20130512, 14:16  #1 
Sep 2005
Berlin
42_{16} Posts 
An Egyptian Fraction
Write as a sum of distinct unit fractions (all numerators are 1), where all denominators are smaller than 10001.

20130512, 15:01  #2 
Aug 2010
Kansas
547 Posts 
1/102+1/10517+1/228264101+1/188377806760266318

20130512, 16:04  #3 
Aug 2006
3×1,993 Posts 
cl10ck3r, that doesn't meet the denominator requirement.
I get 99/10001 = 1/105 + 1/6165 + 1/9198 + 1/9590. Last fiddled with by CRGreathouse on 20130512 at 16:05 
20130512, 16:11  #4 
Aug 2010
Kansas
547_{10} Posts 

20130512, 19:35  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·29·113 Posts 
Many solutions. Here's just a few:
v=[3,14,15];sum(k=1,#v,1/(73*v[k])+1/(137*v[k])) v=[3,10,42,70];sum(k=1,#v,1/(73*v[k])+1/(137*v[k])) v=[5,6,14,30];sum(k=1,#v,1/(73*v[k])+1/(137*v[k])) v=[7,10,14,15,21,35,70];sum(k=1,#v,1/(73*v[k])+1/(137*v[k])) Last fiddled with by Batalov on 20130512 at 19:50 Reason: (add a long one) 
20130512, 21:15  #6  
Aug 2006
3×1,993 Posts 
Quote:


20130512, 21:41  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×29×113 Posts 
Yes, indeed. There was a P.Euler problem in similar vein. (more than one, actually)
Can we count the number of (distinct) solutions of OP? (with d_{1} < d_{2} < ... < d_{n} < 10001) 
20130513, 05:11  #8 
Aug 2006
3×1,993 Posts 
Good question!
It seems very hard, unless the upper bound on the number of terms can be lowered substantially from the naive 98. I don't see why it should be, though. 
20130516, 15:50  #9  
Sep 2005
Berlin
2×3×11 Posts 
Quote:
The "generic" method (which also works for prime denominators) is to choose a smooth number k which factors into small primes and write the numerator in (99k)/(10001k) as a sum of factors of the denominator. This leads to a unit fraction expansion for 99/10001, e.g. 1/247 + 1/292 + 1/949 + 1/1898 + 1/2603 + 1/3562 + 1/5548. An interesting question is to how to find the expansion, where all denominators are bounded by a number as small as possible. 

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