How to prove that if $c$ >$8/3$ then there exist a real number $\theta $ such that $\bigl\lfloor\theta^{c^n}\bigr\rfloor$ is prime

Question

How to prove that if $c$ >$8/3$ then there exist a real number $\theta $ such that $\bigl\lfloor\theta^{c^n}\bigr\rfloor$ is prime for every positive integer $n$?

Answer 1

This is a generalization of Mill's constant (or Mill's Theorem). The proof is really dumb because it has almost nothing to do with primes. You will however need a nontrivial fact about primes.

Let $S=\{p_k\}_{k=1}^\infty$ be a set of numbers $p_k$ which satisfy the following property: There is some fixed $a,b$ with $0<b<1$ such that $(r,r+r^b)$ contains an element of $S$ for each real number $r>a$. So, think of $(r,r+r^b)$ as a sliding window which grows in size as $r$ gets larger. The claim now is that if $c>\min\left(\frac{1}{1-b},2\right)$ then there is an $A$ such that $[A^{c^n}]$ is an element of $S$ for each $n$. In other words $[A^{c^n}]$ is an increasing subsequence of $S$.

To prove this claim, carefully pull out a subsequence of $S$ that satisfies the above criterion. You can find one such proof here.

For your problem, you need to pick $a,b$ such that $(r,r+r^b)$ has a prime for each $r>a$. Now it just depends on what kind of theorems you know about prime gaps. For example, Bertrand's postulate (theorem) says there's a prime between $r$ and $2r$ is not good enough since in this case $b=1$. On the other hand, the prime number theorem guruantees there are about $(r+r^b)/\ln(r+r^b)-(r)/\ln(r)$ primes between $r$ and $r+r^b$. Unfortunately I don't think this is enough to conclude an answer. It seems like the existance of a prime in $[r,r+r^b]$ is related to the Riemann zeta function, according to Steven Stadnicki's comment in this answer. I believe this is related to Zero-Density-Theorems. For comparison, Mills, in his original proof, used $b=6/8$.

Perhaps someone wiser than me will see why $8/3$ in particular comes up, maybe from a known nice result.