Are the factors of high perfect numbers known? - Page 3

Primeinator · 2009-07-22, 23:00

Quote:

Originally Posted by David John Hill Jr

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.

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

Primeinator · 2009-07-23, 01:30

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

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 = 2*p factors

Just something interesting I just noticed.

Kevin · 2009-07-23, 07:20

Quote:

Originally Posted by Primeinator

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

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 = 2*p factors

Just something interesting I just noticed.

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

David John Hill Jr · 2009-07-23, 08:09

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--------

2009-07-23, 08:09	#26
David John Hill Jr Jun 2003 Pa.,U.S.A. 2²×7² Posts	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--------

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2009-07-23, 01:30	#24
Primeinator "Kyle" Feb 2005 Somewhere near M52.. 3·5·61 Posts	Additionally, if we consider $(2^p -1)(2^{p-1})=\sum_{i=0}^{p-1} {2^i} + \sum_{j=0}^{p-2} {2^j(2^p-1)}$ 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 = *2p** factors Just something interesting I just noticed.