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 \(^yx = x^{x^{x^{.^{.^.}}}}\) 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, \({^42}^{^22} = {^32}^{^32} = 2^{64}\).

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 to December in 2019: \begin{eqnarray*} N(m) = \left\{ \begin{array}{ll} 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 \(2^{16 \mod{10}} \mod{10} = 2^6 \mod{10} = 64 \mod{10} = 4\), while the last 1 digit of \(2 \uparrow \uparrow 4 = 2^{16} = 65536\) 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 \(b = 15\) and \(d = 1\), then computes the last 1 digit of \(2 \uparrow \uparrow 4\) in base \(15\) as \(2^{16 \mod{15}} \mod{15} = 2^1 \mod{15} = 2\), while the last 1 digit of \(2 \uparrow \uparrow 4 = 65536\) in base \(15\) 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 and, where \(\phi\) denotes. 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 \((x,b,d) = (2,15,1)\), where \(b^d = 15\) 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.

In particular, the method works when \(x\) is coprime with \(b\) and one of the following holds: If the method works, then the modular exponentiation can be computed very quickly using several algorithms such as. The condition that \(x\) is coprime to \(b\) is important as we have the counterexample \((x,b,d) = (100,10,2)\), although it is often carelessly dropped in this wiki. Further, although it is sometimes believed that the recursive formula works when \(d\) is sufficiently large, \(x = b^d\) is always a counterexample of the recursive formula for any \(b \geq 2\) and \(d \geq 1\) because \(^2 x \mod{b^d} = 0\) and \(x^{x \mod{b^d}} \mod{b^d} = x^0 \mod {b^d} = 1 \mod{b^d} = 1\).
 * \(b = 2^h\) for some \(h \geq 1\).
 * \(b = 2^h \times 3^i\) for some \(h,i \geq 1\).
 * \(b = 2^h \times 5^i\) for some \(h,i \geq 1\), and \(dh \geq 2\).
 * \(b = 2^h \times 3^i \times 5^j\) for some \(h,i,j \geq 1\), and \(dh \geq 2\).
 * \(b = 2^h \times 3^i \times 7^j\) for some \(h,i,j \geq 1\).

Even if \(x\) is not necessarily coprime with \(b\), the last \(1\) digit \(D_b(x,m)\) of \(x^m\) in base \(b\) for an \(m \in \mathbb{N}\) can be quite easily computed in the case \(b\) is squarefree. First, consider the case where \(b\) is a prime number \(p\). Then \(D_p(x,m)\) is computed in the following recursive way by : Since \((S_p^n(m))_{n=0}^{\infty}\) is a decreasing sequences converges to a non-negative integer \(S_p^{\infty}(m)\) smaller than \(p\), the computation of \(D_p(x,m)\) is quickly reduced to the computation of \(D_p(x \mod{p},S_p^{\infty}(m)) \in \{(x \mod{p})^i \mod{p} \mid 0 \leq i < p\}\). Now consider the general case. Then \(D_b(m)\) is computed in the following recursive way by : Be careful that this method does not work when \(b\) is not squarefree or the length \(d\) of digits is greater than \(1\), while such restricted methods are often "extended" to general cases just by dropping conditions without theoretic justification in this community as long as people do not know explicit counterexamples, as we have explained above.
 * 1) Denote by \(S_p(m)\) the sum of all digits of \(m\) in base \(p\).
 * 2) Then \(D_p(x,m) = D_p(x \mod{p},S_p(m))\).
 * 1) Denote by \(p_0,\ldots,p_k\) the enumeration of prime factors of \(b\).
 * 2) If \(k = 0\), then we have done the computation of \(D_b(x,m) = D_{p_0}(x,m)\) by the algorithm above.
 * 3) Suppose \(k > 0\).
 * 4) Put \(b' = p_0 \cdots p_{k-1}\). (Since \(b'\) is also squarefree, \(D_{b'}(x,m)\) is also computed in this algorithm by the induction on \(k\).)
 * 5) Take an \((s,t) \in \mathbb{Z}^2\) such that \(sb' + tp_k = 1\) by . (Such an \((s,t)\) actually exists because \(b'\) is coprime with \(p_k\))
 * 6) Then \(D_b(x,m) = (sb'D_{p_k}(x,m)+tp_kD_{b'}(x,m)) \mod{b}\).

For the case where \(b\) is not squarefree or the length \(d\) of digits is greater than \(1\), we need another method to use a shifted system of representatives of \(\mathbb{Z}/b^d \mathbb{Z}\) other than \(\{0,\ldots,b^d-1\}\). Suppose that \(b^d\) is divisible by the least common multiple of \(\phi(p^d)\) for prime factors \(p\) of \(b\). Let \(\mu \in \mathbb{N}\) denote the maximum of multiplicities of prime factors of \(b\). For an \(n \in \mathbb{N}\), we define \(\textrm{Rep}(n) \in \mathbb{N}\) in the following way: Then even if \(x\) is not necessarily coprime with \(b\), we have \(\textrm{Rep}(x^m) = \textrm{Rep}(x^{\textrm{Rep}(m)})\) unless \(x^{\textrm{Rep}(m)} < \mu d \leq x^m\) by and. Let \(y_0 \in \mathbb{N}\) be a fixed sufficiently small natural number such that \(^{y_0}x\) is known to be greater than or equal to \(\mu d\). Then \(\textrm{Rep}(^yx)\) can be computed in the following way: In particular, the last \(d\) digit of \(^yx\) in base \(b\) can be computed by \(\textrm{Rep}(^yx) \mod{b^d}\) by \((^yx - \textrm{Red}(^yx)) \mod{b^d} = 0\).
 * 1) If \(n < \mu d\), then \(\textrm{Rep}(n) = n\).
 * 2) If \(n \geq \mu d\), then \(\textrm{Rep}(n)\) is the unique \(i \in \{\mu d,\ldots,b^d - 1 + \mu d\}\) such that \(n-i \mod{b^d} = 0\).
 * 1) If \(y < y_0\), then compute \(\textrm{Rep}(^yx)\) in the usual way.
 * 2) If \(y \geq y_0\) and \(x^{\textrm{Rep}(^{y-1}x)} < \mu d\), then \(\textrm{Rep}(^yx) = \textrm{Rep}(x^{\textrm{Rep}(^{y-1}x)}+b^d)\).
 * 3) If \(y \geq y_0\) and \(x^{\textrm{Rep}(^{y-1}x)} \geq \mu d\), then \(\textrm{Rep}(^yx) = \textrm{Rep}(x^{\textrm{Rep}(^{y-1}x)})\).

First digits
Computing the first digits of \(^yx\) in a reasonable amount of time for y > 6 is probably impossible. In base 10:

\[a^b = 10^{b \log_{10} a} = 10^{\text{frac}(b \log_{10} a) + \lfloor b \log_{10} a \rfloor} = 10^{\text{frac}(b \log_{10} a)} \times 10^{\lfloor b \log_{10} a \rfloor}\]

The leading digits of \(^ba\) are then \(10^{\text{frac}(^{b - 1}a \log_{10} a)}\), so the problem is finding the fractional part of \(^{b - 1}a \log_{10} a\). This is equivalent to finding arbitrary base-\(a\) digits of \(^{b - 2}a\) starting at the \(^{b - 2}a\)th place. The most efficient known way to do this is a, which, unfortunately, requires linear time to operate and works only with radixes that are powers of 2. We need an algorithm at least as efficient as \(O(\log^*n)\) (where \(\log^*n\) is the ), and it is unlikely that one exists.

This roadblock ripples through the rest of the hyperoperators. Even if we do find a \(O(\log^*n)\) algorithm, it becomes unworkable at the pentational level. A constant time algorithm is needed, and finding such an algorithm would take a miracle.

For non-integral \(y\)
Mathematicians have not agreed on the function's behavior on \(^yx\) where \(y\) is not an integer. In fact, the problem breaks down into a more general issue of the meaning of \(f^t(x)\) for non-integral \(t\). For example, if \(f(x) := x!\), what is \(f^{2.5}(x)\)? was very interested in the problem of continuous tetration because it may reveal the general case of "continuizing" discrete systems.

Daniel Geisler describes a method for defining \(f^t(x)\) for complex \(t\) where \(f\) is a holomorphic function over \(\mathbb{C}\) using Taylor series. This gives a definition of complex tetration that he calls hyperbolic tetration.

As \(y \rightarrow \infty\)
One function of note is infinite tetration, defined as

\[^\infty x = \lim_{n\rightarrow\infty}{}^nx\]

If we mark the points on the complex plane at which \(^\infty x\) becomes periodic (as opposed to escaping to infinity), we get an interesting fractal. Daniel Geisler studied this shape extensively, giving names to identifiable features.

Examples
Here are some small examples of tetration in action:


 * \(^22 = 2^2 = 4\)
 * \(^32 = 2^{2^2} = 2^4 = 16\)
 * \(^23 = 3^3 = 27\)
 * \(^33 = 3^{3^3} = 3^{27} =\)
 * \(^42 = 2^{2^{2^2}} = 2^{2^4} = 2^{16} = 65,536\)
 * \(^35 = 5^{5^5} \approx 1.9110125979 \cdot 10^{2,184}\)
 * \(^52 \approx 2.00352993041 \cdot 10^{19,728}\)
 * \(^310 = 10^{10^{10}} = 10^{10,000,000,000}\)
 * \(^43 \approx 10^{10^{10^{1.11}}}\)

When given a negative or non-integer base, irrational and complex numbers can occur:


 * \(^2{-2} = (-2)^{(-2)} = \frac{1}{(-2)^2} = \frac{1}{4}\)
 * \(^3{-2} = (-2)^{(-2)^{(-2)}} = (-2)^{1/4} = \frac{1 + i}{\sqrt[4]{2}}\)
 * \(^2(1/2) = (1/2)^{(1/2)} = \sqrt{1/2} = \frac{\sqrt2}{2}\)
 * \(^3(1/2) = (1/2)^{(1/2)^{(1/2)}} = (1/2)^{\sqrt{2}/2}\)

Functions whose growth rates are on the level of tetration include:


 * The Catalan-Mersenne sequence
 * The size of power sets in the von Neumann universe as a function of stage
 * \(f_3\) in the fast-growing hierarchy

Super root
Since ba is perfectly well-defined for non-integer a, we can define a root inverse function as:

\(sr_k(n) = x \text{ such that } ^kx = n\)

Numerical evaluation
The second-order super root can be calculated as:

\(\frac{ln(x)}{W(ln(x))}\)

where \(W(n)\) is the.

Formulas for higher-order super roots are unknown.

Pseudocode
Below is an example of pseudocode for tetration.

function tetration(a, b): result := 1 repeat b times: result := a to the power of result return result