# Prime gaps with respect to the squared primes

Conjecture

If we have two consecutive prime numbers $p_{a}$ and $p_{a+1}$, and two other consecutive primes $p_n$ and $p_{n+1}$,
so that $p_{a} < p_{a+1} < p^2_{n+1}$,
then $p_{a+1} – p_{a} < 2p_{n}$.

Are there any known counter examples and are there any known similar conjectures?

#### Solutions Collecting From Web of "Prime gaps with respect to the squared primes"

No known counterexamples. A slightly stronger conjecture, true as far as anyone has been able to check (up to $4 \cdot 10^{18}$) is that
$$\unicode{x2E2E} \unicode{x2E2E} \unicode{191} \unicode{191} \; p_{a+1} < p_a + 2 \; \sqrt {p_a} \; \unicode{63} ?$$
This is currently unprovable.

What people actually suspect is that,
$$\unicode{191} ¿ \; \mbox{if} \; \; p_a \geq 11, \; \; \mbox{then} \; \; p_{a+1} < p_a + \log^2 {p_a} \; ?$$
Really, really beyond proof.

See Prime pair points slope approaches 1

Note: as far as using the extensive tables of prime gaps, there is a detail involving the fact that $\log^2 x > 2 \sqrt x$ for an interval, roughly $19.6 < x < 187.8.$ So, it was necessary to make a separate confirmation of my version of the conjecture for $p_a < 188.$