A function whose fixed points are the primes If $a(n) = (\text{largest proper divisor of } n)$, let $f:\mathbb{N} \setminus \{ 0,1\} \to \mathbb{N}$ be defined by $f(n) = n+a(n)-1$. For instance, $f(100)=100+50-1=149$. Clearly the fixed points of $f$ are the primes.

Is every number preperiodic? In other words, is $f(f(\ldots(f(n)\ldots))$ eventually prime?

 A: The problem may have connections with 3x+1 problem. To an integer, I can associate $(p_0,p_1,\cdots)$ the sequence of the least prime divisor of the term  of the sequence $u_{n+1}=f(u_n)$ which may have same properties as the parity function in the 3x+1 problem.
For instance : the asymptotic periodicity of $(p_n)$ is equivalent to periodicity of $u_n$ (thus convergence).
I would like to prove that for all composite integer,there exists a prime that appear infinitly many times in the sequence $(p_n)$.  Using some asymptotic estimations, it's not difficult to prove: 
$\forall N>0,\exists p | card \lbrace n \in \mathbb N | p_n=p\geq N\rbrace $.
It makes no use of the fact that $p_n$ is the least divisor of $u_n$.
It would be interesting to have two other theorems "3x+1"-like : 
The sequence $(p_n)$ determines $u_0$ and $\sum_{n=0}^{\infty}\prod_{k\leq n} \frac{p_k}{p_k +1}=u_0$ 
The sum converges since we have the inequality $\sum_{n=0}^{\infty} \prod_{k\leq n} \frac{p_k}{p_k +1}\leq u_0$.
I would be very interested by the links between the choice of a sequence $(p_n)$ and the convergence in an appropriate space of the prime sum. In the "3x+1"-equivalent would be 
$\sum \frac{2^{a_k}}{3^k}$, where $a_k=$ number of even terms before the $k^{th}$ odd term, which converges in $\mathbb{Z}_2$. In the 3x+1 problem generalized to $\mathbb{Z}_2$, the sum establish a bijection between parity functions and initial conditions in $\mathbb{Z}_2$.  
Furthermore is it true that for any k-uplet of prime number $(l_0,\cdots,l_N)$ there exists an initial condition such that $p_i=l_i, i\leq N $. 
