Timeline for Simple/efficient representation of Stirling numbers of the first kind
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 4, 2018 at 4:43 | history | edited | Martin Sleziak |
added (stirling-numbers) tag
|
|
| Sep 4, 2015 at 6:54 | answer | added | Tom Copeland | timeline score: 4 | |
| Nov 26, 2014 at 10:13 | answer | added | David Callan | timeline score: 3 | |
| Oct 2, 2010 at 2:11 | answer | added | Toto | timeline score: -2 | |
| Aug 16, 2010 at 8:31 | comment | added | Roland Bacher | Concerning Fredrik Johannson's comment: Memory requirements of many methods can be made small (at the cost of increasing computing time) by using the Chinese remainder theorem applied with enough small primes ("enough" can be determined for example by computing first an upper bound on the final result, eg. by computing a real approximation). Of course, this is hardly any more a problem on modern computers. | |
| Aug 2, 2010 at 17:08 | comment | added | Fredrik Johansson | Wadim: I'm asking whether there is a formula that does not involve nested Stirling numbers. | |
| Aug 2, 2010 at 6:27 | comment | added | Wadim Zudilin | Fredrik, so what's wrong with Eq. (17) on mathworld.wolfram.com/StirlingNumberoftheFirstKind.html ? (You don't need to compute SNs of the 2nd kind.) In view of your comments to Mariano and Qiaochu, I am trying to understand what is exactly unsatisfactory in all these classical formulae... You can't get something better, because everything is too classical. | |
| Aug 2, 2010 at 1:29 | answer | added | J. M. isn't a mathematician | timeline score: 4 | |
| Aug 2, 2010 at 1:23 | comment | added | Fredrik Johansson | Mariano: yes, for large $n$. Qiaochu: this is a good method, but even expanding the polynomial using a balanced product (I tried it using Sage) is considerably slower for large n than evaluating (1), and of course requires much more memory. I'm interested in whether there exists a formula that does not amount to computing all $k$ numbers. | |
| Aug 1, 2010 at 22:05 | comment | added | Qiaochu Yuan | What's wrong with the first formula in the Wikipedia article? One can easily extract a particular coefficient from it without a recurrence. | |
| Aug 1, 2010 at 20:48 | comment | added | Mariano Suárez-Álvarez | Is it really simpler/faster to use (1) instead of the usual recurrence formula to compute $S_2(n,k)$? | |
| Aug 1, 2010 at 20:36 | history | asked | Fredrik Johansson | CC BY-SA 2.5 |