User blog:Wythagoras/TREE(n)

I've discovered new bounds for TREE(n), using the function TREE(a,b), which combines the powers of TREE and tree: It is the length of the longest sequence using a different brackets. The nth member of the sequence has at most n+b brackets. So TREE(a)=TREE(a,0) and tree(b)=TREE(1,b)

I'll show TREE(n) >> tree_n(n) for n > 2

TREE(2,0) and TREE(2,1)
TREE(2,0)=TREE(2)=3

TREE(2,1)>tree_3(tree_2(tree(8))), using Deedlit11's bound

TREE(2,2)
In my weak sequence, I use the first tree_3(tree_2(tree(8))) trees the same as in Deedlits sequence, with the only difference that each tree has an additional [] bracket, like

3. []

4. [([][])]

5.

So the tree_3(tree_2(tree(8)))th tree is

we can finish with a sequence of more than tree_3(tree_3(tree_2(tree(8)))) but this is a very weak lower bound.

TREE(2,n)
Similiar, TREE(2,n) > tree_3^n(tree_2(tree(8))))