Arrow notation


 * For other arrow notations, see down-arrow notation and chained arrow notation.

Arrow notation or up-arrow notation is a widely used notation for the hyper operators, devised by Donald Knuth in 1976 to represent large numbers. It is defined by the following rules:

\begin{eqnarray} a \uparrow^1 b &=& a^b \\ a \uparrow^n 1 &=& a \\ a \uparrow^{n + 1} (b + 1) &=& a \uparrow^n (a \uparrow^{n + 1} b) \\ \end{eqnarray}

\(a \uparrow^{n} b\) is a shorthand for \(a \uparrow\uparrow\cdots\uparrow\uparrow b\) with n arrows (where n is a positive integer). So, for example, \(a \uparrow^2 b = a \uparrow\uparrow b\). Arrow notation operators are right-associative; \(a \uparrow b \uparrow c\) always means \(a \uparrow (b \uparrow c)\).

Specifically, \(a \uparrow b\) is exponentiation, \(a \uparrow\uparrow b\) is tetration, \(a \uparrow\uparrow\uparrow b\) is pentation, and so forth. In ASCII, these are written a^b, a^^b, a^^^b, ...

Arrow notation has been generalized to other notations. Two notable ones are chained arrow notation and BEAF/BAN.

Examples

 * \(2 \uparrow 3 = 2^3 = 8\)
 * \(5 \uparrow 6 = 5^6 = 15625\)
 * \(10 \uparrow 100 = 10^{100} =\) googol
 * \(3 \uparrow\uparrow 4 = 3 \uparrow 3 \uparrow 3 \uparrow 3 = 3 \uparrow 3 \uparrow 27 = 3^{7625597484987}\)
 * \(5 \uparrow\uparrow 3 = 5 \uparrow 5 \uparrow 5 = 5^{5^5}\)
 * \(2 \uparrow\uparrow\uparrow 2 = 2 \uparrow\uparrow 2 = 2 \uparrow 2 = 2^2 = 4\)
 * \(3 \uparrow\uparrow\uparrow 2 = 3 \uparrow\uparrow 3 = 3 \uparrow 3 \uparrow 3 = 3^{3^3} = 3^{27} = 7625597484987\)
 * \(2 \uparrow\uparrow\uparrow 3 = 2 \uparrow\uparrow 2 \uparrow\uparrow 2 = 2 \uparrow\uparrow 4 = 2 \uparrow 2 \uparrow 2 \uparrow 2 = 2 \uparrow 2 \uparrow 4 = 2 \uparrow 16 = 65536\)
 * \(3 \uparrow\uparrow\uparrow 3 = 3 \uparrow\uparrow 3 \uparrow\uparrow 3 = 3 \uparrow\uparrow 7625597484987 = Tritri\)

Proof of well-definedness
It is straightforward to prove that \(a \uparrow^{n} b\) exists, where all lowercase letters represent positive integers. We use induction on \(n\).

The base case is \(n = 1\), which is true since \(a \uparrow^1 b = a^b\) is well-defined for all \(a,b\).

Suppose \(a \uparrow^{n} b\) exists for all \(a,b\). Then we use induction on \(b\). The base case is \(a \uparrow^{n + 1} 1 = a\). Suppose \(a \uparrow^{n + 1} b\) exists for all \(a\). Then \(a \uparrow^n (a \uparrow^{n + 1} b)\) exists, so \(a \uparrow^{n + 1} (b + 1)\) exists for all \(a\), completing induction on \(b\). Therefore \(a \uparrow^{n + 1} b\) exists for all \(a,b\), completing induction on \(n\). Therefore \(a \uparrow^n b\) for all \(a,b,n\).

Turing machine code
Input form: to represent \(a \uparrow^{c} b\), place a 1's, then c+2 ^'s, then b 1's. For example, 111^^^^^111 computes tritri.

Turing machine code

0 * * r 0 0 _ _ l 1 1 1 _ l 2 2 ^ _ l 3 2 1 1 l 4 3 ^ _ l 3 3 1 _ l 2 4 1 1 l 4 4 ^ 1 l 4' 4 _ 1 l halt 4' ^ ^ l 5 4' 1 1 r 0 5 ^ ^ l 5 5 1 1 r 6 6 ^ x r 7 6 1 y r 9 6 | ^ l 12 7 * * r 7 7 _ | r 8 7 | | r 8 8 * * r 8 8 _ ^ l 11 9 * * r 9 9 | | r 10 10 * * r 10 10 _ 1 l 11 11 * * l 11 11 x ^ r 6 11 y 1 r 6 12 * * l 12 12 ^ ^ l 12' 12' ^ ^ l 12' 12' * * l 13 12' 1 x r 14 13 * * l 13 13 ^ ^ r 20 13 _ _ r 20 13 1 x r 14 14 * * r 14 14 ^ ^ r 15 15 ^ ^ r 15 15 x x r 16 15 1 x l 12 16 x x r 16 16 1 x l 12 16 ^ x r 17 17 ^ ^ r 17 17 1 ^ r 18 18 ^ 1 r 17 18 1 1 r 18 18 _ 1 l 19 19 * * l 19 19 x x l 12 20 x 1 r 20 20 ^ ^ r 21 21 ^ ^ r 21 21 x x r 22 22 x x r 22 22 1 _ r 23 22 ^ ^ l 30 23 1 _ r 23 23 ^ ^ l 24 24 _ ^ r 25 25 ^ ^ r 25 25 1 1 l 26 26 ^ 1 r 27 27 1 1 r 27 27 _ _ l 28 28 1 _ l 29 29 * * l 29 29 _ ^ r 25 29 x 1 l 30 30 x 1 l 30 30 ^ ^ r 31 31 * * r 31 31 _ _ l 32 32 1 _ l 1