I'm wondering whether there's a known way to compare the exponential factorial of n versus the tetration of a fixed number $($ e.g., $3$, since it appears in Graham's number $)$ with the same number of terms.

Here's an example: Find the smallest n such that $n^{!}$ is greater than $^{n-1}3$.

Both of these terms have the same number of exponents $($we ignore $1$ in the factorial exponential, thus $n-1$ terms$)$. A first computation shows that the tetration of $3$ grows much faster:

$4^{!} = 262144,~$ while $~^{3}3 = 3^{27} \approx 7.62 \cdot 10^{12}$

$5^{!} = 5^{262144} \approx 10^{183230},~$ while $~^{4}3 = 3^{3^{27}} \approx 10^{3.6 \cdot 10^{12}}$

Afterwards, numerical evaluation becomes more difficult, but what appears to be clear is that until the basis for the exponent in the exponential factorial becomes much, much larger, the tetration of $3$ will keep growing insanely faster, simply due to the higher last exponent.

Is this a known problem ?

  • $\begingroup$ This is somewhat related: on Googology wiki they say that the exponential factorial satisfies the bound $n^! \leq ^{n-1}n$, although I don't have a proper reference. $\endgroup$ – A.P. Nov 20 '15 at 21:22
  • $\begingroup$ To ease up the computation, repeatedly apply logarithm. $\endgroup$ – Lucian Nov 20 '15 at 21:38
  • $\begingroup$ That does not help at all. After two stages, you end up with ln(ln(ln(....ln(X + b))...), which you can only approximate. X is still huge, but you have transform the sum to make progress, otherwise you're stuck. You can use the identity ln(X+a) = ln(X(1 + a/X)) = ln X + ln(1+a/X), it's not very helpful unless you decide to ignore ln(1+a/X) for X huge, which does not work at all. Applying this logic will tell you that no matter what n grows to, tetration will always win. $\endgroup$ – Olivier Dec 4 '15 at 21:20
  • $\begingroup$ as for $n^{!} \le ^{n-1}n$, it seems totally obvious. You have two towers of the same height, on the left side decreasing from $n$ to 2, on the right side constant at $n$, thus for any $n\ge2$, the inequality will always hold. $\endgroup$ – Olivier Dec 4 '15 at 21:26
  • $\begingroup$ We can derive a very loose minimum bound from the early results. We observe that $3^{27}$ is already much greater than $4^{!}$, and thus can determine what would be the base exponent needed to match the next level of tetration, i.e find $n$ such that $n^{4^{!}}\ge3^{3^{27}}$. $n^Y \ge 3^X \equiv ln(n) \ge (X ln(3)/Y)$. With approximate values for $X \approx 7.62*10^{12}$ and $Y=262144$, we get $N \ge e^{31157912.84...}$. With any smaller $n$, the next level of tetration will be larger. So a very low minimum bound for n is $10^{13879415}$. $\endgroup$ – Olivier Dec 4 '15 at 22:10

In fact, for no $n$ do we have $n^!$ greater than $^{n-1} 3$. For the following, we will use $\log$ to mean logarithm base 3.

We will prove by induction that $(n+2)^{n^!} < \ ^n 3$ for $n \ge 2$.

Base cases: $4^{2^1} = 16 < 27 = \ ^23$.

$5^{3^!} = 5^9 < 3^{27} = \ ^33$

$6^{4^!} = 6^{4^9} < 3^{2* 4^9} < 3^{3^{27}} = \ ^4 3$

Inductive step: Suppose that $(n+2)^{n^!} < \ ^n 3$ for some $n \ge 4$. Then

$\log\log ((n+3)^{(n+1)^!}) = \log((n+1)^! \log (n+3)) = \log((n+1)^!) + \log\log(n+3) = n^! \log(n+1) + \log \log (n+3)$.

Now, I claim that $n^! \log(n+1) + \log \log (n+3) < n^! \log(n+2)$. After rearranging, this is equivalent to $n^! > \frac {\log\log(n+3)}{\log(n+2) - \log(n+1)}$. But in fact

$\frac{\log\log(n+3)}{\log(n+2) - \log(n+1)} = \frac{\log\log(n+3)}{\log(1 + \frac{1}{n+1})} = \frac{\log\log(n+3)}{\frac{\ln(1+\frac{1}{n+1})}{\ln(3)}} < \frac{\log\log(n+3)}{\frac{1}{(n+2)\ln(3)}} = (n+2)\ln(3) \log\log(n+3) < n^9 \le n^!$.

(The last two inequalities follow because $n \ge 4$.)


$\log\log ((n+3)^{(n+1)^!}) < n^! \log(n+2) = \log((n+2)^{n^!}) < \log(^n 3) = \ ^{n-1} 3$

$(n+3)^{(n+1)^!} < \ ^{n+1} 3$, as desired.

So for $n \ge 3$, $n^! < (n+1)^{(n-1)^!} < \ ^{n-1} 3$, and of course one can verify this is true for $n = 2$ as well.

Going in the other direction, obviously $n^! > \ ^{n-2} 3$ for $n \ge 2$, since each term of the first tower is greater or equal to each term of the second, with equality in just one case. So for all $n \ge 2$ we have

$^{n-2} 3 < n^! < \ ^{n-1} 3$.


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