A Stirling number identity

Question

Let $s(n,j)$ denote the signed Stirling numbers of the first kind and $S(n,j)$ the Stirling numbers of the second kind. I need the following (probably trivial) identity $\sum\limits_{j = 0}^n {s(n,j)S(m + j,k)} = 0$ for $k < n$ and $\sum\limits_{j = 0}^n {s(n,j)S(m + j,n)} = {n^m},$ but don't see how to prove it-

Answer 1

Consider the following Spivey like identity $${m+j\brace k}=\sum _{\ell =1}^k\sum _{s=0}^m\binom{m}{s}\ell ^s{j\brace \ell}{m-s\brace k-\ell}.$$ This is stablished by splitting the $m$ objects in the ones that share block with an element of the $j$ objects (This is controlled with $s$) or the ones that are not sharing.

$$(-1)^n\sum _{j=0}^n(-1)^j{n\brack j }\left (\sum _{\ell =1}^k\sum _{s=0}^m\binom{m}{s}\ell^s{j\brace \ell}{m-s\brace k-\ell}\right )=\sum _{\ell =1}^k\sum _{s=0}^m\binom{m}{s}\ell^s{m-s\brace k-\ell}(-1)^n\sum _{j=0}^n(-1)^j{n\brack j }{j\brace \ell},$$ but this last sum is $\delta _{n,\ell}$ by the identity of connecting coefficients in between Stirling numbers.

So we conclude that if $k<n$ the sum is $0.$ If $k=n$ then the only term that survives is $m=s,k=\ell=n,$ giving you $n^m.$

Answer 2

We seek to evaluate (note that this is zero by inspection when $k\gt n+m$):

$$\sum_{j=0}^n (-1)^{n+j} {n\brack j} {m+j\brace k}$$

where $k\le n.$ Using standard EGFs this becomes

$$n! [z^n] \sum_{j=0}^n (-1)^{n+j} \frac{1}{j!} \left(\log\frac{1}{1-z}\right)^j (m+j)! [w^{m+j}] \frac{(\exp(w)-1)^k}{k!} \\ = (-1)^n n! m! [z^n] \sum_{j=0}^n (-1)^{j} {m+j\choose j} \left(\log\frac{1}{1-z}\right)^j \\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+j+1}} \frac{(\exp(w)-1)^k}{k!} \; dw \\ = (-1)^n n! m! [z^n] \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!} \\ \times \sum_{j=0}^n (-1)^{j} {m+j\choose j} \left(\log\frac{1}{1-z}\right)^j \frac{1}{w^j} \; dw.$$

Now $\left(\log \frac{1}{1-z}\right)^j = z^j+\cdots$ so the coefficient extractor $[z^n]$ enforces the upper limit of the sum:

$$(-1)^n n! m! [z^n] \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!} \\ \times \sum_{j\ge 0} (-1)^{j} {m+j\choose j} \left(\log\frac{1}{1-z}\right)^j \frac{1}{w^j} \; dw \\ = (-1)^n n! m! \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!} \\ \times \sum_{j\ge 0} (-1)^{j} {m+j\choose j} \left(\log\frac{1}{1-z}\right)^j \frac{1}{w^j} \; dw \; dz \\ = (-1)^n n! m! \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!} \frac{1}{(1+\frac{1}{w}\log\frac{1}{1-z})^{m+1}} \; dw \; dz \\ = (-1)^n n! m! \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(\exp(w)-1)^k}{k!} \frac{1}{(w+\log\frac{1}{1-z})^{m+1}} \; dw \; dz.$$

Now observe that for the geometric series in $j$ to converge we must have $|\log\frac{1}{1-z}| \lt |w|.$ Note that with $\log \frac{1}{1-z} = z + \cdots$ the image of $|z|=\epsilon$ makes one turn around the origin, a circle of radius $\epsilon$ plus additional lower order fluctuations. We therefore choose $\epsilon$ to shrink this pseudo-circle to be entirely contained in $|w|=\gamma.$ With this choice the pole at $-\log\frac{1}{1-z}$ is inside the contour in $w.$ We thus require

$$\frac{1}{k! \times m!} \left(\sum_{q=0}^k {k\choose q} (-1)^{k-q} \exp(qw)\right)^{(m)} = \frac{1}{k! \times m!} \sum_{q=0}^k {k\choose q} (-1)^{k-q} q^m \exp(qw).$$

Evaluating the integral in $w$ we find

$$(-1)^n \frac{n!}{k!} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \sum_{q=0}^k {k\choose q} (-1)^{k-q} q^m (1-z)^q \; dz$$

which is

$$\bbox[5px,border:2px solid #00A000]{ \frac{n!}{k!} \sum_{q=0}^k {k\choose q} {q\choose n} (-1)^{k-q} q^m.}$$

Now when $k\lt n$ we have ${q\choose n}= 0$ so the entire sum vanishes as claimed. We get just one term when $k=n$ namely

$$\frac{n!}{n!} {n\choose n} {n\choose n} (-1)^{n-n} n^m = n^m$$

also as claimed. This concludes the argument.