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,n)

### 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))))

## TREE(3,n)

### TREE(3,0)

TREE(3,0)>tree_3(tree_2(tree(8)))

### TREE(3,1)

2. {{}}

3. [{}{}]

4. {[[[]]]}

TREE(3,1)>tree_4(tree_3(tree_3(tree_2(tree(8)))))

TREE(3,2)>tree_4(tree_4(tree_3(tree_2(tree(8)))))

TREE(3,n)>tree_4^n(tree_3(tree_2(tree(8)))))

## TREE(n)

TREE(4)>tree_4(tree_3(tree_3(tree_2(tree(8)))))

TREE(5)>tree_5(TREE(4))

TREE(6)>tree_6(TREE(5))

TREE(7)>tree_7(TREE(6))