A Vandermonde Identity for Stirling Numbers?

I'm facing the problem of trying to express a quantity in the simplest possible way (it is, using the least possible number of sum symbols).

$$ \sum_{j=0}^{n} \sum_{\ell=0}^m \frac{1}{j!}\binom{b+j}{j} {j+1 \brack {\ell+1}} {b+2 \brack {m-\ell+1}}$$

Of course, this can be easily written as a convolution between two polynomials (which happen to be more or less simple). I'm pretty sure that approach will not work (at most, one can write the above expression as "the coefficient of $x^m$ in this product [...]", but that is not useful to my purpose).

However, if one explores this sum a little bit, it pretty soon come up the fact that it could be truly useful to, for example, be able to compute this: $$\sum_{\ell=0}^m {j+1\brack{\ell+1}}{b+2 \brack {m-\ell+1}}$$ (which resembles a lot Vandermonde's Identity, but with Stirling numbers instead of binomial coefficients).

I looked up on a couple of books (Concrete Mathematics of Graham-Knuth-Patashnik, and others), and I couldn't find any references pointing to such an identity. Does anybody know something like that? (Perhaps involving other weird numbers as Eulerian or double Eulerian or that kind of stuff?)

Nevertheless, any kind of help simplifying the first double sum would be really appreciated.

We show the following identity is valid for non-negative integers $a,b,M$: \begin{align*} \sum_{k=0}^M {a\brack k}{b\brack M-k}=\sum_{n=0}^{\min{\{a,b\}}}{a+b-n\brack M}\frac{(-1)^na!b!}{n!(a-n)!(b-n)!}\tag{1} \end{align*}

The right-hand sum is regrettably not a simplification. But we have at least one factor as Stirling number of the first kind instead of two (at the cost of some other factors) and we have an upper index containing $a+b$ which comes somewhat close to Vandermonde's identity \begin{align*} \sum_{k=0}^m\binom{a}{k}\binom{b}{M-k}=\binom{a+b}{M} \end{align*}

We show (1) by recalling the generating function: \begin{align*} (1-z)^{-u}=\sum_{n=0}^\infty\sum_{k=0}^n{n\brack k}u^k\frac{z^n}{n!}\tag{2} \end{align*}

At first we derive the left-hand side of (1). We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ and obtain from (2) \begin{align*} \color{blue}{a!b!}&\color{blue}{[u^Mz^at^b](1-z)^{-u}(1-t)^{-u}}\tag{3}\\ &=a!b![u^M]\left([z^a]\sum_{n=0}^\infty\sum_{k=0}^n{n\brack k}u^k\frac{z^n}{n!}\right)\left([t^b]\sum_{n=0}^\infty\sum_{k=0}^n{n\brack k}u^k\frac{t^n}{n!}\right)\\ &=a!b![u^M]\left(\sum_{k=0}^a{a\brack k}u^k\frac{1}{a!}\right)\left(\sum_{l=0}^b{b\brack l}u^l\frac{1}{b!}\right)\\ &=[u^M]\sum_{q=0}^{a+b}\sum_{{k+l=q}\atop{k,l\geq 0}}{a\brack k}{b\brack l}u^q\\ &=[u^M]\sum_{q=0}^{a+b}\sum_{k=0}^q{a\brack k}{b\brack q-k}u^q\\ &\,\,\color{blue}{=\sum_{k=0}^M{a\brack k}{b\brack M-k}}\tag{4} \end{align*}

We take (3) and calculate the coefficient in a somewhat different way.

We obtain \begin{align*} \color{blue}{a!b!}&\color{blue}{[u^Mz^at^b]((1-z)(1-t))^{-u}}\\ &=a!b+t))^{-u}\\ &=a!b![u^Mz^at^b]\left(\sum_{n=0}^\infty\sum_{k=0}^n{n\brack k}u^k\frac{(z(1-t)+t)^n}{n!}\right)\tag{5}\\ &=a!b![z^at^b]\sum_{n=0}^\infty{n\brack M}\frac{(z(1-t)+t)^n}{n!}\tag{6}\\ &=a!b![z^at^b]\sum_{n=0}^\infty{n\brack M}\frac{1}{n!}\sum_{j=0}^n\binom{n}{j}z^j(1-t)^jt^{n-j}\tag{7}\\ &=a!b![t^b]\sum_{n=0}^\infty{n\brack M}\frac{1}{n!}\binom{n}{a}(1-t)^at^{n-a}\tag{8}\\ &=a!b!\sum_{n=0}^{\min\{a,b\}}{n\brack M}\frac{1}{n!}\binom{n}{a}\binom{a}{a+b-n}(-1)^{a+b-n}\tag{9}\\ &=a!b!\sum_{n=0}^{\min\{a,b\}}{n\brack M}\frac{(-1)^{a+b-n}}{(n-a)!(n-b)!(a+b-n)!}\tag{10}\\ &\,\,\color{blue}{=a!b!\sum_{n=0}^{\min{\{a,b\}}}{a+b-n\brack M}\frac{(-1)^n}{n!(a-n)!(b-n)!}}\tag{11} \end{align*} and the claim (1) follows.

Comment:

• In (5) we expand the series according to (2).

• In (6) we select the coefficient of $u^M$.

• In (7) we expand the binomial.

• In (8) we select the coefficient of $z^a$.

• In (9) we select the coefficient of $t^b$.

• In (10) we do some cancellations.

• In (11) we change the order of summation $n\to a+b-n$.