- [I tried to send this email several times these days. I hope this is the good one]

>I posted it, but it didn't go.

First time I see it. Is it a result of yourself? Do you have a proof?

>Look at the formula...

>sigma(abundant#) = 2*sigma(perfect#) + associated Mp + 1

If you have it, I bet it involves even perfect numbers (because of that Mp), so why do you

expect it to be applicable to odd perfects?

The only related result I know says that if N is a perfect or abundant number, then k·N is

abundant, but the converse theorem is not true, if k·N is an abundant number, then N

hasn't to be neither abundant nor perfect (think about 945 and 945/3).

Regards. Jose Brox >> I posted it, but it didn't go.

1) Obviously, if A is even perfect number, it must be of the form

>> Look at the formula...

>> sigma(abundant#) = 2*sigma(perfect#) + associated Mp + 1

>

> First time I see it. Is it a result of yourself? Do you have a proof?

A = 2^(n-1)*M_n, where M_n = (2^n - 1) is a Mersenne prime. The proof is

straightforward, see http://primes.utm.edu/notes/proofs/EvenPerfect.html.

Then, sigma(A) = 2^n*M_n and sigma(2A) = (2^(n+1) - 1)*2^n = 2*sigma(A) +

(2^n-1) + 1.

So, the formula holds for abundant numbers equal to twice-an-even-perfect.

2) There are -way- too many abundant numbers which are NOT of the form

2*perfect number. 18, 20, ... The formula does not (and also -can- not)

hold for them.

> If you have it, I bet it involves even perfect numbers (because of that

In fact... WHAT exactly is "associated Mp" in the case of odd perfect

> Mp), so why do you expect it to be applicable to odd perfects?

numbers?

Peter

--

[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278>1) Obviously, if A is even perfect number, it must be of the form

I was aware of that. I was just asking for a proof of the claim as it was first stated,

>A = 2^(n-1)*M_n, where M_n = (2^n - 1) is a Mersenne prime. The proof is

>straightforward, see http://primes.utm.edu/notes/proofs/EvenPerfect.html.

>Then, sigma(A) = 2^n*M_n and sigma(2A) = (2^(n+1) - 1)*2^n = 2*sigma(A) +

>(2^n-1) + 1.

involving all perfect numbers (even the theoretical odd ones!). I'm also aware that this

proof can't be achieved... it was just a rethorical message, a wait-a-minute-and-think

calling for the author.

>So, the formula holds for abundant numbers equal to twice-an-even-perfect.

Yes, I know :-)

>2) There are -way- too many abundant numbers which are NOT of the form

Any k·even_perfect(n) is an abundant number, in fact.

>2*perfect number. 18, 20, ... The formula does not (and also -can- not)

>hold for them.

Regards. Jose Brox