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$\text{googolplex} = 10^{(10^{100})}$

Is $\text{googolplex}!$ greater than $\text{Graham's number}$? How would this be proven?

If $\text{googolplex}! \le \text{Graham's number}$, (which I expect) then how many cycles of $( \dots(((\text{googolplex}!)!)!) \dots)!$ would be needed to exceed $\text{Graham's number}$?

r.e.s.
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user278589
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2 Answers2

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Graham's number is way bigger.

Basically, any number you construct with a normal-looking number of normal-looking operators acting on a normal amount of normal-sized numbers (terms deliberately not clearly defined) will be far smaller than Graham's number, basically because you're not doing anywhere near enough recursion to approach the recursion done in defining Graham's number.

Chappers
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  • what about the second part. intuition tells me that some n number of cycles will exceed grahams number ? – user278589 Oct 09 '15 at 13:28
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    Way too many factorials just essentially do normal stacking $x!$ is roughly (fine enough for our purposes) $x^x$ so iterating $!$ just gives you $x^{x^{x^{x\cdiag}}}$. This won't be enough to overtake even the first layer of Graham's number. – DRF Oct 09 '15 at 13:36
  • Take a look at the wiki article the $n$ you define might just overtake $g_1$ but I wouldn't bet on it. Actually looking at it again I don't think it will even manage that. – DRF Oct 09 '15 at 13:38
  • Well some $n$ number of cycles certainly will. For example if $n$ is graham's number you're set.:D – DRF Oct 09 '15 at 13:38
  • what fraction of grahams number will do as the number of factorial cycles – user278589 Oct 09 '15 at 13:40
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    Well since $G_{64}$ which is Graham's number is $3\uparrow_{G_{63}}3$ even $G_{63}$ won't be even close to enough. With what is essentially rounding errors you will need $3\uparrow_{G_{63}-1}3$ cycles. Actually now I think about it that will probbably still not be enough, because I think the last arrow increases the power more then just as a tower. – DRF Oct 09 '15 at 13:46
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We derive a dramatic improvement over a comment that suspected even $k= 3\uparrow^{g_{63}-1}3$ is not enough iterations for $f^k(\text{googolplex}) > \text{Graham's number}$, where $f$ is the factorial function.

For convenience in writing Knuth's up-arrows, we use the operator notation $$[a]b = 3\uparrow^a b,$$ so Graham's number $(g_{64})$ is defined by the recursion $$g_0 = 4\\ g_{n+1} = [g_n]3.$$

We show that Graham's number can be written as an exactly specified number of iterations of the tetration operator $([2])$: $$\text{Graham's number } = [2]^t 1 $$ and therefore that $$\text{Graham's number } \gt f^{t-3}(\text{googolplex})$$ because $\ [2]^3 1 > \text{googolplex}\ $ and $\ [2]n > n!\ $ for all positive integers $n$.

Before proceeding to calculate $t$, we state some properties of the hyperoperators $[p]$. Associating from the right, these satisfy the following recursion for $a,b\in Z_{>0}$: $$[a]b = \begin{cases} 3 & \text{if } b=1\\ 3^b & \text{if }a=1\\ [a-1][a](b-1) & \text{otherwise} \end{cases}\tag{E0}$$

which implies the identities

$$\begin{align} [a]b &= [a-1]^b 1 &(a\ge 2,\ b \ge 1)\tag{E1}\\ [a]b &= [a-2]^{[a](b-1)}1 &(a\ge 3,\ b \ge 2)\tag{E2} \end{align}$$

Aside: $(E1)$ is obtained by repeatedly applying $(E0)$ to itself, as $$\begin{align}[a]b &= [a-1][a](b-1)\\ &= [a-1][a-1][a](b-2)\\ &\cdots\\ &= [a-1]^{(b-1)}[a] 1\\ &= [a-1]^{(b-1)}[a-1] 1\\ &= [a-1]^b 1 \end{align}$$ and $(E2)$ is obtained by applying $(E1)$ to $(E0)$ as

$$[a]b = [\underbrace{a-1}_{A}]\underbrace{[a](b-1)}_{B} = [A]B = [A-1]^B 1 = [a-2]^{[a](b-1)}1.$$

General result:

If $p\ge 4\ $ and $\ q \le p-3\ $ are positive integers with opposite parity, then $$[p]3 = [q]^t 1$$ where $$t=[q+2](b_j-1)\\ j={{p-q-1}\over{2}}$$ and $b_j$ is the last term in the very rapidly increasing sequence $b_1 < b_2 < \dots < b_j$, given by the recursion

$$\begin{align} b_1 &= [p-1][p-1]1\\ b_i &= [p-2i+1]^{[p-2i+3](b_{i-1}-1)-1}1\quad (1\lt i\le j).\\ \end{align}$$

Proof-sketch, using repeated application of the previous identity $E2$ until $[p]3$ is reduced to the form $[q]^t 1$: $$\begin{align} [p]3 &= [p-1]([p-1][p-1]1)\\ &= [a_1]b_1,\quad a_1 = p-1,\ b_1=[p-1][p-1]1\\ &= [a_1-2][a_1-2]^{[a_1](b_1-1)-1}1\\ &= [a_2]b_2,\quad a_2 = a_1 - 2=p-3,\ b_2 = [a_2]^{[a_1](b_1-1)-1}1\\ &= [a_2-2][a_2-2]^{[a_2](b_2-1)-1}1\\ &= [a_3]b_3,\quad a_3 = a_2 - 2=p-5,\ b_3 = [a_3]^{[a_2](b_2-1)-1}1\\ &...\\ &= [a_i-2]^{[a_i](b_i-1)},\quad a_i = a_{i-1} - 2=p-2i+1,\ b_i = [a_i]^{[a_{i-1}](b_{i-1}-1)-1}1\\ &...\\ &= [q]^{[q+2](b_j-1)}1,\quad q = a_j-2=p-2\cdot j-1\implies j={{p-q-1}\over{2}}\\ \end{align}$$ QED

Applying the above results to $p=g_{63}$ and $q=2$:

$$\text{Graham's number } = [2]^t 1 > f^{t-3}(\text{googolplex})$$ where $$t=[4](b_j-1)\\ j={{g_{63}-3}\over{2}}$$ and $b_j$ is the last term in the very long and rapidly increasing sequence $b_1 < b_2 < \dots < b_j$ given by the recursion $$\begin{align} b_1 &= [g_{63}-1][g_{63}-1]1\\ b_i &= [g_{63}-2i+1]^{[g_{63}-2i+3](b_{i-1}-1)-1}1\quad (1\lt i\le j).\\ \end{align}$$

NB: We have $t = [4](b_j-1) \gg b_i$ for each $b_i$ in the sequence $b_1 < b_2 < \dots < b_j$, where the number of terms is $j={{g_{63}-3}\over{2}}$, and already the second term is $$b_2 = [g_{63}-3]^{[g_{63}-1]([g_{63}-1]3-1)-1}1.$$ Making some extremely crude simplifications, we have for example $$k \le {[4]^{[6]^{[8]{\cdot^{\cdot^{\cdot^{[g_{63}-3]^Q 1}\cdot}\cdot}\cdot}1}1}1} \ \implies\ f^k (\text{googolplex}) \le \text{Graham's number}$$

where the topmost term is $Q = [g_{63}-1]([g_{63}-1]3-1)-1$ and the height of the "iteration tower" is ${g_{63}-3}\over 2$.

r.e.s.
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