Generating primes via composition of polynomials It is well known that no nonconstant polynomial $f\in \mathbb{Z}[x]$ can assume only prime values at integer arguments. Indeed, if $a\in \mathbb{Z}$ is so large that $|f(a)|>1$, and if $p$ is a prime factor of $f(a)$, then all values of $f(a+pf(x))$ are divisible by $p$.
Question: Can a sequence of the form $f(a),f(f(a)),\ldots$ (for some fixed $a\in \mathbb{Z}$) take on only prime values and tend to infinity? 
If $f$ is linear, then the answer to the question is no. To prove this, assume on the contrary that $f^n(a)$ tends to infinity and takes only prime values. Choose $n$ so large that $p:=|f^n(a)|$ is a prime greater than any coefficient of $f$. Then $f$ is a permutation polynomial mod $p$, therefore $f^m(p)$ is divisible by $p$ for infinitely many $m$, a contradiction.
If $f$ is not linear, and if the sequence in question consists only of primes, then (as Dietrich Burde pointed out) $f$ would be a non-linear polynomial taking infinitely many prime values, something that is not known to exist. 
On the other hand, if the sequence $f(a),f(f(a)),\ldots$ can never take only prime values (which is my guess) then maybe the problem has a simple solution. Does anyone have any ideas or references?
This is an expanded form of a question posted on Math SE here.
 A: Let $d=\deg(f)\ge2$. Since the height of $f^n(\alpha)$ grows like $C^{d^n}$ with $C>1$ (unless the orbit is finite), one would expect the sequence $f^n(\alpha)$ to contain only finitely many primes on probabilistic grounds. So people study other questions. For example, how large is the set of primes that divides at least one term in the sequence? See [1]. Or as joro indicates, one might look at the orbit behavior modulo $p$ for varying $p$; see for example [2] or [3]. Or one might ask how many terms in the sequence have a primitive prime divisor, that is, a prime dividing $f^n(\alpha)$ that does not divide $f^m(\alpha)$ for all $m<n$; see for example [4] and [5].
[1] Hamblenm Spencer and Jones, Rafe and Madhu, Kalyani,
The density of primes in orbits of $z^d + c$, 2013, arXiv:1303.6513.
[2] Akbary, Amir and Ghioca, Dragos, 
Periods of orbits modulo primes, 
J. Number Theory 129 (2009), 2831-2842.
[3] Silverman, Joseph H.,
Variation of periods modulo $p$ in arithmetic dynamics,
New York J. Math. 14 (2008), 601--616.
[4] Ingram, Patrick and Silverman, Joseph H., 
Primitive divisors in arithmetic dynamics, 
Math. Proc. Cambridge Philos. Soc. 146 (2009), 289-302.
[5] Gratton, Chad and Nguyen, Khoa and Tucker, Thomas J., 
ABC implies primitive prime divisors in arithmetic dynamics, 2012,
arXiv:1208.2989.
A: Some comments.
Assuming a plausible, conjecture, one can make
$f(a),f(f(a)),\ldots f^n(a)$ prime for arbitrary large $n$.
Take $f(x)=2 x + 1$.
This is Cunningham chain
From the above link:

It follows from Dickson's conjecture and the broader Schinzel's hypothesis H, both widely believed to be true, that for every $k$ there are infinitely many Cunningham chains of length $k$. There are, however, no known direct methods of generating such chains.

For certain $f$ one can prove this is impossible.
If some $f^n(x)$ is reducible over $\mathbb{Z}[x]$, the rest of the iterates will be reducible too.
This happens for $g(x) = x^2 - x - 1$.
$$g(g(g(x))) = (x^{4} - 3 x^{3} + 4 x - 1) \cdot (x^{4} -  x^{3} - 3 x^{2} + x + 1)$$
By the boxing principle $f^k(x)$ becomes periodic $\mod p$ and if the
periodic part contains zero it is impossible too --
this doesn't work for interesting coprime sequences.
