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Jul 20, 2013 at 13:11 comment added Wenlong DU @Gerry Myerson Yes, thank you very much.
Jul 20, 2013 at 12:43 comment added Gerry Myerson Now posted to m.se, math.stackexchange.com/questions/447999/…
Jun 26, 2013 at 1:57 comment added stankewicz What then does "close to 1" mean? How close?
Jun 26, 2013 at 1:52 history edited Wenlong DU CC BY-SA 3.0
got a real qustion that "Has anyone a clue how to prove or disprove the above conjecture".
Jun 22, 2013 at 1:19 comment added Wenlong DU @Steven Landsburg: $\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$ means that $E=\frac{logN(logN-2loglogN)+2}{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}\approx 1$. "E" is close to 1, but it is not always equal to 1. For example, E=09.5, E=0.98, E=1.02 or E=1.04, and so on. So we can proof that $$\limsup_n \frac{p_{n+1} - p_n}{(\log p_n)^2} \geq 1$$
Jun 21, 2013 at 15:58 comment added Steven Landsburg What does "almost equal to" mean?
Jun 21, 2013 at 12:24 comment added Wenlong DU @Per Alexandersson: Thank you for you advice. It means a lot for me.
Jun 21, 2013 at 12:24 comment added Wenlong DU @Steven Landsburg: This is my first time asking a question in mathoverflow, I am sorry.
Jun 21, 2013 at 12:23 comment added Wenlong DU @The User: "≈" means "almost equal to". But, $\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$ do not mean that $$\lim_{n\rightarrow \infty }\frac{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}{ logN(logN-2loglogN)+2}=1$$.
Jun 21, 2013 at 6:12 comment added Loïc Teyssier Who upvotes this post??
Jun 20, 2013 at 20:56 history closed Steven Landsburg
Andy Putman
Felipe Voloch
Yemon Choi
Zev Chonoles
not a real question
Jun 20, 2013 at 19:42 comment added The User What does ≈ mean?
Jun 20, 2013 at 19:36 comment added Andy Putman This doesn't seem like a real question, just a random speculation backed up by essentially no evidence. I've voted to close.
Jun 20, 2013 at 19:29 answer added user9072 timeline score: 7
Jun 20, 2013 at 19:28 comment added Steven Landsburg How can we possibly tell if your conjecture is a good one or a bad one without any description of the heuristics that led you to it? And where is the mathematical question here?
Jun 20, 2013 at 19:27 comment added Per Alexandersson I can't understand your table. Is B=N? What type of answer do you seek? If it seems to hold for all really large N, then yes, otherwise, clearly no. Mathoverflow is not a place to "publish" conjectures, but to seek advice on problems. A better question would be "has anyone seen a similar estimate before?" or, "This inequality seems to hold, is there any reason why this should be true?"
Jun 20, 2013 at 19:14 history asked Wenlong DU CC BY-SA 3.0