About the Binomial Transform of Incomplete Bell Polynomials $\displaystyle \sum_{j=1}^k(-1)^{j-1}\binom{k}{j} B_{m,j}$?

Question

The binomial transform is the shift operator for the Bell numbers. That is, $$ \sum_{j=0}^k {k\choose j} B_j =B_{k+1} $$ where the $B_n$ are the Bell numbers.

Is there a somewhat similar expressions involving incomplete Bell polynomials: $$ \sum_{j=1}^k(-1)^{j-1}\binom{k}{j} {\hat B}_{m,j}(x_1,x_2,...,x_{m-j+1})=\;? $$

Maybe the definition of incomplete Bell polynomials helps to find an answer: $$ {\hat B}_{m,j}(x_1,x_2,...,x_{m-j+1})=\sum_{{k_0+k_1+\cdots+k_N=j}\atop{k_1+2k_2+\cdots+Nk_N=m}}\binom{j}{k_0,k_1,\ldots,k_N} \prod_{i=1}^Nx_i^{k_i} $$

Answer 1

Hint: We could look at the situation with the help of exponential generating functions.

The following holds: If $$A(z)=\sum_{k=0}^\infty a_k \frac{z^k}{k!}$$ is an exponential generating function of the sequence $(a_k)$, then \begin{align*} A(z)e^z=\sum_{k=0}^\infty \left(\sum_{j=0}^k\binom{k}{j} a_j \right)\frac{z^k}{k!} \end{align*} and applying the differential operator $D_z=\frac{d}{dz}$ we obtain \begin{align*} D_z A(z)=\sum_{k=0}^\infty a_{k+1}\frac{z^k}{k!} \end{align*}

Since the exponential generating function of the Bell numbers is \begin{align*} B(z)=\exp\left({e^z-1}\right)=\sum_{n=0}^\infty B_n\frac{z^n}{n!} \end{align*} the relation \begin{align*} \ \qquad\qquad\sum_{j=0}^k\binom{k}{j}B_j=B_{k+1}\qquad\qquad k\geq 0 \end{align*} translates into \begin{align*} B(z)e^z&=D_z B(z)\\ \end{align*}

We could now try to find an exponential generating function $P(z)$ for the partial ordinary Bell polynomials. The expression \begin{align*} \sum_{j=0}^k(-1)^{j-1}\binom{k}{j}\hat{B}_{m,j}(x_1,x_2,\ldots,m-j+1) \end{align*} looks like the coefficient of \begin{align*} (-1)^{k-1}P(z)e^{-z} \end{align*} We could now check if this series could also be represented using operators like $D_z, e^z$, etc. Here we use the notation $\hat{B}_{m,j}$ in accordance with the Wiki page.

[Add-on 2017-04-30] Some information regarding ordinary generating functions of partial ordinary Bell polynomials. We use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ of a series $A(t)$.

According to the referred Wiki page are the partial ordinary Bell polynomials \begin{align*} \hat{B}_{m,j}(x_1,x_2,\ldots,x_{m-j+1}) \end{align*}

the coefficients of a bivariate generating function $\hat{\Phi}(t,u)$ with

\begin{align*} \hat{\Phi}(t,u)&=\exp\left(u\sum_{l=1}^\infty x_lt^l\right)\\ &=\sum_{k=0}^\infty\left(\sum_{l=1}^\infty x_lt^l\right)^k\frac{u^k}{k!}\\ &=\sum_{k=0}^\infty\left(\sum_{n=k}^\infty\hat{B}_{n,k}\left(x_1,x_2,\ldots,x_{n-k+1}\right)t^n\right)\frac{u^k}{k!} \end{align*}

We can write \begin{align*} \hat{B}_{m,j}(x_1,x_2,\ldots,x_{m-j+1})&=[t^mu^j]\hat{\Phi}(t,u)\\ &=[t^m]\underbrace{\left(\sum_{l=1}^\infty x_lt^l\right)^j}_{:= \hat{B}_j(t)}\tag{1} \end{align*}

We recall the Euler transformation formula of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n\tag{2} \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

We conclude according to (1) and (2) \begin{align*} \sum_{j=0}^m(-1)^{j}\binom{m}{j}\hat{B}_{m,j}(x_1,x_2,\ldots,x_{m-j+1}) &=[t^mu^j]\frac{1}{1-t}\hat{\Phi}\left(\frac{-t}{1-t},u\right)\\ &=[t^m]\frac{1}{1-t}\hat{B}_j\left(\frac{-t}{1-t}\right)\tag{3} \end{align*}

Note: Regrettably there seems to be no simple representation of the coefficients in (3) based on the ordinary generating function. An exponential generating function \begin{align*} \sum_{n=k}^\infty\hat{B}_{n,k}\left(x_1,x_2,\ldots,x_{n-k+1}\right)\frac{t^n}{n!} \end{align*} of the partial ordinary Bell polynomials is not stated at the Wiki-page about Bell polynomials and useful represenations are not known to me.

Answer 2

Let \begin{equation*}%\label{Fall-Factorial-Dfn-Eq} \langle\alpha\rangle_n= \prod_{k=0}^{n-1}(\alpha-k)= \begin{cases} \alpha(\alpha-1)\dotsm(\alpha-n+1), & n\ge1\\ 1,& n=0 \end{cases} \end{equation*} and \begin{equation*} (\alpha)_n=\prod_{\ell=0}^{n-1}(\alpha+\ell) = \begin{cases} \alpha(\alpha+1)\dotsm(\alpha+n-1), & n\ge1\\ 1, & n=0 \end{cases} \end{equation*} denote the falling and rising factorials of $\alpha\in\mathbb{C}$ respectively.

In Theorem 2.1 of the paper [1] below, the formula \begin{equation}\label{Bell-fall-Eq-inv}\tag{3} \sum_{\ell=0}^{k}\frac{B_{n,\ell}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-\ell+1})}{(k-\ell)!} =\frac{\langle\alpha k\rangle_n}{k!} \end{equation} for $n\ge k\ge0$ and $\alpha\in\mathbb{R}$ was established.

As a consequence of \eqref{Bell-fall-Eq-inv}, the formula \begin{equation}\label{Bell-rise-Eq-inv}\tag{5} \sum_{\ell=0}^{k}\frac{B_{n,\ell}((\alpha)_1, (\alpha)_2, \dotsc, (\alpha)_{n-\ell+1})}{(k-\ell)!} =\frac{(\alpha k)_n}{k!} \end{equation} for $\alpha\in\mathbb{C}$ and $n\ge k\ge0$ was derived in Corollary 2.1 of the paper [1] below.

References

1. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Bai-Ni Guo, Closed formulas and identities for the Bell polynomials and falling factorials, Contributions to Discrete Mathematics 15 (2020), no. 1, 163--174; available online at https://doi.org/10.11575/cdm.v15i1.68111.
2. https://mathoverflow.net/a/405003/147732
3. https://math.stackexchange.com/a/4262657/945479
4. https://math.stackexchange.com/a/4261764/945479