This question already has an answer here:

When going through with learning Grahams number, I got stuck at

$$3↑↑↑3$$

Working it through, we have

$$3↑3=3^3$$ $$3↑↑3=3^{3^3}=3↑(3↑3)$$

As such, it would appear to me that

$$3↑↑↑3=3^{3^{3^3}}=3↑(3↑(3↑3))=3↑(3↑↑)$$

Which is incorrect; the correct answer being

$$3↑↑↑3=3↑↑(3↑↑3)$$

What I'm wanting to know is where the error in the way I've worked it through, and how working $3↑↑(3↑↑3)$ through backwards to $3↑↑↑3$ would look?

marked as duplicate by amWhy, Xander Henderson, The Phenotype, Lord Shark the Unknown, JonMark Perry Mar 11 at 2:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $3 \uparrow \uparrow = 7625597484987$ so $3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow (3 \uparrow \uparrow 3) = 3 \uparrow \uparrow 7625597484987$, which is an impossibly large number. Starting with x=1, you iterate $x \mapsto 3 \uparrow x$ 7625597484987 times. Now $3 \uparrow \uparrow \uparrow \uparrow 3$ is even more difficult. Take the number you just got, and start with x=1. That's how many times you need to iterate $x \mapsto 3 \uparrow \uparrow \uparrow x$. – Sheldon L Sep 5 '17 at 20:00
  • Sheldon L I'm not entirely sure I understand what you're expressing here. Are you saying that $3↑↑7625597484987$ is equivalent to $3↑↑↑↑3$, or equivalent to $3↑↑↑3$? – An Entity Sep 6 '17 at 1:17
up vote 2 down vote accepted

You wish to understand Graham's number through these arrows? If so, I'd suggest stepping back down to multiplication and building the way up.

Note that

$$a\times b=\underbrace{a+(a+(\dots+a))}_b$$

For example,

$$3\times3=3+(3+3)=3+6=9$$

And then exponentiation,

$$a^b=a\uparrow b=\underbrace{a\times(a\times(\dots\times a))}_b$$

For example,

$$3\uparrow3=3\times(3\times3)=3\times9=27$$

Now tetration,

$$a\uparrow\uparrow b=\underbrace{a\uparrow(a\uparrow(\dots\uparrow a))}_b$$

For example,

$$3\uparrow\uparrow 3=3\uparrow(3\uparrow3)=3\uparrow27=7625597484987$$

And beyond...

$$a\uparrow\uparrow\uparrow b=\underbrace{a\uparrow\uparrow(a\uparrow\uparrow(\dots\uparrow\uparrow a))}_b$$

$$3\uparrow\uparrow\uparrow3=3\uparrow\uparrow(3\uparrow\uparrow3)=3\uparrow\uparrow7625597484987=\underbrace{3\uparrow(3\uparrow(\dots\uparrow3))}_{7625597484987}=3^{3^{3^{3^{\dots}}}}$$

$$ 3 \uparrow \uparrow n = 3 \uparrow (3 \uparrow \uparrow (n-1))$$ $$ 3 \uparrow \uparrow \uparrow n = 3 \uparrow \uparrow (3 \uparrow \uparrow \uparrow (n-1))$$ $$ 3 \uparrow \uparrow \uparrow \uparrow n = 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow \uparrow (n-1)) ....$$

By definition $$ 3 \uparrow^{n} 0 =1\;\;\;\; 3 \uparrow 1 = 3 $$ One can prove by induction: $$ 3 \uparrow^{n} 1 =3$$ $$ 3 \uparrow^{n} 2 = 3 \uparrow^{(n-1)} 3$$ $$ 3 \uparrow \uparrow 3 = 3 \uparrow {3^3} = 3^{3^3} = 3^{27} = 7625597484987$$ Therefore $$ 3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow (3 \uparrow \uparrow \uparrow 2)= 3 \uparrow \uparrow (3 \uparrow \uparrow 3) = 3 \uparrow \uparrow 7625597484987 $$ $$ 3 \uparrow \uparrow \uparrow 3 = 3^{3^{3^{3...}}}\;\;\; \text{tower height} = 3 \uparrow \uparrow 3 = 7625597484987$$ $$ 3 \uparrow \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3) $$ $$ ^{^{^{...3}3}3}3\;\;\; \text{tower height} = 3 \uparrow \uparrow 7625597484987 $$

  • Okay, so it's because the tower height of $3↑^n$ is equal to $3^{n-1}$, and rather than the power to which $3$ is raised? That explains it, thank-you! – An Entity Sep 6 '17 at 23:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.