Draft:Bowers' array notation

• Comment: Not sourced for notability (and nor is Conway for that matter). ChrysGalley (talk) 14:31, 26 April 2026 (UTC)

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Array Notation is a means of expressing extremely large numbers, created by Jonathan Bowers , vastly exceeding Conway chained arrow notation, a notation designed to produce very large numbers (see Bird's proof).

Definition and rules

Let $a,b,c,\ldots ,a_{1},a_{2},a_{3},\ldots$ denote positive integers and $\#$ denote an arbitrary sequence of positive integers. An entry is a positive integer in an array. An example of a valid array is $\{4,5,1,3,2\}$. The array is evaluated according to the following rules:

1. $\{a\}=a$
2. $\{a,b\}=a^{b}$ (originally $a+b$)
3. $\{\#,1\}=\{\#\}$
4. $\{a,1,\#\}=a$
5. $\{a,b,1,\ldots ,1,c,\#\}=\{a,a,\ldots ,a,\{a,b-1,1,\ldots ,1,c,\#\},c-1,\#\}$ (if the third entry is $1$, all entries before the last $1$ preceding non-$1$ entry $c$ become the first entry, the last of the ones becomes the original array with the second entry decreased by $1$, and the said non-$1$ entry is decreased by $1$)
6. If none of the rules above apply, then $\{a,b,c,\#\}=\{a,\{a,b-1,c,\#\},c-1,\#\}$ $(c>1)$

Examples

1. The array above is evaluated like so: $\{4,5,1,3,2\}=\{4,4,\{4,4,1,3,2\},2,2\}$ (third entry is $1$, rule 5)
2. $\{5,5,1,1,2\}=\{5,5,5,\{5,4,1,1,2\},1\}=\{5,5,5,\{5,4,1,1,2\}\}=$
   $\{5,5,5,\{5,5,5,\{5,3,1,1,2\}\}\}=\{5,5,5,\{5,5,5,\{5,5,5,\{5,2,1,1,2\}\}\}\}=$
   $\{5,5,5,\{5,5,5,\{5,5,5,\{5,5,5,\{5,1,1,1,2\}\}\}\}\}=\{5,5,5,\{5,5,5,\{5,5,5,\{5,5,5,5\}\}\}\}$
3. $\{2,3,3\}=\{2,\{2,2,3\},2\}=\{2,\{2,\{2,1,3\},2\},2\}=\{2,\{2,2,2\},2\}=\{2,\{2,\{2,1,2\},1\},2\}=$
   $\{2,\{2,2,1\},2\}=\{2,\{2,2\},2\}=\{2,4,2\}=\{2,\{2,3,2\},1\}=\{2,\{2,3,2\}\}=$
   $\{2,\{2,\{2,2,2\}\}\}=\{2,\{2,\{2,\{2,1,2\}\}\}\}=\{2,\{2,\{2,2\}\}\}=2^{2^{2^{2}}}=2\uparrow \uparrow 4=65536$
4. $\{a,b,c\}=a\uparrow ^{c}b$

Extended hyperoperators

Bowers also created a set of "extended operators" to visualize his array notation. ${\begin{array}{lcl}a\{c\}b=a\uparrow ^{c}b=\{a,b,c\}\\a\underbrace {\{\{\ldots \{\{} _{d}c\underbrace {\}\}\ldots \}\}} _{d}b=a\{c\}^{d}b=\{a,b,c,d\}\end{array}}$

Properties

Array notation with 4 or more entries is not primitive recursive because the Ackermann function $A(n,n)=(2\{n-2\}(n+3))-3(n\geq 3)$ grows slower than even $\{n,n,n\}=n\{n\}n$

Bird's Proof

Bird’s Proof is a theorem that states that $\forall a\geq 3,b\geq 2,c\geq 1,d\geq 2,\{a,b,c,d\}>\underbrace {a\rightarrow a\rightarrow a\rightarrow \ldots \rightarrow a} _{d}\rightarrow (b-1)\rightarrow (c+1)$ in Conway chained arrow notation. The theorem was proven by Chris Bird and named by Jonathan Bowers. Before proving the main theorem, Bird proved two lemmas:

• $a\rightarrow \#\geq a\forall$ Conway chains.
• $a\rightarrow a\rightarrow \ldots \rightarrow a\rightarrow (y+1)\rightarrow z>(a\rightarrow a\rightarrow \ldots \rightarrow a\rightarrow y\rightarrow z)+1$ ($n$ $a$’s on each side)

and two corollaries:

• $y'>y\geq 1\implies a\rightarrow a\rightarrow \ldots \rightarrow a\rightarrow y'\rightarrow z>a\rightarrow a\rightarrow \ldots \rightarrow a\rightarrow y\rightarrow z$

($n$ $a$’s on each side), $\forall a\geq 3,z\geq 1,n\geq 1$.

• $y'>y\geq 1\implies a\rightarrow a\rightarrow \ldots \rightarrow a\rightarrow y'>a\rightarrow a\rightarrow \ldots \rightarrow a\rightarrow y$

($n$ $a$’s on each side), $\forall a\geq 3,z\geq 1,n\geq 1$. Note that the second corollary is just the first corollary, but with $z=1$.

The main theorem was proven by mathematical induction.

The proof implies that:

1. $\{a,a,a,d\}>\underbrace {a\rightarrow a\rightarrow a\rightarrow \dots \rightarrow a} _{d+2}\forall a\geq 3,d>1$, as $\{a,a,a,1\}=\{a,a,a\}=a\uparrow ^{a}a=a\rightarrow a\rightarrow a$
2. $$\left.{\begin{matrix}\{a,b,1,1,2\}>&a\underbrace {\rightarrow a\rightarrow \ldots \rightarrow } a\\&(a\underbrace {\rightarrow a\rightarrow \ldots \rightarrow } a)+2\\&\underbrace {\qquad \;\;\vdots \qquad \;\;} \\&(a\underbrace {\rightarrow a\rightarrow \ldots \rightarrow } a)+2\\&a+2\end{matrix}}\right\}b{\text{ layers}}$$$\forall a\geq 3$, $b\geq 3$

Graham's Number

Graham's number cannot be easily expressed in array notation but can be bounded by the following: $\{3,65,1,2\}<$ Graham's number $<\{4,65,1,2\}\ll \{3,66,1,2\}$

Proof

Let $f(x)=\{3,3,x\}$. Then, Graham’s Number is $f^{64}(4)$, where the superscript represents iteration. Since $\{3,3,3\}=f(3)$, it follows that $f^{64}(3)=\{3,65,1,2\}<f^{64}(4)=\{3,3,f^{63}(4)\}=$ Graham’s Number $<\{4,4,f^{63}(4)\}<\{4,65,1,2\}\ll \{3,66,1,2\}$

Array of

Bowers created a simple function: $b\&a$ ($b$ array of $a$) $=\underbrace {\{a,a,a,\ldots ,a\}} _{b}$. As one might expect, this grows extraordinarily faster than Conway chained arrow notation. ($a\&a\approx f_{\omega ^{\omega }}(a)$ in the fast-growing hierarchy)

Extended Array Notation

Bowers extended his array notation to more dimensions and even tetrational spaces. It has a growth rate of $f_{\varepsilon _{0}}(a)$ in the fast-growing hierarchy..

Rules

$\#,\#_{1},\ldots$ denote remainders of an array, $\&$ denotes the "array of" operator, $(n),n>0$ denotes a "separator" and $(n_{1})(n_{2})\ldots (n_{m})$ denote separators such that $n_{1}\geq n_{2}\geq \ldots \geq n_{m}>0$. The rules only cover cases up to $f_{\omega ^{\omega ^{\omega }}}(a)$, but can be easily extended to $f_{\varepsilon _{0}}(a)$:

1. 2 entries: $\{a,b\}=a^{b}$
2. Rows ending with 1: $\{\#1(m)\#_{1}1(n)\#_{2}1\}=\{\#(m)\#_{1}(n)\#_{2}\}$
3. Second entry is 1: $\{a,1\#\}=a$
4. Third entry is 1, non-1 entry in same row $\{a,b,1,\ldots ,1,c\#\}=\{a,a,a,\ldots ,a,\{a,b-1,1,\ldots ,1,c,\#\},c-1\#\}$
5. Rules above do not apply: $\{a,b,c\#\}=\{a,\{a,b-1,c\#\},c-1\#\}$
6. 2 entries on main row, next non-1 entry begins a higher structure: $\{a,b(n)\#(k)c\#\}=\{b^{n}\&a(n)\#(k)c-1\#\}$
7. 2 entries on main row, next non-1 entry is preceded by 1's in its row: $\{a,b(n)\#(k)1,\ldots ,1,c\#_{1}\}=\{b^{n}\&a(n)\#(k)a,a,\ldots a,\{a,b-1(n)\#(k)1,1,\dots 1,1,c\#_{1}\},c-1\#_{1}\}$
8. $b^{0}\&a=1\&a=a$
9. $b^{1}\&a=b\&a$
10. $b^{n+1}\&a=\underbrace {b^{n}\&a(n)\ldots (n)b^{n}\&a} _{b}$ for $n\geq 1$