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Oct 22, 2019 at 13:06 comment added Robert Frost I'm going to leave the following nice fact here as there's a non zero chance it's related to the proof of 1 above: Let $f:\Bbb N\to\omega^{<\omega}$ send a natural number to the lengths of consecutive ones and zeroes in its representation, e.g. $f(27)=f(11011_2)=(2,1,2)$ then it can be proven that the representation of the periodic string of $-3^{-n}\in\Bbb Z_2$ is of period $2$ for all $n\in\Bbb N$. For example $-\frac1{27}=\overline{000010010111101101}_2\mapsto(\overline{4,1,2,1,1,\text{ }\color{red}{4,1,2,1,1}})$
S Oct 15, 2017 at 21:15 history suggested Glorfindel CC BY-SA 3.0
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Mar 28, 2012 at 3:10 comment added Gerhard Paseman I don't know how. There is some literature on how close a power of 3 can be to a power of two, but I am unfamiliar with it. The only other idea I have is to consider factors of 2^k + 1, and consider certain combinations of those, but that looks even less likely of an in. Gerhard "Ask Me About System Design" Paseman, 2012.03.27
Mar 27, 2012 at 21:44 answer added GH from MO timeline score: 5
Mar 27, 2012 at 21:21 comment added user6976 @Gerhard: how knowing the last few letters of a word you can prevent the word from being a palindrome?
Mar 27, 2012 at 19:03 answer added user6976 timeline score: 5
Mar 27, 2012 at 18:12 comment added Gerhard Paseman Once 1 is established, 2 follows pretty easily. You might try showing that powers of 3 mod some higher power of 2 provide a block towards being a palindrome. Gerhard "Ask Me About System Design" Paseman, 2012.03.27
Mar 27, 2012 at 18:10 comment added Asterios Gkantzounis computer checking
Mar 27, 2012 at 17:59 history asked Asterios Gkantzounis CC BY-SA 3.0