User blog comment:I am a McCree God/I think I have found how many arrows Graham's Number has./@comment-1605058-20171027080749

This is not even close. Already the number \(g_2\) is \(3\uparrow^{g_1}3\), and the number of arrows is \(g_1=3\uparrow\uparrow\uparrow\uparrow 3\). This evaluates to an enormous power tower of threes, and note that already \(3^{3^{3^{3^3}}}\) is much bigger than the number of arrows you give.

And that's still \(g_2\), let alone \(g_{64}\).