Timeline for Stirling number identity via homology?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2014 at 4:48 | answer | added | Steven Sam | timeline score: 5 | |
| Jul 8, 2012 at 13:38 | comment | added | Patricia Hersh | @John: you are absolutely correct. Mariano's idea will work for alternating sequences summing to 0 if the absolute values form a unimodal sequence, as justified in the following related MO question and answers: mathoverflow.net/questions/90423/… | |
| Jul 6, 2012 at 20:30 | comment | added | John Wiltshire-Gordon | @Mariano Suárez-Alvarez What about 1-5+1-5+13-5=0? These don't seem to fit into an exact sequence. | |
| Jul 3, 2012 at 23:28 | comment | added | David Roberts♦ | It might have a species interpretation, see ncatlab.org/nlab/show/species | |
| Jul 3, 2012 at 20:38 | answer | added | David E Speyer | timeline score: 10 | |
| Nov 9, 2011 at 20:46 | comment | added | Noam D. Elkies | Apropos I.Rivin's comment: a test case is $\sum_{i=0}^n (-1)^i {n \choose i}^3$. The sum is known in closed form, and quite nontrivial for $n$ even. Is there a proof via an Euler-characteristic interpretation? | |
| Nov 9, 2011 at 17:47 | comment | added | Dimitrije Kostic | Most combinatorialists (for example, Enumerative Combinatorics, v.1, by R.P. Stanley, page 18) define the Stirling numbers of the first kind to be $s(k,m) := (-1)^{(k-m)}c(k,m)$. With that definition, you have the identity $\sum_{k \geq 0} S(n,k)s(k,m) = \delta_{n,m}$ (ibid., p. 35). The sum you give does not always yield 0. When $n=2$ and $m=1$, for example, it equals 2. | |
| Oct 27, 2011 at 20:32 | comment | added | Igor Rivin | @Mariano: of course, but this might not necessarily be the most enlightening argument... | |
| Oct 27, 2011 at 20:25 | comment | added | Mariano Suárez-Álvarez | (...a finite alternating sum of positive integers...) | |
| Oct 27, 2011 at 20:24 | comment | added | Mariano Suárez-Álvarez | Igor, if a finite alteranting sum of integers has value zero, you can find an exact complex $X$ of finite dimensional vector spaces which turns that equality to zero into the statement «the Euler characteristic of $X$ is zero». | |
| Oct 27, 2011 at 20:18 | comment | added | Igor Rivin | Can every alternating sum be computed by a homological argument? | |
| Oct 27, 2011 at 20:17 | answer | added | Mariano Suárez-Álvarez | timeline score: 8 | |
| Oct 27, 2011 at 19:34 | history | asked | Gary Kennedy | CC BY-SA 3.0 |