What is the smallest number $n$ , such that $n\uparrow^4 n>3\uparrow^5 3$ holds? 
What is the smallest number $n$, such that $$n\uparrow^4 n>3\uparrow^5 3$$ holds ?

$\uparrow$ stands for Knut's up-arrow-notation and is defined as follows
$a\uparrow b=a^b$
$$a\uparrow \uparrow b=a\uparrow a\uparrow ...\uparrow a\uparrow a$$ with $b$ $a's$
$$a\uparrow^3 b=a\uparrow\uparrow a\uparrow\uparrow...\uparrow\uparrow a\uparrow\uparrow a$$ with $b$ $a's$
$$a\uparrow^4 b=a\uparrow^3 a\uparrow^3 ...\uparrow^3 a\uparrow^3 a$$ with $b$ $a's$ and so on.
We have $$3\uparrow^5 3=3\uparrow^4 3\uparrow^3 3\uparrow\uparrow 3^{27}$$
Note, that every uparrow-expression is calculated from right to left, so
we have $a\uparrow\uparrow a\uparrow\uparrow a=a\uparrow\uparrow (a\uparrow\uparrow a)$
Because of $3\uparrow^5 3=3\uparrow^4 3\uparrow^4 3$ , I guess that $n$ is approximately $3\uparrow^4 3$. Can we calculate $n$ more precisely ?
Notation : Applying Saibian's theorem , we have with
           $S:=3\uparrow^4 3$ :
$$(S-3)\uparrow^4 (S-3)<3\uparrow^5 3<S \uparrow^4 S$$
Proof : $(S-3)\uparrow^4 (S-3)<(3\uparrow^4 3)\uparrow^4 (S-3)<3\uparrow^4 S=3\uparrow^5 3$
The right inequality follows from $3\uparrow^5 3=3\uparrow^4 S<S\uparrow^4 S$
So, the bounds for $n$ are very sharp indeed.
 A: In general, for $k \ge 2$ the smallest $n$ such that $n \uparrow^k n > 3 \uparrow^{k+1} 3$ is $n = (3 \uparrow^k 3) - 2$.
Set $S = 3 \uparrow^k 3$. We will prove:
Lemma: For all $0 \le i \le k$ and $j \ge 2$, $(S-2) \uparrow^i j > 3 \uparrow^i (j+2)$.
Proof by induction on $i$ and then on $j$.
Base case: $i=0$. Using the convention that $a \uparrow^0 b = ab$, we have
$$(S-2)\uparrow^0 j = (S-2)j > 6j \ge 3(j+2) = 3 \uparrow^0 (j+2)$$
Base case: We assume the lemma for $i-1$, and prove it for $i$ and $j=2$.  By the inductive hypothesis, we have $(S-2) \uparrow^{i-1} (S-2) > 3 \uparrow^{i-1} S$.  But then we have
$$(S-2) \uparrow^i 2 = (S-2) \uparrow^{i-1} (S-2) > 3 \uparrow^{i-1} S \ge 3 \uparrow^{i-1} (3 \uparrow^i 3) = 3 \uparrow^i 4$$
Main case: We assume the lemma for $i$ and $j$, and prove it for $i$ and $j+1$.
$$(S-2) \uparrow^i (j+1) = (S-2) \uparrow^{i-1} ((S-2) \uparrow^i j) > 3 \uparrow^{i-1} (3 \uparrow^i (j+2)) = 3 \uparrow^i (j+3)$$
and the lemma is proved.
Setting $i = k$ and $j = S-2$ gets the desired inequality. (Peter has already shown that $n = S-3$ is not sufficient.)
