Timeline for Must bounded sequences be well-distributed to most *composite* moduli?
Current License: CC BY-SA 4.0
18 events
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Sep 10 at 1:22 | history | rollback | Will Sawin |
Rollback to Revision 6
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Sep 9 at 20:52 | history | edited | Will Sawin | CC BY-SA 4.0 |
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Sep 9 at 20:09 | history | edited | Will Sawin | CC BY-SA 4.0 |
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Sep 9 at 19:28 | history | edited | Will Sawin | CC BY-SA 4.0 |
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Sep 9 at 0:16 | comment | added | H A Helfgott | See the answer below. | |
Sep 8 at 12:57 | comment | added | Will Sawin | @HAHelfgott Seems fine to me. | |
Sep 8 at 3:37 | comment | added | H A Helfgott | Given the circumstances, is it socially acceptable to write down the argument on paper, take a picture with my phone and post it below? | |
Sep 8 at 3:35 | comment | added | H A Helfgott | I’ll try to post something soon. (I’m still in Guatemala and my laptop is now refusing to turn on.) | |
Sep 6 at 1:36 | comment | added | Will Sawin | @HAHelfgott It seems that combining the arguments carefully would give that the sum is at most $\epsilon$ for almost all $q$ with at most $C_1 (\epsilon) \sqrt{\log \log N}$ prime factors but can sometimes fail to be at most $\epsilon$ for all $q$ with more than $C_2(\epsilon) \sqrt{\log \log N}$ prime factors for constants $C_1$ and $C_2$ going to $0$ with $\epsilon$. This would be a pretty sharp result and seems worth writing down. | |
Sep 5 at 23:02 | comment | added | H A Helfgott | Ah. Should I give a sketch of the bound for $k=\sqrt{\log \log N}$ below? It still has a wonky step at the beginning, but it can most likely be made rigorous. | |
Sep 5 at 19:43 | comment | added | Will Sawin | @HAHelfgott Thanks! No, what I was trying to express is that the Erdős–Kac argument does not give a good enough bound on the number of large values of $q$ with a small number of prime factors. Of course there are other arguments that estimate that. | |
Sep 5 at 19:30 | comment | added | H A Helfgott | Nice. (I had a piecewise linear function of $\omega(n)$ in my head, but this is simpler.) Your last $>$ should be a $\leq$, right? | |
Sep 5 at 17:48 | history | edited | Will Sawin | CC BY-SA 4.0 |
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Sep 5 at 17:39 | history | undeleted | Will Sawin | ||
Sep 5 at 17:39 | history | edited | Will Sawin | CC BY-SA 4.0 |
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Sep 5 at 17:34 | history | deleted | Will Sawin | via Vote | |
Sep 5 at 17:33 | history | edited | Will Sawin | CC BY-SA 4.0 |
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Sep 5 at 17:26 | history | answered | Will Sawin | CC BY-SA 4.0 |