Are the factors of high perfect numbers known? - Page 2

R.D. Silverman · 2009-07-21, 23:15

Quote:

Originally Posted by alpertron

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

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

Primeinator · 2009-07-22, 05:05

Quote:

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?

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

Quote:

Originally Posted by jasong

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.

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

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 = 496

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))

jasong · 2009-07-22, 05:09

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.

Primeinator · 2009-07-22, 05:16

Quote:

Originally Posted by jasong

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.

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.

wblipp · 2009-07-22, 05:33

Quote:

Originally Posted by jasong

but could a switch from looking for Mersennes to looking for perfect numbers be beneficial?

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.

Mini-Geek · 2009-07-22, 14:05

Quote:

Originally Posted by jasong

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.

I was thinking the same thing, with circular reasoning: I thought if
$2^p-1=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}$ (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.

Primeinator · 2009-07-22, 14:58

Quote:

Originally Posted by Mini-Geek

I was thinking the same thing, with circular reasoning: I thought if
$2^p-1=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}$ (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.

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

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:

$(2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}$

xilman · 2009-07-22, 15:04

Quote:

Originally Posted by alpertron

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

Someone is failing to draw the distinction between "factors" and "distinct factors".

Paul

Mini-Geek · 2009-07-22, 15:04

Quote:

Originally Posted by Primeinator

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

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:

Yes, my mistake.

Quote:

Originally Posted by Primeinator

$(2^p -1)(2^(p-1))=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}$
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)

$(2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}$
Use {} to do what you're trying to do. 2^{p-1} instead of 2^(p-1).

Primeinator · 2009-07-22, 15:08

Quote:

Originally Posted by Mini-Geek

Use {} to do what you're trying to do. 2^{p-1} instead of 2^(p-1).

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

David John Hill Jr · 2009-07-22, 19:59

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.

2009-07-22, 19:59	#22
David John Hill Jr Jun 2003 Pa.,U.S.A. 2²·7² Posts	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.

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2009-07-22, 05:09	#14
jasong "Jason Goatcher" Mar 2005 3×7×167 Posts	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.