User blog:Simply Beautiful Art/3 Arg Array Arsenal (Triple A)

Here's my triple A notation:

$$R(a,n,b) = a-1, ~ a \text{ is an integer}$$

$$R([c,0,e],n,b) = R([c,d,0],n,b) = c$$

$$R([c,d,e],n,b) = [[c,d,R(e,n,b)],R(d,n,b),c] $$

$$R([c,d],0,b) = [0,0,0]$$

$$R([0,d],n,b) = [n,[d],n]$$

$$R([c,d],n,b) = [n,[R(c,n-1,c),d],n]$$

$$R([c],0,b) = [0,0,0]$$

$$R([0],n,b) = [n,n,n]$$

$$R([c],n,b) = [R(b,n-1,b),R(c,n,b)]$$

$$H(0,n) = n$$

$$H(a,n) = H(R(a,n,a),n+1)$$

How large then is $$H([[[[[[[[0]]]]]]]],10)$$, that is, 10 nested single element brackets? The first few expansions are

$$H([[[[[[[[0]]]]]]]],10)$$

$$= H(R([[[[[[[[0]]]]]]]],10,[[[[[[[[0]]]]]]]]),11)$$

$$= H([R([[[[[[[[0]]]]]]]],9,[[[[[[[[0]]]]]]]],R([[[[[[[0]]]]]]],10,[[[[[[[[0]]]]]]]]])],11)$$

And then it becomes kinda "meh" and "bleh" beyond that.