## FANDOM

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For the stronger operators, see Arrow notation.
Down-arrow notation
Type3-argument
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.

## ExamplesEdit

$$3 \downarrow\downarrow 3 = 3^{3^2} = 3^9 = 19,683$$

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

$$3 \downarrow\downarrow\downarrow 3 = (3 \downarrow\downarrow\downarrow 2)\downarrow\downarrow 3 = 19,683 \downarrow\downarrow 3 = 19,683^{19,683^2}$$

Using BCalc, we find $$3 \downarrow\downarrow\downarrow 3 \approx 2.198197017816742204069545026183 \cdot 10^{1,663,618,948}$$

## BoundsEdit

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