There exists n such that number of primes between $(n+1)^{2021}$ and $n^{2021}$ is greater than 1122021140.

Question

To show there exists a natural number $n$ such that there are at least 1122021140 primes between $(n+1)^{2021}$ and $n^{2021}$.

Here is my attempt :

$\Rightarrow$ We define $\pi(x)=\#\{p \leq x \mid p$ is prime $\}$.

Then by prime number theoremn $\pi(x) \sim \frac{x}{\log x}$.

So I need to show $\pi(y)-\pi(x) > 1122021140$ for $y=(n+1)^{2021}$ and $x=n^{2021}$

For large $x$ we have (saw it here) $$ \pi(y)-\pi(x) \sim \frac{y-x}{\log x} \tag1 $$

$$ \begin{aligned} &=\frac{(n+1)^{2021}-n^{2021}}{\log (n)^{2021}} \\ &=\frac{(n+1)^{2021}-n^{2021}}{2021 \log n} \end{aligned} $$

I have two questions :

1. How to prove equation $(1)$ i.e., $\pi(y)-\pi(x) \sim \frac{y-x}{\log x}$ for large $x$.

2. How to show there exists a $n$ such that $\frac{(n+1)^{2021}-n^{2021}}{2021 \log n} > 1122021140$.

Answer 1

More directly, $n=1$ will work. There are many more than $1122021140$ primes between $1$ and $2^{2021}$, and we can prove it.

For example, there is a lower bound on $\pi(2n)$ via the prime factorization of $\binom{2n}{n}$ (for details, see this MSE post) that tells us $$ \pi(2n)\ge \frac{n}{\log_2 (2n)} $$ so in particular, $\pi(2^{2021}) \ge \frac{2^{2020}}{2021}$. This is much larger than $1122021140$: it's a number with over $600$ digits.

Answer 2

Let $\epsilon$, $M$ be positive numbers. Then there exists $n$ such that between $n^{1+\epsilon}$ and $(n+1)^{1+\epsilon}$ there are more than $M$ prime numbers. Indeed, otherwise a simple estimate would show that $$\sum_{p \textrm{ prime}} \frac{1}{p} \le M \cdot \sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon}}< \infty$$ contradiction

$\bf{Added:}$ A similar argument shows that for every increasing sequence $a_n$ with $\sum \frac{1}{a_n} < \infty$ we have $$\lim \sup_{n\to \infty} (\pi(a_{n+1}) -\pi(a_n)) = +\infty$$

Answer 3

From the question you link:

There are unconditional results by Huxley and Heath-Brown showing $(1)$ for $h=y-x$ roughly being $x^{7/12}$.

Since $h\in O(n^{2020})=O(x^{2020/2021})=\omega(x^{7/12})$, there is an unconditional proof of $(1)$. Then $(2)$ follows by simply comparing growth rates of numerator and denominator.