For the stronger operators, see Arrow notation.
Down-arrow notation
Based onExponentiation
Growth rate\(f_{\omega}(n)\)

Down-arrow notation is a left-associative extension for addition, multiplication and exponentiation. Formally it is defined as follows:

\[a \downarrow b = a^b\]

\[a \downarrow^n 1 = a\]

\[a \downarrow^{n + 1} (b + 1) = (a \downarrow^{n +1} b) \downarrow^n a \text{ (otherwise)}\]

\(a \downarrow^n b\) is a shorthand for \(a \downarrow\downarrow\cdots\downarrow\downarrow b\) 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 in the definition of Clarkkkkson.


\(3 \downarrow\downarrow 3 = 3^{3^2} = 3^9 = 19683\)

\(3 \downarrow\downarrow\downarrow 2 = (3 \downarrow\downarrow\downarrow 1)\downarrow\downarrow 3 = 3\downarrow\downarrow 3\)

\(3 \downarrow\downarrow\downarrow 3 = (3 \downarrow\downarrow\downarrow 2)\downarrow\downarrow 3 = 19683 \downarrow\downarrow 3 = 19683^{19683^2}\)

Using BCalc, we find \(3 \downarrow\downarrow\downarrow 3 \approx 2.198197017816742204069545026183 \cdot 10^{1663618948}\)


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

See alsoEdit

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