User blog comment:LittlePeng9/Random Turing machines/@comment-5529393-20130413053946/@comment-1605058-20130413105814

If your count is right, then from 101 to 150 we have 49 states. We can use just 3 of them to simulate (3,3) machine best contestant to leave ([... with >300mln ['s. I think with 5~6 states we can make from this x ([[[...]). Output will be xxxx... and we can make from this x (...) with that many pairs, we can automatize it by adding counter which will repeat this process as many times, as it took first hydra to halt. This probably is well beyond H(3) and Bird's N