A closed form for the sum $\sum_{i=0}^j (-1)^i \binom{j}{i} (j+c-i)^k$ Let $c\in \mathbb{Q}$ be a constant, and let $k\in \mathbb{N}$ be fixed.
Is there a closed form for the sum
$$ L_j := \sum_{i=0}^j (-1)^i \binom{j}{i} (j+c-i)^k, $$
for $j \leq k$ ?
 A: Here is an attempt.  Note that
$$L_j=\sum_{i=0}^j(-1)^i\binom{j}{i}\sum_{r=0}^k\binom{k}{r}c^{k-r}(j-i)^{r}.$$
Therefore,
$$L_j=\sum_{r=0}^kc^{k-r}\binom{k}{r}\sum_{i=0}^j(-1)^i\binom{j}{i}(j-i)^r.$$
Let $s=j-i$.  Then,
$$L_j=\sum_{r=0}^kc^{k-r}\binom{k}{r}\sum_{s=0}^j(-1)^{j-s}\binom{j}{s}s^r.$$
Note that $\displaystyle\frac{1}{j!}\sum_{s=0}^j(-1)^{j-s}\binom{j}{s}s^r$ is the Stirling number $\begin{Bmatrix}r\\j\end{Bmatrix}$.  So, 
$$L_j=j!\sum_{r=0}^k\binom{k}r\begin{Bmatrix}r\\j\end{Bmatrix}c^{k-r}=j!\sum_{r=j}^k\binom{k}r\begin{Bmatrix}r\\j\end{Bmatrix}c^{k-r}.$$
One particular result is when $j=k$, where we have $L_k=k!$.  Also, $L_j=0$ if $j>k$.  For $j\leq k$, $L_j$ is a polynomial in $c$ with non-negative integer coefficients of degree $k-j$.  Some other trivial answers are $L_0=c^k$ (for $k\geq 0$, and $L_1=(c+1)^k-c^k$ (for $k\geq 1$).
A: You can assume that there is no "closed" form.
With a Generating function:
$\displaystyle L_{k,j}(x):=\frac{d^k}{dz^k}e^{xz}\left(e^z-1\right)^j |_{z=0}=\sum\limits_{i=0}^j (-1)^{j-i}{\binom j i}(i+x)^k = {\Delta}^j x^k \enspace$ 
$\hspace{7cm}$ ($j^{\,\text{th}}$ forward difference of $x^k$)
$\Rightarrow \enspace L_j = L_{k,j}(c)$
A: A closed form seems to be out of reach for general $c\in\mathbb{Q}$. We derive an expression which makes this plausible. We use the coefficient of operator $[z^k]$ to denote the coeffcient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
n^k=k![z^k]e^{nz}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^j}&\color{blue}{(-1)^i\binom{j}{i}(j+c-i)^k}\\
&=\sum_{i=0}^j(-1)^{i+j}\binom{j}{i}(c+i)^k\tag{2}\\
&=\sum_{i=0}^j(-1)^{i+j}\binom{j}{i}k![z^k]e^{(c+i)z}\tag{3}\\
&=(-1)^jk![z^k]e^{cz}\sum_{i=0}^j\binom{j}{i}(-e^z)^i\tag{4}\\
&=(-1)^jk![z^k]e^{cz}\left(1-e^z\right)^j\tag{5}\\
&\color{blue}{=k![z^k]e^{cz}\left(e^z-1\right)^j}\tag{6}
\end{align*}
  We conclude from the representation (6) that the expression $e^{cz}\left(e^z-1\right)^j$ cannot be simplified for general $c\in\mathbb{Q}$. Taking the coefficient $[z^k]$ from it will typically lead to more than one term, because we have to do a Cauchy multiplication of the corresponding series. So, we won't get a closed form in general.

Comment:


*

*In (2) we exchange the order of summation $i \to j-i$.

*In (3) we apply the coefficient of operator as we did in (1).

*In (4) we do some rearrangements as preparation for the next step.

*In (5) we apply the binomial theorem.

*In (6) we do some final simplifications.
