How to find the minimum $m$ for a given $n$ in this inequality? For a given $n \in \Bbb N$, how do you find the minimum $m \in \Bbb N$ which satisfies the inequality below?
$$3^{3^{3^{3^{\unicode{x22F0}^{3}}}}} (m \text{ times}) > 9^{9^{9^{9^{\unicode{x22F0}^{9}}}}} (n \text{ times})$$
What I have tried to do so far is decomposing the $9$ on the right side to $3*3$
or to $3^2$, but both ways didn't get me much far and I couldn't find a pattern.
 A: $9^{9^{9^{9^{\dots}}}} n$ times $= 3^{2*3^{2*3^{2*3^2{\dots}}}} $, where the upper most $2$ is $(n+1)^{th}$ power.
$2*3^2 < 3^{2+1} = 3^3 $
$\implies 9^{9^{9^{9^{\dots}}}} n$ times $\lt 3^{3^{3^{3^{\dots}}}} n+1 $ times
So, $ m = n + 1 $
A: I can tell you with certainty that m+n+1 is the minimum value for m. Working it out is the real problem. I'm sure it's correct, though, because if you take the 3's n+1 times and subtract the 9's n times, 
3^3^3^3 (n+1 times) - 9^9^9 (n times) > 0

it's always positive and the difference is increasing. I have not, however, been able to model this with a function that you can take the derivative of and find a minimum value for.
A: Let $a_1=3,b_1=9,a_{n+1}=3^{a_n},b_{n+1}=9^{b_n}$ obviously we have $a_n ,b_n \in N$
$a_1<b_1$ .Suppose $a_n<b_n$ then
$$a_{n+1}=3^{a_n}<3^{b_n}<9^{b_n}=b_{n+1}$$
$a_2>2b_1$ Suppose $a_n>2b_{n-1}$ then
$$a_{n+1}=3^{a_n}\ge3^{2b_{n-1}+1}=3\cdot 9^{b_{n-1}}=3b_n>2b_n$$
Thus by induction $a_n<b_n<\dfrac {a_{n+1}} 2<a_{n+1}$ hence the result follows
