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Cramer's conjecture gives $$p_{n+1}-p_n= O(\log^2 p_n)$$ while Riemann Hypothesis yields just $$p_{n+1}-p_n= O(\sqrt p_n\log^2 p_n).$$

Does Cramer conjecture on prime gaps imply Riemann Hypothesis (or is there a proof that Cramer cannot imply Riemann)?

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  • $\begingroup$ If RH holds, then $$p_{n+1}-p_{n}=O\left(\sqrt{p_{n}}\,\log\left(p_{n}\right)\right).$$ $\endgroup$ – Marco Cantarini Jan 1 '15 at 19:04
  • $\begingroup$ What I have poses no contradiction to what you state. $\endgroup$ – T.... Jan 1 '15 at 21:10
  • $\begingroup$ Cramer's conjecture and the Riemann Hypothesis are essentially unrelated, even though the implications are related. But the Riemann Hypothesis gives results on the error term of the prime number theorem, and Cramer's conjecture gives a regularity result on primes. The primes can be very regular or very irregular and give the same asymptotics in the prime number theorem. $\endgroup$ – davidlowryduda Jan 1 '15 at 22:52
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In direct contradiction with Charles' answer, it is not currently believed that Cramer's conjecture implies the Riemann Hypothesis. It is hard to explain why two things are unrelated. But in a sense, the Riemann Hypothesis concerns global behaviour of the prime-counting function, while Cramer's conjecture concerns local data. Roughly speaking, RH is global and Cramer's is like a statement about derivatives.

More importantly, there seems to be no strong relation between the asymptotics or error term in the prime-counting function and prime gaps. This is directly evidenced by the fact that RH gives such a lousy bound on prime gaps, but encodes all the information about $\pi(n)$ and its error term when compared with $\text{Li}(n)$ or $n/\log(n)$.

Heuristically, this nicely parallels the idea that rough bounds on functions give poor bounds on derivatives of that function. Similarly, rough bounds on derivatives give poor bounds on the function itself.

Unfortunately, this has all been intuition and heuristic. I searched for a definitive resource saying that Cramer's conjecture doesn't imply Riemann. I have two useful articles.

  1. The webpage for the paper Why Philosophers Should Care About Computational Complexity by Scott Aaronsen (a well-known and well-respected mathematician). In it, the paper, the older versions of the paper, and the comments, Joshua Zellinsky points out that Scott Aaronsen had originally claimed that Cramer implies RH with the same handwaving that Charles uses in his answer (i.e. Cramer implies a stronger result, therefore implying RH). But Scott backpedals, revised the paper, apologized for the error and for the excess handwaving, and removes that claim. In particular, footnote 14 of the original version of the paper changes into a much weaker footnote 17 in the revision.

  2. This question at MO from two years ago, which is exactly on this question. The answer there parallels my understanding.

So no, Cramer's conjecture is not thought to imply RH. And there is good reason to suspect that it truly does not imply RH.

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  • $\begingroup$ So you think that this aimath.org/WWN/rh/articles/html/28a is false? (Serious question.) $\endgroup$ – Charles Jan 2 '15 at 15:34
  • $\begingroup$ @charles: no. In fact, I wrote a sketch of the proof in another answer of mine. But I don't see a strong connection with a small error term and strong results on prime gaps. $\endgroup$ – davidlowryduda Jan 2 '15 at 16:14

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