0
$\begingroup$

We can use the following to add the number $2$ to itself $5$ times.

$$f(n,k) = \sum_{x=1}^k n = n\cdot k$$

$$2 + 2 + 2 + 2 + 2 = f(2,5) = \sum_{x=1}^5 2 = 2\cdot 5 = 10$$

We can use a similar strategy for multiplying the number $2$ by itself $5$ times.

$$g(n,k) = \prod_{x=1}^k n = n^k$$

$$2\cdot2\cdot2\cdot2\cdot2 = g(2,5) = \prod_{x=1}^5 2 = 2^5 = 32$$

How could I define $h(n,k)$ such that $h(2,5) = 2^{2^{2^{2^2}}}$?

  • 1
    You just **did** define it. – 2017-02-27
  • 1
    By saying "$h(n,k)$ such that it is $n$ raised to itself $n$ iterations". That's all you need. You could do recursion. "$h(n,0) = 1$ and $h(n,k+1) = n^{h(n,k)};k\ge 0$". Important to not, in case you haven't. This not associative and $3^{(3^3)} \ne (3^3)^3$. This assumes you meant the former not the latter. – 2017-02-27

2 Answers 2

1

Thanks to the tag added in Ivan Neretin's edit, I have found that this operation is formally known as "tetration".

In mathematics, tetration (or hyper-4) is the next hyperoperation after exponentiation, and is defined as iterated exponentiation.

$h(n,k)$ would be defined as follows, using standard tetration format.

$$h(n,k) = ^kn$$

  • 1
    You could "accept" your self-answer, to "close-the-case" for the MSE-readers – 2017-03-01
1

Knuth Arrow notation, best explained here.

It also has accommodations for repeated knuth arrow notations, repeated REPEATED knuth arrows, and so on into the sun.