Timeline for Subsequence of the cubes
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 23, 2021 at 19:33 | comment | added | Notamathematician | More generally, $a\left(\frac{2^{km}-1}{2^k-1}\right)=(a(2^m-1))^{2k-1}$. | |
| Sep 17, 2021 at 15:54 | vote | accept | Notamathematician | ||
| Sep 17, 2021 at 12:50 | history | answered | Peter Taylor | CC BY-SA 4.0 |