While answering this question, I make one question. Define a sequence of polynomials as \begin{align} p_{n}(x)=\sum_{r=0}^{\lfloor (n+1)/2\rfloor} (-1)^{r}\binom{n+1-r}{r} x^{n-r}. \end{align}
I used Wolfram Alpha to find some properties about these polynomials (and solve the original problem, although I used the other method), I find :
Question. (a) For all $n$, $p_{n}(x)$ has only real zeros.
(b) Let $\alpha_{n}$ be a largest zero of $p_{n}(x)$. Then $\alpha_{n}$ is monotone increasing and $\lim_{n\to \infty} \alpha_{n}=4$.
I tried to prove these properties without using the recurrence relation $p_{n+1}(x)=x(p_{n}(x)-p_{n-1}(x))$, but I can't. I need some help.
I suggest checking Theorem 2 in this paper
https://arxiv.org/abs/1601.04381
The generating function provided in this theorem correspond to the recurrence: $H_m=-B(z)H_{m-1}-A(z)H_{m-2}$.
Your question corresponds to A(z)=z and B(z)=-z with a slight modification of the generating function: instead of having 1 in the numerator, we have 1-t:
$\sum_{n=0}^{\infty}P_m(z)t^m=\frac{1-t}{1-zt+zt^2}.$
The numerator does not affect the reality of the roots since we can move to the left side and see $1/(1-t)$ as a constant multiplying $P_m$. Then, the theorem implies that the zeros of $H_m$ are real. What's more, the roots of $P_m$ lie in the interval $(0,4)$ and they are dense on that interval (For each gap in this interval we can find m such that $P_m$ has a zero in this gap).