As in, why does the iteration of the function until $g_{64}$ guarantee this property that defines Graham's number? Why was this number chosen?
If I had to guess (emphasis on guess), I'd say that the Ramsey theoretical problem involving Graham's number involves ${4 \choose 2} = 6$ line segments between four points and two ways to color each, and $2^6 = 64$. But I don't know at all.
It appears that there is no explanation because the 64 in "Graham's number" doesn't come from anywhere! The 64 doesn't appear in Graham and Rothschild's original paper on the topic, “Ramsey's theorem for $n$-parameter sets”; instead the paper has (p.290):
We introduce a calibration function $F(m,n)$ with which me may compare our estimate of $N^*$. This is defined recursively as follows: $$\begin{align} F(1,n)=2^n \qquad F(m,2)=4 &\qquad m\ge 1, n\ge 2, \\ F(m,n) = F(m-1, F(m,n-1)) & \qquad m\ge2, n\ge 3. \end{align} $$ ...
The best estimate we obtain this way is roughly $$N^* \le F(F(F(F(F(F(F(12,3),3),3),3),3),3),3).$$
and according to this post by John Baez:
I asked Graham. And the answer was interesting. He said he made up Graham's number when talking to Martin Gardner! Why? Because it was simpler to explain than his actual upper bound - and bigger, so it's still an upper bound!
