--- In

primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@...>

wrote:

>

>

> ----- Original Message -----

> From: "leavemsg1" <leavemsg1@...>

>

>

> >Sorry,... the formula should have been posted as...

> >(3^(2n-2))*(3^(2n-1)-1)/2.

>

> A = 3^(2n-2), B = (3^(2n-1)-1)/2

> N = A·B

>

> Sigma is a multiplicative function, and B = = 1 (mod 3), which

implies CGD(A,B) = 1, so

>

> Sigma(N) = Sigma(A·B) = Sigma(A)·Sigma(B)

>

> Sigma(A) = 1+3+3^2 + ... + 3^(2n-2) = (3^(2n-1)-1)/2 = B

>

> and as we want Sigma(N) = 2N = 2AB, it follows that we need Sigma

(B) = 2A

>

> Sigma((3^(2n-1)-1)/2) = 2·3^(2n-2), can this be?

>

Forget my attempt to formalize a formula for an odd-perfect number be-

cause by your very sigma definitions... the first perfect number of

any even or odd number must occur at half the number of the first

abundant number found in the list i.e. 6 is the first perfect number

which is half of and less than 12(the first even abundant number).

Hence, by definition... the first odd perfect number would have to

occur before 945/2 which isn't a whole number.

Hence, i believe an odd-perfect number defined by the sigma function

couldn't exist.

Bill

> I thought I had a simple proof for the negative, but it was flawed.

Nevertheless, let's

> try other ways. I don't have the time to prove anyone, but I will

state them:

>

> a) I think that the only exponents that make Sigma(B) = = 2 (mod 4)

are n = 2, 4, 7, 36

> and 52. This could be mistaken easily.

> b) I also believe that Sigma(B) < 2A for every n.

> c) Probably there is yet a simpler way, but I don't see it right

now.

>

> Jose Brox

>