TITLE: Do these polynomials have alternating coefficients?
QUESTION [21 upvotes]: In answering another MathOverflow question, I stumbled across the sequence of polynomials $Q_n(p)$ defined by the recurrence
$$Q_n(p) = 1-\sum_{k=2}^{n-1} \binom{n-2}{k-2}(1-p)^{k(n-k)}Q_k(p).$$
Thus:
$Q_{2}(p) = 1$
$Q_{3}(p) = -p^2
 + 2 p$
$Q_{4}(p) = -2 p^5
 + 9 p^4
 - 14 p^3
 + 8 p^2$
$Q_{5}(p) = 6 p^9
 - 48 p^8
 + 162 p^7
 - 298 p^6
 + 318 p^5
 - 189 p^4
 + 50 p^3$
Numerical calculations up to $n=60$ suggest that:

The lowest-degree term of $Q_n(p)$ is $2n^{n-3}p^{n-2}$.
The coefficients of $Q_n(p)$ alternate in sign.

Are these true for all $n$?
As the title indicates, I'm especially puzzled about 2. Indeed, the original inspiration for the polynomials $Q_n(p)$ comes from a classic paper of E. N. Gilbert (Random graphs, Ann. Math. Stat. 30, 1141-1144 (1959); ZBL0168.40801) where the author studies the sequence of polynomials $P_n(p)$ given by the similar recurrence $$P_n(p) = 1 - \sum_{k=1}^{n-1} \binom{n-1}{k-1}(1-p)^{k(n-k)}P_k(p),$$ which do not have alternating coefficients.

REPLY [12 votes]: To illustrate the suggestion of Richard Stanley about positivity of real parts of zeroes, here are the zeroes of $Q_{20}$. The pattern seems to be the same for all of them.

Another empirical observation: seems that
$$
\frac{Q_n(1-x)}{(1-x)^{n-2}(1+x)}=1+(n-3)x+\left(\binom{n-2}2+1\right)x^2
+\left(\binom{n-1}3+n-3\right)x^3+\left(\binom n4+\binom{n-2}2+1\right)x^4+...+\left(\binom{n+k-4}k+\binom{n+k-6}{k-2}+\binom{n+k-8}{k-4}+...\right)x^k+O(x^{k+1})$$for $n>k+1$

REPLY [7 votes]: To add to the preceding answer: The absolute values of the coefficients appear normal (in particular, unimodal):  Which means that the technology developed, in, eg, 
Lebowitz, J.L.; Pittel, B.; Ruelle, D.; Speer, E.R., Central limit theorems, Lee-Yang zeros, and graph-counting polynomials, J. Comb. Theory, Ser. A 141, 147-183 (2016). ZBL1334.05065.
May be relevant.