Comment by pinkmuffinere
8 months ago
Sure, but the example I'm providing is just meant to illustrate that BB(n) is not greater than arbitrary f(n). I'm not trying to provide the biggest number, I'm trying to illustrate that the sketched-out proof is incorrect. If you want me to provide a bigger number, I suppose another easy example is to define f(n) = BB(BB(n)).
Edit: Oh sorry, I see I misread the direction of your greater than signs. Leaving comments as-is though, hopefully that results in the least confusion
oops, my greater than signs are in the wrong direction. specifically, for any computable function f, there exists some constant c such that f(BB(n))<<BB(n+c)