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Wolfgang
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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.