Timeline for Is the following weak version of second Hardy-Littlewood conjecture already known?
Current License: CC BY-SA 4.0
35 events
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| Dec 8, 2018 at 14:05 | comment | added | user57432 | @RobertFrost: Not really. See Thomas Bloom's response above. | |
| Dec 8, 2018 at 13:33 | comment | added | Robert Frost | I guess we are expecting the full conjecture to be false. | |
| Nov 2, 2018 at 3:19 | vote | accept | CommunityBot | ||
| Nov 1, 2018 at 23:03 | comment | added | GH from MO | See my response below. | |
| Nov 1, 2018 at 23:01 | answer | added | GH from MO | timeline score: 8 | |
| Nov 1, 2018 at 16:01 | comment | added | user57432 | @GHfromMO: Please let me know if there are any more mistakes in the formulation of the question. | |
| Nov 1, 2018 at 16:00 | comment | added | user57432 | @ThomasBloom: Just to be sure. By Montgomery and Vaughan you mean the book Multiplicative Number Theory: I. Classical Theory, right? Also I would be glad if you could comment on the recent version of my post. | |
| Nov 1, 2018 at 15:54 | history | edited | user57432 | CC BY-SA 4.0 |
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| Oct 31, 2018 at 16:27 | comment | added | user57432 | I have edited the question. | |
| Oct 31, 2018 at 16:27 | history | edited | user57432 | CC BY-SA 4.0 |
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| Oct 31, 2018 at 16:26 | history | undeleted | user57432 | ||
| Oct 31, 2018 at 16:19 | history | deleted | user57432 | via Vote | |
| Oct 31, 2018 at 16:18 | history | undeleted | user57432 | ||
| Oct 14, 2018 at 11:02 | history | deleted | user57432 | via Vote | |
| Oct 14, 2018 at 8:23 | comment | added | Jan-Christoph Schlage-Puchta | "For that y_0" does not make sense, as $y_0$ is a fuction of $\varepsilon$. Also the part about $\varepsilon$ follows from any version of the prime number theorem with error term better than $\frac{x}{\log^2 x}$. | |
| Oct 13, 2018 at 23:40 | comment | added | GH from MO | @user170039: It is not clear what is the role of $k$ in the first sentence of your Proposition, since $y_0$ and the first display are independent of $k$. So it seems that the "for all $k>1$" should be the beginning of the second sentence. However, even with that modification, the second statement makes no sense, as it refers to an unspecified $y_0$ (which has nothing to do with $k$). Simply put, the two sentences do not fit. | |
| Oct 13, 2018 at 8:21 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
minor typo
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| Oct 13, 2018 at 3:27 | comment | added | user57432 | Also I don't understand which inequality you were referring. Were you referring to "$y\ge y_0$" or $\pi(ky)+\pi(y)\ge \pi((k+1)y)$? If it is the latter then can you clarify what did you mean by "the last inequality does not contain $\varepsilon"? | |
| Oct 13, 2018 at 3:24 | comment | added | user57432 | @Jan-ChristophSchlage-Puchta: Let me try to state the proposition a but more clearly, "Assume that for all $k>1$ and $\varepsilon\in (0,\ln \sqrt{2}]$ there exists $y_0\in \mathbb{R}$ (depending on $\varepsilon$ only) such that for all $y\ge y_0$ we have, $$\dfrac{y}{\ln y-(1-\varepsilon)}<\pi(y)<\dfrac{y}{\ln y-(1+\varepsilon)}$$Then for that $y_0$ and for all $y\ge y_0$ we have, $$\pi(ky)+\pi(y)\ge \pi((k+1)y)$$ where $\pi$ is the Prime Counting Function." | |
| Oct 12, 2018 at 16:57 | comment | added | Jan-Christoph Schlage-Puchta | But the last inequality does not contain $\epsilon$. | |
| Oct 12, 2018 at 14:49 | comment | added | user57432 | @Jan-ChristophSchlage-Puchta: $y_0$ depends on $\varepsilon$ only and not on $k$. It is the same $y_0$ which was mentioned earlier. | |
| Oct 12, 2018 at 14:09 | comment | added | Jan-Christoph Schlage-Puchta | It is not clear what the last occurrence of $y_0$ in the proposition means. If we assume that $y_0$ may depend on $k$, i.e. $\forall k\exists y_0\forall y>y_0: \pi(ky)+\pi(y)\geq\pi((k+1)y)$, then the statement does not contradict the Prime tuple conjecture. In fact, it follows from any version of the prime number theorem with an error term better than $\frac{x}{\log^2 x}$. | |
| Oct 10, 2018 at 13:09 | comment | added | Thomas Bloom | I suppose not equivalent, but regardless this 'weaker' form of the Hardy-Littlewood conjecture is still enough to contradict the Prime Tuple conjecture (see e.g. Theorem 7.16 in Montgomery and Vaughan) | |
| Oct 10, 2018 at 13:02 | comment | added | Thomas Bloom | As GH says, the claimed Proposition is, thanks to PNT, equivalent to the Hardy-Littlewood conjecture for all large $x,y$. This would contradict the Prime Tuple conjecture, so a) it is certainly not known and b) a proof of it would be very big news. | |
| Oct 10, 2018 at 12:50 | comment | added | user57432 | @GHfromMO: Thank you very much for your suggestions. Indeed, as you wrote the first sentence is meant to be a condition (and I am aware that it follows from PNT). I am also aware that this conjecture is believed to be false by Hensley and Richard's paper. Anyway, please let me know of any other problems that you may find in this post. | |
| Oct 10, 2018 at 12:38 | history | edited | user57432 | CC BY-SA 4.0 |
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| Oct 10, 2018 at 5:39 | history | edited | GH from MO |
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| Oct 10, 2018 at 5:37 | comment | added | GH from MO | It seems that the first sentence in your Proposition is meant to be a condition, so you should start it with "Assume that". However, this condition follows from the Prime Number Theorem and the asymptotic expansion of $\mathrm{li}(y)$. So it seems that you are claiming the second sentence in your Proposition unconditionally, which as you remark is the second Hardy-Littlewood conjecture. Note also that this conjecture is believed to be false, as it contradicts the Hardy-Littlewood conjecture on prime tuples. | |
| Oct 10, 2018 at 5:27 | history | edited | user57432 | CC BY-SA 4.0 |
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| Oct 10, 2018 at 5:21 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Oct 10, 2018 at 5:16 | history | edited | user57432 | CC BY-SA 4.0 |
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| Oct 10, 2018 at 4:52 | comment | added | Nate Eldredge | $\varepsilon$ doesn't appear in the displayed inequality of Proposition 2, so if I can find a $y_0$ that works at all, it's automatically true for every $\varepsilon$. So as written there's no point in including $\varepsilon$ at all. I assume this is not what you meant, and that there is a typo. | |
| Oct 10, 2018 at 4:50 | comment | added | user57432 | @NateEldredge: Sorry, I don't understand. Can you clarify? | |
| Oct 10, 2018 at 4:50 | comment | added | Nate Eldredge | Did you mean for the conclusion of Proposition 2 to actually mention $\varepsilon$? | |
| Oct 10, 2018 at 3:48 | history | asked | user57432 | CC BY-SA 4.0 |