3↑↑↑3= (or near) in power tower of 10 or in ( Knuth ) arrow ↑ notation of 10 to get a sense of it's order of magnitude; I grasp numbers more easily with 10

3↑↑↑3 being the first really huge number in the awesome crescendo of Graham's number, I suspect that it is still within the grasp of imagination, but it would help to get it in terms of 10 instead of 3.

I can't find it on the web, I am pretty sure I'm not the only one to wonder about this....

Please note that I am really talking about the small (?!) 3↑↑↑3 (Sun Tower), not G1=3↑↑↑↑3


!↑↑↑↑! :)

  • $\begingroup$ So as I understand it, you want the value of $3\uparrow \uparrow \uparrow 3$ expressed in scientific notation? $\endgroup$ – user472341 Dec 21 '17 at 3:55
  • $\begingroup$ This answer might give you some idea. Basically you can't write $3\uparrow \uparrow \uparrow 3$ as $10^{\text {something}}$. In theory you can, but that something is unimaginably large. It take many applications of the logarithm to make the number comprehensible. You might enjoy the May, 1982 column of Metamagical Themas by Douglas Hofstadter which you can download here $\endgroup$ – Ross Millikan Dec 21 '17 at 4:19
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    $\begingroup$ Sidenote: the first step to Graham's number is $3\uparrow\uparrow\uparrow\uparrow3$, which is much larger. $\endgroup$ – Simply Beautiful Art Aug 12 '18 at 3:13

For large power towers, what matters for the size is the number of layers, not the numbers inside it. Yes, if you make one of the numbers enormous you can overcome this, but for reasonable sized numbers it doesn't matter which one you pick. $3\uparrow \uparrow \uparrow 3$ is a tower $3^{27}=7625597484987$ layers high. This is about $7.6 \cdot 10^{12}$ layers high, which is how I would write it in terms of $10$.


For starters, I would suggest reading Pentation Notation - How does it work?

Now, we have


One may note that $10^{n/10}10\le3^n<10^n$ for $n>2$, hence,


where the upper bound is trivial and the lower bound may be observed inductively by proving $(10\uparrow\uparrow n)10<3\uparrow\uparrow(n+1)$ for $n>1$.

The base case:


The inductive case:

\begin{align}(10\uparrow\uparrow(n+1))10&=(10\uparrow(10\uparrow\uparrow n))10\\&<(10\uparrow((3\uparrow\uparrow(n+1))/10))10\\&<3\uparrow(3\uparrow\uparrow(n+1))\\&=3\uparrow\uparrow(n+2)\end{align}

as claimed.

If this is easier to read, the end result is essentially:



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