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 :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

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