I have a question on how the number of the first layer of the Graham's number ($g_1$) is computed.

From Wikipedia:


$g_1 = 3\uparrow\uparrow\uparrow\uparrow3 $

enter image description here

As I understand it, this means that the number of the first layer of the Graham's number $g_1$ is a tetration ($\uparrow\uparrow$) in the form:

$$g_1 = 3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}} = \,{}^{n}3$$

Where $n$ is:

$3\uparrow\uparrow(3\uparrow\uparrow3) = \,{}^{7625597484987}3$

Thus, the height of the tower, is this what Wikipedia says?

If so, now, if:

$$3\uparrow\uparrow3 = \,{}^{3}3 = 3^{3^{3}}$$

$$3\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow(3\uparrow\uparrow3)= \,{}^{3\uparrow\uparrow3}3 = \,{}^{7625597484987}3$$

Why g1 is (as in the posted link):

$$g_1 = 3\uparrow\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow3)$$

And not:

$$g_1 = 3\uparrow\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow(3\uparrow\uparrow\uparrow3)$$

??? Anyway, why:

$\,\,3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow3)\,\,\,\,\,$ is $\,\,\,\,\,3\uparrow\uparrow(3\uparrow\uparrow(3\uparrow\uparrow ... (3\uparrow\uparrow3)...))$

  • $\begingroup$ A discussion of this can also be found at mpmueller.net/reihenalgebra.pdf (However I don't know whether this is really helpful for you, since your notations seem to be very similar to that what I remember from an early version of the article) $\endgroup$ – Gottfried Helms Mar 15 '15 at 21:06
  • $\begingroup$ Thanks for the link! After rereading the example on Wikipedia I understood the principle: $3\uparrow\uparrow\uparrow\uparrow3$ refers to a recursive-recursive tetration repeated 3 times ($3\uparrow\uparrow\uparrow(3\uparrow\uparrow\uparrow3) = 3\uparrow\uparrow\uparrow(3\uparrow\uparrow(3\uparrow\uparrow3))$ which in turn means a recursive tetration repeated $3\uparrow\uparrow(3\uparrow\uparrow3)$ times, where $3\uparrow\uparrow(3\uparrow\uparrow3)$ is not the height of the tower, but the number of tetrations which overwrap one after each other, leading to an even bigger number. $\endgroup$ – user3019105 Mar 17 '15 at 8:24
  • $\begingroup$ Possibly of interest: Graham's Number : Why so big? $\endgroup$ – MJD Mar 30 '15 at 20:09

The equality $3\uparrow\uparrow\uparrow\uparrow3=3\uparrow\uparrow(3\uparrow\uparrow3)$ is wrong there.

Hyperoperation (from tetration and so on) written in Knuth's notation satisfy the relation: $a\uparrow^nb=a\uparrow^{n-1}a\uparrow^{n-1}a\uparrow^{n-1}\dots \uparrow^{n-1}a$ where $n$ is the number of arrows, and $\uparrow^{n-1}$ is iterated b times.

So the first layer which is $3\uparrow\uparrow\uparrow\uparrow\uparrow2$ or (more usually) $3\uparrow\uparrow\uparrow\uparrow3$ is equal to:


So after you exponentiate 3 to itself 7625597484987 times you get how many times you have to tetrate 3 to itself, and that's only the first layer!

  • $\begingroup$ Yeah, that's a really big thing, can't even imagine it with a number... $\endgroup$ – user3019105 Mar 31 '15 at 7:43
  • 1
    $\begingroup$ We can't imagine it, it has more digit than particles in the whole universe! If you try to compute that with your mind you'll probably collapse in a blackhole. (I don't remember where i read this). $\endgroup$ – Renato Faraone Mar 31 '15 at 9:30
  • $\begingroup$ And trying to compute the Ackermann's function of two Graham's number (xkcd -> xkcd.com/207) as parameters will probably make the whole universe collapse in a blackhole. Not saying about what's behind of it, if anything. $\endgroup$ – user3019105 Mar 31 '15 at 13:21
  • $\begingroup$ Your 1st sentence states an inequality, which is true :). $\endgroup$ – dan Jan 1 '16 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.