20190423, 23:41  #1 
Feb 2019
2^{4} Posts 
What is this called? repitend of 1/95
Forgive me, I'm an amateur. I call this front/back stack. The repeating decimal of 1/95 is 0.01052631578947368421, which can be found by stacking the sum of 5^n from left to right or the sum of 2^n from right to left as shown in the attached image. Thanks.

20190424, 08:21  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,281 Posts 
the right lower corner of your summation is = 
20190424, 11:27  #3 
Feb 2019
2^{4} Posts 
Thanks! Is it known that all repitends are in this form?

20190424, 13:09  #4 
Feb 2017
Nowhere
5×691 Posts 
The following geometric series expression may also be illustrative.
1/95 = 1/(100  5) = (1/100)/(1  5/100) = (1/100)[1 + 5/100 + (5/100)^2 + (5/100)^3 + ...] = .01 + .0005 + .000025 + .00000125 + ... Obviously, this sort of expression depends on being close to a power of 10. Of course, repeating decimals are closely related to geometric series. You might not be able to get a small common ratio in a geometric series for a given fraction, but in some cases you can, with some manipulation. A wellknown example is 1/7 = 14/98 = (14/100) * [1 + 2/100 + (2/100)^2 + ...] = 14/100 * [1 + .02 + .0004 + .000008 + .00000016 + ...] = .14 28 57... (the "7" in "57" is due to a carry from 7 x 16.) Last fiddled with by Dr Sardonicus on 20190424 at 13:10 Reason: fixing typos 
20190424, 19:22  #5 
Feb 2019
10000_{2} Posts 
Fascinating. I’ll look into it more. You are right about it being close to a power of 10. Other front/back combinations I have found are more complicated. For example 1/93 is a front stack of 7^n and a back stack of 143^n (again, please excuse my jargon). What is this area of study called?

20190425, 15:05  #6  
Feb 2017
Nowhere
6577_{8} Posts 
Quote:
1/93 = (1/100)/(1  7/100) = (1/100)* [1 + (7/100) + (7/100)^2 + ...] The "back stacking" also appears to be related to summation formulas for a geometric series like the one in Batalov's post above, although I confess I don't see 143 jumping off the screen as being related to 1/93. An unusual sort of "front stacking" involving the Fibonacci sequence instead of the powers of an integer is described here. 

20190428, 14:08  #7  
Feb 2019
2^{4} Posts 
Quote:


20190429, 11:41  #8 
"Jane Sullivan"
Jan 2011
Beckenham, UK
EA_{16} Posts 
Have you read Samuel Yates, Repunits and Repetends, Star Publishing, 1982?

20190429, 17:09  #9 
Feb 2019
10000_{2} Posts 

20190504, 20:42  #10  
Feb 2019
2^{4} Posts 
Quote:


20190507, 13:04  #11  
Feb 2017
Nowhere
110101111111_{2} Posts 
Quote:
1/7 + 100/7^2 + 100^2/7^3 + ... + 100^(k1)/7^k = ((100/7)^k  1)/93. The numerators of the fractions for the partial sums are what you're seeing. ? t=1/7;s=0;for(i=1,20,s+=t;t*=100/7;print(s)) 1/7 107/49 10749/343 1075243/2401 107526701/16807 10752686907/117649 1075268808349/823543 107526881658443/5764801 10752688171609101/40353607 1075268817201263707/282475249 107526881720408845949/1977326743 10752688172042861921643/13841287201 1075268817204300033451501/96889010407 107526881720430100234160507/678223072849 10752688172043010701639123549/4747561509943 1075268817204301074911473864843/33232930569601 107526881720430107524380317053901/232630513987207 10752688172043010752670662219377307/1628413597910449 1075268817204301075268694635535641149/11398895185373143 107526881720430107526880862448749488043/79792266297612001 

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