All natural numbers have a unique factorization into primes. I'm interested in a set $Q$ for which all natural numbers have a unique factorization into distinct elements of $Q$. This leads inductively to $$Q = \{p^{2^a} \mid p \textrm{ is prime}, a \in \mathbb{N}\} = \{2,3,4,5,7,9,11,13,16,17,19,23, \ldots \}$$
Let $I$ index the elements of $Q$ in ascending order: $I(2) = 0$, $I(3) = 1$, $I(4) = 2$, etc.
Now consider the function $f : \mathbb{Z}^+ \to \mathbb{N}$ which factors $n$ into distinct elements $\{q_0, q_1, \ldots, q_k \} \subset Q$ and yields $\sum_{i=0}^k 2^{I(q_i)}$. For example, $24 = 2 \times 3 \times 4$, so $f(24) = 2^0 + 2^1 + 2^2 = 7$.
Iterating $f$ may give loops. E.g. $f(8) = 2^{I(2)} + 2^{I(4)} = 2^0 + 2^2 = 5$, and $f(5) = 2^{I(5)} = 2^3 = 8$.
My question is, is there a finite number of loops? Since all numbers can be reached by only one number, all chains are closed loops, except for the one that reaches $0$. It's feasible that the chain of numbers that goes to $0$ ($0 \leftarrow 1 \leftarrow 2 \leftarrow 3 \leftarrow 6 \leftarrow 12 \leftarrow 20 \leftarrow 28 \leftarrow \cdots$) will pass through nearly all numbers. Does it?