 mersenneforum.org the scary fraction of one seventh
 Register FAQ Search Today's Posts Mark Forums Read 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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post MattcAnderson MattcAnderson 9 2021-12-12 20:44 jvang jvang 12 2019-01-14 04:08 Yamato Puzzles 8 2013-05-16 15:50 tinhnho Miscellaneous Math 4 2005-01-17 19:45 Cyclamen Persicum Miscellaneous Math 9 2003-04-13 14:56

All times are UTC. The time now is 17:32.

Thu Aug 11 17:32:43 UTC 2022 up 35 days, 12:20, 3 users, load averages: 1.26, 1.26, 1.25