[sligocki]

Yes, it seems that BB(3, 4) >>> BB(5, 2) (BB(5) = BB(5, 2)). This is not too surprising since BB(3, 4) has 12 transitions in it's table (3*4), while BB(5, 2) only has 10. But it seems that also BB(3, 4) >> BB(6, 2) (which both have the same number of transitions, so it appears that having more symbols is valuable to these little TMs.

[tgv]

I think I can explain that. If you only have two symbols, you can encode the same information as having four symbols in two cells, so performing the same operation on them requires an extra state to read the first of the two and then move to one of two other states to handle the second. Less symbols requires more states.

This is related to the linear speed-up theorem, which roughly states that you can speed up any TM by expanding its alphabet. And speed-up is not what BB is about.

So actually, it would make sense to limit the the busy beavers to BB(n, 2) only.