What is known about Goldbach's conjecture, that every even integer greater than 4 is the sum of two odd...

In article <9124.25db4280@amherst.bitnet> ksbolduan@amherst.bitnet writes:
> 2. The Goldbach Conjecture
>         The hyptheses that every even integer, i, where i >=6
> could be represented as the sum of two odd primes, and following from
> this that every odd number j, where j>=9 could be represented as the sum
> of three odd primes.  This was proposed in the 16th century by a
> conjecture: 9=3+3+3, 13=3+5+5, 21=15+3+3.  The first conjecture
> has been verified up to 100,000.

[ Editor: in the quoted text, 15 is not prime (21 = 7+7+7 would do),
and Goldbach made the conjecture in a letter to Euler in 1742, the 18th
century. ]

Again, let me correct some of this.

Goldbach's conjecture has been verified to 10^8.

Chen Jing-Run has shown that all large enough even integers are the sum
of a prime and a number having at most two prime factors.

The conjecture of Hardy & Littlewood gives the number of representations
N2(n) of an even number n as the sum of two primes as

2cn/(log^2 n) product( (p-1)/(p-2))

where 2c ~ 1.3203  and the product is taken over all odd prime divisors of n.

Stein and Stein have calculated N2(n) up to 10^5.

I also refer the reader to:

H.L. Montgomery & R.C. Vaughan [Editor: spelled "Vaughn" in the
original], "The exception set in Goldbach's problem"
Acta Arithmetica V. 27 (1975)

It is known that the exception set has asymptotic density 0.

> Another interesting way to think about the first part of this problem is as
> follows:
>
> Every even integer >6 can be thought of as 2n, and 2n = (n+x) + (n-x).
> Therefore, if we can find prime numbers equidistant (on a numberline) from any
> n, then those two primes fit Goldbach's conjecture. E.g. if we're trying to
> make 36 from two primes, we can look at 18 and find a number (in this case 1)
> when added to 18 gives a prime (19) and subtracted from 18 gives another prime
> (17). After some work on this with a computer, I found that Goldbach's
> conjecture works this way for all even n up to 2000.

It's much faster to generate the set of primes, then use hash tables
to verify a particular n.

> ... two primes which can be added together to make an even number, but
> ones that happen to be equidistant from (1/2)n.

This is true of ANY two numbers that add up to n. If a+b = n, and n
is even, then a and b are equidistant from n/2.

> For some numbers, x must be rather large, but it seems that we can then
> rephrase Goldbach's Conjecture in this manner, which, if nothing else is a
> fresh way of seeing things. (Of course someone may have already come up with
> this, and if so, could someone give me a reference?)

It's not very useful.

-- Bob Silverman <bs@linus.mitre.org>