# 高德納箭號表示法

## 簡介

$3 \uparrow \uparrow 2 = {^{2}3} = 3^3 = 27$
$3 \uparrow \uparrow 3 = {^{3}3} = 3^{3^3} = 3^{27} = 7,625,597,484,987$
$3 \uparrow \uparrow 4 = {^{4}3} = 3^{3^{3^3}} = 3^{7625597484987} \approx 1.2580143\times 10^{3638334640024}$
$3 \uparrow \uparrow 5 = {^{5}3} = 3^{3^{3^{3^3}}} = 3^{3^{7625597484987}} \approx 3^{1.2580143\times 10^{3638334640024}}$

$3 \uparrow\uparrow \uparrow 2 = 3 \uparrow\uparrow 3 = {^{3}3} = 3^{3^3} = 3^{27} = 7,625,597,484,987\,\!$

$3 \uparrow\uparrow \uparrow 3 = 3 \uparrow\uparrow 3 \uparrow\uparrow 3 = {^{^{3}3}3} = {^{7625597484987}3} = \begin{matrix} \underbrace{ 3^{3^{3^{...}}} } \\ 7625597484987\end{matrix}$

## 使用指數來解釋高德納箭號表示法

$a \uparrow \uparrow b$代表重複的冪，或迭代冪次，例如： $a \uparrow \uparrow 4 = a \uparrow (a \uparrow (a \uparrow a)) = a^{a^{a^a}}$

$a \uparrow \uparrow b = \underbrace{a^{a^{.^{.^{.{a}}}}}}_{b}$

$a \uparrow \uparrow \uparrow 2 = a \uparrow \uparrow a = \underbrace{a^{a^{.^{.^{.{a}}}}}}_{a}$
$a \uparrow \uparrow \uparrow 3 = a \uparrow \uparrow (a \uparrow \uparrow a) = \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{a} }$
$a \uparrow \uparrow \uparrow 4 = a \uparrow \uparrow (a \uparrow \uparrow (a \uparrow \uparrow a)) = \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{a} }}$

$a \uparrow \uparrow \uparrow b = \left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} b$

$a \uparrow \uparrow \uparrow \uparrow 2 = a \uparrow \uparrow \uparrow a = \left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} a$
$a \uparrow \uparrow \uparrow \uparrow 3 = a \uparrow \uparrow \uparrow (a \uparrow \uparrow \uparrow a) = \left.\left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} a$
$a \uparrow \uparrow \uparrow \uparrow 4 = a \uparrow \uparrow \uparrow (a \uparrow \uparrow \uparrow (a \uparrow \uparrow \uparrow a)) = \left.\left.\left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} a$

$a \uparrow \uparrow \uparrow \uparrow b = \underbrace{ \left.\left.\left. \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{a^{a^{.^{.^{.{a}}}}}}_{ \underbrace{\vdots}_{a} }} \right\} \cdots \right\} a }_{b}$

## 一般化

$\begin{matrix} a\uparrow^n b & = & \mbox{hyper}(a,n+2,b) & = & a\to b\to n \\ \mbox{(Knuth)} & & & & \mbox{(Conway)} \end{matrix}$

## 定義

 $a\uparrow^n b= \left\{ \begin{matrix} 1, \\ a^b, \\ a\uparrow^{n-1}(a\uparrow^n(b-1)), \end{matrix} \right.$ 若$b=0$； 若$n=1$； 其他。