Timeline for Sets with both additive and multiplicative gaps
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 25, 2022 at 21:36 | comment | added | Ivan Meir | @WillSawin Thanks you that is extremely enlightening, very much appreciated. | |
| Jul 24, 2022 at 21:08 | comment | added | Will Sawin | It makes sense to cut at a string of zeroes, but which one? I saw that choosing a long string of zeroes would be helpful, so why not the longest? I tried that and convinced myself that it works, then worked out the details. | |
| Jul 24, 2022 at 21:08 | comment | added | Will Sawin | @IvanMeir I started with the solution when $p=5$. I then wondered if the solution could e extended to other $p$ by choosing a lift $c_a$ of each $a$ and looking at $c_a$ mod $5$. The question is how to choose a lift so that the proportion of elements where the lifting isn't locally consistent goes to $0$. I tried a few things before deciding to look at the Mersenne prime case and think about the periodic binary digit expansions. I tried to see what makes a good lifting and saw that it amounted to cutting the loop of the binary digit expansion. | |
| Jul 24, 2022 at 13:11 | comment | added | Ivan Meir | @WillSawin Very cool and interesting example for the lower bound! Can you give some indication how you came up with it? | |
| Jul 22, 2022 at 16:22 | vote | accept | Seva | ||
| S Jul 21, 2022 at 16:21 | history | suggested | Anurag Sahay | CC BY-SA 4.0 |
fixing a mathematical typo.
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| Jul 21, 2022 at 15:59 | review | Suggested edits | |||
| S Jul 21, 2022 at 16:21 | |||||
| Jul 21, 2022 at 15:57 | comment | added | Anurag Sahay | That last paragraph should say $M v_\alpha = v_{\alpha/2} + 2 \cos (2\pi /p) v_\alpha + v_{2\alpha}$ not $M v_\alpha = v_{\alpha/2} + 2 \cos (2\pi \alpha/p) + v_{2\alpha}$, right? | |
| Jul 21, 2022 at 14:55 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 2 characters in body
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| Jul 21, 2022 at 14:51 | comment | added | Ivan Meir | @WillSawin Should $5 |A| \leq p$ be $5 |A| \leq 2p$ in your second paragraph? | |
| Jul 21, 2022 at 14:40 | history | edited | Will Sawin | CC BY-SA 4.0 |
edited body
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| Jul 21, 2022 at 14:28 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 2596 characters in body
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| S Jul 21, 2022 at 14:02 | history | suggested | mathworker21 | CC BY-SA 4.0 |
fixed some typos
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| Jul 21, 2022 at 14:01 | comment | added | Will Sawin | @mathworker21 But one can never connect $5z+1$ to $z$ only using the operations $z \mapsto z+1, z-1, 2z, z/2$ without using a congruence modulo a specific prime $p$. Local means any connected component of a finite subgraph one can form without using congruences mod $p$. | |
| Jul 21, 2022 at 13:56 | review | Suggested edits | |||
| S Jul 21, 2022 at 14:02 | |||||
| Jul 21, 2022 at 13:54 | comment | added | mathworker21 | $+1$, though the "local" argument you gave isn't that local. If you had something like $5z+1$ in your subgraph, then the argument wouldn't work if $n$ is a multiple of $5$, since it's then not the case that $5z+1$ runs over $\mathbb{Z}/n\mathbb{Z}$ as $z$ does. | |
| Jul 21, 2022 at 13:00 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 687 characters in body
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| Jul 21, 2022 at 12:53 | history | answered | Will Sawin | CC BY-SA 4.0 |