15}{8}} \lt N^3 +N^2 \lt (N + 1)^3 - 1$.
Let $P_0$ be a prime greater than $K^8$. Then by the lemma we can construct an infinite sequence of primes, $P_0,P_1, P_2,\cdots$, such that
$P_n^3 \lt P_{n+1} \lt (P_n +1)^3 - 1$.
Let
3.$ u_n = P_n^{3^{-n}} , v_n = (P_n + 1)^{3^{-n}}$.
Then
4.$ v_n \gt u_n, u_{n+1} = P_{n+1}^{3^{-n-1}} \gt P_n^{3^{-n}} = u_n$,
5.$ v_{n+1} = (P_{n+1} + 1)^{3^{-n-1}} \lt (P_n + 1)^{3^{-n}} = v_n$.
It follows at once that the $u_n$ form a bounded monotone increasing sequence. Let $ A = \lim_{n \to \infty} u_n$.
$A^{3^n}$ is a prime-representing function.
From (4) and (5) it follows that $u_n \lt A \lt v_n$, or $P_n \lt A^{3^n} \lt P_n+1$ .
Therefore $\left [A^{3^n}\right ] = P_n$ and $\left [A^{3^n}\right ]$ is a prime-representing function.