Date: Mar 31, 2000 6:38 AM
Author: Milo Gardner
Subject: Unitary (Egyptian) fractions

Tapio's two part question, how do you write 4/733 and 183/733

as 3-term series, is easily answered in terms of

4/733 = 1/184 + 1/67436 + 1/134872

which means that

732/733 = 183/184 + 183/67436 + 183/134872

still a very difficult situation.

3/733 = 1/367 + 1/733 + 1/269011

also implies that

732/733 = 244/367 + 244/733 + 244/269011,

another very difficult problem to write in fewer than 7-terms.

What I was referencing was that an additive selection of

n/733 members be considered to find the 732/733 'optimal'

series,

such as,

367/733 = 1/2 + 1/1466

184/733 = 1/4 + 1/1466 + 1/2932

or,

551/733 = 1/2 + 1/4 + 1/733 + 1/2932

leaving

181/733 = 1/a + 1/b

which I have not been able to find.

Yes, this is a very difficult problem is the shortest series

is the only goal. However, considering the smallest last term

issue, the problem even becomes more interesting.

Regards to all,

Milo