Timeline for A balls-and-colours problem
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2011 at 21:24 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Oct 14, 2010 at 0:08 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Oct 13, 2010 at 21:41 | comment | added | JBL | Barring arithmetic errors, we can also rewrite your final expression as $\frac{(2j - 1)(n - 1)(n - j)}{n} - (n - 1)(n - j) \sum_{k = 1}^{j - 1} \frac{1}{n - k}$, which does indeed look fairly simple in terms of harmonic numbers, as A. Rex remembered. | |
| Oct 13, 2010 at 19:16 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Oct 13, 2010 at 18:49 | history | answered | David E Speyer | CC BY-SA 2.5 |