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A couple days ago I was bored and thought of the following problem and am hoping someone can either answer my question or point me in the right direction.

Is there an easy way to compute the number of digits of the following? Follow up is, can the number of digits be presented in an easily presentable/human readable way?

googolplex knuth's up-arrow googolplex

References:

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closed as off-topic by Namaste, Daniel W. Farlow, user91500, Claude Leibovici, hardmath Mar 17 '17 at 10:50

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  • $\begingroup$ No, not if you want to write it out fully. $\endgroup$ – Simply Beautiful Art Mar 16 '17 at 23:27
  • $\begingroup$ With uparrows it is easy to make numbers that have so many digits that the log is inadequate to attack it. More uparrows is much more powerful than big numbers between the arrows. You might see my answer here where I count the number applications of log it takes to get a number handy and find that even that is inadequate for even $3 \uparrow \uparrow \uparrow 3$, which is much larger than your number. $\endgroup$ – Ross Millikan Mar 16 '17 at 23:51
  • $\begingroup$ Another two applications of the log will bring the number of digits down to about $100$ $\endgroup$ – Ross Millikan Mar 17 '17 at 0:04
  • $\begingroup$ Please clarify how many arrows you want by the way. As is, I'll have to close it for being unclear. $\endgroup$ – Simply Beautiful Art Mar 17 '17 at 0:11
  • $\begingroup$ You might also enjoy this question which deals with how powerful different ways to make large numbers are. $\endgroup$ – Ross Millikan Mar 17 '17 at 4:54
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One uparrow in Knuth's notation is just regular exponentiation, so you are asking about $N=(10^{10^{100}})^{10^{10^{100}}}=10^{(10^{100}\cdot {10^{10^{100}}})}$ which has $10^{100}\cdot {10^{10^{100}}}+1$ digits.

This is actually not such a large number. If we take $\log_{10} (\log_{10} (\log_{10} N)))$ we get about $100$. In this answer I consider $3 \uparrow \uparrow \uparrow 3$ and find it takes $7625597484985$ applications of the $\log$ to make the number handy. We define $\log^*$ as the number of applications of $\log$ to make a number handy, so we have $\log^* N=4$, but $\log^*3 \uparrow \uparrow \uparrow 3=7625597484985$ so it takes $\log \log^*3 \uparrow \uparrow \uparrow 3$ to get handy. Bigger numbers will need $\log^{**}$, the number of times you need to apply $\log^*$ and so on.

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  • 2
    $\begingroup$ But isn't that googol to the googol, not googolplex to the googolplex? $\endgroup$ – WB-man Mar 16 '17 at 23:49
  • $\begingroup$ @WB-man: you are right. Fixed. $\endgroup$ – Ross Millikan Mar 17 '17 at 0:02
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At the risk of being snarky, there is a simple way to write out the number of digits:

$$ 1 + \lfloor \log_{10}\left( \text{googolplex} \uparrow \uparrow \text{googolplex}\right) \rfloor $$

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  • $\begingroup$ Up voted because its true and funny. But curious if a way to represent it without referencing itself :) $\endgroup$ – VenomFangs Mar 16 '17 at 23:32
  • $\begingroup$ @VenomFangs Its how you get to it, but as I've said, you cannot write it out fully. $\endgroup$ – Simply Beautiful Art Mar 16 '17 at 23:33
  • $\begingroup$ OP only used one uparrow, which is just regular exponentiation. $\endgroup$ – Ross Millikan Mar 16 '17 at 23:41
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If you just mean a single up arrow, then this is the same as computing a googol plex to the power of a googol plex. If I let $g$ denote the ordinary googol ($=10^{100}$), and $G$ denote the googol plex ($=10^{10^{100}}$), then the number of digits is the log base $10$ of $G^G$ plus $1$ (since both $g$ and $G$ are powers of $10$): \begin{align} \mathrm{Num\ Digits} &= 1 + \log_{10} \left(G^G \right) \\ &= 1 + G \log_{10} G \\ &= 1 + G \log_{10} \left[10^\left(10^{100} \right) \right] \\ &= 1 + G \cdot 10^{100} \\ &= 1 + Gg \end{align} So the number of digits is a googol plex times a googol, plus $1$.

Or to put it another way: the number of digits is one plus $1$ followed by a googol zeros and another hundred zeros.

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