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If we define $G_0 = 10^{100}$, and $G_n = 10^{G_{n-1}}$ (hence $G_0$ is a googol, $G_1$ is a googolplex, $G_2$ is a googolplexian), for what first value of n will $G_n$ exceed Graham's number?

Peter
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prestokeys
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  • +1 Great question. I'm fairly sure that the answer here is not going to be $n = 5$. I think the the answer is more going to be $n = G_5 $ in all likelihood. There is a great numberphile video that describes just how big it is that I would recommend. https://www.youtube.com/watch?v=XTeJ64KD5cg – Leonhard Euler Feb 17 '18 at 16:47
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    @stuartstevenson I think your overestimating $G_n$ a bit here. – Arthur Feb 17 '18 at 16:50
  • @ Arthur This is the fastest growing function I can think of right now. As long as n is expressible in closed form (e.g. as a $G_n$ number like stuartstevenson suggests), I'm content, because I can then finally sense the size of Graham's number. – prestokeys Feb 17 '18 at 16:52
  • @Arthur Do you mean to say that I have overshot Graham's number with what I've put? – Leonhard Euler Feb 17 '18 at 16:58
  • @stuartstevenson No, I don't think you're anywhere close to Graham's number. I think $G_n$ is comparable to $10\uparrow\uparrow n$, and chaining $G$'s together as one another's indices $n$ tones is comparable to $10\uparrow^3n$. You need to step up your game to get close to $g_0$, much less $g_{634}$. – Arthur Feb 17 '18 at 17:02
  • Oh, I see. My point was just that the answer was not going to be easily expressible. – Leonhard Euler Feb 17 '18 at 17:04
  • Actually, the answer "no such n can be expressed", if such an answer is correct, will succeed in making me understand the enormity of Graham's number. – prestokeys Feb 17 '18 at 17:07
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    I sincerely doubt we can come anywhere even close to finding the exact $n$ where it happens. But my gut says that it happens somewhere in the (relative) vicinity of what you get if you take Graham's number $3\uparrow\cdots\uparrow3$ and remove a single arrow, then use that as $n$. That's a lot smaller than Graham's number, to be sure, but it's just as difficult to comprehend. – Arthur Feb 17 '18 at 17:33
  • Even faster is something like $G_n = G_{n-1}^{G_{n-2}}$ where we define $G_1, G_2$ to be large. – MathematicsStudent1122 Feb 17 '18 at 19:20
  • Power towers cannot compete with Graham's number! – Peter Feb 17 '18 at 22:54
  • For googology-standards, this function is growing quite slow, just tetrational growth rate. This is the very beginning for creating large numbers. Graham's number is in a completely other league! It is so big that (if we denote Graham's number with $N$) $G_N$ is not much larger than $N$ itself. – Peter Feb 17 '18 at 23:01

1 Answers1

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Well, let's take a look at the definition of Graham's number.

$$g_n = \begin{cases} 3\uparrow\uparrow\uparrow\uparrow 3, &n=1\\ 3\uparrow^{g_{n-1}}3, &n\geq 2,n\in\mathbb{N}\end{cases}$$

Specifically, Graham's number is $g_{64}$. To clarify, the up arrows are Knuth's up-arrow notation:

$$a\uparrow^nb = \begin{cases} a^b, &n=1\\ 1, &n\geq1,b=0\\ a\uparrow^{n-1}(a\uparrow^n(b-1)), &\mathrm{otherwise}\end{cases}$$

A shortcut I will also use is tetration notation: ${^33}=3^{3^{3}}$. So let's look at some examples of the values of the numbers on the way to Graham's number, like $g_1$:

$$\begin{align}g_1 &= 3\uparrow\uparrow\uparrow\uparrow3\\ &= 3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow\uparrow2)\\ &=3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow\uparrow1))\\ &=3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow3))\\ &=3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow(3\uparrow\uparrow\uparrow2)))\\ &=3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow(3\uparrow\uparrow(3\uparrow\uparrow\uparrow1))))\\ &=3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow(3\uparrow\uparrow(3\uparrow\uparrow3)))\\ &=3\uparrow\uparrow\uparrow\left(3\uparrow\uparrow\uparrow\left(3\uparrow\uparrow3^{3^{3}}\right)\right)\\ &=3\uparrow\uparrow\uparrow\left(3\uparrow\uparrow\uparrow{^{3^{3^{3}}}3}\right)\end{align}$$

Now, I'm not going to bother reducing it further, because the reduction of $\left(3\uparrow\uparrow\uparrow{^{3^{3^{3}}}3}\right)$ requires over ${^{7625597484987}3}$ iterations. That is, 3 raised to itself 7625597484987 times. This is nested four additional times- so you find 3 raised to itself 7625597484987 times, take 3 and raise it to itself that many times, and repeat this process twice more, to get the number of iterations required to reduce this fully.

To note, your function is roughly equivalent to $10\uparrow\uparrow n$, or to raise 10 to the power of itself n times. As you can tell, it's vastly dominated even by $g_1$; any $n$ such that your function even comes close to $g_{64}$ will probably be impossible to write in any closed notation, much like Graham's number itself.

If I had to put an approximate guess as to what the value of n would be, I would put it at either $g_{63}$ or $g_{62}$.

LuuBluum
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