1:02 AM
Woo, I'm back and bring with me a cool upper/lower bound to a particular case in the fast growing hierarchy!
$$n\operatorname{Tet}(2^n,n,1)\le f_3(n)\le\operatorname{Tet}(2\sqrt[n]n,n,n)$$
where $$\operatorname{Tet}(a,b,c)=\underbrace{a\widehat{} a\widehat{}\dots\widehat{}a\widehat{}}_bc=\begin{cases}c, &b=0\\a\widehat{}\operatorname{Tet}(a,b-1,c) ,&b>0\end{cases}$$
(the little arrows are supposed to be exponent thingies)
@user21820 @StevenH. you guys might be interested in the above.