User blog comment:GamesFan2000/Cubic graphing function (extremely powerful, I think)/@comment-30754445-20190217081315

This is weaker than tetration. In fact, G(n) < n↑↑4 for n>3.

The number of available points is 8xn3x103n which is less than 104n (for n>3)

Each of these points can either be in our graph or not, so the total possible number of point-sets is 2 to the power of the previous number. That's 210 4n < 1010 4n.

For a point-set of N points, the number of possible graphs is exactly 2N(N-1)/2 (see here) which is less then 2N 2.

So an upper bound of G(n) would be:

1010 4n x2(10 4n)2 = 1010 4n x210 8n <10108n < n↑↑4 for n>3