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