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...

[QUOTE=alpertron;182118]Except when a=b. Let's get the factors of 9. 1 is paired with 9, and 3 is paired with...[/QUOTE]

3 is paired with 3.......

[QUOTE=jasong;181666]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?[/QUOTE]

Mini-Geek makes a good point, as shown by axn.

[QUOTE=jasong;182024]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'll let others worry about maybe testing the other numbers.
[/QUOTE]

M(5) = 2^5 -1 = 31. The associated perfect number is (2^4)(31) = 496

Using axn's approach, which is a direct form of what you posted yourself, the factors of 496 are :

a) [B]1[/B], [B]2[/B], 2^2 = [B]4[/B], 2^3 = [B]8[/B], 2^4 = [B]16[/B]
b) 2^5 -1 = [B]31[/B], 2(2^5 -1) = [B]62[/B], 2^2 * (2^5 -1) = [B]124[/B], 2^3 * (2^5 -1) = [B]248[/B], 2^4 * (2^5 -1) = [B]496[/B]

If you want to check: Simply construct a factor tree or, since this is a perfect number, add the factors except for the perfect number itself.

1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = [B]496[/B]

It checks out. I'm sure there is a relatively elementary proof for this that is undoubtedly closely tied to the form of a perfect number of (2^n-1)(2^(n-1))

I'm sure this has been covered before, but could a switch from looking for Mersennes to looking for perfect numbers be beneficial? We may be doing this ass-backwards, maybe searching for perfect numbers is the better plan.

[QUOTE=jasong;182187]I'm sure this has been covered before, but could a switch from looking for Mersennes to looking for perfect numbers be beneficial? We may be doing this ass-backwards, maybe searching for perfect numbers is the better plan.[/QUOTE]

I don't think so...because look at the form of the associated perfect number. It still requires knowing 2^n to find a perfect number, defeating the whole purpose of looking for mersenne associated perfect numbers to find a mersenne prime. One could consider using the factor scheme to construct such a number, but then you still don't have the means of knowing if 2^n -1 is prime without testing, once again defeating the whole purpose. It is a nice idea though.

[QUOTE=jasong;182187]but could a switch from looking for Mersennes to looking for perfect numbers be beneficial?[/QUOTE]

Nobody knows how to do that. The only reasonable method known to find even perfect numbers is to first find Mersenne primes. It's unlikely a better approach will be found because it would automatically give an approach to testing Mersenne primes that is better than the Lucas-Lehmer test - a very tough standard.

[quote=jasong;182187]I'm sure this has been covered before, but could a switch from looking for Mersennes to looking for perfect numbers be beneficial? We may be doing this ass-backwards, maybe searching for perfect numbers is the better plan.[/quote]
I was thinking the same thing, with circular reasoning: I thought if
[tex]2^p-1=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex] (a mathematical form of the series of all factors given previously, caring only about the sum it produces)
then 2^p-1 must be prime, but in fact the sum of all of the factors of 2^p-1 is only represented by the above equation when 2^p-1 is prime (the equation is always true, what you might take it to imply isn't). Like others have said, there is no easy way to find perfect numbers directly.

[QUOTE=Mini-Geek;182225]I was thinking the same thing, with circular reasoning: I thought if
[tex]2^p-1=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex] (a mathematical form of the series of all factors given previously, caring only about the sum it produces)
then 2^p-1 must be prime, but in fact the sum of all of the factors of 2^p-1 is only represented by the above equation when 2^p-1 is prime (the equation is always true, what you might take it to imply isn't). Like others have said, there is no easy way to find perfect numbers directly.[/QUOTE]

Shouldn't this be: [tex]Pn of Mp=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]

aka the associated perfect number of any mersenne prime is denoted by the sum of all its proper factors as shown by the sum of the two series. Alternatively...you could represent it this way as well:

[tex](2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]

[QUOTE=alpertron;182118]Except when a=b. Let's get the factors of 9. 1 is paired with 9, and 3 is paired with...[/QUOTE]Someone is failing to draw the distinction between "factors" and "distinct factors".

Paul

[quote=Primeinator;182229]Shouldn't this be: [tex]Pn of Mp=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]

aka the associated perfect number of any mersenne prime is denoted by the sum of all its proper factors as shown by the sum of the two series. Alternatively...you could represent it this way as well:[/quote]
Yes, my mistake.
[quote=Primeinator;182229][tex](2^p -1)(2^(p-1))=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]
Though here it should be known that on the second function on the left side of the equation, the entire part p-1 is exponentiated, but it is hard to tell for some reason (I'm not very good with Tex)[/quote]
[tex](2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]
Use {} to do what you're trying to do. 2^{p-1} instead of 2^(p-1).

[QUOTE=Mini-Geek;182231]
Use {} to do what you're trying to do. 2^{p-1} instead of 2^(p-1).[/QUOTE]

Thanks, that does make a big difference! I need to get into the habit of using tex instead of just typing things out.

A case where the perfect rendition might be faster
 

I had been meaning to post this elsewhere , but this looks like a better place.

Of very rare occurence and usage.

Begin at (p-1)!+1= K*p (AS befitting Wilson's theorem.)

Perform K+1, then as given 2^p-1 and a corresponding 2^(p-1), then
2^(p-1) | K+1

In reverse , as an integer multiple of 2^(p-1) subtract 1, and one has K, the
Wilson's theorem proof coefficient.

Two cases where this appears to hold(and possibly on up by induction)
is 2^3-1 and 2^7-1.(and should therefor really be included in the given statement)


As 'academic' to the whole picture that this may appear, they are(exist) cases
where going the perfect way should be a lot faster than full scale wilson calculation.

[QUOTE=David John Hill Jr;182267]I had been meaning to post this elsewhere , but this looks like a better place.

Of very rare occurence and usage.

Begin at (p-1)!+1= K*p (AS befitting Wilson's theorem.)

Perform K+1, then as given 2^p-1 and a corresponding 2^(p-1), then
2^(p-1) | K+1

In reverse , as an integer multiple of 2^(p-1) subtract 1, and one has K, the
Wilson's theorem proof coefficient.

Two cases where this appears to hold(and possibly on up by induction)
is 2^3-1 and 2^7-1.(and should therefor really be included in the given statement)


As 'academic' to the whole picture that this may appear, they are(exist) cases
where going the perfect way should be a lot faster than full scale wilson calculation.[/QUOTE]

Doesn't this still require knowing the value of p?

Additionally, if we consider
[tex](2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]

then the number of distinct factors for the associated perfect number of any Mersenne prime can be given by:

factors = 1 + (p-1) + 1 + (p-2) + 1 = (2p-3) + 3 = [B]2*p[/B] factors

Just something interesting I just noticed.

[QUOTE=Primeinator;182298]Additionally, if we consider
[tex](2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}[/tex]

then the number of distinct factors for the associated perfect number of any Mersenne prime can be given by:

factors = 1 + (p-1) + 1 + (p-2) + 1 = (2p-3) + 3 = [B]2*p[/B] factors

Just something interesting I just noticed.[/QUOTE]

That's pretty easy to see directly. Assuming 2^{p}-1 is prime, all of your factors are either one of $p$ powers of 2 (2^0,2^1,/ldots, 2^{p-1}), or one of $p$ powers of two times the associated Mersenne prime ((2^0*(2^{p}-1),2^1*(2^{p}-1),/ldots, 2^{p-1}*(2^{p}-1)), for $2p$ distinct factors.

/too drunk to bother with the tex tags

In response to 2nd to last entry
 

To clarify, I was not suggesting a shortcut for a search for a p in general. The examples are just that , as occuring.For examples much higher the numbers are too great to verify with my software, and so , to verify a pattern
of the division by some 2^(p-1), as going this way.
I was simply noting the occurence of the +1 to the K of Wilson and division
by the 2^(p-1) as being another whole number, and further as far as computing it might be a faster way around.


As to the original thread post, back to the last entry--------