The other day I was asked how to represent really big numbers. I half-jokingly replied to just take the logarithm repeatedly:

$$\log \log \log N$$ makes almost any number $N$ handy. (Assume base 10).

But applying this to Graham's Number probably does zip to make it handy (a handy number is defined as a number less than my or your age in years.) Then just keep applying logarithms. Can anyone estimate or even calculate how often to apply a $\log$ (of base 10, or 3 or $e$ or any handy base) to arrive at a handy number?

  • $\begingroup$ So, you want $log^*$ of Graham's number? $\endgroup$ – Dennis Meng Oct 31 '13 at 19:42
  • $\begingroup$ @DennisMeng I'm unfamiliar with that notation; if it means $\log\log\log ...(k \ times) N$ = handy number, then yes, I'm interested in $k$. $\endgroup$ – Jens Oct 31 '13 at 19:46
  • $\begingroup$ I'm used to $log^*$ being defined colloquially as "the number of times you need to take the log", so the little mini-example there would have $log^* N = k$. $\endgroup$ – Dennis Meng Oct 31 '13 at 19:48
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    $\begingroup$ Is this handy number concept your own creation? I like it. $\endgroup$ – Patrick Oct 31 '13 at 19:55
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    $\begingroup$ @Patrick Yes, as far as I know. I made it up myself but as so often in science, I may just have reinvented something. A minute ago I asked a search engine. This was futile because "handy number" in German is "cell phone number". :-) $\endgroup$ – Jens Oct 31 '13 at 20:05

Just getting $3 \uparrow\uparrow\uparrow 3$, which is a power tower of $3 \uparrow 3 \uparrow 3=3^{27}\ \ =7625597484987\ \ 3$'s to be a handy number takes $7625597484985$ applications of the $\log$ to get to $3^3=27$. The logarithm is woefully inadequate for this purpose.

The concept of $\log^*$ is a step in the right direction, but still not enough. We have $\log^* 3 \uparrow\uparrow\uparrow 3=7625597484985$, which isn't handy, but $\log \log^* 3 \uparrow\uparrow\uparrow 3=27$ is. Unfortunately we have a lot more uparrows to go. We probably need to define $\log^{**}$ as the number of times you apply $\log^*$ to get handy, then $\log^{***}$, etc. I suspect we need another (several) layers-define $\log^\&$ as the number of stars you have to put on $\log$ to get a handy number in one go. I have no idea how to do the computation, or even what sort of data structure is appropriate.

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    $\begingroup$ +1 That makes sense. To make such numbers handy a more powerful tool is required. What about an operation that cancels an arrow... a down arrow $\downarrow$ could be the symbol and the operation could be called logarrowthm. I shouldn't drink and post. :-) $\endgroup$ – Jens Oct 31 '13 at 20:58
  • $\begingroup$ $\log^{\&}(n)$ grows about as fast as the inverse Ackermann function $\alpha(n)$. We have $\log^{\&}(G_1)=3$, since $$\log^{**} G_1 = \log^{**} 3 \uparrow^4 3 =\log^{**} 3 \uparrow^3 (3 \uparrow^3 3) \approx 3 \uparrow^3 3$$, which is still too large. On the other hand, $\log^{***} (G_1 )=2$ (or 3, I'm not entirely sure). But $\log^{\&}(G_2)\approx G_1$, which isn't handy. $\log^{\&}(\log^{\&}(G_2))$ is handy, on the other hand. In general, we need to apply $\log^{\&}$ $n$ times to get $G_n$ handy. (This can be proven using induction. It's not difficult). Note: $ \uparrow^k$ is $k$ arrows. $\endgroup$ – wythagoras Dec 4 '16 at 14:24
  • $\begingroup$ If we define $\log^{\&*}$ as the number of times that $\log^{\&}$ has to be applied, then $\log^{\&*}(G_{64})=64$. $\endgroup$ – wythagoras Dec 4 '16 at 14:26
  • $\begingroup$ I absolutely love it! Logarrowthm. $\endgroup$ – return true Mar 12 at 23:17

Can anyone estimate or even calculate how often to apply a log (of base 10, or 3 or e or any handy base) to arrive at a handy number?

Graham's number $G$ can be written as an exteremely tall exponential tower of $3$s:

$$G \ =\ 3\uparrow\uparrow height $$

With $3$ as the base of the logarithms, you're asking for the $height$ of this tower (or rather $height - 2$, for a handy number of $27$). Now, a formula for the exact height can be found using the following property of Knuth arrows:

$$3\uparrow^x y = 3\uparrow^{x-1}(3\uparrow^x(y-1)) \ \ \ \ (x \ge 2, y \ge 2)$$

Applying this repeatedly gives

$$3\uparrow^x 3 \\ =3\uparrow^{x-1}(3\uparrow^{x}2) \\ =3\uparrow^{x-2}(3\uparrow^{x-1}(3\uparrow^{x}2 - 1))\\ =3\uparrow^{x-3}(3\uparrow^{x-2}(3\uparrow^{x-1}(3\uparrow^{x}2 - 1) - 1)) \\ \cdot\\ \cdot\\ =3\uparrow^2( 3\uparrow^3 (3\uparrow^4 ( \ \cdots \ (3\uparrow^{x-1}(3\uparrow^{x}2 - 1) - 1) \cdots -1 ) ) ) $$

Thus, the handy number 27 is the result of starting with $G$ and applying the base-3 $\log$ exactly $height-2$ times, where $$height = 3\uparrow^3 (3\uparrow^4 ( \ \cdots \ (3\uparrow^{g_{63}-1}(3\uparrow^{g_{63}}2 - 1) - 1) \cdots -1 )) $$

NB: A very crude lower bound on $height$ is given by
$$3\uparrow^{g_{63}} 3 \ = \ 3\uparrow^{g_{63}-1}(3\uparrow^{g_{63}-1}3) \ \ggg \ 3\uparrow^{2}(3\uparrow^{g_{63}-1}3)$$


$$height \ \ggg \ 3\uparrow^{g_{63}-1}3 $$

which has only one less arrow than $G$ itself!


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