Convolution Identity for Stirling Numbers

Suppose we seek to verify that

$${n\brace m} = \sum_{k=m}^n {k\choose m} \sum_{q=0}^k (-1)^{n+q} {n+q-m\brace k} (-1)^{k+q} \left[k\atop q\right] {n\choose n+q-m}$$

which is

$${n\brace m} = (-1)^n \sum_{k=m}^n {k\choose m} (-1)^k \sum_{q=0}^k {n+q-m\brace k} \left[k\atop q\right] {n\choose m-q}$$

where presumably $n\ge m.$ We need for the second binomial coefficient that $m\ge q$ so this is

$${n\brace m} = (-1)^n \sum_{k=m}^n {k\choose m} (-1)^k \sum_{q=0}^m {n+q-m\brace k} \left[k\atop q\right] {n\choose m-q}.$$

Observe that the Stirling number of the second kind vanishes when $k\gt n$ so we may extend the summation to infinity, getting

$${n\brace m} = (-1)^n \sum_{k\ge m} {k\choose m} (-1)^k \sum_{q=0}^m {n+q-m\brace k} \left[k\atop q\right] {n\choose m-q}.$$

Recall that

$$\left[k\atop q\right] = [w^q] k! \times {w+k-1\choose k}.$$

Starting with the inner sum we obtain

$$n! \sum_{q=0}^m \frac{1}{(m-q)!} [z^{n+q-m}] (\exp(z)-1)^k [w^q] {w+k-1\choose k} \\ = n! \sum_{q=0}^m \frac{1}{q!} [z^{n-q}] (\exp(z)-1)^k [w^{m-q}] {w+k-1\choose k}.$$

Now when $q\gt m$ the coefficient extractor in $w$ yields zero, hence we may extend the sum in $q$ to infinity:

$$n! \sum_{q\ge 0} \frac{1}{q!} [z^{n-q}] (\exp(z)-1)^k [w^{m-q}] {w+k-1\choose k}.$$

We thus obtain

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

Preparing the outer sum we obtain

$$\sum_{k\ge m} {k\choose m} (-1)^k (\exp(z)-1)^k {w+k-1\choose k} \\ = \sum_{k\ge m} {k\choose m} (-1)^k (\exp(z)-1)^k [v^k] \frac{1}{(1-v)^w}.$$

Note that for a formal power series $Q(v)$ we have

$$\sum_{k\ge m} {k\choose m} (-1)^{k-m} u^{k-m} [v^k] Q(v) = \frac{1}{m!} \left.(Q(v))^{(m)}\right|_{v=-u}.$$

We get for the derivative in $v$

$$\left(\frac{1}{(1-v)^w}\right)^{(m)} = m! {w+m-1\choose m} \frac{1}{(1-v)^{w+m}}.$$

Substituting $u=\exp(z)-1$ yields

$$m! {w+m-1\choose m} \exp(-(w+m)z).$$

Returning to the double integral we find

$$\frac{(-1)^n\times n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (\exp(z)-1)^m (-1)^m \\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1}} \exp(zw) {w+m-1\choose m} \exp(-(w+m)z) \; dw \; dz \\ = \frac{(-1)^n\times n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (\exp(z)-1)^m (-1)^m \exp(-mz) \\ \times \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{m+1}} {w+m-1\choose m} \; dw \; dz \\ = \frac{(-1)^n\times n!}{2\pi i \times m!} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (\exp(z)-1)^m (-1)^m \exp(-mz) \; dz \\ = \frac{(-1)^n\times n!}{2\pi i \times m!} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1-\exp(-z))^m (-1)^m \; dz \\ = \frac{(-1)^n\times n!}{2\pi i \times m!} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (\exp(-z)-1)^m \; dz.$$

Finally put $z=-v$ to get

$$-\frac{(-1)^n\times n!}{2\pi i \times m!} \int_{|v|=\epsilon} \frac{(-1)^{n+1}}{v^{n+1}} (\exp(v)-1)^m \; dv \\ = \frac{n!}{2\pi i \times m!} \int_{|v|=\epsilon} \frac{1}{v^{n+1}} (\exp(v)-1)^m \; dv.$$

This is

$$n! [v^n] \frac{(\exp(v)-1)^m}{m!} = {n\brace m}$$

and we have the claim.

Note: This is a rather challenging identity. The following is inspired by an in-depth analysis of the answer of @MarkoRiedel which was of great help to overcome some difficulties.

This answer is based upon the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}

Here we use D. Knuth's notational convention and write $\left[n\atop k\right]$ for the unsigned Stirling numbers of the first kind and ${n\brace k}$ for the Stirling numbers of the second kind.

From generating functions we get \begin{align*} \left[k\atop i\right]=[z^k]n!\binom{z+n-1}{n}\qquad\text{and}\qquad {n\brace k}=n![z^n]\frac{\left(e^z-1\right)^k}{k!}\\ \end{align*}

The following is valid for $0\leq m\leq l$

\begin{align*} {l\brace m}=\sum_{k=m}^l\binom{k}{m}\sum_{i=0}^k(-1)^{l+k}{l+i-m\brace k}\left[k\atop i\right]\binom{l}{l+i-m} \end{align*}

Note that here is the sign $(-1)^{l+k}$ whereas in OPs formula the sign is $(-1)^{l+i}$. This is due to $s(k,i)=(-1)^{k+i}\left[k\atop i\right]$.

We obtain

\begin{align*} \sum_{k=m}^l&\binom{k}{m}\sum_{i=0}^k(-1)^{l+k}{l+i-m\brace k}\left[k\atop i\right]\binom{l}{l+i-m}\\ &=\sum_{k=m}^l\binom{k}{m}\sum_{i=0}^m(-1)^{l+k}{l-i\brace k}\left[k\atop m-i\right]\binom{l}{i}\tag{1}\\ &=\sum_{k\geq m}\binom{k}{m}\sum_{i\geq 0}(-1)^{l+k}(l-i)![z^{l-i}]\frac{\left(e^z-1\right)^k}{k!} [u^{m-i}]k!\binom{u+k-1}{k}\binom{l}{i}\tag{2}\\ &=\sum_{k\geq m}\binom{k}{m}\sum_{i\geq 0}(-1)^{l+k}[z^{l-i}]\left(e^z-1\right)^k [u^{m-i}]\binom{-u}{k}(-1)^k\frac{l!}{i!}\tag{3}\\ &=(-1)^ll!\sum_{k\geq m}\binom{k}{m}[z^{l}]\left(e^z-1\right)^k [u^{m}][t^k](1+t)^{-u}\sum_{i\geq 0}\frac{(zu)^i}{i!}\tag{4}\\ &=(-1)^ll![z^{l}][u^{m}]e^{zu}\sum_{k\geq m}\binom{k}{m}\left(e^z-1\right)^k [t^k](1+t)^{-u}\tag{5}\\ &=(-1)^ll![z^{l}][u^{m}]e^{zu}\left(e^z-1\right)^m\frac{1}{m!}D_t^m\left.(1+t)^{-u}\right|_{t=e^z-1}\tag{6}\\ &=(-1)^ll![z^{l}][u^{m}]e^{zu}\left(e^z-1\right)^m(-1)^m\binom{u+m-1}{m}\left.(1+t)^{-(u+m)}\right|_{t=e^z-1}\tag{7}\\ &=(-1)^{l}l![z^{l}]\left(e^z-1\right)^me^{-mz}(-1)^m[u^{m}]\binom{u+m-1}{m}\tag{8}\\ &=(-1)^ll![z^{l}]\left(e^{-z}-1\right)^m\frac{1}{m!}\tag{9}\\ &=l![z^{l}]\frac{\left(e^{z}-1\right)^m}{m!}\tag{10}\\ &={l\brace m} \end{align*} and the claim follows.

Comment:

• In (1) we set the upper limit of the inner sum to $m$ since $k\geq m$ and therefore $\binom{l}{l+i-m}=0$ for $i\geq m$. We also change the order of the inner sum by $i\rightarrow m-i$.

• In (2) we apply the coefficient of operator twice to the Stirling numbers of the first and second kind. We also set the upper limit of both sums to $\infty$ without changing anything since we are adding zeros only.

• In (3) we do some cancellation and use the binomial identity $$\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$$

• In (4) we do some rearrangements and apply the rule \begin{align*} [z^{p-q}]A(z)=[z^p]z^qA(z) \end{align*}

• In (5) we do some rearrangements and expand the exponential series.

• In (6) we recall the substitution rule of the coefficient of operator \begin{align*} A(z)=\sum_{k\geq 0} a_k z^k=\sum_{k\geq 0} z^k [t^k]A(t) \end{align*} Using the differential operator $D_z:=\frac{d}{dz}$ and apply it $m$ times we obtain \begin{align*} D_z^mA(z)&=\sum_{k\geq m} k(k-1)\cdots (k-m+1)a_k z^{k-m}\\ &=m!\sum_{k\geq m}\binom{k}{m}a_kz^{k-m}\\ &=m!\sum_{k\geq m}\binom{k}{m}z^{k-m}[t^k]A(t) \end{align*}

It follows using the substitution $t:= e^z-1$ \begin{align*} \sum_{k\geq m}\binom{k}{m}\left(e^z-1\right)^{k-m}[t^k](1+t)^{-u} =\frac{1}{m!}D_t^m\left.(1+t)^{-u}\right|_{t=e^z-1} \end{align*}

• In (7) we do the differentiation.

• In (8) we substitute and simplify

• In (9) we select the coefficient of $u^m$ from the polynomial $\binom{u+m-1}{m}$ in $u$ and simplify the expression.

• In (10) we substitute $z\rightarrow -z$.