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Take some 4 consecutive primes $p_n,p_{n+1},p_{n+2},p_{n+3}$ where $p_n \geq 5$.

Now form two products: $p_n \cdot p_{n+3}$ and $p_{n+1} \cdot p_{n+2}$.

Is there always at least one prime in the interval $[p_n \cdot p_{n+3},p_{n+1} \cdot p_{n+2}]$ (it does not matter if we have that $p_n \cdot p_{n+3}>p_{n+1} \cdot p_{n+2}$, if that is the case then just reverse the endpoints of the interval)?

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    $\begingroup$ No. Gerhard "Sometimes The Interval Is Empty" Paseman, 2017.04.11. $\endgroup$ – Gerhard Paseman Apr 11 '17 at 18:41
  • $\begingroup$ @GerhardPaseman How can it be empty? It only can be that the left endpoint is bigger than the right one but that does not matter, then we can just reverse the endpoints. $\endgroup$ – Paladin Apr 11 '17 at 18:49
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    $\begingroup$ Consider $n=26$. Gerhard "And Sometimes It's Too Small" Paseman, 2017.04.11. $\endgroup$ – Gerhard Paseman Apr 11 '17 at 18:52
  • $\begingroup$ Current data suggest that in such examples, the sum of the middle two primes is the same as the sum of the outer two primes. However, proving that those are the only cases would advance the state of the art. Gerhard "Prime Gaps Could Be Big" Paseman, 2017.04.11. $\endgroup$ – Gerhard Paseman Apr 11 '17 at 18:56
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    $\begingroup$ The real q. is is $p_{n+1} < p_n + p_n^{\frac 12}$ for infinitely many $n$?--it must be still open. I feel that the Q. above is unnecessarily complicated in the comparison to the classical q. $\endgroup$ – Włodzimierz Holsztyński Apr 11 '17 at 21:34
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It is very likely that there are infinitely many counterexamples. More precisely, by a special case of the prime tuple conjecture (due to Dickson and Hardy-Littlewood), there are infinitely many consecutive primes of the form $p<p+2<p+6<p+8$. The corresponding difference then equals $$ (p+2)(p+6)-p(p+8)=12, $$ and the requirement is that there is a prime between $p(p+8)$ and $p(p+8)+12$. The expected prime gap size around $p(p+8)$ is $\sim2\log p$, which is much larger than $12$, so we have every reason to believe that the requirement fails for most of the above $p$'s.

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I find about 29,000 cases where this fails with all four primes below 10^8.

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  • $\begingroup$ Are there any of these 29000 cases (besides the first two) where $p_{n+3} - p_{n+2}$ is different from $p_{n+1} - p_n$? Gerhard "It Does Make A Difference" Paseman, 2017.04.11. $\endgroup$ – Gerhard Paseman Apr 11 '17 at 21:00
  • $\begingroup$ @GerhardPaseman No. Wouldn't that make for huge intervals? $\endgroup$ – Charles Apr 11 '17 at 21:12
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    $\begingroup$ It would. However, experiment would confirm the observation, or reveal a bug. Gerhard "It's A Win Either Way" Paseman, 2017.04.11. $\endgroup$ – Gerhard Paseman Apr 11 '17 at 21:40

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