Timeline for Every possible number of partitions by restricting parts?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2019 at 14:02 | vote | accept | Brian Hopkins | ||
| Sep 10, 2019 at 14:02 | comment | added | Brian Hopkins | I appreciate the edits you made; I get it now. (I knew (**) is essential; I was hung up on details in the three lines above it.) Thanks very much for this solution. | |
| Sep 10, 2019 at 13:08 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
added 140 characters in body
|
| Sep 10, 2019 at 13:05 | comment | added | Ilya Bogdanov | (If the question is `why do we do what we are doing', then the answer is: if $(**)$ were not corect, then for some value of $k$ the algorithm would fail; so this is mostly what we need to check, in order to decide validity of the algorithm.) | |
| Sep 10, 2019 at 13:03 | comment | added | Ilya Bogdanov | Sorry, I do not understand the question. I clarified a bit how $S_m$ are defined; does it help? | |
| Sep 10, 2019 at 13:03 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
added 140 characters in body
|
| Sep 10, 2019 at 0:43 | comment | added | Brian Hopkins | Ok, I agree that this works. There's one step I'd like to understand better. In the (*) proof, the "otherwise" case is when $m$ is not removed, so it must be that $p_{S_m} (n) < k$, i.e., $p_{S_m} (n) \le k - 1$. And $S_m = S_{m-1}$. Certainly removing $m$ cannot increase the number of partitions, but why move to $p_{S_{m-1}\setminus \{m\}}$? | |
| S Sep 8, 2019 at 14:08 | history | suggested | Brian Hopkins | CC BY-SA 4.0 |
added initial S value
|
| Sep 8, 2019 at 13:53 | review | Suggested edits | |||
| S Sep 8, 2019 at 14:08 | |||||
| Sep 7, 2019 at 6:30 | comment | added | Ilya Bogdanov | Yes, and then going through the nimbers in increasing order. | |
| Sep 7, 2019 at 2:36 | comment | added | Brian Hopkins | Thanks, working through this. So you're starting from $S=[n]$, right? | |
| Sep 6, 2019 at 15:34 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |