Positivity of a finite sum involving Stirling numbers In my research in theoretical physics, I have arrived at some coefficients $a_{n,m}$ depending on two integers, $n\geq 1$ and $0\leq m\leq n$:
$$
a_{n,m}=\sum_{j=0}^{n-1} {2j \choose j+m} \left(\frac{n}{4}\right)^j s(n,j+1)
$$
where $s(n,j+1)$ are Stirling numbers of the first kind. Although this expression does not make it manifest (to me), these coefficients are zero when $n+m$ is even. For physical reasons, I am convinced that when $m+n$ is odd, $a_{n,m}>0$, but I haven't been able to prove it. 
I would like to find a proof that $a_{n,m}\geq 0$.
I have found various ways to rewrite these coefficients. For instance, in terms of power series involving modified Bessel functions of the first kind $I_m$:
$$
a_{n,m}= \left. \frac{d^{n-1} \,\left( (1-z)^{\frac{n-2}{2}} \, I_m\left(-\frac{n}{2} \log (1-z)\right) \right)}{d z^{n-1}} \right|_{z=0}
$$
but I am not able to conclude that $a_{n,m}\geq 0$ from this expression either.
Alternatively, when $n$ is a multiple of 4, these coefficients are integers, so I was hoping that there might be a combinatorial argument for this particular case. However, I have been unable to produce it.
Any help would be greatly appreciated.
 A: The numbers $a_{n,m}$ are in fact the Fourier coefficients of the polynomial
$$P_n(x)=\prod_{j=1}^{n-1} \Big( \frac{nx}{2} + \frac{n}{2}-j\Big) $$
with respect to the Chebyshev measure $d\sigma:=(1-x^2)^{-1/2}dx$ on $[-1,1]$, and its orthogonal bases of the Chebyshev polynomials of the first kind. Precisely,
for $0\le m\le n$
$$a_{n,m}=\frac{1}{\pi}  \int_{-1}^1 P_n(x)T_m(x)d\sigma  \ .$$
Changing variable, we have a trigonometric version:
$$a_{n,m}=\frac{1}{\pi}\int_0^\pi P_n(\cos \theta)\cos (m\theta)d\theta \ .$$
Note that the polynomials  $P_n$ and $T_m$ are odd resp. even, according to the parity of $n-1$, respectively $m$, so the integrand $P_n(x)T_m(x)$ has the same parity of $n+m-1$. On the other hand, the Chebyshev measure is symmetric, which explains the vanishing property $a_{n,m}=0$ whenever $n+m$ is even. Moreover, for odd $n+m$ the integrand is positive and concentrated about $\pm 1$; this should hopefully yield to the desired estimate $a_{n,m}>0$. I'll try some computation and in case add details later.
To compute the integral we may also use the Chebyshev-Gauss Quadrature formula on $N$ nodes, which is exact on polynomials of degree less than $2N$. Therefore, for  $2N\ge n+m$ we have
$$a_{n,m}=\frac{1}{N}\sum_{k=1}^{N} P_n\Big(\cos\big( \frac{2k-1}{2N}\pi \big) \Big)\ \cos\big( m\frac{2k-1}{2N}\pi\big) \ .$$
