Timeline for What is the maximal number of distinct values of the product of n permuted ordinals
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 29, 2014 at 16:16 | vote | accept | Gérard Lang | ||
| Aug 28, 2014 at 14:36 | answer | added | Joseph Van Name | timeline score: 13 | |
| Aug 27, 2014 at 22:46 | comment | added | Gérard Lang | The paper of Wakulicz (pages 255-266) is written in french. The formula valid for n>20 is obtained by writing all ordinals in their Cantor normal form (as a pseudo-polynomial of powers of ω) and analyzing very precisely the consequences of the fact that every term with a power of ω written at the left of another term with a strictly greater power of ω can be eliminated of the sum. So that I do not think that this method can be used in the case of the product. Gérard Lang | |
| Aug 27, 2014 at 14:18 | comment | added | David E Speyer | To save other people hunting, Wakulicz paper is matwbn.icm.edu.pl/ksiazki/fm/fm36/fm36126.pdf . The answer for addition (once $n > 20$) is $81^t 193^r$ where $n-1 = 5t+6r$ and $0 \leq r \leq 4$. This is of size roughly $81^{n/5} \approx 2.41^n$, beating the value $2^{n-1}$ obtained by using $\omega$, $\omega^2$, $\omega^3$, ..., $\omega^n$. I have not yet understood Wakulicz's proof. | |
| Aug 27, 2014 at 10:26 | answer | added | Joel David Hamkins | timeline score: 9 | |
| Aug 27, 2014 at 10:10 | history | asked | Gérard Lang | CC BY-SA 3.0 |