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

From Wikipedia:

http://en.wikipedia.org/wiki/Graham%27s_number#Magnitude

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

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  • $\begingroup$ 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, 2015 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, 2015 at 8:24
  • $\begingroup$ Possibly of interest: Graham's Number : Why so big? $\endgroup$
    – MJD
    Mar 30, 2015 at 20:09

1 Answer 1

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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:

$3\uparrow\uparrow\uparrow\uparrow3=3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3=3\uparrow\uparrow\uparrow(3\uparrow\uparrow3\uparrow\uparrow3)=3\uparrow\uparrow\uparrow(3\uparrow\uparrow7625597484987)$

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!

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  • $\begingroup$ Yeah, that's a really big thing, can't even imagine it with a number... $\endgroup$
    – user3019105
    Mar 31, 2015 at 7:43
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    $\begingroup$ 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). $\endgroup$
    – AlienRem
    Mar 31, 2015 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, 2015 at 13:21
  • $\begingroup$ Your 1st sentence states an inequality, which is true :). $\endgroup$
    – dan
    Jan 1, 2016 at 13:00

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