Timeline for Number of squares in a grid under certain conditions
Current License: CC BY-SA 3.0
2 events
| when toggle format | what | by | license | comment | |
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| Feb 12, 2015 at 1:01 | comment | added | user66997 | Yes, that's the answer. $E(6)$ and $E(7)$ must be fixed. We have $E(n) = \sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}(n+1-2k)^2 = \frac{(n-1)n(n+1)}{6}$. | |
| Feb 12, 2015 at 0:57 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |