counting Misiurewicz points

Question

I enumerated the number of Misiurewicz points using SageMath to factor into irreducible polynomials over $\mathbb{Z}$, where the degree (after discarding factors corresponding to lower (pre)periods) gives the number of roots over $\mathbb{C}$ (which are guaranteed to have multiplicity $1$). Then I played around trying to fit a function to the data in Haskell, and came up with this $M(q,p)$ that counts the number of Misiurewicz points in the Mandelbrot set:

$$ \begin{aligned} H :\mathbb{N^+} &\to \mathbb{N} \\ H (p) &= \sum_{d \mid p} \mu\left(\frac{p}{d}\right) 2 ^ {d - 1} \\ M :\mathbb{N^+} \times \mathbb{N^+} &\to \mathbb{N} \\ M (q, p) &= \left(2 ^ {q - 1} - \mathbf{1}_{q \equiv 1 \mod p}\right) H (p) \end{aligned} $$

$\mu$ is the Moebius function. $\mathbf{1}$ is the indicator function. H(p) counts hyperbolic components with preperiod $0$. $q$ is the desired preperiod, $p$ is the desired period.

However while this fits all the data I have available ($q + p \le 16$), and I know that $H$ is correct, I have no proof of correctness of $M$. Any ideas how to go about proving this? In all my tests I ended up with just $1$ irreducible polynomial, with multiplicity $1$, after discarding lower (per)period factors. But tests are not a proof.

Answer 1

Misiurewicz Points for Polynomial Maps and Transversality by Benjamin Hutz, Adam Towsley

Corollary 3.3. The number of $(m,n)$ Misiurewicz points for $f_{d,c}$ is $$ M_{m,n} = \begin{cases} \sum_{k \mid n} \mu\left(n \over k \right) d^{k-1} & m = 0 \\ (d^m - d^{m-1} - d + 1) \sum_{k \mid n} \mu\left(n \over k \right) d^{k-1} & m \ne 0 \text{ and } n \mid (m - 1) \\ (d^m - d^{m-1}) \sum_{k \mid n} \mu\left(n \over k \right) d^{k-1} & \text{otherwise} \end{cases} $$

There is a typo in the PDF statement of the corollary ($k \mid d$ in place of $k \mid n$), but its proof uses $k \mid n$, so the above is not a verbatim quote.

They have $f_{d,c}(z) = z^d + c$, so this result is more general than the case $d = 2$ I asked about.