On 07/24/2012 10:52 AM, Ben Bacarisse wrote: > mluttgens<email@example.com> writes: > <snip> >> Thank you! You are of course right. >> >> But my aim was to show that a sum s? = a + b of two uneven numbers, at >> least one of them not being a prime, could easily be transformed into >> a sum of two primes, simply by adding and subtracting some even number >> from its terms: >> >> The chosen example was: >> >> s? = 13 + 15 = (13-8) + (15+8) = 5+23 >> = (13-2) + (15+2) = 11+17 >> >> It has been claimed that such transformation could sometimes not be >> possible. >> I am wondering about which terms a and b should be chosen to justify >> that claim. >> Till now, I did not find a clue in the litterature, but you have >> perhaps a reference? > > Your transformation is possible if GC it true and false otherwise. > Every counter-example to GC (of which none are known, of course) would be > an example of what you seek with s = 1 + (s-1). Computers have checked > GC up to about 10^18, but since almost everyone thinks GC is true, why > would you go searching for a counter-example? > > Every reference in the literature about GC is a reference that will > help you in your quest, because your statement about transforming > non-prime sums into prime sums is exactly the same as GC. >
Kevin Brown has a rather unique kind of presence on the Web. His math pages rarely mention the name of the author (himself). I read that there are no links going to other web-sites there ...
In any case, Kevin Brown is listed in the Numericana Hall of Fame along with other distinguished web-authors:
In his essay "Evidence for Goldbach", Brown tries to compensate the number of prime partitions of an even number 2n for/(according to) the residue class (modulo 3) of 2n, with a logical argument. There's further compensation [justified probabilistically] for 2n (modulo p) for all larger odd primes p.
The end result, quoting K.B., << If we plot the log of this function divided by the log of n we find that the scatter is reduced almost entirely to a single line as shown below: >>