[seqfan] Re: Egyptian fractions over the primes A165558

[franktaw at netscape.net]

For any sum e_i/p_i with distinct primes p_i, if a particular p_i does 
not divide e_i, there will be a p_i in the denominator.  Thus to get an 
integer sum, all the terms must be integers.  This gives us two classes 
of solutions: p_1/p_1 + p_2/p_2, and (2p_1)/p_1.

Franklin T. Adams-Watters


-----Original Message-----
From: Richard Mathar <mathar at strw.leidenuniv.nl>

For those who like puzzles, here is one tightly related to extending
http://research.att.com/~njas/sequences/A165558 non-trivially:
How can the number 2 be decomposed into a sum over distinct weighted 
inverse
primes?
I.e., find solutions to
2 = e_1/p_1 + e_2/p_2 +e_3/p_3 +...
where p_i are a set of distinct primes, and e_i >=1 is a set of 
associated
integers (which ultimately play the role of prime powers in A165558).
The case 2 = 4/2 with p_1=2 and e_1=4 is so far the only case shown in 
A165558
which
is not of the "trivial" kind of selecting e_i=p_i, i=1,2, for an 
arbitrary set
of 2 distinct primes.
(This may already have been answered somewhere else, for example in the 
OEIS.. )