Is $\liminf \frac{\sigma_{k}(({2}^{m-1})({2^m-1}))}{\phi_{k}(({2}^{m-1})({2^m-1}))}$ finite for every $k$? I would like to check if this limit :$$\liminf \frac{\sigma_{k}(({2}^{m-1})({2^m-1}))}{\phi_{k}(({2}^{m-1})({2^m-1}))}$$ finite for every $k$?
where :$\phi_{k}$ is iterating Euler - totient function and $\sigma_{k}$ is iterating sum divisor .
note(01) : Here :$\sigma_{k}(n)=\sigma(\sigma(\sigma(\dots n)))$ is the $k$-th iterate of the sum of divisors function.
and :$\phi_{k}(n)=\phi(\phi(\phi(\dots n)))$ is the $k$-th iterate of the euler totiont  function.
Note(02) :I tried to evaluate the recent limit I accrossed this problem :can I  write  :$${\phi_{k}(({2}^{m-1})({2^m-1}))}=({2}^{m-1-k})\phi_{k}({2^m-1})$$ ? I know only that is true iff gcd $({2}^{m-1},{2^m-1})=1$ for $m\geq 1$ and $k=1$ ?
Thank you for any help 
 A: The problem is probably beyond current knowledge. It is well possible that $\liminf\omega(2^m-1)=\infty$, where $\omega(n)$ denotes the number of distinct prime divisors of $n$. If this was the case, one could expect that for each small prime $p$ there is some $m_0$, such that for $m>m_0$ we have that $2^m-1$ has a prime divisor $q\equiv 1\pmod{p}$. If this was the case, then $\frac{\varphi(\varphi(2^m-1))}{2^m-1}\rightarrow 0$, thus for $k=2$ the $\liminf$ in question would be infinite.
Excluding these possibilities is probably not much easier then proving that there are infinitely many Mersenne primes, which is certainly difficult.
A: For $k=1$ and $m$ a not too small prime, there are fewer than $f=m/(\log_2 m)$ prime factors of $2^m-1$, all of them at least as large as $m$, and your ratio is $O((m/(m-2))^f)$, which is bounded, so the answer is yes the lim inf is finite for $k=1$.
For $k \gt 1$ fixed, we have no nice estimates on the number or size of distinct prime factors of either numerator or denominator of your fraction, and at present we don't have a good understanding of the dynamics of the $\sigma$ or the $\phi$ functions to  give a good guess.  I am looking at a concept with the working title "factoral abundance" to address questions like this.
Getting some help with the estimates raised in  How to get a good upper bound on $\sum_{1 \lt d, d|n} \phi(d)/\log d$? would push us nearer to such answers.
Gerhard "And Help On Other Questions" Paseman, 2015.12.29
