Timeline for Number of partitions of a number on a combinatorial bracelet
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 23, 2011 at 16:15 | comment | added | Douglas Zare | It looks like your bijection for that case is a restriction of my bijection. | |
| Jun 23, 2011 at 16:15 | comment | added | user9072 | Thank you for the clarification, I guess I should have seen myself what you mean. I guess I also need coffee. | |
| Jun 23, 2011 at 16:03 | comment | added | Gerhard Paseman | For each bead labeled b, I replace it with b white beads, and separate each group with a black bead. This should be a bijection. However, Douglas Zare is a professional at this, so I may be speaking in haste. Gerhard "Really Needs Coffee For This" Paseman, 2011.06.23 | |
| Jun 23, 2011 at 15:34 | comment | added | user9072 | Sorry, I do not understand the second paragraph. OK, you fix a size k, which seems like an interesting idea, and I assume your n is the N. But then how do you get what your write? | |
| Jun 23, 2011 at 15:23 | comment | added | Gerhard Paseman | Of course, for n < 5, the number of bracelets is the number of partitions of n if 0 is not a color. If 0 is a color, again there are infinitely many bracelets. Gerhard "Shouldn't Do This Sans Coffee" Paseman, 2011.06.23 | |
| Jun 23, 2011 at 15:14 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |