User blog comment:Vel!/Ordinal Hyper-E 2.0/@comment-25418284-20130409173638/@comment-5150073-20130409183500

I wonder to know that isomorphism exactly means, but if I understand it correctly, E^ can look similar with BEAF only for ordinals < e_0. After that we reach difficult problem (although I have idea how to avoid it!) that we can't express X^^(X+1) and #^^## structures naturally. Take for example, a^^b and a^^(b+1). What's n such that (a^^b)^n = a^^(b+1)? I was tried to find it out, and get to n = (a^^b)/(a^^(b-1)). The problem that arrays closed only positive rank operations: addition, multiplication and exponentiation. So division and subtraction are illegal when we speak about X structures.

I decided to go to the Bowers' website and read that he speak about pentational arrays (these ones cover cases X^^(X+1), X^^(X*2), X^^(X^2), and so on). He proposed an idea: we can separate parts of power towers using curly brackets. For example, for X^^X structures are written as {X^X^X...X^X^X} (with X X's). Then:

X^^(X+1) = {X^X^X...X^X^X}^X (line of X's and single X's) X^^(X+2) = {X^X^X...X^X^X}^X^X (line of X's and two X's) X^^(X*2) = {X^X^X...X^X^X}^{X^X^X...X^X^X} (two rows of X's) X^^(X*3) = {X^X..X^X}^{X^X..X^X}^{X^X..X^X} (three rows of X's) X^^(X^2) = {{X^X..X^X}^{X^X..X^X}...{X^X..X^X}^{X^X..X^X}} (now there X rows, each contains X X's. In other words, it is a tetrational plane.)

You can feel the tension. The reasonable question is: how to solve it? Well, we can think about it as separators enclosed in parenthesis (I described that form of separators in my blog post about formal definition of tetrational arrays). For example, X^^(X+1) & a[b] is {a,b (X^^(X+1)) 2} = {a,b ({X^X..X^X}^X) 2}. Then we start to transform this into array. Let a=3, b=2:

{3,2 ({X^X}^X}) 2} = {X^X}^X & 3[2] = {X^X}^2 & 3[2] = {X^(X*X)} & 3[2] = {X^(X*2)} & 3[2] = {X^(X+X)} & 3[2] = {X^(X+2)} & 3[2] = {{X^(X+1)} & 3[2] ({X^(X+1)}) {X^(X+1)} & 3[2]} = {{X^X} & 3[2] ({X^X}) {X^X} & 3[2] ({X^(X+1)}) {X^X} & 3[2] ({X^X}) {X^X} & 3[2]} = {3,3 (X) 3,3 ({X^X}) 3,3 (X) 3,3 ({X^(X+1)}) 3,3 (X) 3,3 ({X^X}) 3,3 (X) 3,3} = {3,N ({X^X}) 3,3 (X) 3,3 ({X^(X+1)}) 3,3 (X) 3,3 ({X^X}) 3,3 (X) 3,3}

Now notice: since {X^X} separator enclosed with curly brackets, there can be any number of X's. So, {X}, {X^X}, {X^X^X}, {X^X^X^X} are equivalent separators, all these equivalent to X^^X separator. Therefore, {3,N ({X^X}) 3 #} = {N^^N & 3[N] ({X^X}) 2 #}. That is, imagine N^N^N...N^N^N & 3[N] (with N N's from the left to the & operator) and how long time need to solve it into the form {3,M}.

Is that's useful anyways?