156 个页面

BEAF，或鮑爾斯爆炸數陣函數，是由喬納森·鮑爾斯發明的大數函數。這篇文章會直觀的描述它，並慢慢的介紹。

## 箭號表示法

$$a\uparrow b = a^b$$
$$a\uparrow\uparrow b = \underbrace{a\uparrow a\uparrow\ldots\uparrow a\uparrow a}_b = \underbrace{a^{a^{a^{.^{.^.}}}}}_b$$。大數研究者稱之為迭代冪次。箭號應從右到左運算。
$$a\uparrow\uparrow\uparrow b = \underbrace{a\uparrow\uparrow a\uparrow\uparrow\ldots\uparrow\uparrow a\uparrow\uparrow a}_b$$。又稱五級運算

$$a\uparrow^n b = \underbrace{a\uparrow^{n-1} a\uparrow^{n-1}\ldots\uparrow^{n-1} a\uparrow^{n-1} a}_b$$

$$3\uparrow 4 = 3^4 = 81$$（3的4次方）
$$2\uparrow\uparrow 4 = 2\uparrow 2\uparrow 2\uparrow 2 = 2\uparrow 2\uparrow 4 = 2\uparrow 16 = 65536$$
$$4\uparrow\uparrow 4 = 2361022671...5261392896$$，大約有$$8.0723\cdot 10^{153}$$位數
$$3\uparrow\uparrow\uparrow 3 = 3\uparrow\uparrow 3\uparrow\uparrow 3 = 3\uparrow\uparrow 3^{3^3} = 3\uparrow\uparrow 7625597484987$$

## 運算記號

$$a\ \{n\}\ b = a\uparrow^n b$$

$$a\ \{\{1\}\}\ b = \underbrace{a\ \{a\ \{\ldots a\ \{a}_b\}\ a\ldots\}\ a\}\ a$$

$$3\ \{\{1\}\}\ 3 = 3\ \{3\ \{3\}\ 3\}\ 3 = 3\ \{3\ \{2\}\ 7625597484987\}\ 3$$

$$a\ \{\{2\}\}\ b = \underbrace{a\ \{\{1\}\}\ a\ \{\{1\}\}\ \ldots\ \{\{1\}\}\ a\ \{\{1\}\}\ a}_b$$

n的值更高的情況：

$$a\ \{\{3\}\}\ b = \underbrace{a\ \{\{2\}\}\ a\ \{\{2\}\}\ \ldots\ \{\{2\}\}\ a\ \{\{2\}\}\ a}_b$$（powerexpansion
$$a\ \{\{4\}\}\ b = \underbrace{a\ \{\{3\}\}\ a\ \{\{3\}\}\ \ldots\ \{\{3\}\}\ a\ \{\{3\}\}\ a}_b$$（expandotetration
expandopentation，expandohexation，等。

$$a\ \{\{\{1\}\}\}\ b = \underbrace{a\ \{\{a\ \{\{\ldots a\ \{\{a\}\}\ a\ldots\}\}\ a\}\}\ a}_b$$
$$a\ \{\{\{2\}\}\}\ b = \underbrace{a\ \{\{\{1\}\}\}\ a\ \{\{\{1\}\}\}\ \ldots\ \{\{\{1\}\}\}\ a\ \{\{\{1\}\}\}\ a}_b$$（multiexplosion
$$a\ \{\{\{3\}\}\}\ b = \underbrace{a\ \{\{\{2\}\}\}\ a\ \{\{\{2\}\}\}\ \ldots\ \{\{\{2\}\}\}\ a\ \{\{\{2\}\}\}\ a}_b$$（powerexplosion
$$a\ \{\{\{4\}\}\}\ b = \underbrace{a\ \{\{\{3\}\}\}\ a\ \{\{\{3\}\}\}\ \ldots\ \{\{\{3\}\}\}\ a\ \{\{\{3\}\}\}\ a}_b$$（explodotetration
explodopentation，explodohexation，等。

$$a\ \{1\}\ b = a^b$$
$$a\ \{1\}^d\ b = \underbrace{a\ \{a\ \{\ldots a\ \{a\}^{d - 1}\ a\ldots\}^{d - 1}\ a\}^{d - 1}\ a}_b$$
$$a\ \{c\}^d\ b = \underbrace{a\ \{c - 1\}^d\ a\ \{c - 1\}^d\ \ldots\ \{c - 1\}^d\ a\ \{c - 1\}^d\ a}_b$$

## 線性數陣記號

• $$\{a, b, 1, 1\} = a^b$$
• $$\{a, b, 1, d\} = \underbrace{\{a, a, \{a, a, \ldots \{a, a, a, d - 1\} \ldots, d - 1\}, d - 1\}}_b$$ if $$d > 1$$
• $$\{a, b, c, d\} = \underbrace{\{a, \{a, \ldots \{a, a, c - 1, d\} \ldots, c - 1, d\}, c - 1, d\}}_b$$ if $$c > 1$$

1被認為是默認值，所以數陣結尾的1可以被去掉。如$$\{a, b, 1, 1\}$$可以簡化成$$\{a, b\} = a^b$$。

$$\{a, b, 1, d\} = \underbrace{\{a, a, \{a, a, \ldots \{a, a, a, d - 1\} \ldots, d - 1\}, d - 1\}}_b$$
$$= \{a, a, \underbrace{\{a, a, \ldots \{a, a, a, d - 1\} \ldots, d - 1\}}_{b - 1}, d - 1\} = \{a, a, \{a, b - 1, 1, d\}, d - 1\}$$
$$\{a, b, c, d\} = \underbrace{\{a, \{a, \ldots \{a, a, c - 1, d\} \ldots, c - 1, d\}, c - 1, d\}}_b$$
$$= \{a, \underbrace{\{a, \ldots \{a, a, c - 1, d\} \ldots, c - 1, d\}}_{b - 1}, c - 1, d\} = \{a, \{a, b - 1, c, d\}, c - 1, d\}$$

$$\{a, 1, c, d\} = a$$

### 五項以上的數陣

• $$\{a, b, 1, 1\} = \{a, b\} = a^b$$
• $$\{a, 1, c, d\} = a$$
• $$\{a, b, 1, d\} = \{a, a, \{a, b - 1, 1, d\}, d - 1\}$$, $$b,d > 1$$
• $$\{a, b, c, d\} = \{a, \{a, b - 1, c, d\}, c - 1, d\}$$, $$b,c > 1$$

• $$\{a, b\} = a^b$$
• $$\{a, 1, c, d, e\} = a$$
• $$\{a, b, 1, 1, e\} = ???$$
• $$\{a, b, 1, d, e\} = \{a, a, \{a, b - 1, 1, d, e\}, d - 1, e\}$$, $$b,d > 1$$
• $$\{a, b, c, d, e\} = \{a, \{a, b - 1, c, d, e\}, c - 1, d, e\}$$, $$b,c > 1$$

$$\{a, b, 1, 1, e\} = \{a, a, a, \{a, b - 1, 1, 1, e\}, e - 1\}$$

• $$\{a, b\} = a^b$$
• $$\{a, 1, c, d, e, f\} = a$$
• $$\{a, b, 1, 1, 1, f\} = \{a, a, a, a, \{a, b - 1, 1, 1, 1, f\}, f - 1\}$$, $$b,f > 1$$
• $$\{a, b, 1, 1, e, f\} = \{a, a, a, \{a, b - 1, 1, 1, e, f\}, e - 1, f\}$$, $$b,e > 1$$
• $$\{a, b, 1, d, e, f\} = \{a, a, \{a, b - 1, 1, d, e, f\}, d - 1, e, f\}$$, $$b,d > 1$$
• $$\{a, b, c, d, e, f\} = \{a, \{a, b - 1, c, d, e, f\}, c - 1, d, e, f\}$$, $$b,c > 1$$

1. 基礎規則：如果沒有駕駛員（即指數後的項均為1），則$$v(A) = b^p$$。
2. 指數規則：如果指數為1，$$v(A) = b$$。
3. 災難規則：否則，
1. 將副駕駛換成原數陣，但其指數減去1。
2. 駕駛員減1。
3. 所有乘客變為$$b$$。
（定義結束）

### 例子

\begin{eqnarray*} \{3,3,1,1,1,3\} &=& \{3,3,3,3,\{3,2,1,1,1,3\},2\} \\ &=& \{3,3,3,3,\{3,3,3,3,3,2\},2\} \\ &=& \{3,3,3,3,\{3,\{3,2,3,3,3,2\},2,3,3,2\},2\} \\ &=& \{3,3,3,3,\{3,\{3,\{3,1,3,3,3,2\},2,3,3,2\},2,3,3,2\},2\} \\ &=& \{3,3,3,3,\{3,\{3,3,2,3,3,2\},2,3,3,2\},2\} \end{eqnarray*}

• $$\{a,a,a\} = \{a,2,1,2\}$$
• $$\{a,a,\{a,a,a\}\} = \{a,3,1,2\}$$
• $$\{a,a,\{a,a,\{a,a,a\}\}\} = \{a,4,1,2\}$$等

• $$\{a,a,a,a\} = \{a,2,1,1,2\}$$
• $$\{a,a,a,\{a,a,a,a\}\} = \{a,3,1,1,2\}$$
• $$\{a,a,a,\{a,a,a,\{a,a,a,\{a,a,a,\{a,a,a,a\}\}\}\}\} = \{a,6,1,1,2\}$$

• $$\{a,a,a,a,a\} = \{a,2,1,1,1,2\}$$
• $$\{a,a,a,a,\{a,a,a,a,a\}\} = \{a,3,1,1,1,2\}$$
• $$\{a,a,a,a,\{a,a,a,a,\{a,a,a,a,a\}\}\} = \{a,4,1,1,1,2\}$$

## 多維數陣

$$b \& a = \underbrace{\{a, a, ..., a, a\}}_b$$

1. 基礎規則：如果沒有駕駛員（即指數後的項均為1），則$$v(A) = p \& b$$。

1. 基礎規則：如果沒有駕駛員、且$$r = 1$$，則$$v(A) = b^p$$。
2. 階規則：如果沒有駕駛員、且$$r > 1$$，則$$v(A) = p \&_{r - 1} b$$。
3. 指數規則和災難規則不變。

$$\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}}$$

$$\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}_{._{._.}}}}$$（每個數陣有$$a$$項）

$$\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}_{._{._.}},2},2}$$

$$\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}_{\{a, a, \ldots, a, a\}_{._{._.}},n - 1},n - 1}$$

$\left\{ \begin{matrix} b,p \\ 1,1,2 \end{matrix} \right\}$

$$\{b, p (1) 1, 1, 2\} = \{b, b, b, \ldots (1) b, \{b, p - 1 (1) 1, 1, 2\}\}$$

• 底數為第一項。
• 指數為第二項。
• 駕駛員是指數後的第一個非1項。
• 副駕駛是駕駛員的前一項。如果駕駛員是第二列的開始，那麼它不存在。
• 前項代表在指定項那列、在指定項前面的項。（因此，在$$\{a, b (1) c, d\}$$中，$$c$$是$$d$$的前項，而$$a$$和$$b$$不是。）
• 列的指數塊包含p項。
• 飛機是駕駛員和的它前項。如果駕駛員在第二列，飛機還包括前一列的指數塊。
• 乘客代表飛機中，除了駕駛員和副駕駛（如果有）外的所有項。

\begin{eqnarray*} \{3,3,3 (1) 2\} &=& \{3,\{3,2,3 (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,1,3 (1) 2\},2 (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,3,2 (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,2,2 (1) 2\},1 (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,2,2 (1) 2\} (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,\{3,1,2 (1) 2\},1 (1) 2\} (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,3 (1) 2\} (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,3,3 (1) 1\} (1) 2\},2 (1) 2\} \\ &=& \{3,\{3,\{3,3,3\} (1) 2\},2 (1) 2\} \end{eqnarray*}

### 更多列

• 飛機是駕駛員和它的前項，和前列的指數塊。

### 三維

• 列的指數塊包含$$p$$項，平面的指數塊包含成方形的$$p \times p$$項（或$$p$$列）。
• X的前項是在X那列、在X前的項。X的前列是在X的平面中、在X前的列。
• 飛機包含駕駛員、駕駛員的前項、和駕駛員前結構的指數塊。

### 四維和更多

• 結構是項、列、平面、領域、福潤、5維空間等。
• 列的指數塊為$$p$$項；平面的指數塊為$$p \times p$$的方形；領域的指數塊為$$p \times p \times p$$方塊，等。整體來說，列的指數塊為$$p$$項，$$X^n$$結構的指數塊為$$p$$ $$X^{n-1}$$結構。
• A的前結構為跟A在同個$$X^{n + 1}$$結構、在A前的$$X^n$$結構。具體來說，A的前項是跟A同列、在A前的項，A的前列是跟A同平面、在A前的列，等。
• 飛機包含駕駛員、前項、前結構的指數塊。

• 底數$$b$$為第一項。
• 指數$$p$$為第二項。
• 駕駛員為指數後第一個非1項。
• 副駕駛是駕駛員的前一項。如果駕駛員位在它那列的第一項，則副駕駛不存在。
• 結構是項、列、平面、領域、福潤、5維空間等。
• 列的指數塊為$$p$$項；平面的指數塊為$$p \times p$$的方形；領域的指數塊為$$p \times p \times p$$方塊，等。整體來說，列的指數塊為$$p$$項，$$X^n$$結構的指數塊為$$p$$ $$X^{n-1}$$結構。
• A的前結構為跟A在同個$$X^{n + 1}$$結構、在A前的$$X^n$$結構。具體來說，A的前項是跟A同列、在A前的項，A的前列是跟A同平面、在A前的列，等。
• 飛機包含駕駛員、前項、前結構的指數塊。
• 乘客代表飛機中，除了駕駛員和副駕駛（如果有）外的所有項。

1. 基礎規則：如果沒有駕駛員（指數後的項均為1）則$$v(A) = b^p$$。
2. 指數規則；如果指數為1，$$v(A) = b$$。
3. 災難規則：否則，
1. 將副駕駛換成原數陣，但其指數減去1。
2. 駕駛員減1。
3. 所有乘客變為$$b$$。
(定義結束)

## 數陣運算符

X以上的結構是X+1,X+2,X+3,...,2X,2X+1,2X+2,2X+3,...,3X,4X,...,$$X^2$$，等。X+m可被認為是由原本的列和含m項的第二列組成的，2X為兩列，3X為三列，nX+m由n列和它下面有m項的列組成。下面是剛剛的記號和一般數陣間的對比：

\begin{eqnarray*} 1\ \&\ n &=& \{n\} = n \\ 2\ \&\ n &=& \{n,n\} = n^n \\ 3\ \&\ n &=& \{n,n,n\} = n \uparrow^{n} n \\ 4\ \&\ n &=& \{n,n,n,n\} = n \underbrace{ \{\{\cdots\{\{n\}\}\cdots\}\} }_{n \{\}'s} n \\ m\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{m n's}}\} &=& \{n,m (1) 2\} \\ X\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}}\} &=& \{n,n (1) 2\} \\ X+1\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) n\} \\ X+2\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) n,n\} \\ X+3\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) n,n,n\} \\ X+m\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) \underbrace{n,n,n,\cdots,n,n,n}_{\text{m n's}}\} \\ 2X\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) \underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}}\} \\ 3X\ \&\ n &=& \{\underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) \underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}} (1) \underbrace{n,n,n,\cdots,n,n,n}_{\text{n n's}}\} \\ mX\ \&\ n &=& \{\underbrace{X\ \&\ n(1)X\ \&\ n(1)X\ \&\ n(1)\cdots(1)X\ \&\ n(1)X\ \&\ n(1)X\ \&\ n}_{\text{m X's}}\} &=& \{n,m (2) 2\} \\ X^2\ \&\ n &=& \{\underbrace{X\ \&\ n(1)X\ \&\ n(1)X\ \&\ n(1)\cdots(1)X\ \&\ n(1)X\ \&\ n(1)X\ \&\ n}_{\text{n X's}} &=& \{n,n (2) 2\} \end{eqnarray*}

## 迭代冪次數陣

$$\{b, p (0, 1) 2\}$$

Technical note: the space described by an $$X^X$$ structure is called a dimensional group. In a dimensional group, each coordinate is described by a row of coordinates (order type $$\omega$$) rather than a fixed finite number.

$$\{b, p (1, 1) 2\}$$

This is an $$X^{X + 1}$$ structure; its prime block is $$p^{p + 1}$$. Generally, an $$(n, m)$$ separator creates an $$X^{mX + n}$$ structure. It is a row of dimensional groups.

Linear separator arrays can describe all possible $$X^P$$, where $$P$$ is a polynomial in $$X$$. (Here and in the rest of the article, we restrict "polynomial" to have nonnegative integer coefficients.) They do this by supplying a list of coefficients from degree 0 on:

$$(a_0, a_1, a_2, a_3, \ldots)$$ describes a structure of type $$X^{a_0 + a_1X + a_2X^2 + a_3X^3 + \cdots}$$

So, for example, (0, 0, 1) describes $$X^{X^2}$$ (dimensional gang), and (1, 0, 6, 1) describes $$X^{1 + 6X^2 + X^3}$$. You will notice that now zeroes are the default rather than 1's; this is one of BEAF's more annoying properties.

$$\{b, p (0 (1) 1) 2\} = \{b, p ((1) 1) 2\}$$

Of course, we have no reason to stop at linear arrays for separators. This structure is $$X^{X^X}$$, a superdimensional group. (Vectors in a superdimensional group have order type $$\omega^\omega$$.) In general for the second row,

$$(a_{00}, a_{01}, a_{02}, \ldots (1) a_{10}, a_{11}, a_{12}, \ldots )$$ describes a structure of type $$X^{a_{00} + a_{01}X + a_{02}X^2 + a_{03}X^3 + \cdots + a_{10}X^X + a_{11}X^{X + 1} + a_{12}X^{X + 2} + \ldots}$$.

The third row ((1)(1)1) describes $$X^{X^{2X}}$$ (no, we're not at $$X^{X^{X^X}}$$ yet!), and the (n + 1)-th row describes $$X^{X^{nX}}$$. The second plane ((2)1) is $$X^{X^{X^2}}$$, the second realm ((3)1) is $$X^{X^{X^3}}$$, and so on and so forth. We finally reach $$X^{X^{X^X}} = X \uparrow\uparrow 4$$ at the second dimensional group, ((0, 1) 1). The inner pair of parentheses describes a polynomial $$P$$ in $$X$$, with the outer one describing $$X^{X^{X^P}}$$.

$$\{b, p (((1) 1) 1) 2\}$$

This is $$X^{X^{X^{X^X}}} = X \uparrow\uparrow 5$$.

$$\{b, p (((0, 1) 1) 1) 2\}$$ describes $$X \uparrow\uparrow 6$$
$$\{b, p ((((1) 1) 1) 1) 2\}$$ describes $$X \uparrow\uparrow 7$$
$$\{b, p ((((0, 1) 1) 1) 1) 2\}$$ describes $$X \uparrow\uparrow 8$$
etc.

The limit of all this is the $$X \uparrow\uparrow X$$ structure. If we stop here, we have tetrational array notation.

### Attempt at definition

Bowers never bothered to formalize tetrational arrays, but we'll give it a try. The only thing holding us back right now is the definition of prime blocks, which is specifically tailored for dimensional arrays only. Let's look at our current inductive definition:

• The prime block of an entry is an entry.
• The prime block of an $$X^{n + 1}$$ structure is the prime blocks of its first $$p$$ $$X^n$$-structures.

(There is a slight change here, but it can be seen that it has minimal effect.) This definition doesn't say what happens when our structure is an $$X^X$$ structure, because it's not a structure of the form $$X^{n + 1}$$.

Let's try to extend this definition to allow all $$X^P$$ for degree-1 polynomials $$P$$.

• The prime block of an entry is an entry.
• The prime block of an $$X^{P+1}$$ structure is the prime blocks of its first $$p$$ $$X^P$$-structures, where $$P$$ is a polynomial in $$X$$.
• The prime block of an $$X^{P+X}$$ structure is the prime block of its first $$X^{P + p}$$-structure.

This breaks down at $$X^{X^2}$$ as expected, but there is a pattern emerging.

• The prime block of an entry is an entry.
• The prime block of an $$X^{P + 1}$$ structure is the prime blocks of its first $$p$$ $$X^P$$-structures, where $$P$$ is a polynomial in $$X$$.
• The prime block of an $$X^{P + X}$$ structure is the prime block of its first $$X^{P + p}$$-structure.
• The prime block of an $$X^{P + X^2}$$ structure is the prime block of its first $$X^{P + pX}$$-structure.
• The prime block of an $$X^{P + X^3}$$ structure is the prime block of its first $$X^{P + pX^2}$$-structure.
• In general, the prime block of an $$X^{P + X^{n + 1}}$$ structure is the prime block of its first $$X^{P + pX^n}$$-structure.

This works up until $$X^{X^X}$$. At this rate we'll never get to $$X \uparrow\uparrow X$$!

### Second attempt

Let's generalize the concept of "structure" to help out our definition.

• $$0$$ is a structure.
• If $$\alpha$$ is a structure, $$X^\alpha$$ is also a structure.
• If $$\alpha$$ and $$\beta$$ are structures, $$\alpha + \beta$$ is also a structure.

This allows things like $$2X$$ and $$4X^X + X + 1$$, and works up until our limit of $$X \uparrow\uparrow X$$. Furthermore, let's define a limit structure as one of the form $$X^\alpha$$ for $$\alpha > 0$$ or $$\alpha+\beta$$ for $$\alpha$$ and $$\beta$$ being limit structures, and a successor structure as one of the form $$\alpha + 1$$.

Now we'll recursively define the prime block of a structure $$\alpha[p]$$.

• $$0[p] = \{\}$$
• $$(\alpha + 1)[p] = \alpha[p] \cup \{\text{the last entry of }\alpha + 1\}$$
• $$X[p] = p$$ and $$X^{\alpha + 1}[p] = X^{\alpha} p$$
• $$X^{\alpha}[p] = X^{\alpha[p]}$$ if and only if $$\alpha$$ is a limit structure.
• $$(X^{\alpha_1} + X^{\alpha_2} + \cdots + X^{\alpha_{k - 1}} + X^{\alpha_k})[p] = (X^{\alpha_1} + X^{\alpha_2} + \cdots + X^{\alpha_{k - 1}} )[p] \cup X^{\alpha_k}[p])$$ where $$\alpha_k$$ is the smallest $$\alpha_i$$
• $$X\uparrow\uparrow X[0] = 1$$ and $$X\uparrow\uparrow X[p] = X^{X\uparrow\uparrow X[p-1]}$$

If you are confused by now, skip it. All you need an intuitive conception of how this works.

There's one final step. Notice how we used the word "smallest" in the definition above, but structures aren't numbers and we haven't yet defined an ordering. This is easy:

• $$\alpha < \alpha + X^0$$
• $$X^\alpha < X^\beta$$ iff $$\alpha < \beta$$
• $$\alpha + \gamma < \beta + \gamma$$ iff $$\alpha < \beta$$

And we're done! Tetrational arrays.

But can we make this simpler?

Some of our more experienced readers may notice the parallels to ordinal hierarchies. $$X$$ functions a lot like $$\omega$$, and our prime block definition looks like that of the Wainer hierarchy — we picked the $$\alpha[p]$$ notation for a reason. We have developed a notation for ordinals up to $$\varepsilon_0$$, which is also the strength of BEAF up to this point. It seems reasonable to rewrite our notation using ordinals, and we'll do just that.

We'll define an array as a function $$A : \varepsilon_0 \mapsto \mathbb{N}_+$$, where the number of outputs greater than 1 is finite. (This prevents infinite arrays like {6, 6, 6, ...}.) Let $$b = A(0)$$, $$p = A(1)$$, $$\pi = \min\{\alpha > 1: A(\alpha) > 1\}$$ (the position of the pilot), and $$\kappa = \pi - 1$$ (the position of the copilot).

Define the prime block $$\Pi(\alpha)$$:

• $$\Pi(0) = \{\}$$
• $$\Pi(\alpha + 1) = \{\alpha\} \cup \Pi(\alpha)$$
• $$\Pi(\alpha) = \Pi(\alpha[p])$$ if $$\alpha$$ is a limit ordinal

(This looks a lot like the slow-growing hierarchy.) Define the passengers as $$S = \Pi(\pi) \backslash \{\pi, \kappa\}$$.

And the three rules:

1. Base rule. If $$\pi$$ does not exist, $$v(A) = b^p$$.
2. Prime rule. If $$p = 1$$, $$v(A) = b$$.
3. Catastrophic rule. Define $$A'$$ as $$A$$ with the following modifications:
• Define $$B$$ as $$A$$, but with $$B(1) = p - 1$$.
• If $$\kappa$$ exists, $$A'(\kappa) := v(B)$$.
• $$A'(\pi) = A(\pi) - 1$$.
• $$A(\sigma) := b$$ for $$\sigma \in S$$.
• $$v(A) = v(A')$$.

Although less accessible, this is a far simpler definition than above!

One more thing; let's define some fundamental sequences for ordinals below $$\varepsilon_0$$.

• $$\omega^\alpha[n] = \omega^{\alpha[n]}$$ for limit ordinals $$\alpha$$
• $$\omega^{\alpha + 1}[n] = \omega^\alpha n$$
• $$(\omega^{\alpha_1} + \omega^{\alpha_2} + \cdots + \omega^{\alpha_k})[n] = \omega^{\alpha_1} + \omega^{\alpha_2} + \cdots + \omega^{\alpha_k}[n]$$ where $$\alpha_1 \geq \alpha_2 \geq \cdots \geq \alpha_k$$
• $$\omega \uparrow\uparrow 0 = 1$$
• $$(\omega \uparrow\uparrow (\alpha + 1))[n] = \omega^{\omega \uparrow\uparrow \alpha}[n]$$

We threw in a definition of $$\omega \uparrow\uparrow n$$ to ensure that we properly mirror Bowers' $$X$$ hierarchy. Indeed, after $$\varepsilon_0$$ the fundamental sequences depart a bit from the usual.

## Pentational arrays

Pentational arrays are a bit of a mindblow. By far the worst part is that no notation has yet been devised to support them! Bowers never bothered to put one together; he just used the array of operator.

Remember the array of operator?

$$a \& b = \{b, a (1) 2\}$$

We left it behind long ago in the single-row era. Turns out, & has an extension that allows use beyond this level.

$$a^2 \& b = \{b, a (2) 2\}$$

It's not hard to see that the right-hand side is an a by a array of b's, or an a2 array of b's. Note that the expression "$$a^2$$" should not be solved first; rather it is a blueprint for the dimensions of the array.

$$a^k \& b = \{b, a (k) 2\}$$

In general, the array-of operator lets us specify many dimensions, and even tetrational arrays such as this one:

$$a^{a^a} \& b = a \uparrow\uparrow 3 \& b = \{b, a ((1) 1) 2\}$$

For those who understood the ordinal definition above, we can formally define $$f(a) \& b = \{0 \mapsto b, 1 \mapsto a, f(\omega) \mapsto 2\}$$.

The smallest pentational array is this one:

$$(a \uparrow\uparrow a) \& b = \{b, a (???) 2\}$$

There is no notation for this array (an $$X \uparrow\uparrow X$$ structure), but it's certainly definable with ordinals, where $$\omega \uparrow\uparrow \omega = \varepsilon_0$$ using the fundamental sequence $$\varepsilon_0[n] = \omega \uparrow\uparrow n$$. We'll take a moment to apologize to the readers who don't get ordinals yet. Skip ahead to the legion arrays if you're one of them.

At this point you may be wondering what $$X \uparrow\uparrow\uparrow X$$ is, but we can't move on to that until we've defined $$X \uparrow\uparrow (X2)$$ or the fundamental sequence of "$$\omega \uparrow\uparrow (\omega2)$$". Our current definition of $$\uparrow\uparrow$$ over the ordinals would make $$\omega \uparrow\uparrow (\omega2) = \omega \uparrow\uparrow \omega$$, so we need to repair that.

If we amend our definition to $$\omega \uparrow\uparrow (1 + \alpha) = \omega^{\omega \uparrow\uparrow \alpha}$$, we can clearly see that $$\omega \uparrow\uparrow (1 + \omega) = \omega \uparrow\uparrow \omega$$ is the fixed point of the function $$\alpha \mapsto \omega^\alpha$$.

$$\omega \uparrow\uparrow (\omega2)$$ should be the next fixed point, i.e. $$\varepsilon_1$$. One particularly clean definition of $$\varepsilon_1$$ is the limit of $$\varepsilon_0, \varepsilon_0^{\varepsilon_0}, \varepsilon_0^{\varepsilon_0^{\varepsilon_0}}, \ldots$$, so why not make $$\varepsilon_1 = \varepsilon_0 \uparrow\uparrow \omega = (\omega \uparrow\uparrow \omega) \uparrow\uparrow \omega$$? This seems slightly odd at first, since in general $$a \uparrow\uparrow a2 \neq (a \uparrow\uparrow a) \uparrow\uparrow a$$. Ordinals work in mysterious ways.

The next fixed point is $$\omega \uparrow\uparrow (\omega3) = \varepsilon_1 \uparrow\uparrow \omega = \varepsilon_2$$, and in general $$\omega \uparrow\uparrow (\omega n) = \varepsilon_{n - 2} \uparrow\uparrow \omega = \varepsilon_{n-1}$$ for finite $$n > 0$$. Understandably, the limit of all these is $$\omega \uparrow\uparrow (\omega^2) = \varepsilon_\omega$$.

Yada yada yada. Nothing too interesting here until $$\varepsilon_{\varepsilon_0} = \omega \uparrow\uparrow (\omega \uparrow\uparrow \omega) = \omega \uparrow\uparrow\uparrow 3$$ an $$\varepsilon_{\varepsilon_{\varepsilon_0}} = \omega \uparrow\uparrow\uparrow 4$$. The triple arrows are quite promising, and indeed the limit of all this is $$\zeta_0 = \omega \uparrow\uparrow\uparrow \omega$$. Boom. Done.

### Formal definition

Good news: formalizing this is just a matter of defining some functions over the ordinals and the fundamental sequences they create. Arrow notation is not part of standard ordinal arithmetic, so we have to define it ourselves:

• $$\alpha \uparrow\uparrow 0 = 1$$
• $$\alpha \uparrow\uparrow (n + 1) = \alpha^{\alpha \uparrow\uparrow n}$$ for $$n < \omega$$
• $$\alpha \uparrow\uparrow \beta = \bigcup\{\gamma < \beta : \alpha \uparrow\uparrow \gamma\}$$ for limit ordinals $$\beta$$
• $$\alpha \uparrow^{k} (\beta + \omega) = lim\{(\alpha \uparrow^k \beta),(\alpha \uparrow^k \beta)^{(\alpha \uparrow^k \beta)},(\alpha \uparrow^k \beta)^{(\alpha \uparrow^k \beta)^{(\alpha \uparrow^k \beta)}},...\}$$ for $$\beta \geq \omega$$
• $$(\alpha \uparrow\uparrow \beta)[n] = \alpha \uparrow\uparrow \beta[n]$$

## Further sublegion arrays

We ended the last section with $$\zeta_0 = \omega \uparrow\uparrow\uparrow \omega$$. We can also write $$\omega \uparrow\uparrow\uparrow \omega = \{\omega, \omega, 3\}$$, and what's to stop us from defining $$\{\omega, \omega, 4\}$$? Or $$\{\omega, \omega, 1, 2\}$$? Or $$\{\omega, \omega (0, 1) 2\}$$?

The only barrier is formality: we need to define BEAF for ordinals. We'll start off with arrow notation for a finite number of arrows:

• $$\alpha \uparrow \beta = \alpha^\beta$$
• $$\alpha \uparrow^k 0 = 1$$
• $$\alpha \uparrow^{k + 1} (n + 1) = \alpha \uparrow^k (\alpha \uparrow^{k + 1} n)$$ for $$n < \omega$$
• $$\alpha \uparrow^k \beta = \sup\{\gamma < \beta : \alpha \uparrow^k \gamma\}$$ for limit ordinals $$\beta$$
• $$\alpha \uparrow^{k} (\beta + \omega) = lim\{(\alpha \uparrow^k \beta),(\alpha \uparrow^k \beta)\uparrow^{k-1}(\alpha \uparrow^k \beta),(\alpha \uparrow^k \beta)\uparrow^{k-1}(\alpha \uparrow^k \beta))\uparrow^{k-1}(\alpha \uparrow^k \beta),...\}$$ for $$\beta \geq \omega$$
• $$(\alpha \uparrow^k \beta)[n] = \alpha \uparrow^k \beta[n]$$

It's not hard to see that ordinal arrow notation is similar to the Veblen function, and the differences are mostly in the fundamental hierarchies. Analogous to the Veblen unction, $$\alpha \mapsto \omega \uparrow^{k + 1} (\omega + \alpha)$$ enumerates the fixed points of $$\alpha \mapsto \omega \uparrow^k \alpha$$.

Already this definition is mildly annoying due to the sheer number of rules. Anyways, let's try $$k = \omega$$:

• $$\alpha \uparrow^\omega \beta = \sup\{k < \omega : \alpha \uparrow^k \beta\}$$
• $$(\alpha \uparrow^\omega \beta)[n] = \alpha \uparrow^n \beta$$

Take note that $$\omega \uparrow^\omega \omega = \phi_\omega(0)$$. Our existing framework works for successor ordinals $$k$$, but we need some limits here:

• $$\alpha \uparrow^\kappa \beta = \sup\{\gamma < \kappa : \alpha \uparrow^\gamma \beta\}$$
• $$(\alpha \uparrow^\kappa \beta)[n] = \alpha \uparrow^{\kappa[n]} \beta$$

Putting it all together:

• $$\alpha \uparrow \beta = \alpha^\beta$$
• $$\alpha \uparrow^\kappa 0 = 1$$
• $$\alpha \uparrow^{\kappa + 1} (n + 1) = \alpha \uparrow^\kappa (\alpha \uparrow^{\kappa + 1} n)$$ for $$n < \omega$$
• $$\alpha \uparrow^\kappa (\beta + 1) = (\alpha \uparrow^\kappa \beta) \uparrow^\kappa \alpha$$ for $$\beta \geq \omega$$
• $$\alpha \uparrow^\kappa \beta = \sup\{\gamma < \kappa : \alpha \uparrow^\gamma \beta\}$$ for limit ordinals $$\kappa$$
• $$\alpha \uparrow^\kappa \beta = \sup\{\gamma < \beta : \alpha \uparrow^\kappa \gamma\}$$ for limit ordinals $$\beta$$
• $$(\alpha \uparrow^\kappa \beta)[n] = \alpha \uparrow^{\kappa[n]} \beta$$ for limit ordinals $$\kappa$$
• $$(\alpha \uparrow^\kappa \beta)[n] = \alpha \uparrow^\kappa \beta[n]$$ for limit ordinals $$\beta$$

This definition works okay, but the last two pairs of rules have identical conditions. In the case of $$(\omega \uparrow^\omega \omega)[n]$$, we'd rather have $$\omega \uparrow^n \omega$$, so if $$\kappa$$ and $$\beta$$ are limit ordinals, we should focus on $$\kappa$$ first.

In the meantime, let's write this out as three-entry array notation. This is ordinary three-entry notation with some limit ordinal cases. To simplify things, we won't allow any of the entries to be 0 as with ordinary array notation.

• $$\{a, b, 1\} = a^b$$
• $$\{a, 1, c\} = a$$
• $$\{a, b + 1, c + 1\} = \{a, \{a, b, c + 1\}, c\}$$ for $$b < \omega$$
• $$\{a, b + 1, c\} = \{\{a, b, c\}, a, c\}$$ for $$b \geq \omega$$
• $$\{a, b, c\} = \sup\{x < c: \{a, b + 1, x\}\}$$ for $$c \in \text{Lim}$$
• $$\{a, b, c + 1\} = \sup\{x < b: \{a, b + 1, c + 1\}\}$$ for $$b \in \text{Lim}$$
• $$\{a, b, c\}[n] = \{a, b, c[n]\}$$ for $$c \in \text{Lim}$$
• $$\{a, b, c + 1\}[n] = \{a, b[n], c + 1\}$$ for $$b \in \text{Lim}$$

Where $$\text{Lim}$$ is set of all limit ordinals.

We can also remove the fourth and fifth rules if we define them as the limits of the sixth and seventh rules, respectively.

Now for four entries:

• $$\{a, b, 1, 1\} = a^b$$
• $$\{a, 1, c, d\} = a$$
• $$b < \omega:$$
• $$\{a, b + 1, 1, d + 1\} = \{a, \{a, b, 1, d + 1\}, 1, d\}$$
• $$\{a, b + 1, c + 1, d\} = \{a, \{a, b, c + 1, d\}, c, d\}$$
• $$b > \omega:$$
• $$\{a, b + 1, c, d\} = \{\{a, b, c, d\}, a, c, d\}$$
• $$\{a, b, c, d\}[n] = \{a, b, c, d[n]\}$$ for $$d \in \text{Lim}$$
• $$\{a, b, c, d + 1\}[n] = \{a, b, c[n], d + 1\}$$ for $$c \in \text{Lim}$$
• $$\{a, b, c + 1, d + 1\}[n] = \{a, b[n], c + 1, d + 1\}$$ for $$b \in \text{Lim}$$

Generalization is easy from here.

Linear ordinal array notation

Definitions are unchanged — the first entry is the base $$b$$, the second entry is the prime $$p$$, the first non-1 entry after the prime is the pilot, the entry before it is the copilot, and the passengers are all entries before the copilot. In addition, we'll define the limiter $$l$$ as the last limit ordinal after the base; it may not exist if there are only successor ordinals from the prime on.

All entries are between zero and $$\omega_1$$ exclusive.

1. Base rule. If there is no pilot, $$v(A) = b^p$$.
2. Prime rule. If $$p = 1$$, $$v(A) = b$$.
3. Catastrophic rule. If there is no limiter and $$p < \omega$$...
1. Replace the copilot with a copy of the array with the prime decreased by 1. (The prime must be a successor ordinal, otherwise there would be a limiter.)
2. Decrease the pilot by 1.
3. Set all the passengers to $$b$$.
4. Infinite catastrophic rule. If there is no limiter and $$p \geq \omega$$...
1. Replace the base with a copy of the array with the prime decreased by 1. (The prime must be a successor ordinal, otherwise there would be a limiter.)
2. Set the prime to the original value of $$b$$.
5. Limit rule. Otherwise...
1. Let $$v(A)[n]$$ be the value of the array with the limiter changed to $$l[n]$$.
2. $$v(A) = \sup\{1 \leq n < \omega : v(A)[n]\}$$.
(end of definition)

Of course, we have no reason to stop at linear arrays. Our tetrational BEAF easily extends to ordinals; we just need to throw in the limit rule, and expand the domain of $$A$$ from $$\varepsilon_0$$ to $$\omega_1$$, the first ordinal without a fundamental sequence with countable length. We'll prohibit ourselves from just plugging in any large countable ordinal, since we're building from the ground up and using BEAF to define the ordinals as we go along.

By the way, what ordinals have we defined? Quite a few.

• $$\Gamma_0 = \{\omega, \omega, 1, 2\}$$ (expandal arrays)
• $$\vartheta(\Omega^2) = \{\omega, \omega, 1, 1, 2\}$$
• $$\vartheta(\Omega^\omega) = \{\omega, \omega (1) 2\}$$
• $$\vartheta(\Omega^\Omega) = \{\omega, \omega, 2 (1) 2\}$$
• $$\vartheta(\varepsilon_{\Omega + 1}) = \{\omega, \omega (\varepsilon_0) 2\}$$

The limit of our notation at the moment is $$\vartheta(\Omega_\omega)$$. This is quite an accomplishment!

## Legion arrays

Thank you, non-ordinal readers, for waiting this long! Up to now we have what could be called "sublegion BEAF." Bowers called everything up to this point "L-space."

We're resurrecting the array of operator & again. As our notations get more and more powerful, so does &: by now we have a good definition for things such as {3, 3, 3}&3 (triakulus), a pentational array. The array of operator will always be able to diagonalize over our most powerful notations.

Note that we can also write triakulus as (3&3)&3, and we'll make the operator left-associative so we can get rid of parentheses as in 3&3&3. This naturally invites an operator, bp = b&b&...&b&b p times. This operator expresses the limit of sublegion BEAF; it's the most powerful function we've defined so far. Note that the ♦ notation hasn't been used by Bowers himself.

$$b♦p$$ reminds us a bit of $$b^p$$, so let's use the same trick we did far back in two-row arrays. We can make this "second-order sublegion BEAF" by redefining the base rule to $$b♦p$$.

As we did with two-row array notation, we could keep expanding higher-order sublegion BEAF until we need to merge the order into the array. We'll cut to the chase and immediately put the order into the array.

$$\{b, p / 2\} = b♦p$$

The slash separator marks off a new kind of space called legion space. The pilot 2 is in the second legion, in the same way that we moved the pilot to the second row way back in planar array notation. The prime block of a legion is $$b♦p$$, so for example $$\{b, p / 3\} = \{b♦p / 2\}$$.

$$\{b, p / 1, 2\} = \{b♦p / \{b, p - 1 / 2\}\}$$

Legions form a larger space than rows, planes, or any structure we've defined so far, so we can put multiple entries in the second legion.

$$\{b, p / 1 / 2\} = \{b♦p / b♦p\}$$

We can also throw in more than two legions...

$$\{b, p (/1) 2\} = \{b♦p / b♦p / \ldots / b♦p / b♦p\}$$ with $$p$$ legions
$$\{b, p (/0,1) 2\} = \{b♦p / b♦p / \ldots / b♦p / b♦p\}$$ where the legions form a $$p^p$$ structure

...and the legions themselves can take on dimensional structures of any size.

Another notation for $$\{b, p (/1) 2\} = \{b♦p / b♦p / \ldots / b♦p / b♦p\}$$ is $$p \&\& b$$, or p legion array of b. Similarly, $$\{b, p (/(1)1) 2\} = p^p \&\& b$$, so the && operator works similarly to & but with legions instead of entries. We'll use b♦♦p to mean b&&b&&..&&b&&b p times, and we'll introduce "legion legion space" using the separator $$//$$.

$$\{b, p // 2\} = b♦♦p = b \&\& b \&\& \ldots \&\& b \&\& b$$
$$\{b, p (//1) 2\} = \{b♦♦p / b♦♦p / \ldots / b♦♦p / b♦♦p\} = b \&\&\& p$$
$$\{b, p /^n 2\} = b♦^np = b \&^n b \&^n \ldots \&^n b \&^n b$$
$$\{b, p (/^n1) 2\} = \{b♦^np / b♦^np / \ldots / b♦^np / b♦^np\} = b \&^{n + 1} p$$

Extension is straightforward.

$$\{b, p (1)/ 2\} = \{b, p /^p 2\} = b (1)\& p$$
$$\{b, p ///(1)///(1)/// 2\}$$

Why not give the legions a multidimensional structure themselves? Trouble is, we need to define this. Let L be a legion structure, which is greater than all structures up to but not including legions. (So L is a legion in the same way that X is a row.) We could say that L = {X, X / 2}, but that's uncomfortably circular. Anyways, the prime block of a legion is $$b♦p$$.

Two legions is a $$2L$$ structure, a row of legions is an $$XL$$ structure, a legion legion // is an $$L^2$$ structure. From here we can put L inside an array the same way we wrote things such as {X, X, X}. We can write $$\{L, L\}$$ or $$\{L, L, L\}$$ or $$\{L, X (1) 2\} = X\&L$$. Unfortunately, by this time we have no notation for separators, but Bowers supplied the notation $$\{L, L\}_{b, p}$$ (say) to mean an $$\{L, L\}$$ structure solved with b as the base and p as the prime.

We continue to drop L into larger and larger arrays until we reach the landmark $$\{L, L / 2\}$$, or lugion space. Bowers also writes this $$L2$$ using the separator \ and the operator $$a@b$$, or a legiattic array of b.

### Formalization

We will quickly catch up on the definition of legion arrays with ordinals. First, we'll notice that the array-of operator has three parts: a base, prime, and structure type. For example, $$3\&4$$ has base 4, prime 3, and structure type X, and $$3 \uparrow\uparrow\uparrow 4 \& 5$$ has base 5, prime 3, and structure type $$X \uparrow\uparrow\uparrow 4$$. Bowers' minor abuse of notation can be mended with a three-argument array-of operator:

$$\lambda(b, p, \xi) = \{(0, b), (1, p), (\xi, 2)\}$$

That is, base $$b$$, prime $$p$$, and the number 2 in position $$\xi$$. Normally $$\xi$$ is a limit ordinal, but our new function allows any $$\xi > 1$$.

The ordinal for legion space is the unnamed ordinal $$\vartheta(\Omega_\omega) = \{\omega, \omega / 2\}$$, and its fundamental sequence is $$\vartheta(\Omega_\omega)[1] = \omega,\,\vartheta(\Omega_\omega)[2] = \lambda(\omega, \omega, \omega),\,\vartheta(\Omega_\omega)[3] = \lambda(\omega, \omega, \lambda(\omega, \omega, \omega)),$$ and so and so forth. This defines the prime block of a $$\vartheta(\Omega_\omega)$$ structure as defined before.