Timeline for Doubling for Sumset of the same set
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2013 at 12:26 | comment | added | Seva | I would expect that no characterisation can be given and, in fact, you have $\sigma(2^kA)\le\sigma(2^{k-1}A)$ for all "non-pathological" cases. As just one example, consider the situation where $A$ is a sufficiently dense subset of a subgroup. | |
| Nov 18, 2013 at 9:54 | comment | added | yue | and non-coset $A$ | |
| Nov 18, 2013 at 9:50 | comment | added | yue | I noticed this special case as well. Thanks for reminding. I should have mentioned it. But I'd love to see whether there is a non-group $A$ satisfying the requirement. | |
| Nov 18, 2013 at 9:32 | comment | added | Seva | If $A$ is a subgroup, then $\sigma(mA)=1$ for every $m\ge 1$. Similarly, taking $A$ to be an arithmetic progression yields $\sigma(mA)=2+o(1)$. Does this answer your question? | |
| Nov 18, 2013 at 6:13 | review | First posts | |||
| Nov 18, 2013 at 6:19 | |||||
| Nov 18, 2013 at 6:10 | history | edited | yue | CC BY-SA 3.0 |
typo
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| Nov 18, 2013 at 5:58 | comment | added | Brendan McKay | Is there a typo? $s=2$? | |
| Nov 18, 2013 at 5:54 | history | asked | yue | CC BY-SA 3.0 |