greedy computation of pseudoperfect numbers?

[franktaw at netscape.net]

Since the other question has been answered, let me say something
about the first one.  First of all, it is possible to have a set of 
divisors of
n summing to n whose lcm is not n.  One such is:

210, 140, 105, 84, 70, 60, 42, 35, 30, 28, 21, 15;

this sums to 840, but the lcm is only 420.

Looking at this question is perhaps easier if we divide the terms
through by n.  Now we are looking for a set of unit fractions,
excluding 1, which sum to 1, have denominators all dividing n,
and have gcd not equal to 1.  (Egyptian fractions!)  While it is
certainly possible to do this, it seems unlikely that it can be done
in such a way that one cannot improve the set of denominators
to get rid of the common factor.  In other words, the "greedy
with backtracking" algorithm should always produce a set with
gcd = 1 (so lcm = n in the original problem).

The greedy algorithm, when it finds a solution, will certainly
always find one with gcd = 1.  If p and q are the two smallest
prime divisors of n (no prime power can be abundant), and p^k
is the largest power of p < q, then the first terms will be

1/p, 1/p^2, ..., 1/p^k, 1/q.

(Even when p = 2, the sum of these terms is < 1.)  Now
p and q are relatively prime, so the gcd is 1.

----
A not-unrelated question: what is the least upper bound of
sigma(n)/n - 2 for n a weird number?  For the numbers in
b006037, the maximum is 74/1043 = .070949..., for
a(8) = 10430.  (It is clear that the lub is not obtained,
since any weird number n times any prime > sigma(n)
produces another weird number, and this will have a
larger value for this expression.  The primes required grow
quite rapidly when this operation is iterated, however.)

The maximum for any primitive weird number (A002975)
appears to be 2/35 = .0571428..., for 70, the very first
weird number.

Franklin T. Adams-Watters

-----Original Message-----
From: Joerg Arndt <arndt at jjj.de>

For both sorts of pseudoperfect numbers, allowing divisor one A005835,
and forbidding divisor one A136446, a greedy method _seems_ to always
work for deciding whther a given n in in the sequence:

...

The computation is obviously extremely fast.

Further, the lcm of all divisors taken for the sum has always lcm==1.
(cf. the line if ( l!=n,  error(" lcm=!=n ") ); which is never 
executed).

Is there any information available about whether the greedy approach
always works, and whether the lcm property holds?
I do not know whether the questions are easy or very hard.