
Unitary (Egyptian) fractions
Posted:
Mar 31, 2000 6:38 AM


Tapio's two part question, how do you write 4/733 and 183/733 as 3term 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 7terms.
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

