2020-11-13, 02:20 | #1 |
Nov 2020
1_{2} Posts |
Perfect number pass test?
I'm new to this, and I suck at math, so call me a lunatic for asking...
But I'm looking and I'm not seeing anything anywhere on what seems to be to be an obvious step. Everything seems to be working off working fully on the prime suspects directly, but there's two points of potential verification for any potential prime candidate. The potential prime itself, and the implied number that would be the corresponding perfect number. The corresponding perfect number candidate has to obey a bunch of rules to pass as a valid candidate, but more importantly it needs to have it's factors add up to itself; meaning there's a potential target for a coarse invalidation pass and as more factors are found the potential number of remaining factor pairs for a valid solution drops as the known factors eat into the factor summation headroom remaining to get a 'perfect' solution. As invalidating the perfect number candidate is as good as finding a factor in the prime candidate, I'm left wondering if this is already being done, or if I'm missing something or what. Seems like if you're going to brute force the space for all possible factors, it should be overall quicker if you have a target number that invalidates the operation if found midway to be unreachable. Am I crazy? Someone enlighten me. |
2020-11-13, 02:58 | #2 |
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×3^{2}×167 Posts |
I'm not sure what your actual question is.
Are just trying to find perfect numbers? If so, then all Mersenne primes can be used to construct an even perfect number. This is easy. A more difficult problem is to find an odd perfect number. Much work has been done already, but so far none have been identified. |
2020-11-13, 03:48 | #3 |
"Curtis"
Feb 2005
Riverside, CA
2^{2}·13·89 Posts |
I think you are asking if we could rule out primality of a Mersenne candidate by showing its corresponding perfect-number-candidate isn't perfect. But, to do that you'd have to completely factor that number.
Factoring numbers is many orders of magnitude harder than testing for primality. Numbers with special form (where the form helps us factor faster) can be factored up to about 300 digits. Beyond that size, you have to hope all the factors are small- like under 70 digits. And, you only know you're done factoring when the factors you've broken the candidate into are tested to be prime- which means a primality test on th Mersenne candidate is a component of your hope of finding a perfect number directly! |
2020-11-13, 04:09 | #4 |
Jan 2012
Toronto, Canada
110100_{2} Posts |
It has already been proven that an even number N is a perfect number if and only if N = 2^(k-1)*(2^k-1) and 2^k-1 is prime: https://primes.utm.edu/notes/proofs/EvenPerfect.html
As retina has mentioned, there has been much effort dedicated to finding odd perfect numbers and it has been shown that no such odd number exists below 10^1500 (and in all honesty is unlikely to exist). Current efforts there are focused on factoring "roadblock" numbers that can be many hundreds of digits long and even the smallest ones take months to years for a single machine to factor. We have a thread for that too: https://mersenneforum.org/showthread.php?t=19066 |
Thread Tools | |
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
I found the primality test, there seems to be no composite numbers that pass the test | sweety439 | sweety439 | 7 | 2020-02-11 19:49 |
my own Integer-based LL-test not pass 3 Mprimes!? | ktpn2011 | Software | 56 | 2019-01-17 06:11 |
Is this a Perfect Number ? | Godzilla | Miscellaneous Math | 8 | 2016-09-05 05:56 |
Composites that pass Mathematica's pseudoprime test | grandpascorpion | Math | 15 | 2012-03-23 02:52 |
please help me pass the test. | caliman | Hardware | 2 | 2007-11-08 06:12 |