User blog comment:TySkyo/The Super-duper Big and Huge Number/@comment-5529393-20160803173437

Actually, your number is smaller than Graham's number. Let's break it down.

$$10^{10^{10^{10^{10^{246.6198748165387}}}}}$$ is less than $$10 \uparrow\uparrow 7$$. The double factorial is a bit faster than exponential, but much slower than doubly exponential, so taking the double factorial of that number gets you a number less than $$10 \uparrow\uparrow 8$$. Same with the superfactorial; we wind up with a number less than $$10\uparrow\uparrow 9$$.

Then, you apply $$A(n_0, n_0\uparrow^{n_0} n_0)$$. $$n_0\uparrow^{n_0} n_0$$ is less than $$A(n_0 + 2, 2n_0)$$, so we have $$n_1 = A(n_0, n_0\uparrow^{n_0} n_0) < A(n_0, A(n_0 + 2, 2n_0)) < A(n_0+3, n_0 + 3) = A(n_0+3)$$. (Here I am defining A(n) = A(n,n).). So $$n_1 < A(10\uparrow\uparrow 9)$$.

Then we take the factorial of n_1, which makes basically no difference to a number expressed in Ackermann notation, as $$A(n)! < A(n+1)$$. So $$n_1 ! < A(10\uparrow\uparrow 9)$$ Then we apply the Ackermann function to it, so $$SBAH < A(A(10\uparow\uparrow 9))$$

Now, compare this to the construction of Graham's number: $$G_1 = 3\uparrow^4 3 > 10 \uparrow\uparrow 9$$. Then for each subsequent $$G_i$$ we apply $$3 \uparrow n \uparrow 3 > A(n)$$, so $$G_3 > A(A(10\uparrow\uparrow 9)) > SBAH$$. Graham's number is $$G_{64}$$, so it's much greater than SBAH.

TREE(3) is in a completely different universe in terms of size, and requires more sophisticated recursion techniques even to get to the best lower bound.