Formula for gaps between primes [closed]

Question

The twin prime conjecture refers to:

$$ \liminf_{n\to \infty}\; p_{n+1} - p_{n} = 2. $$

By reasoning I arrive at the following simple formula for gaps between primes:

\begin{align} p_{n+1}-p_{n} & = \frac{p_{n+1}^{2}-p_{n}^{2}}{p_{n+1}+p_{n}}=2 \\ & \Rightarrow (p_{n}+h)^{2}-p_{n}^{2} =2(2p_{n}+h)\\ & \Rightarrow h^{2}+2p_{n}h =2h+4p_{n}\\ & \Rightarrow \frac{h^{2}-2h}{2}+p_{n}h =2p_{n}\\ & \Rightarrow p_{n} < p_{n}+h < 2p_{n} \le p_{n}h \le\frac{h^{2}-2h}{2}+p_{n}h\\ & \Rightarrow p_{n} < p_{n}+h < 2p_{n} \le \frac{h^{2}-2h}{2}+p_{n}\\ & \Rightarrow \frac{h^{2}-2h}{2} \le p_{n}-2\\ & \Rightarrow (h-\sqrt{2})^{2}\le 2p_{n}-2\\ & \Rightarrow (h-\sqrt{2})^{2}< 2p_{n} \\ & \Rightarrow h < \sqrt{2p_{n}}+\sqrt{2} \end{align}

so: $$ p_{n+1} - p_{n} < \sqrt{2p_{n}}+\sqrt{2} $$

or: $$ p_{n+1} < p_{n} + \sqrt{2p_{n}}+\sqrt{2}. $$

It's a simple conclusion, and one that is stronger and more valid than the Bertrand's postulate.

For example, we often use $[\sqrt{x},x]$ as a judgment and determination of prime number interval, and the above inference can directly give the answer.

I want to ask, is this conclusion correct?

By the way, this formula only holds if the twin prime conjecture holds.

Answer 1

Such conjectures are known in the literature. For example, Andrica's conjecture.

It is known that under the Riemann Hypothesis one has $p_{n+1} - p_n = O(\sqrt{p_n} \log p_n)$. The best unconditional result of this sort is due to Baker, Harman, and Pintz, who showed that there is always a prime in the interval $[x, x + x^{0.525})$ for $x$ sufficiently large.