----- 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?

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

wrote:>

implies CGD(A,B) = 1, so

>

> ----- 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

>

(B) = 2A

> 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

>

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

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

>

are n = 2, 4, 7, 36

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

> and 52. This could be mistaken easily.

now.

> 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

>

> Jose Brox

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

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

>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.

Why is that so? If you have a sound argument for that idea, i.e., that the first perfect

number must occur at half of the first abundant number... then you could gain fame and

fortune instantly! The Odd Perfect's Conjecture is considered to be a hard one.

Regards. Jose Brox