Can you confirm the positivity of a quantity involving the Stirling numbers of the first kind

Question

Let $s(m,n)$ denote the Stirling numbers of the first kind. For $m,n\in\mathbb{N}$, define \begin{equation} \mathcal{Q}(m,n)=(-1)^n\sum_{\ell=0}^{2n} \binom{m+\ell-1}{m-1} s(m+2n-1,m+\ell-1)\biggl(\frac{m+2n-2}{2}\biggr)^{\ell}. \end{equation} I knew the identities \begin{align*} \mathcal{Q}(1,n)&=\biggl[\frac{(2n-1)!!}{2^{n}}\biggr]^2,\\ \mathcal{Q}(2,n)&=\biggl[\frac{(2n)!!}{2^{n}}\biggr]^2=(n!)^2,\\ \mathcal{Q}(3,n)&=\biggl[\frac{(2n+1)!!}{2^{n}}\biggr]^2 \sum_{\ell=0}^{n}\frac{1}{(2\ell+1)^2},\\ \mathcal{Q}(4,n)&=\biggl[\frac{(2n+2)!!}{2^{n}}\biggr]^2 \sum_{\ell=0}^{n}\frac{1}{(2\ell+2)^2}, \end{align*} and \begin{equation*} \mathcal{Q}(5,n)= \frac{1}{2}\biggl[\frac{(2n+3)!!}{2^{n}}\biggr]^2 \Biggl[\Biggl(\sum_{\ell=0}^{n+1}\frac{1}{(2\ell+1)^2}\Biggr)^2 -\sum_{\ell=0}^{n+1}\frac{1}{(2\ell+1)^4}\Biggr] \end{equation*} for $n\in\mathbb{N}$.

I guess that

the quantity $\mathcal{Q}(m,n)$ for $m,n\in\mathbb{N}$ is positive.

Can you confirm the correctness of this guess? Thank you very much.

This problem comes from the main results in the following references.

References

1. B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), no. 2, 427--466; available online at https://doi.org/10.2298/AADM210401017G.
2. F. Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstr. Math. 55 (2022), no. 1, 710--736; available online at https://doi.org/10.1515/dema-2022-0157.

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One more problem. For given $m\ge1$ and $n\ge0$, is the quantity $$ \mathfrak{Q}_{m,n}(\ell)=\frac{\mathcal{Q}(m,n+\ell+2)}{\mathcal{Q}(m,n+\ell+1)}\frac{(m+2 n+2 \ell+2)!}{(m+2 n+2 \ell+4)!}(\ell+1) $$ an increasing sequence in $\ell\ge0$? Equivalently, is the sequence $$\tag{OO} \frac{t(m+2\ell+2n+2,m)}{t(m+2\ell+2n,m)} \frac{(m+2\ell+2n)!}{(m+2\ell+2n+2)!} \ell $$ decreasing in $\ell\in\mathbb{N}$ for fixed integers $n\ge0$ and $m\in\mathbb{N}$? where $t(m,n)$ stands for the first kind of central factorial numbers.

Answer 1

(This is not an answer, it is just too long for a comment. What follows is mostly taken from John Riordan, Combinatorial Identities (1968), Chapter 6.)

Let $x^{[n]}=x\prod_{k=1}^{n-1}(x+n/2-k)$. The central factorial number (of the first kind) is the coefficient of $x^k$ in the expansion of $x^{[n]}$; that is, $x^{[n]}=\sum_{k=0}^nt(n,k)x^k$. The central factorial number (of the second kind) is similarly defined by $x^n=\sum_{k=0}^nT(n,k)x^{[k]}$ (notice the analogy to the Stirling numbers).

To prove your claim, it would suffice to show that $$\mathcal Q(2m,n)=(-1)^nt(2m+2n,2m),\tag{1}$$ $$\mathcal Q(2m+1,n)=(-1)^nt(2m+2n+1,2m+1),\tag{2}$$ $$\sum_{k=1}^nt(2n,2k)x^{2k-2}=\prod_{k=1}^{n-1}(x^2-k^2),\tag{3}$$ and $$\sum_{k=0}^nt(2n+1,2k+1)x^{2k}=4^{-n}\prod_{k=1}^n(4x^2-(2k-1)^2).\tag{4}$$ The identities (3) and (4) follow from the definitions of the central factorial numbers. Perhaps the easiest way to prove (1) and (2) would be to show that they satisfy the definitions directly, or to check that $\mathcal Q$ satisfies the recurrence $$t(n,k)=t(n-2,k-2)-\frac{1}{4}(n-2)^2t(n-2,k),\tag{5}$$ but I do not have the time to check this right now unfortunately.

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Edit (15 October 2023). We prove that $t(2n,2k)$ has sign $(-1)^{n-k}$. Let us fix some notation first. We write $[n]=\{1,\dots,n\}$ for the set of the first $n$ positive integers, ${[n]\choose k}=\{S\subset[n]:|S|=k\}$ for the set of all $k$-element subsets of $[n]$, and $[x^k]f(x)$ for the coefficient of $x^k$ in the expansion of $f(x)$. The main idea we will use is that $\prod_{i=1}^n(a_i+b_i)=\sum_{S\subset[n]}\prod_{i\in S}a_i\prod_{i\not\in S}b_i$, as each term in the expansion of this product arises from choosing $a_i$ or $b_i$ in the $i$-th factor; the set $S$ then runs over all possible choices.

Since $\sum_{k=1}^nt(2n,2k)x^{2k-2}=\prod_{i=1}^{n-1}(x^2-i^2)$, we may compute \begin{align*} t(2n,2k) &=[x^{2k-2}]\prod_{i=1}^{n-1}(x^2-i^2)\\ &=[x^{2k-2}]\sum_{S\subset[n-1]}\prod_{i\in S}x^2\prod_{i\in[n-1]\setminus S}(-i^2)\\ &=\sum_{S\in{[n-1]\choose k-1}}\prod_{i\in[n-1]\setminus S}(-i)^2\\ &=(-1)^{n-k}\sum_{S\in{[n-1]\choose k-1}}\prod_{i\in[n-1]\setminus S}i^2. \end{align*} The sum in the last term is positive, and so the claim follows.