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From Wikipedia, Knuth's up-arrow notation begins at exponentiation and continues through the hyperoperations:

$a \uparrow b = a^b$

$a \uparrow\uparrow b = {\ ^{b}a} = \underbrace{a^{a^{.^{.^{.^{a}}}}}}_b$ (the tetration of a and b; an exponentiation tower of a, b elements high)

This already produces numbers much larger than the number of Planck volumes in the observable universe with very small numbers; $3 \uparrow\uparrow 3$ is a relatively modest 7.6 trillion, but $3 \uparrow\uparrow 4 = 3^{7.6t} = 10^{3.6t}$.

Then there is pentation ($a\uparrow\uparrow\uparrow b = a\uparrow^3b$) and hexation ($a\uparrow\uparrow\uparrow\uparrow b = a\uparrow^4b$). The pentation of 3 and 3 is $\underbrace{3^{3^{.^{.^{.^{3}}}}}}_{\ ^{3}3}$, an exponentiation tower of 3s 7.6 trillion elements in height. Hexation is an exponentiation tower of 3s equal in height to the value of the pentation of 3 and 3. And that is just $g_1$, the first layer of calculation necessary to compute Graham's number, $g_{64}$, where $g_n = 3\uparrow^{g_{n-1}}3$.

I'm having considerable, and I hope understandable, difficulty simply wrapping my head around a number of this magnitude. So, the question is, is there value in understanding the scope of numbers produced by Knuth's up-arrow notation, or is this simply a way for mathematicians to make each others' heads explode?

If it's the latter, I leave you with the following:

$A(g_{64},g_{64})$

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The only realistic answer to the title question is no. –  Brian M. Scott Jun 1 '12 at 21:18
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Isn't the last number from xkcd? [Yes, it is.] –  Asaf Karagila Jun 1 '12 at 21:19
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In my humble opinion it is a bit pointless trying to get any sort of feel for the real size of numbers of these types, and better to stick with the idea that they form increasing sequences of "enormity" in which each stage is ridiculously tiny compared to the next ones in the sequence. Its a bit like trying to visualise inter-stellar and inter-galactic distances in the universe. –  Old John Jun 1 '12 at 21:55
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Yes, I know of Graham's number as the question text indicates. The question is, what practical value does Graham's Number have? –  KeithS Jun 1 '12 at 22:18

1 Answer 1

up vote 3 down vote accepted

Yes, see "Enormous Integers in Real life" by Harvey Friedman.

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