That link reveals that a certain heuristic but extremely plausible assymptotic formula is highly accurate at least up to $10^{10}$. That information tells us what a graph over a larger range would look like. So one knows certain patterns are there even if the graph is not drawn. (Maybe someone has drawn it, I don't know).
Roughly, the expected number of pairs is about $0.66\frac{n}{{\log}^2(n)}$ for a large number which is twice a prime. Multiply that by $\prod_{p|n}\frac{p-1}{p-2}$ to get the estimate for any even n (the product over odd prime divisors of $n$).
This explains these patterns (do you see others?): There should be a lower half very roughly hitting at $10^6$ from 3460 to 6700 and then an upper half from 6920 to 13400 about twice the lower part. The lower part is for 2 or 4 mod 6 and the upper for mutiples of 6. Numbers which are or are not multiples of 5 and/or 7 should create 4 bands in each half (I can't see much beyond that). Those patterns (including with more primes taken into account)continue very faithfully as far as calculated.
added information
The name "Goldbach's Comet" is probably not useful in finding extended computations. "Goldbach partitions" might do better. As far as I can tell the state of the art is
MR1850627 (2002g:11142) Richstein, Jörg . Computing the number of Goldbach partitions up to $5\cdot 10^8$. Algorithmic number theory (Leiden, 2000), 475--490, Lecture Notes in Comput. Sci., 1838, Springer, Berlin, 2000.
That paper does include a graph for the range $500500000 \le n \le 500660160$. The author no doubt has the data and could generate other graphs, but may not see any reason to. There are a number of observations in that paper about patterns. You might be better served by looking at color coded plots over short ranges (say according to congruence class mod 30 or 210). for example a plot over a modest range colored mod 6 shows that the top is all of the $0 \mod 6$ and nothing else (of course) BUT it also shows that the very bottom is boundry is heavily in favor of $2 \mod 6$ leading me to pose this question.
