Closed form of $ \sum_{i=1}^{n-k} {n-1-i\choose k-1}i^a + \sum_{i=1}^k {n-1-i\choose n-1-k}$

Question

For all natural numbers $a$, is there a known closed form of $ \sum_{i=1}^{n-k} {n-1-i\choose k-1}i^a + \sum_{i=1}^k {n-1-i\choose n-1-k}$, where $k$ is fixed?

For example, letting $k=1$ gives the basic powersum of which the closed form is Faulhaber's formula for $a$. Is there a Faulhaber-like formula for all the rest of the possible fixed $k$ values?

I'm asking this because this corresponds to the values in Pascal triangles where the left diagonal is the sequence $m^a$ and the right diagonal is 1.

Answer 1

The second sum is equal to $\binom{n-1}{n-k}$. For each fixed $a$ the first sum can be evaluated but I don't think there's a nice general formula. For example, for $a=1$, Maple gives $$-\frac{{\binom{n -1-k}{k}} \left(k^{2}+n \right)}{k +1}+{\binom{n}{k +1}} $$

Answer 2

If we assume that $a$ is fixed, then the first sum can be rewritten in a closed form with $a+1$ terms: $$\sum_{j=0}^a \left\langle a\atop j\right\rangle \binom{n+a-1-j}{k+a},$$ where $\left\langle a\atop j\right\rangle$ are Eulerian numbers. Correspondingly, the whole expression simplifies to $$\binom{n-1}{n-k} + \sum_{j=0}^a \left\langle a\atop j\right\rangle \binom{n+a-1-j}{k+a}.$$