
Re: Goldbach conjecture
Posted:
Nov 3, 1997 9:59 PM


In article <EJ3G2n.4Ep@syd.csa.com.au>, warwickp@syd.csa.com.au wrote:
> Euler saw that n^2 + n + 41 was prime for n = 0, 1, 2, ..., 39 and > conjectured that it always produced prime values:
I can assure you that Euler did no such thing.
> ...you've missed the point of my post: although there's "so damn many" > primes they end up being very sparsely distributed. Verifying a conjecture > for a large number of examples might make the conjecture "obvious" to the > lay person, but it shouldn't for the serious mathematician.
Then we're even, as you've missed the point of my post. It is not just that the conjecture is verified for a large number of examples: it is that a very much stronger conjecture is verified for as far out as any lay person or serious mathematician can go. I stick to my position that there is no serious doubt about, and no reason to doubt, the truth of the Goldbach conjecture.
The primes are not sparsely distributed, not for the purposes of Goldbach; they're just off from being all the numbers by a factor of log. Here's yet another way to think of things: there are about 1200 primes below 10000. If you chose 1200 odd numbers below 10000 at random (without replacement),  let's call them, Usenetprimes  what is the probability that you would be unable to express 10000 as a sum of two Usenetprimes?
Well, there are 5000 odd numbers to choose from, so the total number of ways of choosing is (5000 choose 1200). Now, if you pair the odd numbers off, like so: (1, 9999); (3, 9997); (5, 9995); etc., then in order to avoid two Usenetprimes adding to 10000, you have to pick at most one number from each pair. So, first pick 1200 pairs  (2500 choose 1200) ways to do this  then pick one from each pair  2tothe1200 ways to do that. So, the probability of falsifying the Usenet/Goldbach conjecture is
(2tothe1200) times (2500 choose 1200) over (5000 choose 1200).
That's about 1 in (4 times 10tothe83). That's what I meant when I said, a few posts ago, that it would take a massive conspiracy among the primes to falsify Goldbach. And it gets worse, as you go to higher numbers.
There is no similar argument in favor of Li(x) vs Pi(x), or x^2 + x + 41.
Gerry Myerson (gerry@mpce.mq.edu.au)

