# How much bigger is 3↑↑↑↑3 compared to 3↑↑↑3?

3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?

• No, it is much much much much much much much much much much much much much much much much much much much much much much much much much much much larger. – Yves Daoust Oct 22 '18 at 21:29

By definition, $$3\uparrow\uparrow\uparrow\uparrow 3 = 3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow 3))$$. $$\begin{matrix} 3\uparrow\uparrow\uparrow 3= & \underbrace{3^{3^{3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}}}}} \\ & \mbox{7,625,597,484,987 copies of 3} \end{matrix}$$

which should make it a little clearer how much bigger $$3\uparrow\uparrow\uparrow\uparrow 3$$ is than $$3\uparrow\uparrow\uparrow 3$$. It's hard to even give a more concrete answer because the numbers of powers of $$3$$ get unreasonably large to compute or type.

• What's the name of the operator? – 0x90 Oct 22 '18 at 22:16
• I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation. – kcborys Oct 23 '18 at 16:25
• Correction: $3\uparrow^k3=3\uparrow^{k-1}3\uparrow^{k-1}3$, and generally $a\uparrow^kb=a\uparrow^{k-1}a\uparrow^{k-1}\dots\uparrow^{k-1}a$ with $b$ many $a$'s. – Simply Beautiful Art Jan 9 at 21:25

For any integers $$B>A>1,$$ to gauge "how much bigger is $$B$$ compared to $$A,$$" consider the following questions :

1. How many $$A$$s do we need to add together to reach $$B$$?

Ans: The least $$x$$ such that $$\underbrace{A+A+...+A}_{x\ A\text{s}}\ge B$$, i.e., $$A\times x \ge B: \quad x=\left\lceil{B\over A}\right\rceil.$$

2. How many $$A$$s do we need to multiply together to reach $$B$$?

Ans: The least $$x$$ such that $$\underbrace{A\times A\times...A\times A}_{x\ A\text{s}}\ge B$$, i.e., $$A\uparrow x\ge B:\quad x=\left\lceil{\log B\over \log A}\right\rceil.$$

3. More generally, for any operation $$@$$ in the hyperoperation sequence $$(+,\times,\uparrow,\uparrow^2,...)$$, we can ask for the least $$x$$ such that $$\underbrace{A@ A@...A@A}_{x\ A\text{s}}\ge B$$, i.e., $$A@'x\ge B,$$ where $$@'$$ is the next hyperoperation after $$@$$.

Now in the case of $$A=3\uparrow^3 3$$ and $$B=3\uparrow^4 3=3\uparrow^3 A,$$ note the following:

\begin{align}B &=3\uparrow^4 3\\ &=3\uparrow^3 3\uparrow^3 3\\ &=3\uparrow^3 A\\ &=3\uparrow^2 3\uparrow^3 (A-1)\\ &=3\uparrow^1 3\uparrow^2 (3\uparrow^3 (A-1) - 1)\\ &=3\uparrow^1 3\uparrow^1 3\uparrow^2 (3\uparrow^3 (A-1) - 2)\\ &\\ A&=3\uparrow^3 3\\ &=3\uparrow^2 3\uparrow^3 2\\ &=3\uparrow^1 3\uparrow^2 (3\uparrow^3 2 - 1)\\ &=3\uparrow^1 3\uparrow^1 3\uparrow^2 (3\uparrow^3 2 - 2)\\ \end{align}

Then the answer to (1) is $$x$$ such that $$A\times x=B$$: \begin{align}x={B\over A}={3^{3\uparrow^2 (3\uparrow^3 (A-1) - 1)} \over 3^{3\uparrow^2 (3\uparrow^3 2 - 1)}}=3^{3\uparrow^2 (3\uparrow^3 (A-1) - 1) - 3\uparrow^2 (3\uparrow^3 2 - 1)}. \end{align}

Similarly, the answer to (2) is $$x$$ such that $$A\uparrow x=B$$: \begin{align}x&= {\log B\over \log A}={3^{3\uparrow^2 (3\uparrow^3 (A-1) - 2)} \over 3^{3\uparrow^2 (3\uparrow^3 2 - 2)}}=3^{3\uparrow^2 (3\uparrow^3 (A-1) - 2) - 3\uparrow^2 (3\uparrow^3 2 - 2)}. \end{align}

Furthermore, we can give a lower bound on the least $$x$$ such that $$A\uparrow^3 x\ge B$$, as we have by Saibian's inequality, $$(3\uparrow^3 3)\uparrow^3 (3\uparrow^3 3\,-\,3)\ < \ (3)\uparrow^3 (3\uparrow^3 3\,-\,3\,+\,3)\ =\ 3\uparrow^4 3;$$ that is, $$A\uparrow^3 (A-3)=\underbrace{A\uparrow^2 A\uparrow^2 ...A\uparrow^2 A}_{(A-3)\ A\text{s}}< B.$$ Therefore, $$A\uparrow^2 A\uparrow^2 ...A\uparrow^2 A\ge B$$ requires more than $$(A-3)$$ $$A$$s on the left-hand side.

(Of course, this is also a lower bound on the least $$x$$ such that $$\underbrace{A\uparrow A\uparrow ...A\uparrow A}_{x\ A\text{s}}=A\uparrow^2 x\ge B.$$)

Knuth's up-arrow notation is defined so that we have

$$a\uparrow^kb=\underbrace{a\uparrow^{k-1}a\uparrow^{k-1}\dots\uparrow^{k-1}a}_{b\text{ many }a\text{'s}}$$

where $$\uparrow^k=\underbrace{\uparrow\uparrow\dots\uparrow\uparrow}_k~$$ and we evaluate from right to left (e.g. $$a\uparrow b\uparrow c=a\uparrow(b\uparrow c)\ne(a\uparrow b)\uparrow c$$ in general).

This means we have:

$$3\uparrow\uparrow\uparrow\uparrow3=3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3$$

and we also have

$$3\uparrow\uparrow\uparrow3=3\uparrow\uparrow3\uparrow\uparrow3\uparrow\uparrow3$$

and

$$3\uparrow\uparrow3=3\uparrow3\uparrow3=3^{3^3}=7,625,597,484,987$$

Using this result, we may further calculate $$3\uparrow\uparrow\uparrow3$$ as

$$\underbrace{3^{3^{3^{.^{.^.}}}}}_{7,625,597,484,987}$$

which is pretty large. Keep in mind that this is $$3\uparrow\uparrow\uparrow3$$, or $$3\uparrow\uparrow3\uparrow\uparrow3$$. If we took this result and did $$\uparrow\uparrow$$ that many times, we'd basically arrive at $$3\uparrow\uparrow\uparrow\uparrow3$$, which is equivalent to:

$$3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3=\underbrace{3\uparrow\uparrow3\uparrow\uparrow\dots\uparrow\uparrow3}_{3\uparrow\uparrow\uparrow3}$$

which is much much much larger compared to the relatively puny

$$3\uparrow\uparrow\uparrow3=\underbrace{3\uparrow\uparrow3\uparrow\uparrow3}_3$$