I have been studying odd perfect numbers, and I have noticed that there
cannot be an odd perfect number below 45047 (no, I didn't try them all).
Then I proved that there must be at least nine factors by some very limited
trial and error (only about 10 cases were necessary). It seems quite likely
that if there is an odd perfect number, then it will have a number of
factors that comes close to what I call a triangular choose, meaning, for
example, 4C1+4C2+4C3+4C4 for 4. It would probably be slightly less. If it
helps.
SJRS
----- Original Message -----
From: Pa'l La'ng <[EMAIL PROTECTED]>
To: Fred W. Helenius <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Sunday, May 14, 2000 6:23 AM
Subject: Re: Mersenne: To crazy to be true, but it is....
> "Fred W. Helenius" wrote:
>
> > No, this only works one way: Given a perfect number, the sum of the
> > reciprocals
> > of its divisors (excluding 1) is 1. It doesn't work the other way
around;
>
> 1) Really, I cocentrated only to the words" weird"and " off topic" and
> unforgiveably I
> was unexact. The word "general" must have been replaced by the word
"special".
> I apologize.
>
> > 1/2 + 1/4 + 1/6 + 1/12 = 1, but the LCM, 12, isn't perfect.
>
> 2) This is not counter-example, because multiplied by 2 we receive
>
> 1 + 1/2 +1/3+1/6 = 2 and 1/2 +1/3 +1/6 = 1 (This is the
> base-equation, the LCM is perfect,
> as the number "6" , too. This procedure could be repeated by any other
perfect
> number)
> >
> >
> > > 3) So the existance (or non-existance) of producing the unity
as the sum
> > >of a finite number of reciprocals of odd numbers is equivalent with the
> > >existance (or non-existance)
> > >of odd perfect numbers.
>
> 3) To the word "special". The unity as a sum of odd reciprocals has the
> following
> special necessery and sufficient conditions to be equivalent of the
existance or
> non-existance
> of odd perfect numbers ( according to Euler):
> 31) The number of the summands must be odd
> 32) The reciprocals must be either those of squares, or those
of
> such other
> prime-powers being not the divisors of any square(the exponents of this
prime
> should be
> continued from the number "1" to an odd one).
>
> > 1/3 + 1/5 + 1/7 + 1/9 + 1/15 + 1/21 + 1/27 + 1/35 + 1/63 + 1/105 + 1/135
= 1,
> > but you can't get an odd perfect number out of it.
>
> Among these denominators you find no squares, so I really can't get a
perfect
> number from this sum.
> And to find one fulfilling Euler's conditions -- , it is not easy at all.
>
> Best regards
>
> Pa'l
> La'ng , Budapest, Hungary
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