Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind?

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 $$$$\label{QGWSID1}\tag{SID1} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j-1,j,-\frac{j}2\bigr)}{\binom{2m+j-1}{j}}=0, \quad k,m\in\mathbb{N}$$$$ and $$$$\label{QGWSID2}\tag{SID2} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}}=0, \quad k>m\ge1$$$$ should be valid.

Stronger but simpler than \eqref{QGWSID1}, the identity $$$$\label{QGWSID3}\tag{SID3} R\biggl(2m+j-1,j,-\frac{j}2\biggr) =\frac{(-1)^j}{j!}\sum_{\ell=0}^j(-1)^{\ell}\binom{j}{\ell}\biggl(\ell-\frac{j}{2}\biggr)^{2m+j-1} =0$$$$ for $$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 paper [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 and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Applicable Analysis and Discrete Mathematics 18 (2024), no. 1, 92–115; available online at https://doi.org/10.2298/AADM230902020Q.
• you will want to add some motivation (you "guessed" this ?), to promote a helpful response. Commented Apr 11, 2022 at 8:11
• Yes, I guessed them without explicit and short motivation. The motivation is in my brain. Commented Apr 11, 2022 at 10:47
• Using the software Mathematica, I numerically verified these two identities, so it is better to prove their validity mathematically or combinatorially. Commented Apr 11, 2022 at 10:53
• @CarloBeenakker The motivation you required is at arxiv.org/abs/2204.05612. For this, I edited my question again. Commented Apr 13, 2022 at 6:01
• @YemonChoi -- indeed, I will delete the mistaken comment. Commented Apr 16, 2022 at 6:48

$$R(n, k, -\tfrac k2)$$ 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. 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 $$\begin{eqnarray*} \textrm{LHS} &=& \sum_{j=1}^k (-1)^j \binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}} \\ &=& \sum_{j=1}^k (-1)^j \frac{k!(2m)!}{(k-j)! (2m+j)!} T(2m+j,j) \\ &=& \frac{(-1)^k}{2^{2m} \binom{2m+k}{k}} \sum_{j=1}^k (-1)^{k-j} \binom{2m+k}{2m+j} \operatorname{A136630}(2m+j, j) \\ \end{eqnarray*}$$

• Thank Peter Taylor for your proof of the identity (SID3), and then (SID1), in my question. Could you please give another reference to the relation $R(n, k, r)=h_{n-k}(r, r+1, r+2, \dotsc, r+k)$ in your answer? Commented Apr 14, 2022 at 9:11
• @qifeng618, only partially. If we have a triangle $T(n,k) = h_{n-k}(s_1, s_2, \ldots, s_{k+1})$ where $h_0(\vec{s}) = 1$ then it's easy to derive $T(n+1,k) = T(n,k-1) + s_{k+1} T(n,k)$ and conversely to show that this recurrence with base case $T(0,k) = [k=0]$ gives $h_{n-k}(\cdots)$. You may be able to find a reference for this in a textbook: I haven't looked. Then Carlitz, Weighted Stirling numbers of the first and second kind, II, Fibonacci Quart. 18 (1980), no. 3, 242-257, eqn (1.4) as corrected reads $R(n+1,k,\lambda) = R(n,k-1,\lambda)+(k+\lambda)R(n,k,\lambda)$. Commented Apr 14, 2022 at 12:46
• @qifeng618, I've found a reference. Andrei Z. Broder, The r-Stirling numbers, Discrete Math. 49, 241-259 (1984). The correspondence between Broder's notation and Carlitz's is ${n \brace k}_r = R(n-r, k-r, r)$. Unless I've overlooked it, none of the identities in Broder's paper is immediately equivalent to SID2. Commented Apr 16, 2022 at 15:55
• Thank Peter Taylor for your hard endeavor to find a reference. You are a real mathematician. How can I contact with you via e-mail? My e-mail address is qifeng618@...... Commented Apr 17, 2022 at 3:51
• I will throw out for anyone who's interested that I conjecture a stronger result than the theorem proved, namely replacing the binomial coefficient $\binom{2m+k}{2m+j}$ with $\binom{3m+1}{k-j}$. This would allow proving by induction that the same holds for $\binom{n}{k-j}$ if $3m+1 \le n \le 2m+k$. It may have a similar combinatorial proof, but I think it's likely to be somewhat less elegant. Commented Apr 25, 2022 at 22:42