Timeline for Letting $S(m)$ be the digit sum of $m$, then $\lim_{n\to\infty}S(3^n)=\infty$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2023 at 22:07 | answer | added | Marek Kryspin | timeline score: 1 | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jan 11, 2014 at 18:25 | answer | added | Shahrooz | timeline score: 0 | |
| Oct 23, 2013 at 9:26 | answer | added | Alexey Ustinov | timeline score: 3 | |
| Sep 27, 2013 at 14:32 | vote | accept | mathlove | ||
| Sep 27, 2013 at 1:01 | comment | added | so-called friend Don | See this paper of Blecksmith, Filaseta, and Nicol for some closely related results: math.sc.edu/~filaseta/papers/blockpaper.pdf | |
| Sep 26, 2013 at 18:40 | comment | added | Pietro Majer | Btw this sequence is A004166 in the OEIS oeis.org/A004166. It has no much information however; maybe you could add there the link to this question... | |
| Sep 26, 2013 at 15:36 | answer | added | Vesselin Dimitrov | timeline score: 38 | |
| Sep 26, 2013 at 15:35 | comment | added | mathlove | @TheMaskedAvenger: Thank you for pointing it out. | |
| Sep 26, 2013 at 15:28 | comment | added | The Masked Avenger | I like your argument. Note you can replace 4m by ceil(mlog_2 10), giving a slightly tighter location of nonzero digits. For 3, maybe you can argue that a large string of 0's near the least significant bits gives that many nonzeros further up. | |
| Sep 26, 2013 at 14:54 | history | asked | mathlove | CC BY-SA 3.0 |