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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $\underbrace{A\uparrow A\uparrow ...A\uparrow A}_{x\ A\text{s}}=A\uparrow^2 x\ge B.$)

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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

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

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