Splitting $r^k (r+n)!$ as a sum of factorials

Solution 1:

We are given that

$$r^k (r+n)! = \sum_{m=0}^k \lambda_m (r+n+m)!$$

and seek to determine the $\lambda_m$ independent of $r.$ We claim and prove that

$$\lambda_m = (-1)^{k+m} \sum_{p=0}^{k-m} {k\choose p} {k+1-p\brace m+1} n^p.$$

With this in mind we re-write the initial condition as

$$r^k = \sum_{m=0}^k \lambda_m m! {r+n+m\choose m}.$$

We evaluate the RHS starting with $\lambda_m$ using the EGF of the Stirling numbers of the second kind which in the present case says that

$${k+1-p\brace m+1} = \frac{(k+1-p)!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2-p}} \frac{(\exp(z)-1)^{m+1}}{(m+1)!} \; dz.$$

We obtain for $\lambda_m$

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

The inner term vanishes when $p\ge k+2$ but in fact even better it also vanishes when $p\gt k-m$ which implies $m+1\gt k+1-p$ because $(\exp(z)-1)^{m+1}$ starts at $[z^{m+1}]$ and we are extracting the term on $[z^{k+1-p}].$

Hence we may extend $p$ to infinity without picking up any extra contributions to get

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

This is

$$(-1)^{k+m} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} \frac{(\exp(z)-1)^{m+1}}{(m+1)!} ((k+1)-nz) \exp(nz) \; dz.$$

Substitute this into the outer sum to get

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

We have

$${r+n+m\choose m} \frac{1}{m+1} = {r+n+m\choose m+1} \frac{1}{r+n}$$

and hence obtain

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

We may extend $m$ to $m\gt k$ in the remaining sum because the term $(\exp(z)-1)^{m+1}$ as before starts at $[z^{m+1}]$ which would then be $\gt k+1$ but we are extracting the coefficient on $[z^{k+1}],$ which makes for a zero contribution.

Continuing we find

$$-\sum_{m\ge 0} {r+n+m\choose r+n-1} (-1)^{m+1} (\exp(z)-1)^{m+1} \\ = 1 - \frac{1}{(1-(1-\exp(z)))^{r+n}} = 1 - \exp(-(r+n)z).$$

We get two pieces on substituting this back into the main integral, the first is

$$\frac{(-1)^{k}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(nz) \; dz \\ = \frac{(-1)^{k}}{r+n} (k+1)! \frac{n^{k+1}}{(k+1)!} - \frac{(-1)^k}{r+n} k! n \frac{n^{k}}{k!} = 0.$$

and the second is

$$\frac{(-1)^{k+1}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(nz) \exp(-(r+n)z) \; dz \\ = \frac{(-1)^{k+1}}{r+n} \frac{k!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+2}} ((k+1)-nz) \exp(-rz) \; dz \\ = \frac{(-1)^{k+1}}{r+n} (k+1)! \frac{(-r)^{k+1}}{(k+1)!} - \frac{(-1)^{k+1}}{r+n} k! n \frac{(-r)^k}{k!} \\ = \frac{1}{r+n} (k+1)! \frac{r^{k+1}}{(k+1)!} + \frac{1}{r+n} k! n \frac{r^k}{k!} \\ = \frac{1}{r+n} r^{k+1} + \frac{1}{r+n} n r^k = r^k.$$

This concludes the argument.

Addendum Nov 27 2016. Markus Scheuer proposes the identity

$$\lambda_m = (-1)^{m+k} \sum_{p=m}^k {p\brace m} {k\choose p} (n+1)^{k-p}.$$

To see that this is the same as what I presented we extract the coefficient on $[n^q]$ to get

$$(-1)^{m+k} \sum_{p=m}^k {p\brace m} {k\choose p} {k-p\choose q}.$$

Now we have

$${k\choose p} {k-p\choose q} = \frac{k!}{p! q! (k-p-q)!} = {k\choose q} {k-q\choose p}.$$

We get

$$(-1)^{m+k} {k\choose q} \sum_{p=m}^k {p\brace m} {k-q\choose p}.$$

We now introduce with deployment of the Egorychev method in mind

$${k-q\choose p} = {k-q\choose k-q-p} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-q-p+1}} (1+z)^{k-q} \; dz.$$

This certainly vanishes when $p\gt k-q$ so we may extend $p$ to infinity, getting for the sum

$$(-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-q+1}} (1+z)^{k-q} \sum_{p\ge m} {p\brace m} z^p \; dz.$$

Using the OGF of the Stirling numbers of the second kind this becomes

$$(-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-q+1}} (1+z)^{k-q} \prod_{l=1}^m \frac{z}{1-lz} \; dz.$$

Now put $z/(1+z) = w$ to get $z = w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$ to obtain

$$(-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q}} \frac{1-w}{w} \frac{1}{(1-w)^2} \prod_{l=1}^m \frac{w/(1-w)}{1-lw/(1-w)} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+1}} \frac{1}{1-w} \prod_{l=1}^m \frac{w}{1-w-lw} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+1}} \frac{1}{1-w} \prod_{l=1}^m \frac{w}{1-(l+1)w} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+2}} \frac{w}{1-w} \prod_{l=2}^{m+1} \frac{w}{1-lw} \; dw \\ = (-1)^{m+k} {k\choose q} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k-q+2}} \prod_{l=1}^{m+1} \frac{w}{1-lw} \; dw \\= (-1)^{m+k} {k\choose q} {k-q+1\brace m+1}.$$

This is the claim and we are done.

Solution 2:

This answer is based upon an analysis of @MarkoRiedel's instructive answer. Although it looks somewhat different it is essentially the same.

We consider OPs identity in the form \begin{align*} r^k=\sum_{m=0}^k\lambda_m \binom{n+r+m}{m}m!\tag{1} \end{align*}

The essence

The binomial identity (1) in terms of corresponding generating functions is the relationship \begin{align*} e^{-rz}=\frac{e^{(n+1)z}}{e^{(n+1+r)z}}\tag{2} \end{align*}

We will see that the generating function of $r^k$ is essentially the LHS of (2) whereas the generating function of the sum of the RHS in (1) corresponds essentially to the coefficients of $z^n$ of the RHS of (2).

The claim

The following is valid \begin{align*} \lambda_m=(-1)^{m+k}\sum_{p=m}^k {p\brace m}\binom{k}{p}(n+1)^{k-p}\qquad\qquad 0\leq m\leq k \end{align*}

In the following we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} \binom{r}{k}=[z^k](1+z)^r\qquad\text{and}\qquad r^k=k![z^k]e^{rz} \end{align*}

We obtain \begin{align*} r^k&=(-1)^kk![z^k]e^{-rz}\\ &=(-1)^kk![z^k]\frac{e^{(n+1)z}}{e^{(n+1+r)z}}\tag{3}\\ &=(-1)^kk![z^k]\frac{e^{(n+1)z}}{\left(1-\left(e^z-1\right)\right)^{n+1+r}}\\ &=(-1)^kk![z^k]e^{(n+1)z}\sum_{m=0}^\infty\binom{-(n+1+r)}{m}(e^z-1)^m\tag{4}\\ &=(-1)^kk![z^k]e^{(n+1)z}\sum_{m=0}^k\binom{n+r+m}{m}(-1)^m(e^z-1)^m\tag{5}\\ &=(-1)^kk![z^k]e^{(n+1)z}\sum_{m=0}^k\binom{n+r+m}{m}(-1)^mm!\sum_{p=m}^\infty{p\brace m} \frac{z^p}{p!}\tag{6}\\ &=(-1)^kk!\sum_{m=0}^k\binom{n+r+m}{m}(-1)^mm!\sum_{p=m}^k{p\brace m} \frac{1}{p!}[z^{k-p}]e^{(n+1)z}\tag{7}\\ &=(-1)^kk!\sum_{m=0}^k\binom{n+r+m}{m}(-1)^mm!\sum_{p=m}^k{p\brace m} \frac{1}{p!}\cdot\frac{(n+1)^{k-p}}{(k-p)!}\tag{8}\\ &=\sum_{m=0}^k\binom{n+r+m}{m}m!\left((-1)^{m+k}\sum_{p=m}^k{p\brace m}\binom{k}{p}(n+1)^{k-p}\right)\tag{9}\\ \end{align*} and the claim follows.

Comment:

• In (3) we make an extension which looks simple but carries some power. In the following line we can rewrite the denominator to prepare for a binomial series expansion and it also contains the series for the Stirling numbers of the second kind.

• In (4) we apply the binomial series expansion.

• In (5) we use the binomial identity \begin{align*} \binom{-p}{q}=\binom{p+q-1}{q}(-1)^q \end{align*} We also set the upper limit of the inner sum to $k$ by noting that the expansion of $(e^z-1)^m$ starts with $z^m$ and since we are looking for the coefficient of $z^k$ we can restrict the limit to $m=k$.

• In (6) we the use the following exponential series representation of the Stirling numbers of the second kind. \begin{align*} \sum_{p=m}^\infty{p\brace m} \frac{z^p}{p!}=\frac{(e^z-1)^m}{m!} \end{align*}

• In (7) we do some rearrangements, apply the coefficient of operator rule \begin{align*} [z^{p-q}]A(z)=[z^{p}]z^qA(z) \end{align*} and restrict the upper limit of the inner sum to $k$ since the exponent $k-p$ has to be non-negative.

• In (8) we select the coefficient of $z^{k-p}$ of $e^{(n+1)z}$.

• In (9) we do some more rearrangements and collect terms to easier see $\lambda_m$.

Solution 3:

The exposition herewith is not going to add anything substantial to the valuable answers already provided by Markus and by Marko, but is intended just to show a couple of other ways to prove the assertion.
Clearly, for $r \in R$, $(r+n)!$ and $(r+n+k)!$ make sense only if represented through the $\Gamma$ function. So, as already shown, we actually are to prove that

$$ \begin{gathered} r^{\,q} = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {\lambda _{\,q,\;j} \left( {r + n + j} \right)^{\,\underline {\,j\,} } } = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {\lambda _{\,q,\;j} \left( {r + n + 1} \right)^{\,\overline {\,j\,} } } = \hfill \\ = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {j!\lambda _{\,q,\;j} \left( \begin{gathered} r + n + j \\ j \\ \end{gathered} \right)} = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {\left( { - 1} \right)^{\,j} j!\lambda _{\,q,\;j} \left( \begin{gathered} - r - n - 1 \\ j \\ \end{gathered} \right)} \quad \quad \left| \begin{gathered} \;r \in \mathbb{R}\,,\quad n \in \mathbb{Z} \hfill \\ \;0 \leqslant q \in \mathbb{Z}\; \hfill \\ \;\lambda _{\,q,\;j} \;\text{indep}\text{.}\,\text{from}\;r \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$

This is just an identity between polynomials, expressed in different basis, so it is assured that a linear combination, with coefficients independent from $r$, exists and is unique.

• 0) the straight one $$ \begin{array}{l} r^{\,q} = \sum\limits_{0\, \le \,j\, \le \,q} {\lambda _{\,q,\;j} \left( {r + n + 1} \right)^{\,\overline {\,j\,} } } \quad \Rightarrow \\ \left( {s - \left( {n + 1} \right)} \right)^{\,q} = \sum\limits_{0\, \le \,j\, \le \,q} {\lambda _{\,q,\;j} s^{\,\overline {\,j\,} } } = \\ = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left\{ \begin{array}{c} q \\ k \\ \end{array} \right\}\left( {s - \left( {n + 1} \right)} \right)^{\,\overline {\,k\,} } } = \\ = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left\{ \begin{array}{c} q \\ k \\ \end{array} \right\}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( \begin{array}{c} k \\ j \\ \end{array} \right)\left( { - \left( {n + 1} \right)} \right)^{\,\overline {\,k - j\,} } s^{\,\overline {\,j\,} } } } = \\ = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left\{ \begin{array}{c} q \\ k \\ \end{array} \right\}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left( { - 1} \right)^{\,k - j} \left( \begin{array}{c} k \\ j \\ \end{array} \right)\left( {n + 1} \right)^{\,\underline {\,k - j\,} } s^{\,\overline {\,j\,} } } } \quad \Rightarrow \\ \end{array} $$

$$ \lambda _{\,q,\;j} = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}\left( \begin{gathered} k \\ j \\ \end{gathered} \right)\left( {n + 1} \right)^{\,\underline {\,k - j\,} } } = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\frac{{k!}} {{j!}}\left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}\left( \begin{gathered} n + 1 \\ k - j \\ \end{gathered} \right)} \tag{0} $$

• 1) : development of $r^q ==> (-r-n-1+n+1)^q$
  The last of the four expressions formulated above suggests that we might try and develop $r^q$ as follows $$ \begin{gathered} r^{\,q} = \left( { - 1} \right)^{\,q} \left( { - r} \right)^{\,q} = \left( { - 1} \right)^{\,q} \left( { - r - n - 1 + n + 1} \right)^{\,q} = \left( { - 1} \right)^{\,q} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered} q \\ k \\ \end{gathered} \right)\left( {n + 1} \right)^{\,q - k} \left( { - r - n - 1} \right)^{\,k} } = \hfill \\ = \left( { - 1} \right)^{\,q} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered} q \\ k \\ \end{gathered} \right)\left( {n + 1} \right)^{\,q - k} \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,k} \right)} {j!\left\{ \begin{gathered} k \\ j \\ \end{gathered} \right\}\left( \begin{gathered} - r - n - 1 \\ j \\ \end{gathered} \right)} } \hfill \\ \end{gathered} $$ which compared with the starting identity gives:

  $$ \lambda _{\,q,\;j} = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered} q \\ k \\ \end{gathered} \right)\left\{ \begin{gathered} k \\ j \\ \end{gathered} \right\}\left( {n + 1} \right)^{\,q - k} } = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( { - 1} \right)^{\,q - k} \left( \begin{gathered} q \\ k \\ \end{gathered} \right)\left( {n + 1} \right)^{\,q - k} \left( { - 1} \right)^{\,k - j} \left\{ \begin{gathered} k \\ j \\ \end{gathered} \right\}} \tag{1} $$

which is the answer anticipated by Markus Scheuer.

• 2) Eulerian Numbers and Worpitzky identity
  The target identity is quite similar to the Worpitzky's Identity, except for having the binomial diagonally shifted. So it is not difficult to convert into the standard Worpitzky form $$ r^{\,q} = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered} q \hfill \\ k \hfill \\ \end{gathered} \right\rangle } \left( \begin{gathered} r + k \\ q \\ \end{gathered} \right) = \; \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered} q \hfill \\ k \hfill \\ \end{gathered} \right\rangle } \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered} - n + k - 1 - j \\ q - j \\ \end{gathered} \right)\left( \begin{gathered} r + n + j \\ j \\ \end{gathered} \right)} $$ where the Eulerian Numbers can be expressed in different ways, among which $$ \begin{gathered} \left\langle \begin{gathered} q \\ m \\ \end{gathered} \right\rangle = \sum\limits_{0\, \leqslant \,k\, \leqslant \,m} {\left( { - 1} \right)^{\,k} \left( \begin{gathered} q + 1 \\ k \\ \end{gathered} \right)\left( {m + 1 - k} \right)^{\,q} } = \hfill \\ = \sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,q - m} \right)\,} {\left( { - 1} \right)^{\,q - m + k} \left( \begin{gathered} q + 1 \\ m + 1 + k \\ \end{gathered} \right)\,k^{\,q} } = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( { - 1} \right)^{\,q - k - m} \left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}\left( \begin{gathered} q - k \\ m \\ \end{gathered} \right)\,\;k!} \hfill \\ \end{gathered} $$ So this leads to express $\lambda _{\,q,\;j} $ as

  $$ \lambda _{\,q,\;j} = \frac{1} {{j!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered} q \hfill \\ k \hfill \\ \end{gathered} \right\rangle } \left( \begin{gathered} - n + k - 1 - j \\ q - j \\ \end{gathered} \right) = \frac{1} {{j!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left\langle \begin{gathered} q \hfill \\ k \hfill \\ \end{gathered} \right\rangle } \left( { - 1} \right)^{\,q - j} \left( \begin{gathered} q + n - k \\ q - j \\ \end{gathered} \right) \tag{2} $$

In conclusion let's note that the identities above could be conveniently put into matricial notation, whereby one can more easily get the properties of $\lambda$. For instance identity (1) can be written as:

$$ \mathbf{\Lambda }(n) = \mathbf{B}^{\, - \,\mathbf{(n + 1)}} \;\mathbf{St}_{\,\mathbf{1}} ^{\, - \,\mathbf{1}} = \;\left( {\mathbf{St}_{\,\mathbf{1}} \,\mathbf{B}^{\,\,\mathbf{(n + 1)}} } \right)^{\, - \,\mathbf{1}} \quad \Rightarrow \quad \Lambda _{\,q,\,m} (n) = \left. {\frac{1} {{q!}}\nabla _{\,x} ^m \,x^{\,q} } \right|_{\,x\, = \, - \,(n + 1)} \tag{3} $$

where all matrices are LT (indexed from $0$), and
$\mathbf{B}$ is the Pascal matrix
$\mathbf{St}_{\,\mathbf{1}}$ is the matrix of the unsigned Stirling N. of 1st kind
$\nabla_{\,x}$ is the backward Delta
so that the last identity indicates the connection with the Newton backward development of $r^q$, that comes of no surprise.
Given the wide range of properties of the Pascal matrix, the formula above allows to transfer them to $\mathbf{\Lambda }(n) \mathbf{St}_{\,\mathbf{1}}$ and $ \mathbf{St}_{\,\mathbf{1}} \mathbf{\Lambda }(n)$.

Finally, it shall be remarked that the exposition made extends quite evenly to $r$ and $n$ complex.

• ---- Addendum -----*

Might be interesting to signalize that, for $n$ non-negative integer, $\lambda$ can be expressed in terms of the so called r-Stirling Numbers ( re. e.g. to this paper by A. Z. Broder ), denoted as $\left\{ \begin{gathered} n \\ m \\ \end{gathered} \right\}_r $.
Accordingly, formula (1) can be rewritten as

$$ \begin{gathered} \lambda _{\,q,\;j} (n) = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,q} \right)} {\left( \begin{gathered} q \\ k \\ \end{gathered} \right)\left\{ \begin{gathered} k \\ j \\ \end{gathered} \right\}\left( {n + 1} \right)^{\,q - k} } \quad \left| {\;\;0 \leqslant n \in \;\;\mathbb{Z}\,} \right.\quad = \hfill \\ = \left( { - 1} \right)^{\,q - j} \left\{ \begin{gathered} q + n + 1 \\ j + n + 1 \\ \end{gathered} \right\}_{n + 1} = \left( { - 1} \right)^{\,q - j} \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n + 1} \right)} {\left( { - 1} \right)^{\,k} \left[ \begin{gathered} n + 1 \\ n + 1 - k \\ \end{gathered} \right]\left\{ \begin{gathered} q + n + 1 - k \\ n + 1 + j \\ \end{gathered} \right\}} \hfill \\ \end{gathered} \tag{a} $$

thereby getting an additional expression in terms of standard Stirling Numbers. ($\left[ \begin{gathered} n \\ m \\ \end{gathered} \right]$ being the unsigned Stirling N. of 1st kind).