Let $p_n$ be the $n$th Mandelbrot polynomial, so $p_1:=c$ and $p_{n+1}:=p_n^2+c$. Then each $p_n$ is divisible by all $p_k$ for which $k$ is a proper divisor of $n$; to get rid of those factors, we can recursively divide out all the zeroes that were introduced in those previous polynomials:
$$q_{n}:=\frac{p_n}{\prod_{k\mid n,k<n}q_k}=\prod_{k\mid n}p_k^{\mu(n/k)}.$$
For example, the first few such polynomials are: \begin{align*} %p_1&=c&q_1&=c\\ %p_2&=c^2+c&q_2&=c+1\\ %p_3&=c^4+2c^3+c^2+c&q_3&=c^3+2c^2+c+1\\ %p_4&=c^8+4c^7+6c^6+6c^5+5c^4+2c^3+c^2+c&q_4&=c^6+3c^5+3c^4+3c^3+2c^2+1 %p_1&=c&&=q_1\\ %p_2&=c^2+c&&=c\cdot(c+1)=q_1q_2\\ %p_3&=c^4+2c^3+c^2+c&&=c\cdot(c^3+2c^2+c+1)=q_1q_3\\ %p_4&=c^8+4c^7+6c^6+6c^5+5c^4+2c^3+c^2+c&&=c\cdot(c+1) p_1&=c&&=q_1\\ p_2&=c\cdot(c+1)&&=q_1q_2\\ p_3&=c\cdot(c^3+2c^2+c+1)&&=q_1q_3\\ p_4&=c\cdot(c+1)\cdot(c^6+3c^5+3c^4+3c^3+2c^2+1)&&=q_1q_2q_4\\ p_5&=c\cdot(c^{15}+8c^{14}+...+c+1)&&=q_1q_5\\ p_6&=c\cdot(c+1)\cdot(c^3+2c^2+c+1)\cdot(c^{27}+13c^{26}+...-c+1)&&=q_1q_2q_3q_6. \end{align*} Now my question is, are those already the irreducible factors of $p_n$ over $\mathbb Q$?
I suspect they are, but haven't been able to find any proof or counterexample of this so far. It does hold for the first 13 polynomials, but due to the exponentially growing degree I can't really check many more. Reduction modulo two also doesn't help, as $q_4=(c^2+c+1)\cdot(c^4+c+1)\in\mathbb F_2[c]$ is already not irreducible there, and I don't know what else I could try to proof or disproof this. I also haven't been able to find any information about this online; if one of you is more lucky there or has any advice for how else this could possibly be solved, that would be greatly appreciated.
This seems to be a result advertised in the abstract of version 1 of this preprint:
There are two later versions of the paper, I'm not sure what the differences are.