## RE odd perfect number

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• [I tried to send this email several times these days. I hope this is the good one] ... First time I see it. Is it a result of yourself? Do you have a proof? If
Message 1 of 3 , Sep 8, 2006
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[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.
>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?

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
• ... 1) Obviously, if A is even perfect number, it must be of the form A = 2^(n-1)*M_n, where M_n = (2^n - 1) is a Mersenne prime. The proof is straightforward,
Message 2 of 3 , Sep 9, 2006
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>> I posted it, but it didn't go.
>> 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?

1) Obviously, if A is even perfect number, it must be of the form
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
> Mp), so why do you expect it to be applicable to odd perfects?

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

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• ... I was aware of that. I was just asking for a proof of the claim as it was first stated, involving all perfect numbers (even the theoretical odd ones!). I m
Message 3 of 3 , Sep 9, 2006
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>1) Obviously, if A is even perfect number, it must be of the form
>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.

I was aware of that. I was just asking for a proof of the claim as it was first stated,
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
>2*perfect number. 18, 20, ... The formula does not (and also -can- not)
>hold for them.

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

Regards. Jose Brox
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