Busy Beavers: quadratic relation of output size and running time? 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.
It's true, all current busy beaver candidates have a quadratic relationship between output size and steps. As Ross points out, this has to do with how they act. They move back and forth across much of the output every time they increase the output size by some constant. That means that to get to size N they must run for approximately 1 + 1 + 2 + 2 + 3 + 3 + ... = N(N+1) steps.

However, there are many other types of Turing machines. For example, there are ones often called counters which move back and forth across a section of output, each time incrementing a binary (or other base) counter. These take exponential time to increase the size of the output. One interesting one is Brady's "Surprise in a Box" http://www.drb.insel.de/~heiner/BB/simAB3Y_SB.html which expands quadratically for a little while and then moves completely within the "box" until it halts.

There are also machines which have a sort of recursive "back and forth" behavior so that it takes them N^2 steps to increase the size of the tape. They have a cubic relationship between output size and steps.

I suspect that for larger values busy beavers will not have this relation because they will get more complex in their behavior, but I don't know of any proofs.