User blog comment:Ikosarakt1/Introduction to pentational arrays/@comment-5529393-20130413060024/@comment-5150073-20130413104838

The problem that you can't come naturally from X^^X to X^^(X+1) using only addition, multiplication and exponentiation. For example, we want to move from n^^m to n^^(m+1), what is p such that (n^^m)^p = n^^(m+1)? I found p = (n^^m)/(n^^(m-1)) (you can verify me on the calculator). On arrays, division is undefined, so we can't use it.

In your blog post, you recommend to take X^^(X+1) = (X^^X)^p = X^^X*X^^X...X^^X*X^^X (p X^^X's). However, it doesn't stacks correctly with the main rule of "array of" operator: the number of entries before applying main rules should be equivalent to the solved expression from the left to the & operator. To obtain that expression we replace all X to p's and solve it normally. By that, X^^(X+1) = (X^^X)^p. However, if we take, say, p=3:

X^^(X+1) => 3^^(3+1) = 3^^4 = 3^7625597484987

(X^^X)^p => (3^^3)^3 = (3^27)^3 = 3^81

Therefore, we really need the special notation such as I proposed.