My question is whether there exists $(a_0, a_1, \ldots, a_{2n-1}) \in \mathbb{R}_{+}^{2n}$ such that

(1). $a_{2k} + a_{2k+1} = \binom{3n-1}{3k} + \binom{3n-1}{3k+1} + \binom{3n-1}{3k+2}$ for all $0 \le k \le n-1$,

(2). The $(2n-1)$-degree polynomial $ P(x): = a_0 + a_1 x + \cdots + a_{2n-1}x^{2n-1}$ has all real roots.

The problem is motivated from 1). proving some central limit theorems for determinantal processes and 2). approximating Poisson binomial distributions. Here are a few ways that I tried:

(a). There is a necessary and sufficient condition for a real-coefficient polynomial to have all real roots: the Toeplitz matrix $(a_{j-i})$ is totally positive. But it seems hard to apply in my problem, since there is no obvious candidate for $a_{2k}$ and $a_{2k+1}$.

(b). Newton's inequality gives a necessary condition: $a_{k-1}a_{k+1} \ge (1+\frac{1}{k})(1+\frac{1}{2n-k})a_k^2$. A simple consequence is that for each fixed $k$, $a_k = \Theta(n^{1/2 + 3k/2})$. I wonder if there is any combinatorial sequence with $n^{3k/2}$ growth rate. If so, there might be some candidates for $a_{2k}$ and $a_{2k+1}$, and root interlacing argument might be possible.

(c). There are sufficient conditions: $a_{k-1}a_{k+1} \ge 4 a_k^2$ by Hutchinson, Kurtz, Handelman. But this condition is too strong for the problem.

(d). A polynomial has all real roots is equivalent to the fact that the leading coefficients of its Sturm's sequence are all nonnegative. The latter can be represented by the subresultants of the Sylvester matrix of $P$ and $P'$. So the problem is equivalent to whether a semi-algebraic set is empty or not. There is a famous Stengle's Positivstellensatz, but I don't see how to apply in my case.

The answer to the question is positive for $n = 1,2,3,4,5$, but I wonder if it holds for large $n$.

My question is whether there exists $(a_0, a_1, \ldots, a_{2n-1}) \in \mathbb{R}_{+}^{2n}$ such that

(1). $a_{2k} + a_{2k+1} = \binom{3n-1}{3k} + \binom{3n-1}{3k+1} + \binom{3n-1}{3k+2}$ for all $0 \le k \le n-1$,

(2). The $(2n-1)$-degree polynomial $ P(x): = a_0 + a_1 x + \cdots + a_{2n-1}x^{2n-1}$ has all real roots.

The problem is motivated from 1). proving some central limit theorems for determinantal processes and 2). approximating Poisson binomial distributions. Here are a few ways that I tried:

(a). There is a necessary and sufficient condition for a real-coefficient polynomial to have all real roots: the Toeplitz matrix $(a_{j-i})$ is totally positive. But it seems hard to apply in my problem, since there is no obvious candidate for $a_{2k}$ and $a_{2k+1}$.

(b). Newton's inequality gives a necessary condition: $a_k^2 \ge (1+\frac{1}{k})(1+\frac{1}{2n-k})a_{k-1}a_{k+1}$. A simple consequence is that for each fixed $k$, $a_k = \Theta(n^{1/2 + 3k/2})$. I wonder if there is any combinatorial sequence with $n^{3k/2}$ growth rate. If so, there might be some candidates for $a_{2k}$ and $a_{2k+1}$, and root interlacing argument might be possible.

(c). There are sufficient conditions: $a_k^2 \ge 4 a_{k-1}a_{k+1}$ by Hutchinson, Kurtz, Handelman. But this condition is too strong for the problem.

(d). A polynomial has all real roots is equivalent to the fact that the leading coefficients of its Sturm's sequence are all nonnegative. The latter can be represented by the subresultants of the Sylvester matrix of $P$ and $P'$. So the problem is equivalent to whether a semi-algebraic set is empty or not. There is a famous Stengle's Positivstellensatz, but I don't see how to apply in my case.

The answer to the question is positive for $n = 1,2,3,4,5$, but I wonder if it holds for large $n$.