Theory

RichardB · 2010-04-09, 17:14

If the perfect numbers are all 2^n-1x(2ⁿ-1) and all mersenne numbers are 2^n-1 and there is a clear (admittedly on and off) binary pattern for perfect numbers (110, 11100, 111110000) why is there so much trial and error as I am told "The chance that the exponent you are testing will yield a Mersenne prime is about 1 in 421010. " ?

Sorry I don't understand and thanks in advance for any replies. :)

Mini-Geek · 2010-04-09, 17:33

All perfect numbers in binary are indeed 11...1100...00 with p 1's following by p-1 0's, (which is a direct result of the form being $(2^p-1)*(2^{p-1})$ ) but unfortunately this doesn't help any more than noting that all Mersenne numbers (not only the primes, but all numbers 2^n-1) in binary are 11...11 with p 1's (which is a direct result of the form being $2^p-1$ ).

Since all Mersenne numbers (and so all potential Mersenne primes) in binary are 11...11, and all associated potential perfect numbers are 11...1100...00, this is essentially a useless observation. We still need to determine if the Mersenne number is prime.

xilman · 2010-04-09, 18:23

Quote:

Originally Posted by Mini-Geek

All perfect numbers in binary are indeed 11...1100...00 with p 1's following by p-1 0's, (which is a direct result of the form being $(2^p-1)*(2^{p-1})$ ) but unfortunately this doesn't help any more than noting that all Mersenne numbers (not only the primes, but all numbers 2^n-1) in binary are 11...11 with p 1's (which is a direct result of the form being $2^p-1$ ).

Correction: all even perfect numbers have that form.

It is not yet known whether any odd perfect numbers exist. Many mathematicians appear believe that they do not exist. Some mathematicians believe they may exist. What is known is that if they do exist, they must be at least several hundred decimal digits long and that their prime factorization is severely constrained.

The last constraint, a very particular prime factorization, is also true for even perfect numbers, of course.

Paul

Jens K Andersen · 2010-04-09, 22:12

Quote:

Originally Posted by RichardB

If the perfect numbers are all 2^n-1x(2ⁿ-1) and all mersenne numbers are 2^n-1 and there is a clear (admittedly on and off) binary pattern for perfect numbers (110, 11100, 111110000) why is there so much trial and error as I am told "The chance that the exponent you are testing will yield a Mersenne prime is about 1 in 421010. " ?

Only few of the Mersenne numbers are prime numbers, and only the prime Mersenne numbers lead to a perfect number. The computational challenge is to determine which of the Mersenne numbers are prime. With the best known methods it can take weeks or months to test a single Mersenne number that hasn't been tested before. And the estimated chance that the tested number turns out to be prime may be about 1 in 420000 (depending on the exact size and factorization effort).

RichardB · 2010-04-10, 02:11

OK thanks everyone :) got it

Just a quickie, can I take the 100 cpu limit off the program, I have a dual core (200 cpu points) iMac so it could run alot faster.

Mini-Geek · 2010-04-10, 14:13

Quote:

Originally Posted by RichardB

Just a quickie, can I take the 100 cpu limit off the program, I have a dual core (200 cpu points) iMac so it could run alot faster.

Recent versions (25.x) of Prime95/mprime support multithreading (i.e. letting you use more than one core).
You can get it from this page:
http://www.mersenne.org/freesoft/
Here's the link for Mac OS X:
http://mersenneforum.org/gimps/Prime95-MacOSX-2511.zip
Once you're running a version that supports multiple cores, it should automatically configure itself to run on all available cores. If not, go to Test > Worker Windows and configure it there. You'd want to run 2 worker windows, with each using one core.

RichardB · 2010-04-10, 18:39

Thanks, found it, I was a bit confused with all the menus, I had looked in preferences and 'CPU' option windows.

2010-04-09, 17:14	#1
RichardB Apr 2010 England 2×7 Posts	Theory If the perfect numbers are all 2^n-1x(2ⁿ-1) and all mersenne numbers are 2^n-1 and there is a clear (admittedly on and off) binary pattern for perfect numbers (110, 11100, 111110000) why is there so much trial and error as I am told "The chance that the exponent you are testing will yield a Mersenne prime is about 1 in 421010. " ? Sorry I don't understand and thanks in advance for any replies. :)

2010-04-09, 17:33	#2
Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 7×13×47 Posts	All perfect numbers in binary are indeed 11...1100...00 with p 1's following by p-1 0's, (which is a direct result of the form being $(2^p-1)(2^{p-1})$ ) but unfortunately this doesn't help any more than noting that all Mersenne numbers (not only the primes, but all numbers 2^n-1) in binary are 11...11 with p 1's (which is a direct result of the form being $2^p-1$ ). Since all Mersenne numbers (and so all potential Mersenne primes) in binary are 11...11, and all associated potential perfect numbers are 11...1100...00, this is essentially a useless observation. We still need to determine if the Mersenne number is prime. Last fiddled with by Mini-Geek on 2010-04-09 at 17:40*

2010-04-10, 02:11	#5
RichardB Apr 2010 England 2×7 Posts	OK thanks everyone :) got it Just a quickie, can I take the 100 cpu limit off the program, I have a dual core (200 cpu points) iMac so it could run alot faster. Last fiddled with by RichardB on 2010-04-10 at 02:12

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2010-04-10, 18:39	#7
RichardB Apr 2010 England 2×7 Posts	Thanks, found it, I was a bit confused with all the menus, I had looked in preferences and 'CPU' option windows.