Heuristic argument on Goldbach's conjecture

Question

If $p$ is prime number $p \not = 2,3$ then $p= \pm 1 \bmod 6$

If $n$ is even:

case I

$n= 0 \bmod 6$

$n=(+1+6 \cdot a)+(-1+6 \cdot b)=p_a+p_b$

$p_a= 1 \bmod 6,p_b= -1 \bmod 6$

case II

$n= 2 \bmod 6$

$n=(+1+6 \cdot a)+(+1+6 \cdot b)=p_a+p_b$

$p_a,p_b= 1 \bmod 6$

and

$n=3+(-1+6 \cdot c)=3+p_c$

$p_c= -1 \bmod 6$

case III

$n= -2 \bmod 6$

$n=(-1+6 \cdot a)+(-1+6 \cdot b)=p_a+p_b$

$p_a,p_b= -1 \bmod 6$

and

$n= 4 \bmod 6=3+(+1+6 \cdot c)=3+p_c$

$p_c= 1 \bmod 6$

Example case III

for $ n=100$

In this graph the pairs of numbers $n=p_a+p_b$

$p_a,p_b= -1 \bmod 6$

Note the symmetry with respect to $n/2$.

and

$n= 3+ 97$

$97= 1 \bmod 6$

Question

Considering for example $n= -2 \bmod 6$ with $n-3$ composite. Considering that for large values of $n$ the distance between two consecutive primes increases. Is it possible to say that it is very likely that there is a large $ n $ such that for every prime number $ p $ less than or equal $ n / 2 $ corresponds to a composite number $ (n-p) $?

Answer 1

Heuristic arguments don't prove anything, but they can provide an indication of what's likely to be true. In this case, heuristic reasoning seems to favor Goldbach:

For $n=6k-2$ there are $k$ numbers congruent to congruent to $-1$ mod $6$. Roughly $k/2$ of them are less than $n/2$ and $k/2$ of them are greater than $n/2$. Of the ones less than $n/2$, roughly $(k/2)/\ln(k/2)\approx k/(2\ln k)$ are prime (using $\ln(k/2)=\ln k-\ln2\approx\ln k$ when $k$ is large). For each such prime $p$, the probability that $n-p$ is prime is (very) roughly $1/\ln n\approx1/\ln(6k)\approx1/\ln k$, which means the probability it's composite is (again very) roughly $1-1/\ln k$. This means the probability that $n-p$ is composite for all the relevant primes $p$ is (very!) roughly

$$\left(1-{1\over\ln k}\right)^{k/(2\ln k)}=\left(\left(1-{1\over\ln k}\right)^{\ln k}\right)^{k/(2(\ln k)^2)}\approx e^{-k/(2(\ln k)^2}\lt{1\over k^2}$$

so the expected number of counterexamples to Goldbach for numbers of the form $n=6k-2$ with $k\ge K$ (which is easy to make fairly large) is, heuristically, less than

$$\sum_{k=K}^\infty{1\over k^2}\approx{1\over K}$$

The important thing here really is that the series representing the expected number of counterexamples is convergent, suggesting that the number of counterexamples is finite.