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: $$ {\hat B}_{m,j}(x_1,x_2,...,x_{m-j+1})=\sum_{{k_0+k_1+\cdots+k_N=j}\atop{k_1+2k_2+\cdots+Nk_N=m}}\binom{j}{k_0,k_1,\ldots,k_N} \prod_{i=1}^Nx_i^{k_i} $$

  • $\begingroup$ The incomplete Bell polynomials $B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$ become Stirling numbers of the second kind $S(n,k)$ if we put all $x_i=1$. Therefore the Bell numbers come from the complete Bell polynomials by putting all $x_i=1$. The shift formula for Bell numbers is thus more directly related to the formula $$\sum_{i=0}^{n}\binom{n}{i}B_{n-i}(x_1,x_2,\ldots,x_{n-i})x_{i+1}=B_{n+1}(x_1,x_2,\ldots,x_{n+1})$$ for complete Bell polynomials, see the wiki link for all the details. $\endgroup$ – N. Shales Apr 24 '17 at 13:29

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.

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.

  • $\begingroup$ ok, I think I start to see what you mean. I also checked the wiki page again and found $\Phi (t,u)=\exp \left(u\sum _{r=1}^{\infty }x_{r}{\frac {t^{r}}{r!}}\right)=\sum _{n,k\geq 0}B_{n,k}(x_{1},\ldots ,x_{n-k+1}){\frac {t^{n}}{n!}}u^{k}$. Putting your suggestions this becomes $\Phi (t,u)e^{-t}=\sum _{n,k\geq 0}\left(\sum_{j=0}^k(-1)^{j-1}\binom{k}{j}B_{n,j}(x_1,x_2,\ldots)\right)u^{k}{\frac {t^{n}}{n!}}$, where I'm not sure if I've chosen the correct variable $t$ or $u$... $\endgroup$ – draks ... Apr 26 '17 at 13:46
  • $\begingroup$ You mean using this: $\frac{1}{1-z}A\left(\frac{z}{1-z}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^n\binom{n}{j}a_j\right)z^n$? But where do the $x_k$'s come from then? And $m$ looks like a free parameter so it should be $A_m(z)$ then right? $\endgroup$ – draks ... Apr 27 '17 at 10:28
  • $\begingroup$ Thanks a lot so far. Cheers, $\endgroup$ – draks ... Apr 27 '17 at 14:35
  • $\begingroup$ My comment on the summation was crap. I deleted it. You also deleted one. May I ask why? $\endgroup$ – draks ... Apr 29 '17 at 18:59
  • 1
    $\begingroup$ @MarkusScheuer, nope. Just made the comment so others would not be confused and become aware of a difference. Both forms are rather important. $\endgroup$ – Tom Copeland May 1 '17 at 19:48

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