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Down-arrow notation
Type3-argument
Based onExponentiation
Growth rate\(f_{\omega}(n)\)

Down-arrow notation refers to one of the two functions:

  1. A left-associative extension for addition, multiplication and exponentiation. It is a weaker version of arrow notation, but comparable with arrow notation with the index of the arrow halved.[1]
  2. The logarithmic inverse of Knuth's up-arrow notation. It is a very slow-growing function and extends the idea of iterated logarithm (log*). [2]

We explain both definitions in this article.

Formulation 1[]

Definition[]

Formally, \(a \downarrow \cdots \downarrow b\) for positive integers \(a\) and \(b\) is defined as follows: \begin{eqnarray*} a \downarrow^n b = \left\{ \begin{array}{cl} a^b & (n = 1) \\ a & (n > 1, b = 1) \\ (a \downarrow^n (b-1)) \downarrow^{n-1} a & (n > 1, b > 1) \end{array} \right. \end{eqnarray*} Here, \(a \downarrow^n b\) for a positive integer \(n\) is a shorthand for \(a \downarrow\cdots\downarrow b\) with n down-arrows.

When \(n = 2\), \(a \downarrow\downarrow b = a^{a^{b-1}}\). The inequality \(a \downarrow\downarrow a = a^{a^{a-1}} < a^{a^a} = a \uparrow\uparrow 3\) is useful when bounding down-arrows in terms of up-arrows.

It can be shown that \(a \downarrow^{2n-1} b \ge a \uparrow^n b\) for \(a, b, n \ge 1\).

Down-arrow notation is not as important in googology as the up-arrow notation, but it is used to distinguish the two possible meanings of fuga- and in the definition of Clarkkkkson. Alistair has assigned fuga- to the down-arrow form, and agreed to rename the larger up-arrow form as megafuga-.

Examples[]

  • \(n \downarrow\downarrow n\) = fuga-n, in contrast to \(n \uparrow\uparrow n\) = megafuga-n (Alistair Cockburn)
  • \(3 \downarrow\downarrow 3 = ((3 \downarrow\downarrow 1)\downarrow 3)\downarrow 3 = (3^3)^3 = 3^9 = 19,683\)
  • \(3 \downarrow\downarrow\downarrow 2 = (3 \downarrow\downarrow\downarrow 1)\downarrow\downarrow 3 = 3\downarrow\downarrow 3\)
  • \(3 \downarrow\downarrow\downarrow 3 = (3 \downarrow\downarrow\downarrow 2)\downarrow\downarrow 3 = 19,683 \downarrow\downarrow 3 = 19,683^{19,683^2} = 3^{3^{20}}\)

Using BCalc[dead link], we find \(3 \downarrow\downarrow\downarrow 3 \approx 2.198197017816742204069545026183 \cdot 10^{1,663,618,948}\)

Bounds[]

  • \(a\downarrow\downarrow\downarrow b\approx a\uparrow\uparrow(b+1)\)
  • \(a\downarrow^4 b\approx a\uparrow\uparrow(a(b-1)+1)\)
  • \(a\downarrow^5 b\approx(a\uparrow\uparrow)^{b-1}(a(a-1)+1))>a\uparrow\uparrow\uparrow b\)
  • \(a\downarrow^6 b\approx(a\uparrow\uparrow)^{(a-1)(b-1)}(a(a-1)+1)>a\uparrow\uparrow\uparrow((a-1)(b-1)+1)\)

Formulation 2[]

Definition[]

Another formulation of down-arrow notation is given as an "inverse" of Knuth's up-arrow notation in some sense. [2]

For example, \(e \downarrow n\) is defined as \(\log n\), and \(e \downarrow \downarrow n\) is defined as the iterated logarithm \(\log^* n\).

See also[]

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