Common roots of recursive defined polynomial I have a series of polynomials $P_j(x)$ given by the recursive formula
$$P_{j+1}=\frac{e_j}{c_j}xP_{j}-\frac{f_j}{c_j}P_{j-1}
$$
with $P_{-1} \equiv 0$, $P_0 \equiv 1$, where
$$c_j = (j+1)(j+2\kappa+1),\\
e_j = (2j+2\kappa+1)(j+\kappa+1),\\
f_j = (j+\kappa)(j+\kappa+1),\\
j=0, \dots, N-1.$$
For $\kappa=\dfrac{1}{2}$ numerical results indicate that the roots of the polynomial $P_i$ are a subset of the roots of $P_{2i+1}$. E.g. The most simple case:
$$P_1(x)=\frac{3}{2}x, \qquad x_1=0\\
P_2(x)=\frac{5}{2}x^2-\frac{5}{8}, \qquad x_{1,2}=\pm\frac{1}{2},\\
P_3(x)=\frac{35}{8}x^3-\frac{35}{16}x \qquad x_1=0, \ x_{2,3}=\pm-\frac{1}{\sqrt{2}}.$$
How can this be proved or disproved?

Define $Q_n(x) = \dfrac{(2n)!!}{(2n - 1)!!} P_{n - 1}(x)$ for $n \geqslant 0$, where $(-1)!! = 0!! = 1$, then $Q_0(x) = 0$, $Q_1(x) = 2$, and for $n \geqslant 0$,$$
(n + 1)(n + 2) P_{n + 1}(x) = (2n + 2)\left( n + \frac{3}{2} \right) xP_n(x) - \left( n + \frac{1}{2} \right)\left( n + \frac{3}{2} \right) P_{n - 1}(x)\\ \Longrightarrow Q_{n + 2}(x) = 2xQ_{n + 1}(x) - Q_n(x).$$
To prove that $P_n(x) \mid P_{2n + 1}(x)$ for $n \geqslant -1$, it is equivalent to prove that $Q_n(x) \mid Q_{2n}(x)$ for $n \geqslant 0$. Now it suffices to restrict the domains of all $Q_n$'s to $\mathbb{R} \setminus (-1, 1)$. Solving the recurrence relation,$$
Q_n(x) = \frac{1}{\sqrt{x^2 - 1}} ((x + \sqrt{x^2 - 1})^n - (x - \sqrt{x^2 - 1})^n), \quad \forall n \geqslant 0
$$
then for any $n \geqslant 0$,$$
\frac{Q_{2n}(x)}{Q_n(x)} = (x + \sqrt{x^2 - 1})^n + (x - \sqrt{x^2 - 1})^n \in \mathbb{R}[x] \Longrightarrow Q_n(x) \mid Q_{2n}(x).
$$
Therefore, $P_n(x) \mid P_{2n + 1}(x)$ for any $n \geqslant -1$, which implies the set of all roots of $P_n$ (multiplicity counted) is a subset of those of $P_{2n + 1}$.
