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Topic: unitary (Egyptian) fractions
Replies: 17   Last Post: Mar 31, 2000 6:38 AM

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Milo Gardner

Posts: 1,105
Registered: 12/3/04
unitary (Egyptian) fractions
Posted: Mar 22, 2000 5:13 AM
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Dear David, Joe, and others:

I can agree that with the first partitions 1/2, 1/3, 1/7 that four
additional partitions are needed to solve the 732/733 problem. The
algebraic identities needed to solve this problem include onlu
366/733 = 1/2 - 1/1466
245/733 = 1/3 + 2/2199

which leaves,

121/733 to be paritioned by two terms,

as well as one other term smaller than 1/733.

That is, I will now go onto the 1/2, 1/4, 1/8 first terms and get
back with you all, especially David that listed an 'optimal'
solution for 1/2, 1/4. If these fail, I will search a little
longer using highly composite first partitions that allow the

n/p - 1/A = (nA -p)/Ap


n/p = 1/A + (nA -p)/Ap

method to work, as Fibonacci himself understood in 1202 AD.

Yes, it takes a little time to directly solve this problem
even using Fibonacci's indeterminate equation method - that
I do not think that I have seen listed among Eppstein's < 40

Regards to all,

Milo Gardner

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