Timeline for Is the sum $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0?$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2018 at 0:09 | history | edited | Pietro Majer | CC BY-SA 3.0 |
m
|
| Mar 4, 2014 at 15:51 | history | edited | Pietro Majer | CC BY-SA 3.0 |
minor edit
|
| S Nov 9, 2013 at 9:02 | history | suggested | Abhimanyu Pallavi Sudhir | CC BY-SA 3.0 |
,,,,,,,,,,,,,,,,,,,,
|
| Nov 9, 2013 at 7:49 | review | Suggested edits | |||
| S Nov 9, 2013 at 9:02 | |||||
| Jun 23, 2010 at 23:23 | comment | added | Jacques Carette | That last series / rational function can also be expressed as $Li_{-n}(x)$, a polylogarithm of negative order. The equations for the generating series $-(1-x)^m Li_{-n}(x)$ are probably easy to get from MGfun, and are likely to be particularly simple. | |
| Jun 23, 2010 at 15:28 | history | edited | Pietro Majer | CC BY-SA 2.5 |
edited body
|
| Jun 23, 2010 at 15:16 | vote | accept | Alexandra Seceleanu | ||
| Jun 23, 2010 at 15:00 | history | edited | Pietro Majer | CC BY-SA 2.5 |
added 1362 characters in body
|
| Jun 23, 2010 at 6:52 | comment | added | Wadim Zudilin | I didn't expect that the original sequence is so easily expressed thru the Eulerian numbers. A very nice and elementary solution! +1 | |
| Jun 23, 2010 at 0:27 | history | answered | Pietro Majer | CC BY-SA 2.5 |