In the paper [1] below, among other things, Carlitz introduced weighted Stirling numbers of the second kind $R(n,k,r)$. He also proved that the numbers $R(n,k,r)$ can be generated by
\begin{equation\*}%\label{S(n,k,x)-dfn}
\frac{(\textrm{e}^z-1)^k}{k!}\textrm{e}^{\lambda z}=\sum\_{n=k}^\infty R(n,k,\lambda)\frac{z^n}{n!}
\end{equation\*}
and can be explicitly expressed by
\begin{equation\*}%\label{S(n,k,x)-satisfy-eq}
R(n,k,r)=\frac1{k!}\sum\_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n
\end{equation\*}
for $r\in\mathbb{R}$ and $n\ge k\ge0$.
Specially, when $\lambda=0$, the quantity $R(n,k,0)$ becomes the Stirling numbers of the second kind $S(n,k)$.


I guess that the identities
\begin{equation}\label{QGWSID1}\tag{SID1}
\sum\_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j-1,j,-\frac{j}2\bigr)}{\binom{2m+j-1}{j}}=0, \quad k,m\in\mathbb{N}
\end{equation}
and
\begin{equation}\label{QGWSID2}\tag{SID2}
\sum\_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}}=0, \quad k>m\ge1
\end{equation}
should be valid.


Stronger but simpler than \eqref{QGWSID1}, the identity
\begin{equation}\label{QGWSID3}\tag{SID3}
R\biggl(2m+j-1,j,-\frac{j}2\biggr)
=\frac{(-1)^j}{j!}\sum\_{\ell=0}^j(-1)^{\ell}\binom{j}{\ell}\biggl(\ell-\frac{j}{2}\biggr)^{2m+j-1}
=0
\end{equation}
for $k,m\in\mathbb{N}$ should be true.


These guesses \eqref{QGWSID1}, \eqref{QGWSID2}, and \eqref{QGWSID3} are related to series expansions at $x=0$ of the functions
$$
\biggl(\frac{\sin x}{x}\biggr)^r
\quad\text{and}\quad
\biggl(\frac{\sinh x}{x}\biggr)^r
$$
for real number $r\in\mathbb{R}$. For details, please read the arXiv preprint [2] below.


Could you please confirm or deny these identities \eqref{QGWSID1}, \eqref{QGWSID2}, and \eqref{QGWSID3} involving the weighted Stirling numbers of the second kind $R(n,k,r)$?


References


1. L. Carlitz, *Weighted Stirling numbers of the first and second kind, I*, Fibonacci Quart. **18** (1980), no. 2, 147--162.
2. Feng Qi, *Series expansions for any real powers of (hyperbolic) sine functions in terms of weighted Stirling numbers of the second kind*, arXiv (2022), available online at <https://arxiv.org/abs/2204.05612> or <https://doi.org/10.48550/arXiv.2204.05612>.



Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind?
$R(n, k, -\tfrac k2)$ is just the central factorial number $T(n, k)$. (Given the definition of the central factorial numbers, it may be more natural to use them in your context than $R$).


Consider [A136630](https://oeis.org/A136630). We have $\operatorname{A136630}(n, k) = 2^{n-k} T(n,k)$ (this may be stated explicitly by Comtet, Riordan, or Charalambides; if not, it will follow from the egf) and the combinatorial interpretation that $\operatorname{A136630}(n, k)$ counts set partitions of $n$ elements into $k$ odd-sized blocks. This immediately gives (SID3), since an even-sized set cannot be partitioned into an odd number of odd-sized blocks, nor an odd-sized set partitioned into an even number of odd-sized blocks.


(SID2) is equivalent to the following theorem: if $1 \le m < k$ then $$\sum\_{j=1}^k (-1)^{k-j} \binom{2m+k}{2m+j} \operatorname{A136630}(2m+j, j) = 0$$


This theorem has a combinatorial proof: consider set partitions of $2m+k$ elements into $k$ odd-sized blocks where blocks of size $3$ or greater are coloured red and singleton blocks can be coloured red or blue. Then the sum counts such set partitions weighted by $(-1)^{\textrm{number of blue partitions}}$. ($j$ is the number of red partitions). Observe that partitions containing at least one singleton can be paired with the partition which differs only in the colour assigned to the singleton with the smallest element, so that the sum counts the number of partitions of $2m+k$ elements into $k$ odd-sized blocks of at least $3$ elements each. But if $k > m$ there are no such partitions, proving the theorem.


For the equivalence to (SID2) we rearrange $$\begin{eqnarray\*}
\textrm{LHS} &=& \sum\_{j=1}^k (-1)^j \binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}} \\
&=& \sum\_{j=1}^k (-1)^j \frac{k!(2m)!}{(k-j)! (2m+j)!} T(2m+j,j) \\
&=& \frac{(-1)^k}{2^{2m} \binom{2m+k}{k}} \sum\_{j=1}^k (-1)^{k-j} \binom{2m+k}{2m+j} \operatorname{A136630}(2m+j, j) \\
\end{eqnarray\*}$$


