Timeline for Proof that $3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2019 at 21:15 | comment | added | Max Alekseyev | @GottfriedHelms: I do not quite follow your comment. Could you please elaborate? | |
| Nov 17, 2019 at 20:29 | comment | added | Gottfried Helms | Hmm, I didn't yet work through this, but isn't this just the cycle in the negative integers? (Another would be the $17,...,17$-cycle of I think seven or eleven steps, see wikipedia). Or did I overlook something trivial? | |
| Nov 10, 2019 at 15:47 | comment | added | Max Alekseyev | @ReverseFlow: Fixed, thanks! | |
| Nov 10, 2019 at 15:47 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
typo corrected
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| Nov 10, 2019 at 15:45 | comment | added | ReverseFlowControl | I think you have a typo, did you mean $2^{a_k}$ instead of $a^{a_k}$? | |
| Nov 9, 2019 at 22:37 | history | answered | Max Alekseyev | CC BY-SA 4.0 |