What is this operator called?

If $x \cdot 2 = x + x$

and $x \cdot 3 = x + x + x$

and $x^2 = x \cdot x$

and $x^3 = x \cdot x \cdot x$

Is there an operator $\oplus$ such that:

$x \oplus 2 = x^x$

and $x \oplus 3 = {x^{x^x}}$?

Also, is there a name for such a set of operators ops where...

Ops(2) is multiplication

Ops(3) is exponentiation

Ops(4) is $\oplus$

...and so on

Also, is there a branch of math who actually deals with such questions? Have these questions already been answered like 2000 years ago?

• You may be interested in the Ackermann function. Makes for fun problems in recursion theory and such. – Jyrki Lahtonen Jul 5 '16 at 9:48
• FYI $(x^x)^x = x^{(x^2)} \ne x^{(x^x)}$ so the operation is not associative so the notation ${x^x}^x$ is ambiguous. We usually take it to mean $x^{(x^x)}$ because $(((x^x)^x)^x).... = x^{(x^m)}$ and can be expressed that way. – fleablood Jul 5 '16 at 17:34
• Also of note: the "Ops" are known as Hyperoperations. – mbomb007 Jul 5 '16 at 18:56
• i think fleablood got the convention backward. $a^{b^c}$ is understood as $a^{(b^c)}$ because $(a^b)^c = a^{bc}$. – Anton Sherwood Jul 6 '16 at 0:52

This operation ${\rm Ops}(4)$ is called tetration, from the greek root tetra meaning four; it's also sometimes called a "power tower". There are also many further generalizations of this type of sequence; Knuth's up-arrow notation gives $a^{a^{a^a}}=a\uparrow\uparrow4$, so that $a\uparrow\uparrow n$ is the tetration operation. By adding more arrows you get pentation and so on, and the Conway chained arrow notation generalizes this still further.

FYI, for "to the power of-ation" the word you're looking for is exponentiation.

• Also that is similar to graham number. en.wikipedia.org/wiki/Graham%27s_number – Takahiro Waki Jul 5 '16 at 9:19
• Aha, and the answer to the second question is hyperoperation: en.m.wikipedia.org/wiki/Hyperoperation – uzilan Jul 5 '16 at 12:19
• Ah, the whims of the internet. This does not deserve to be my highest-voted answer. – Mario Carneiro Jul 6 '16 at 4:03
• It's the bikeshed, Mario; you get used to it. – J. M. is a poor mathematician Jul 6 '16 at 12:36
• The problem with this answer is that a↑↑n is not defined to be tetration operation. It is a consequence of a↑n being defined as $a^n$ and then further arrows is defined to be n number of iterations of itself minus one arrow on a. Much prefer @Curd 's answer – Ariana Jul 8 '16 at 17:59

A more general function that combines all those operators has been defined by Ackermann:

$\varphi(m,n,p) = \begin{cases} \varphi(m, n, 0) = m + n \\ \varphi(m, 0, 1) = 0 \\ \varphi(m, 0, 2) = 1 \\ \varphi(m, 0, p) = m &\text{ for } p > 2 \\ \varphi(m, n, p) = \varphi(m, \varphi(m, n-1, p), p - 1) &\text{ for } n > 0 \text{ and } p > 0. \end{cases}$

So for $p = 0, 1, 2$ you get

$\phi(m, n, 0) = m + n$
$\phi(m, n, 1) = m \cdot n$
$\phi(m, n, 2) = m ^ n$

and

$\phi(m, n, 3) = \overbrace{{{m ^ m} ^ m} ^ {...}}^{n}$

It is known as a tetration, and it is normally written as $^na$ where n is the height of the power tower. It is the forth hyperoperation.

The zeroth hyperoperation is the successor function, and the first is the zeroth hyperoperation iterated, and so on

A more general way to define the nth hyperoperation is, using the notation, $H_n(a,b)$ where n is the nth hyperoperation,

$H_{n}(a,b)={\begin{cases}b+1&{\text{if }}n=0\\a&{\text{if }}n=1{\text{ and }}b=0\\0&{\text{if }}n=2{\text{ and }}b=0\\1&{\text{if }}n\geq 3{\text{ and }}b=0\\H_{n-1}(a,H_{n}(a,b-1))&{\text{if }n\in\mathbb{N},n>3}\end{cases}}$

Some notations for hyperoperations are(for $H_n(a,b)$:

1. Square bracket notation: $a[n]b$
2. Box notation: $a{\,{\begin{array}{|c|}\hline {\!n\!}\\\hline \end{array}}\,}b$
3. Nambiar's notation : $a\otimes ^{n-1}b$
4. Knuth's up arrow notation: $a\uparrow^{n-2}b$
5. Goodstien's notation: $G(a,b,n)$
6. Conway's chained arrow notation: $a\rightarrow b\rightarrow (n-2)$
7. Bowers exploding array function: $\{a,b,n,1\}$
8. Original Ackermann function: ${\begin{matrix}\phi (a,b,n-1)\ {\text{ for }}1\leq n\leq 3\\\phi (a,b-1,n-1)\ {\text{ for }}n\geq 4\end{matrix}}$
• Wow, Arianna Grande as a mathematician and stackexchange user. Exciting, – Chisko Jul 6 '16 at 17:26
• @Cheskos Ariana – Ariana Jul 6 '16 at 17:40
• Is any of these 8 extending Peter Hurfords $C(a,\dots)$ (see Extending the Extensions)? – Rudi_Birnbaum Mar 11 '18 at 21:47
• @R_Berger The extension, I believe, grows faster than every function in the 8 listed – Ariana Mar 12 '18 at 10:24