Knuth's up-arrow notation extends the concept of exponentiation in a recursive manner.
Euler's theorem can be restated in the arrow notation as: $a \uparrow \phi(m) \equiv 1 \mod m$, whenever $(a, m) = 1$. I was wondering what function $f_k(m)$ would ensure that $a \uparrow^k f_k(m) \equiv 1 \mod m$ and under what conditions can we ensure that such a $f_k(m)$ exists. It is elementary to note that, when $\phi(m) | a$, for all values $n, k >= 2$, we have: $a \uparrow^k n \equiv 1 \mod m$. I could neither find such a $f_k(m)$ nor prove that no such number exists when $\phi(m) \not | a$
As Euler's $\phi(n)$ doesn't always equal the order of $n$ in $m$, I'm only looking for a function $f_k(m)$ that ensures that $a \uparrow^k f_k(m)$ is congruent to $1$ modulo $m$ (under appropriate conditions); not necessarily the smallest such number.