## FANDOM

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In this blog post I'll introduce my vision to how pentational arrays work. I shall start with structures only a bit larger that tetrational, then, when the main gist will be explained, I'll go to pentationals and probably even beyond.

First, remember my blog post: formal definition of tetrational arrays. There I proposed to enclose structures into separators: $$A \&\ b[p] = \{b,p (A) 2\}$$, where A is the any structure. It allows to make it more natural rather than use arrays in arrays.

So, what is the $$(X \uparrow\uparrow X)$$ separator? I decided to use Bowers' ideas and used curly brackets. Let $$(X \uparrow\uparrow X) = (\{X\}) = (\{X^X\}) = (\{X^{X^X}\}) = (\{X^{X^{X^X}}\}) = (\{\underbrace{X^{X^{\cdots^{X^X}}}}_{n\text{ }X's}\})$$ In otherwords, there is no matter how high the power tower enclosed in curly brackets, because X can represent the arbitrary number of them. To distinguish it, however, from the other similar separators in the form $$(X \uparrow\uparrow n)$$, I've used curly brackets. So, $$(\{X^X\}) = (X \uparrow\uparrow X)$$, but $$(X^X) = (X \uparrow\uparrow 2)$$. When the array in the form $$\{b,p (\{X^{X^{\cdots^{X^X}}}\}) 2\}$$, it is equal to $$X \uparrow\uparrow X \&\ b[p]\}$$.

This is the sketch of the new rule for my pentational arrays. All rules for tetrational arrays should be applied ignoring curly brackets.

For example:

$$\{3,3 (\{X^{X+1}\}) 2\} = \{3,3 (\{X^X\}*X\}) 2\} = \{3,3 (\{X^X\}*3\}) 2\} = \{3,3 (\{X^X\}+\{X^X\}+\{X^X\}) 2\}$$

$$= \{\{X^X\} \&\ 3 (\{X^X\}) \{X^X\} \&\ 3 (\{X^X\}) \{X^X\} \&\ 3\} = \{X \uparrow\uparrow X \&\ 3 (\{X^X\}) X \uparrow\uparrow X \&\ 3 (\{X^X\}) X \uparrow\uparrow X \&\ 3\}$$. Imagine that when it, eventually, after really long time, becomes $$\{3,N (\{X^X\}) \#\}$$, we should construct $$X \uparrow\uparrow N \&\ 3$$ (by my new rule), and N is very, very large. And we're barely scratch the surface of pentational arrays.

The next natural question is: what is $$X \uparrow\uparrow (X+1)$$ structure? Well, we can think about $$X+1$$ as a row with X items (arbitrary number of them), and 1 item above (or below) it. Thus, we can use curly brackets as a row delimiters and $$(X \uparrow\uparrow (X+1)) = ({\{X^{X^{\cdots^{X^X}}}\}}^X)$$. That is, we can still apply the same rules (including my new, that I don't defined formally, but I believe that you understood how it work).

For example:

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

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

$$= \{3,3 (X) 3,3 (\{X^X\}) 3,3 (X) 3,3 (\{X^{X+1}\}) 3,3 (X) 3,3 (\{X^X\}) 3,3 (X) 3,3\}$$.

Using my rules, we can notate and solve the further structures:

$$X \uparrow\uparrow (X+2) = {{\{X^{X^{\cdots^{X^X}}}\}}^X}^X$$

$$X \uparrow\uparrow (X+3) = {{{\{X^{X^{\cdots^{X^X}}}\}}^X}^X}^X$$

$$X \uparrow\uparrow (X*2) = {\{X^{X^{\cdots^{X^X}}}\}}^{\{X^{X^{\cdots^{X^X}}}\}}$$

$$X \uparrow\uparrow (X*2+1) = {\{X^{X^{\cdots^{X^X}}}\}}^{{\{X^{X^{\cdots^{X^X}}}\}}^X}$$

$$X \uparrow\uparrow (X*3) = {\{X^{X^{\cdots^{X^X}}}\}}^{{\{X^{X^{\cdots^{X^X}}}\}}^{\{X^{X^{\cdots^{X^X}}}\}}}$$

Now I really need to introduce some new type of separator that separates planes of X's: let normal curly brackets $$\{\}$$ become $$\{\}^X$$, and the new type should be $$\{\}^{X^2}$$. Now $$X \uparrow\uparrow (X^2) = \{\cdots\cdots\}^{X^2}$$, where $$\cdots\cdots$$ just means $$X \times X$$ grid of X's, X sections of X's separated by the normal curly brackets.

For example:

$$X \uparrow\uparrow X^2 \&\ 3[2] = \{3,2 ({\{X^X\}}^{X^X}) 2\} = \{3,2 ({\{X^X\}}^{X^2}\}) 2\} = \{3,2 ({\{X^X\}}^{X*X}\}) 2\} = \{3,2 ({\{X^X\}}^{X*2}\}) 2\}$$, etc.

Then we can make $$\{\}^{X^3}$$-typed brackets that separates realms, $$\{\}^{X^4}$$-typed brackets that separates flunes, and so on. Then you may ask, how to transform $$X \uparrow\uparrow A$$ structure into a separator? Below is the process that shows how to convert $$X \uparrow\uparrow X^2 \&\ 3[2]$$ into $$\{3,2 ({\{X^X\}}^{X^X}) 2\}$$:

$$X \uparrow\uparrow X^2 \&\ 3[2] = X \uparrow\uparrow (X*X) \&\ 3[2] = X \uparrow\uparrow (X*2) \&\ 3[2]$$

$$= X \uparrow\uparrow (X+X) \&\ 3[2] = X \uparrow\uparrow (X+2) \&\ 3[2] = X^{X \uparrow\uparrow (X+1)} \&\ 3[2]$$

$$= {\{X^X\}}^{X \uparrow\uparrow X} \&\ 3[2] = {\{X^X\}}^{X \uparrow\uparrow 2} \&\ 3[2] = {\{X^X\}}^{X^X} \&\ 3[2]$$

$$= \{3,2 ({\{X^X\}}^{X^X}) 2\}$$.

In other words, we can simply evaluate all X's after $$\uparrow\uparrow$$ to p's, solve the expression, make the power tower of that many X's, and then separate each $$p^A$$ X's by $$\{\}^A$$-typed brackets.

Notice that C can be also tetrational structure itself, for instance $$X \uparrow\uparrow (X \uparrow\uparrow (X+1))$$, and we eventually can come to the point where one type of brackets expresses another. This is the limit of pentational arrays.

We can also recall $$\{\}^A = \{\}^{X \uparrow\uparrow A}$$ like I did similar replaces for tetrational arrays, and continue my hierarchy even further, for example, if $$X \uparrow\uparrow\uparrow X = \{X \uparrow\uparrow X \uparrow\uparrow X \cdots X \uparrow\uparrow X \uparrow\uparrow X\}^{X \uparrow\uparrow\uparrow X}$$, then we can take $$X \uparrow\uparrow\uparrow (X+1) = \{X \uparrow\uparrow X \uparrow\uparrow X \cdots X \uparrow\uparrow X \uparrow\uparrow X\}^{X \uparrow\uparrow\uparrow X} \uparrow\uparrow X$$. And we get larger and larger numbers.