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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.

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2 Answers 2

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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 $$-\frac{{\binom{n -1-k}{k}} \left(k^{2}+n \right)}{k +1}+{\binom{n}{k +1}} $$

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  • $\begingroup$ What about if k was fixed and a was free variable? $\endgroup$ Commented Jun 15, 2021 at 3:28
  • $\begingroup$ 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. $\endgroup$
    – Ira Gessel
    Commented Jun 15, 2021 at 19:15
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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 $$\binom{n-1}{n-k} + \sum_{j=0}^a \left\langle a\atop j\right\rangle \binom{n+a-1-j}{k+a}.$$

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