Evidence for Goldbach

Goldbach's famous conjecture is that every natural number can be
expressed as a sum of three primes, which, as Euler pointed out, is
equivalent to the statement that every even number can be expressed
as a sum of two odd primes.  There has been a great deal of work on
this question, and many strong results have been proven, although
we still don't have a complete proof of the conjecture.

Of course, even though we are unable to prove that every even number
is expressible in at least one way as a sum of two odd primes, the
empirical evidence is that most even numbers are expressible in MANY
ways as a sum of two odd primes.  In fact, if we let f(n) denote the
number of distinct ways in which n can be expressed as a sum of primes,
then empirically it appears that f(2n) is greater than the square root
of 2n for all 2n greater than 15788.  Also, an upper bound on f(2n) 
seems to be given by (2n)^(2/3), although perhaps the exponent should 
be a bit greater.

Assuming a relation of the form f(2n) = (2n)^k, the value of the 
exponent k is given by k = ln(f(2n)) / ln(2n), so we can plot this 
ratio to see how it behaves.  A plot of these exponents for 2n up 
to 45000 is shown below.

  

This shows that there are two distinct populations, depending on 
whether or not n is a multiple of 3.  The reason for this is simply
that if 2n=0 (mod 3) then two primes p,q (greater than 3) such that 
p+q=2n can be either p=1,q=2 or p=2,q=1 (mod 3), whereas if 2n=1 
(mod 3) we must have p=q=2 (mod 3), and if 2n=2 (mod 3) we must have
p=q=1 (mod 3).  Hence there are twice as many "chances" for finding
suitable pairs of primes when 2n is a multiple of 3 than when it 
isn't.  If we divide each f(2n) by 2 whenever 2n is a multiple of
3, the upper band in the above figure drops down to coincide with
the lower band, cutting the width of the band in half.  (Notice that
cutting f(2n) in half before computing log(f(2n))/log(n) is equivalent
to subtracting log(2)/log(n) from the ratio.)

Similarly if 2n=0 (mod 5) we have four possibilities

          p=1,q=4    p=2,q=3    p=3,q=2    p=4,q=1

whereas if 2n=1 (mod 5) we have only three possibilities

               p=2,q=4    p=3,q=3   p=4,q=2

The other residue classes also give three possibilities each, so we
find that there are 4/3 as many chances for finding suitable pairs
of primes when 2n is a multiple of 5 than when it isn't.  So, to place
them all on a common footing, we can multiply each f(2n) by 3/4 if
2n is a multiple of 5.

Likewise for every odd prime p we can apply a factor of (p-2)/(p-1) 
to the function f(2n) if 2n is divisible by p.  This effectively
normalizes the function, and we arrive at the function

                                 p - 2
           F(m)  =   f(m)  PROD  -----
                            p|m  p - 1

In other words, F(m) is defined as f(m) multiplied by a factor of
(p-2)/(p-1) for each prime p that divides m.  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:

     

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