Parity of TREE(3)? The number TREE(3) is somewhat famous for being incomprehensible big. But since it's just a finite number it must have a parity. Is the parity of TREE(3) known?
 A: It seems difficult to determine the parity of TREE(3) without finding the exact sequence in question.  
The one thing I can say is, if one takes the "obvious" well-ordering of maximal order type for labelled trees, and then chooses the sequence by using the greedy strategy (always choose the next tree to be the one highest in the ordering that is allowed), then the sequence will have odd length, for any TREE(n). The reason is that one will eventually be down to just one label, and then eventually be forced to choose just paths for the rest of the sequence, as paths of label 1 are the lowest in the well-ordering I mentioned above.  At that point, if we are at the $n$th tree in the sequence, clearly the best option is to let $T_n$ be a path of $n$ vertices, $T_{n+1}$ be a path of $n-1$ vertices, and so on, until $T_{2n-1}$ is just the root, and we are done.
The same argument will hold for any sequence that leaves either monolabelled paths or monolabelled trees of height one for last.  As this seems like a natural thing to do, I think the odds are pretty good that TREE(n) is always odd, but of course this is very far from a proof.
