Tetration

tetration also known as (**hyper4**, **superpower**, **superexponentiation**, **superdegree**, **powerlog**, or **power tower**, is a binary mathematical operator (that is to say, one with just two inputs), defined as yx=xxx⋅⋅⋅ with y copies of x. In other words, tetration is repeated exponentiation. Formally, this is

\[^0x=1\]

\[^{n + 1}x = x^{^nx}\]

where \(n\) is a nonnegative integer.

Tetration is the fourth hyper operator, and the first hyper operator not appearing in mainstream mathematics. When repeated, it is called pentation.

If \(c\) is a non-trivial constant, the function \(a(n) = {}^nc\) grows at a similar rate to \(f\_3(n)\) in FGH.

Daniel Geisler has created a website, IteratedFunctions.com (formerly tetration.org), dedicated to the operator and its properties.

## Basis

Addition is defined as repeated counting:

\[x + y = x + \underbrace{1 + 1 + \ldots + 1 + 1}\_y\]

Multiplication is defined as repeated addition:

\[x \times y = \underbrace{x + x + \ldots + x + x}\_y\]

Exponentiation is defined as repeated multiplication:

\[x^y = \underbrace{x \times x \times \ldots \times x \times x}\_y\]

Analogously, tetration is defined as repeated exponentiation:

\[^yx = \underbrace{x^{x^{x^{.^{.^.}}}}}\_y\]

But since exponentiation is not an associative operator (that is, \(a^{b^{c}}\) is generally not equal to \(\left(a^b\right)^c = a^{bc}\)), we can also group the exponentiation from bottom to top, producing what Robert Munafo calls the **hyper4 operator**, written \(x\_④y\). \(x\_④y\) reduces to \(x^{x^{y - 1}}\), which is not as mathematically interesting as the usual tetration. This is equal to \(x \downarrow\downarrow y\) in down-arrow notation.

### Notations

Tetration was independently invented by several people, and due to lack of widespread use it has several notations:

* \(^yx\) is pronounced "to-the-\(y\) \(x\)" or "\(x\) tetrated to \(y\)." The notation is due to Rudy Rucker, and is most often used in situations where none of the higher operators are called for.
* Robert Munafo uses \(x^④y\), the *hyper4 operator*.
* In arrow notation it is \(x \uparrow\uparrow y\), nowadays the most common way to denote tetration.
* In chained arrow notation it is \(x \rightarrow y \rightarrow 2\).
* In array notation it is \(\{x, y, 2\}\) or \(x\ \{2\}\ y\).
	+ The latter of these also represents tetration in X-Sequence Hyper-Exponential Notation.
* In Hyper-E notation it is E[x]1#y (alternatively x^1#y).
* In plus notation it is \(x ++++ y\).
* In star notation (as used in the Big Psi project) it is \(x \*\*\* y\).
* An exponential stack of *n* 2's was written as E\*(n) by David Moews, the man who held Bignum Bakeoff.
* Harvey Friedman uses \(x^{[y]}\).

## Properties

Tetration lacks many of the symmetrical properties of the lower hyper-operators, so it is difficult to manipulate algebraically. However, it does have a few noteworthy properties of its own.

### Power identity

It is possible to show that \({^ba}^{^ca} = {^{c + 1}a}^{^{b - 1}a}\):

\[{^ba}^{^ca} = (a^{^{b - 1}a})^{(^ca)} = a^{^{b - 1}a \cdot {}^ca} = a^{^ca \cdot {}^{b - 1}a} = (a^{^ca})^{^{b - 1}a} = {^{c + 1}a}^{^{b - 1}a}\]

For example, \({^42}^{^22} = {^32}^{^32} = 2^{4J}\).

### Moduli of power towers

Be careful that **there were many unsourced (and perhaps wrong) statements on last digits of tetration in this wiki.** For example, the last \(d\) digits \(N(y)\) of \(^yx\) in base \(b\) was believed to be computed by the following **wrong** recursive formula in this community up may 2020
\begin{eqnarray\*}
N(m) = \left\{ \begin{array}{l1} 1 & (m=0) \\ x^{N(m-1)} \mod{b^d} & (m>0) \end{array} \right.
\end{eqnarray\*}
Here, \(s \mod{t}\) denotes the remainder of the division of \(s\) by \(t\). For example, we have the following counterexamples:

* If \(b = 10\) and \(d = 1\), then the formula computes the last 1 digit of \(2 \uparrow \uparrow 4\) in base \(10\) as \(2^{13 \mod{10}} \mod{10} = 2^6 \mod{10} = 64 \mod{10} = 4\), while the last 1 digit of \(2 \uparrow \uparrow 4 = 2^{13} = 23XX3\) in base \(10\) is \(6\).
* If \(b = 10\) and \(d = 2\), then the formula computes the last 2 digits of \(100 \uparrow \uparrow 2\) in base \(10\) as \(100^{100 \mod{10^2}} \mod{10^2} = 100^0 \mod{10^2} = 1\) (i.e. \(01\)), while the last 2 digits of \(100 \uparrow \uparrow 2 = 100^{100} = 1000 \cdots 000\) in base \(10\) is \(00\).
* If \(b = 12\) and \(d = 1\), then computes the last 1 digit of \(2 \uparrow \uparrow 4\) in base \(12\) as \(2^{13 \mod{12}} \mod{12} = 2^1 \mod{12} = 2\), while the last 1 digit of \(2 \uparrow \uparrow 4 = 23XX3\) in base \(12\) is \(1\).

The method works when \(x\) is coprime with \(b\) and \(b^d\) is divisible by the least common multiple of \(\phi(p^d) = (p-1)p^{d-1}\) for prime factors \(p\) of \(b\) by Template:W and Template:W, where \(\phi\) denotes Template:W. For example, when \(b = 10\) and \(d \geq 3\), then the condition holds. It is obvious that it does not hold when \(b\) is an odd prime number. Even if \(b\) is not a prime number, the method does not necessarily hold as we have seen the counterexamples \((x,b,d) = (2,10,1)\), where \(b^d = 10\) is not divisible by the least common multiple \(4\) of \(\phi(2^d) = 1\) and \(\phi(5^d) = 4\), and \((x,b,d) = (2,12,1)\), where \(b^d = 12\) is not divisible by the least common multiple \(4\) of \(\phi(3^d) = 2\) and \(\phi(5^d) = 4\). In general, the condition never holds when \(b\) is an odd number.