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 2021-11-15, 07:53 #1 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 5·233 Posts the scary fraction of one seventh Okay, it is not that scary. Some of us know that 1/7 can be represented as an infinite repeating decimal. The six digits are 1,4,2,8,5, and7 So here it is with those digits twice, because we all want to see it twice :) 1/7 = 0.142857 142857 Regards, Matt
 2021-11-15, 09:04 #2 Dobri   "Καλός" May 2018 11×31 Posts Here is a link to more fun facts about 142857, see https://en.wikipedia.org/wiki/142,857.
 2021-11-15, 09:14 #3 kruoli     "Oliver" Sep 2017 Porta Westfalica, DE 2·3·181 Posts Edit: This first part is also seen on the page Dobri provided. I was slower with my writeup than he with a link. The arithmethic progression shown there based on 3 is also interesting. You can also see how $\frac{1}{7}=\sum^{\infty}_{n=1}{\frac{2^n\cdot{}7}{10^{2n}}}$ by "visually inspecting" the terms: Code: x | value of x-th term -------------- 1 | 0.1400000000 2 | 0.0028000000 3 | 0.0000560000 4 | 0.0000011200 5 | 0.0000000224 ... ----------------- = 0.142857142... Of course this is not a proof, this is only a neat visualisation, similar to the one for $\frac{1}{89}=\sum^{\infty}_{n=0}{\frac{F_n}{10^{n+1}}}$ where $$F_n, n \in \mathbb{N}_0$$ are the Fibonacci numbers. This latter one can be generalised: For $$\frac{1}{89}$$, we have "one digit per $$F_n$$". If you want to have $$d$$ digits, you should take $\frac{1}{10^{2d}-10^d-1}=\sum^{\infty}_{n=0}{\frac{F_n}{10^{d\cdot{}n+1}}}$ instead, e.g. $$d=4$$: $\frac{1}{10^{2d}-10^d-1}=\frac{1}{10^8-10^4-1}=\frac{1}{99989999}=0.\ 0000\ 0001\ 0001\ 0002\ 0003\ 0005\ 0008\ 0013\ 0021\ 0034\ 0055\ 0089\ 0144 \dots$ Last fiddled with by kruoli on 2021-11-15 at 09:20 Reason: Crosspost.
 2021-11-25, 03:49 #4 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 22158 Posts The two previous posts are very cool.

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