Timeline for Is there a formula/name for the sum of all possible products of $i$ distinct terms in the first $k$ integers?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2017 at 3:04 | review | Close votes | |||
| Feb 28, 2017 at 11:30 | |||||
| Feb 23, 2017 at 12:55 | comment | added | Pat Devlin | Welcome to mathOverflow! By the way, whenever I encounter any integer sequence, the first thing I do is type it into oeis.org. That tells me more about the thing roughly 90% of the time (and when the thing isn't in there, I'll add an entry about it!). That's a good resource to have in your back pocket. | |
| Feb 23, 2017 at 9:14 | answer | added | JMP | timeline score: 1 | |
| Feb 23, 2017 at 9:02 | comment | added | Alice J. | @GerryMyerson Thank you, that's exactly what I needed ! | |
| Feb 22, 2017 at 22:15 | comment | added | Gerry Myerson | Your $10,35,50,24$ are Stirling numbers of the first kind. See oeis.org/A094638 or oeis.org/A008276 or oeis.org/A054654 | |
| Feb 22, 2017 at 16:22 | comment | added | Alice J. | @GHfromMO I wanted to express it as a sum because it was allowing me to cancel other terms of my formula but I think this expression is actually better, it's more understandable. Thanks a lot :) | |
| S Feb 22, 2017 at 15:49 | history | suggested | Alex M. | CC BY-SA 3.0 |
MathJax formatting
|
| Feb 22, 2017 at 15:48 | review | Close votes | |||
| Feb 22, 2017 at 16:47 | |||||
| Feb 22, 2017 at 15:42 | review | Suggested edits | |||
| S Feb 22, 2017 at 15:49 | |||||
| Feb 22, 2017 at 15:37 | comment | added | GH from MO | You can calculate your product directly: it equals $2^k\left(\binom{n}{2}-1\right)!/\left(\binom{n}{2}-k-1\right)!$. | |
| Feb 22, 2017 at 15:30 | history | edited | Alice J. | CC BY-SA 3.0 |
added 15 characters in body; edited title
|
| Feb 22, 2017 at 15:27 | review | First posts | |||
| Feb 22, 2017 at 15:43 | |||||
| Feb 22, 2017 at 15:22 | history | asked | Alice J. | CC BY-SA 3.0 |