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
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| Jan 11, 2014 at 19:52 | comment | added | Vesselin Dimitrov | @WillSawin: Your are right, it is enough to take $S = \{2,3,5\}$. (I was thinking of accomodating the prime factors of every digit in $\{1,\ldots,9\}$, but this is unnecessary since e.g. we may split $7 \cdot 10^k \cdot 3^{-m}$ into $7$ new $\{2,3,5\}$-unit variables $= 10^k \cdot 3^{-m}$.) | |
| Jan 11, 2014 at 18:51 | comment | added | Will Sawin | What is $7$ achieving in $S$? Is $\{2,3,5\}$ insufficient? | |
| Sep 27, 2013 at 14:32 | vote | accept | mathlove | ||
| Sep 26, 2013 at 23:26 | comment | added | Gerry Myerson | The only problem with copying my references is that you copied my typo. I can't fix it in the comments at 38971, but I did fix it here. Incidentally, Senge & Straus also published their result in another paper of the same title, Period. Math. Hungar. 3 (1973) 93–100, MR0340185 (49 #4941). | |
| Sep 26, 2013 at 23:22 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
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| Sep 26, 2013 at 17:47 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Sep 26, 2013 at 16:23 | comment | added | mathlove | Again, thank you for great information. | |
| Sep 26, 2013 at 16:03 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Sep 26, 2013 at 15:56 | comment | added | Vesselin Dimitrov | Actually, see here: mathoverflow.net/questions/38971/… | |
| Sep 26, 2013 at 15:54 | comment | added | Vesselin Dimitrov | You are welcome! And actually, upon looking at a paper of Stewart from 1980, I realize that I was wrong about effectivity: an explicit lower bound is actually possible with Baker's method. I will edit. | |
| Sep 26, 2013 at 15:36 | history | answered | Vesselin Dimitrov | CC BY-SA 3.0 |