When glancing at the running times and output sizes of some Busy Beaver candidates on this page, I find that there seems to be a rough (up to some small power of 10) quadratic relation between output size and running time. Do you know a heuristic argument or even proof for this? Is it related to the computational model of Turing machines, or is it some more general principle?
Because (under the assumption of a computational model where the binary description of the algorithm and the algorithms' length can be read out and operated on by the program) no program can decide the halting problem for all programs a little bit longer than itself and access to the busy beaver program of some length entails ability to solve the Halting problem for that length by checking whether a given program runs longer than the busy beaver, it can't be that this is somehow a test (for said programs with efficient access to memory) that will vastly constrain the possible candidates for all sizes, as then a smaller number of supplied bits would suffice to generate the busy beaver and thee would be a program could solve the halting problem for somewhat bigger programs. If you sharpen the notion of "vastly", this reasoning will also apply to Turing machines.