We denote for an integer $n>1$ its square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,\tag{0}$$ with the definition $\operatorname{rad}(1)=1$. You can see this definition and the properties of this arithmetic function from the Wikipedia Radical of an integer. Then I wondered about a variant of two problems showed in last paragraphs of section B 41 of Guy's book [1].
Definition. For each fixed integer $n\geq 1$, I consider the arithmetic function $$f(n)=f^1(n):=\operatorname{rad}(n)+1,\tag{1}$$ and we create the iterated $$f^{k+1}=f(f^{k}(n))\tag{2}$$ until $f^{T}(n)$ reaches a prime number, here thus $T=T(n)$ is a function depending on $n$ (that is the number of steps in our iteration that we need to wait until the sequence of iterated $f^{k}(n)$ reaches a prime number for the first time). We can call stopping time to the arithmetic function $T(n)$ (defined thus for integers $n\geq 1$ and taking positive values).
Example 1. Let $n=1$, then in a step the sequence of iterations reaches a prime number since $f(1)=f^1(1)=\operatorname{rad}(1)+1=1+1=2$ is a prime number. The same argument works with the composite number $n=8$, because $f(8)=f^1(8)=\operatorname{rad}(8)+1=2+1=3$ a prime number. Thus in our table $T(1)=1$ iteration, and also in this example we've seen that the corresponding sequence defined for the integer $8$ has stopping time $T(8)=1$.
Example 2. We work about the calculation of $T(33)$. We iterate while the result is a composite integer. Here $\operatorname{rad}(33)=33$, thus $f(33)=34$ and since $34=2\cdot 17$ is also square-free then $f^2(33)=f(f(33))=34+1=35$. Again our last iterated, that is $35=5\cdot 7$ has no repeated prime factors, thus we write $f^3(33)=f(35)=36$. Finally $f^4(33)=\operatorname{rad}(36)+1=6+1=7$ that results a prime number. Thus we conclude that $T(33)=4$.
Computational fact. I did few experiments/calculations and seems that the arithmetic function $T(n)$ is erratic, thus it makes sense to study averages of consecutive values of this stopping time.
Question. I would like to know if it is possible to deduce a tighter upper bound for $$\frac{1}{N}\sum_{1\leq n\leq N}T(n)\tag{3}$$ as $N$ grows (if you are able to provide an elaborated statement about the asymptotic behaviour of $(3)$ as $N\to\infty$ feel free to do it). Many thanks.
Final remarks. My motivation was to propose a similar variant of the mentioned problems in Guy's book, now for other interesting arithmetic function $\operatorname{rad}(n)$. I've no idea about a strategy to attack the problem to get idea about the size of the average means of the $T(n)$.
References:
[1] Richard K. Guy, Unsolved Problems in Number Theory, Volume I, Second Edition, Springer (1994).
Note that $f(n)=n+1$ if $n$ is square-free, which can happen at most three times in a row when iterating $f$, namely one of $n,n+1,n+2,n+3$ must be a multiple of $4$. And if $n$ is not square-free, then $2\le f(n)\le \frac n2+1$. Thus it takes up to four iterations to take $n$ to a number $\le \frac{n+3}2+1$. Thus if we start with $n\le 5+2^m$, we arrive at a number $\le 5+2^{m-1}$ after at most $4$ iterations. Thus after at most $4\log_2(n-5)$ we are at a prime or at $4$ or $6$. This gives the (correct, but probably very pessimistic) upper bound $$ T(n)\le 1+4\log_2(n-5)\sim \frac 4{\ln2}\cdot \ln n\approx 5.77\ln n$$ (which then holds in similar form for the averaged $T$).
A more heuristic appoach is that the probability of $n$ beinf squarefree is $\frac 6{\pi^2}$, hence on average, we should be able to replace the factor $4$ above with $\frac{\pi^2}6\approx1.64$. We might also note that when $n$ is a multiple of $4$, its is a multiple of $8$ or higher about half the time etc. Thus we should then not step from $\sim 2^m$ to $\sim 2^{m-1}$, but on average more like to $2^{m-1-\frac12-\frac14-\ldots}=2^{m-2}$. Boldly combining both ideas, we get a heuristic estimate of $\sim \frac{\pi^2}6\log_4n\approx 1.19\ln n$. And there is still room for impovement by stopping t other primes than $2,3,5,7$.