This question is related to Strange (or stupid) arithmetic derivation. The original question whether an unbounded sequence of iterates exists is still unanswered. $$n=\prod_{i=1}^{k}p_i^{\alpha_i} \quad f(n):= \prod_{i=1}^{k}\alpha_ip_i^{\alpha_i-1}\quad f^{(k)}(n):= \underbrace{f(\ldots f(f}_{k \text{ times}}(n))\ldots)$$

Suppose we have an unbounded sequence: $$\lim_{k\rightarrow \infty}f^{(k)}(n)= \infty$$ Is it true that there is a prime, such that its exponent is unbounded in this sequence?

**Additional information**: We think that this true but we are unable to prove it. If not, there is a sequence, such for every prime $p$ there is a constant $c_p$ such that no element of the sequence $\{f^{(k)}(n)\}_{k=1}^{\infty}$ is divisible by $p^{c_p}$. It is easy to see that $c_p$ must depend on $p$.

It is particularly annoying that we can not disprove the existence of case where the exponent of $2$ is maximized at an even number (thus it is stuck there). Therefore no exponent (larger than one) can decrease anymore!

On the other hand, we manged to show that no divergent sequence exists with bounded number of prime divisors. The proof of this and some other related claims about the derivation can be found here: http://www.math.bme.hu/~kovacsi/Pub/arithmetic_derivation_v04.pdf