Certain totient chain growth assuming Carmichael's Totient Function Conjecture

Question

Assume the Carmichael's Totient Function Conjecture.

Consider the totient chain

$$n_0=\phi(3^2)\rightarrow n_1=\phi((\phi^{-1}(n_0))^2)\rightarrow n_2=\phi((\phi^{-1}(n_1))^2)\rightarrow\dots$$

where $\phi^{-1}(\phi(x))$ is the smallest second integer $y\neq x$ such that $\phi(x)=\phi(y)$.

(1) Does this chain grow doubly exponentially? (Shown below by Greg Martin)

(2) At every $i$ is there a prime $p_{i+1}$ with $p_{i+1}|n_{i+1}$ and $gcd(p_{i+1},\prod_{j=1}^in_j)=1$? What is the size of this prime?

What if $3^2$ is replaced by $a^k$ where $a,k+1\geq3$ and exponent in $\phi$ is $k$?

Answer 1

The suggested dynamic is more clearly written (using $\phi(m^2)=m\phi(m)$) as $$n_{j+1}= n_j\phi^{-1}(n_j) \gt (n_j)^2,$$

which clearly shows superexponential growth in $j$. Using $k$ in place of $2$ gets an inequality like $n_{j+1} \gt n_j^k$. Since it is unclear how $\phi^{-1}$ acts as a map, it should be tantamount to the Carmichael conjecture to ask for a primitive prime factor for each $n_j$.

Gerhard "Remember Jacobi Says 'Always Invert!'" Paseman, 2016.06.07.

Answer 2

The answer to (1) is yes. Using the increasing lower bound $$ \phi(n) > \frac n{4\log\log n} \quad\text{for }n\ge7 $$ (which follows from known bounds), we get $$ n_{j+1} = \phi(\phi^{-1}(n_j)^2) > \frac{\phi^{-1}(n_j)^2}{4\log\log \phi^{-1}(n_j)^2} > \frac{n_j^2}{4\log\log n_j^2}, $$ which certainly implies $n_j \gg \exp((2-\varepsilon)^j)$ for any $\varepsilon>0$.