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Old 2018-12-04, 00:08   #11
goldbug
 
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Dec 2018

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Let me know if this holds water. The numbers I am searching for satisfy the following property. Maybe there is an easier way to search besides looking each order k=2,3,4,... separately?



Given an even number 2n there exists some subset of the prime non-divisors of n 2<p1<p2<p3<...<pk<n such that (2n-p1)(2n-p2)(2n-p3)...(2n-pk) only has p1,p2,p3,...,pk as factors.


So if any of the 2n-p are prime it fails or if 2n-p has a factor besides the k primes it fails.



It would seem that Goldbach is true for numbers which do not have this property. Imagine a process where one starts with one prime non-divisor p. In this case 2n-p must have another as prime non-divisor as a factor if it is not prime. Then, given the two prime non-divisors p1 and p2, we can find another factor from (2n-p1)(2n-p2) and so on. But this process cannot continue forever so 2n-p must be prime for one of the prime non-divisors. This would be a proof by infinite decent I think, but of course only for numbers without the property above.



However, even if the above conjecture is true it is not clear to me how many numbers with this property there are. If the order 2 search is any indication it may be that there are very few maybe none beyond 2200?

Last fiddled with by goldbug on 2018-12-04 at 00:20
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