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?

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ What's the name of the operator? $\endgroup$ – 0x90 Oct 22 '18 at 22:16
  • $\begingroup$ I believe it's hyperexponentiation. The symbol is from Knuth's up-arrow notation. $\endgroup$ – kcborys Oct 23 '18 at 16:25
  • 1
    $\begingroup$ 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. $\endgroup$ – 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:


and we also have




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


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:


which is much much much larger compared to the relatively puny



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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