Are the factors of high perfect numbers known?
 

I accept that when you find a Mersenne prime that (2^n-1)(2^(n-1)) is a perfect number, but is there a simple way to calculate the factors of the perfect number. For instance, is there a systematic way to calculate the factors of the perfect number related to M47?

The factors will either be a) powers of 2 or b) powers of 2 times Mp.

a) 1, 2, 2^2, ... 2^(n-1)
b) 2^n-1, 2*(2^n-1), 2^2*(2^n-1), ... 2^(n-1)*(2^n-1)

[quote=jasong;181666]I accept that when you find a Mersenne prime that [B](2^n-1)(2^(n-1))[/B] is a perfect number, but is there a simple way to calculate the factors of the perfect number. For instance, is there a systematic way to calculate the factors of the perfect number related to M47?[/quote]
You just answered your question. Think for a moment about the factorization of each of these sides: the left is the Mersenne prime, and so by definition is prime. The right is a power of 2, which obviously has no factors besides 2. So the prime factorization of an even perfect number is always 2^(p-1)*(2^p-1)

I think Jason was asking about all of the factors, not just the prime factors. The answer of axn is correct. His last factor in part b is the number itself, not a proper factor.

[quote=philmoore;181684]I think Jason was asking about all of the factors, not just the prime factors. The answer of axn is correct.[/quote]
Oh, ok. Then yeah looks like axn's answer is correct.
[quote=philmoore;181684]His last factor in part b is the number itself, not a proper factor.[/quote]
He also lists 1 as the first factor in part a, so he's apparently including the trivial factors (1, itself) as well.

In general, if you know the prime factorization of a number, then you can trivially enumerate all of its factors - they are precisely the products of its prime factors raised to powers no higher than in the factorization.

[QUOTE=Mini-Geek;181686]He also lists 1 as the first factor in part a, so he's apparently including the trivial factors (1, itself) as well.[/QUOTE]1 + 2 + 3 = 6

You would not include 1?

[QUOTE=axn;181671]The factors will either be a) powers of 2 or b) powers of 2 times Mp.

a) 1, 2, 2^2, ... 2^(n-1)
b) 2^n-1, 2*(2^n-1), 2^2*(2^n-1), ... 2^(n-1)*(2^n-1)[/QUOTE]
Just because I'm anal, let's test it. :) Since only (a) includes 1, I'll assume that both a and b are supposed to be used together.

The first Mersenne prime is 3, 2^2-1. This makes the first perfect number (2^2-1)(2^(2-1))=3*2=6.

So for a this gives us 1 and 2. For b you get 3.

I'm addicted to Runescape, so I'll let others worry about maybe testing the other numbers.

Btw, I'm biguglydude3, and I hang out on world 98. :)

[QUOTE=jasong;182024]The first Mersenne prime is 3, 2^2-1. This makes the first perfect number (2^2-1)(2^(2-1))=3*2=6.

So for a this gives us 1 and 2. For b you get 3.[/QUOTE]

No, for b you get 3 and 6.

[QUOTE=jasong;182024]Just because I'm anal, let's test it. :)[/QUOTE]

Isn't it obvious?

[QUOTE=lavalamp;181739]1 + 2 + 3 = 6

You would not include 1?[/QUOTE]

Factors come in PAIRS; N = ab.

If you list 1, then you should N.

[QUOTE=R.D. Silverman;182049]Factors come in PAIRS; N = ab.

If you list 1, then you should N.[/QUOTE]

Except when a=b. Let's get the factors of 9. 1 is paired with 9, and 3 is paired with...