 mersenneforum.org An Egyptian Fraction
 Register FAQ Search Today's Posts Mark Forums Read 2013-05-12, 14:16 #1 Yamato   Sep 2005 Berlin 2×3×11 Posts An Egyptian Fraction Write as a sum of distinct unit fractions (all numerators are 1), where all denominators are smaller than 10001.    2013-05-12, 15:01 #2 c10ck3r   Aug 2010 Kansas 10001000112 Posts 1/102+1/10517+1/228264101+1/188377806760266318   2013-05-12, 16:04 #3 CRGreathouse   Aug 2006 22×33×5×11 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 2013-05-12 at 16:05   2013-05-12, 16:11   #4
c10ck3r

Aug 2010
Kansas

54710 Posts Quote:
 Originally Posted by CRGreathouse cl10ck3r, that doesn't meet the denominator requirement. I get 99/10001 = 1/105 + 1/6165 + 1/9198 + 1/9590.
Ah, glanced over that part.   2013-05-12, 19:35 #5 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 25·7·41 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 2013-05-12 at 19:50 Reason: (add a long one)   2013-05-12, 21:15   #6
CRGreathouse

Aug 2006

22×33×5×11 Posts Quote:
 Originally Posted by Batalov 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]))
Using (99/10001)/(1/73+1/137) = 33/70. Convenient, but not good if you're trying to minimize terms (as I was). Also interesting would be minimizing the maximum term, maximizing the minimum term, and minimizing the maximum ratio (or difference) between terms.   2013-05-12, 21:41 #7 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 25·7·41 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 d1 < d2 < ... < dn < 10001)   2013-05-13, 05:11 #8 CRGreathouse   Aug 2006 10111001101002 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.   2013-05-16, 15:50   #9
Yamato

Sep 2005
Berlin

4216 Posts Quote:
 Originally Posted by CRGreathouse Using (99/10001)/(1/73+1/137) = 33/70. Convenient, but not good if you're trying to minimize terms (as I was). Also interesting would be minimizing the maximum term, maximizing the minimum term, and minimizing the maximum ratio (or difference) between terms.
Thx for your postet solutions, this is an interesting approach to solve this problem.

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.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Raman Math 7 2009-02-27 18:01 tinhnho Miscellaneous Math 4 2005-01-17 19:45 Cyclamen Persicum Miscellaneous Math 9 2003-04-13 14:56

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