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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?

Brad Graham
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    You might want to highlight the usage of $n$ here - it took me three readings to be sure that the $p_{n+1}^2$ term in the inequality wasn't a typo. – Steven Stadnicki Oct 27 '13 at 19:33

1 Answers1

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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 $p_a < 188.$

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Will Jagy
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  • Interesting thank you. If $p_{a}<p_{a+1}<p^2_{n+1}$, then in most cases $p^2_{n}<p_{a}<p_{a+1}<p^2_{n+1}$, so $p_{a+1}<p_{a}+2\sqrt{p_a}%=$ would be implied. This conjecture would only be stronger in the case that $p_{a}<p^2_{n}<p_{a+1}<p^2_{n+1}$ – Brad Graham Oct 27 '13 at 22:24
  • i.e my conjecture is stronger because $p_{a+1} -p_{a} < 2p_{n}$ so $p_{a+1} -p_{a} < 2\sqrt{p_{a}}$ implying your conjecture $p_{a+1} < p_{a} + 2\sqrt{p_{a}}$ – Brad Graham Aug 02 '15 at 02:40