Are the factors of high perfect numbers known?

[jasong]

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?

[axn]

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)

[Mini-Geek]

Originally Posted by jasong

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?

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)

[philmoore]

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.

[Mini-Geek]

Originally Posted by philmoore

I think Jason was asking about all of the factors, not just the prime factors. The answer of axn is correct.

Oh, ok. Then yeah looks like axn's answer is correct.

Originally Posted by philmoore

His last factor in part b is the number itself, not a proper factor.

He also lists 1 as the first factor in part a, so he's apparently including the trivial factors (1, itself) as well.

[Mr. P-1]

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.

[lavalamp]

Originally Posted by Mini-Geek

He also lists 1 as the first factor in part a, so he's apparently including the trivial factors (1, itself) as well.

1 + 2 + 3 = 6

You would not include 1?

[jasong]

Originally Posted by axn

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)

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. :)

[CRGreathouse]

Originally Posted by jasong

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.

No, for b you get 3 and 6.

Originally Posted by jasong

Just because I'm anal, let's test it. :)

Isn't it obvious?

[R.D. Silverman]

Originally Posted by lavalamp

1 + 2 + 3 = 6

You would not include 1?

Factors come in PAIRS; N = ab.

If you list 1, then you should N.

[alpertron]

Originally Posted by R.D. Silverman

Factors come in PAIRS; N = ab.

If you list 1, then you should N.

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