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

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.

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 $$-\frac{{\binom{n -1-k}{k}} \left(k^{2}+n \right)}{k +1}+{\binom{n}{k +1}}$$
• If you multiply the first sum by $\frac{x^a}{a!}y^nz^k$ and sum on $a$, $n$, and $k$ (starting the sum with $i=0$ to make things a little simpler), you get $$\frac{1-y}{(1-y-yz)(1-ye^x)}$$ from which you can derive whatever explicit formulas exist. When you expand in powers of $x$ you will get Stirling numbers of the second kind. Commented Jun 15, 2021 at 19:15
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 $$\binom{n-1}{n-k} + \sum_{j=0}^a \left\langle a\atop j\right\rangle \binom{n+a-1-j}{k+a}.$$