But of course BB can get dwarfed by other uncomputable functions, like Rayo's:
f(n)=n (n Knuth up arrows) n
Coming back to Rayo, my number is instead f(10^100) rather than Rayo(10^100), and given the series of lower bounds presented in that article, this should easily dwarf that, unless Rayo's will end up growing Knuth arrows faster than linear. I'm not smart enough to evaluate that though. If it would, wonder if this would beat it:
f(n)=n (n Knuth up arrows) n (n Knuth up arrows) n
Or n copies of n Knuth up arrows...
At any rate, it's known that BB(18) is larger than Graham's number [2]. In fact, the same person has a bunch of long-lasting Turing machines for values around 18 [3], giving the lower bounds:
BB(15) > f_omega(2046) > 2046 (2046 up arrows) 2046
BB(16) > f_omega(f_omega(5))
BB(17) > f_(omega+1)(10) BB(18) > f_(omega+1)(7e57) > Graham's number
BB(19) > f_(omega+1)^3(9) > f_(omega+2)(3)
BB(20) > f_(omega+1)^4(7e57) > f_(omega+2)(4)
BB(21) > f_(omega+2)(f_(omega+1)^2(4))
BB(22) > f_(omega+2)^3(f_(omega+1)^2(4))
And for larger values again [2]:
BB(38) > f_(omega2)(167)
BB(64) > f_(omega^2)(4098)
BB(85) > f_(epsilon_0)(1907)
Of course, we expect the Busy Beaver function to grow a lot faster than these bounds.
[1] https://en.wikipedia.org/wiki/Fast-growing_hierarchy [2] https://googology.wikia.org/wiki/Busy_beaver_function#Graham... [3] https://googology.wikia.org/wiki/User_blog:Wythagoras/The_ni...!