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Original question is a duplicate of this question. Please refer to after the edit

Original question:

I was investigating formulas of the form $\sum_{k=0}^n k^a \binom{n}{k}$ after noticing on Wikipedia that $\sum_{k=0}^n k \binom{n}{k} = n2^{n-1}$ and $\sum_{k=0}^{n} k^2 \binom{n}{k} = (n+n^2)2^{n-2}$.

Wolfram Alpha notes that for $a=3$, $\sum_{k=0}^n k^3 \binom{n}{k} = n^2 (n + 3)2^{n-3}$ but for $a=4,5$ then switches to $\sum_{k=0}^n k^4 \binom{n}{k} = n\ _4 F_3(2, 2, 2, 1 - n;1, 1, 1;-1)$ and $\sum_{k=0}^n k^5 \binom{n}{k} = n\ _5 F_4(2, 2, 2, 2, 1 - n;1, 1, 1, 1;-1)$.

My conjecture is that, in general, $$\sum_{k=0}^n k^a \binom{n}{k} = n\ _a F_{a-1}(2, 2, ..., 2, 1 - n;1, 1, ..., 1; -1)$$ Which is some kind of polynomial since one of the numerators there is non-positive (and I suspect Wolfram Alpha just spat this function out so as to avoid calculating the polynomial outright), but I don't know how to use this function for computing with a given $n$ and $a$ (though I begin to suspect there isn't an easy way to do so).

I'm also wondering if there's a formula at all for $\sum_{k=0}^n k^n \binom{n}{k}$.

Any suggestions or advice welcome.

EDIT

Since my original question already has an answer, I'd like to pivot to asking how the resulting identity works. $$\begin{align} \newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} \sum_{k=0}^n k^a \binom{n}{k} &= n\ _a F_{a-1}(2, 2, ..., 2, 1 - n;1, 1, ..., 1; -1)\\ &=2^{n-m}\sum_{j=0}^m\binom{n}{j}2^{m-j}\stirtwo{m}{j}j!\\ \end{align}$$

Sherlock9
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    Check these: https://math.stackexchange.com/q/1400588/42969, https://math.stackexchange.com/q/923115/42969, https://math.stackexchange.com/q/355262/42969 – all found with Approach0 – Martin R Sep 20 '23 at 08:06
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    If $p_a(x)=x(x-1)\cdots(x-a+1),$ the "falling factorial," then $\sum_{k=0}^n p_a(k)\binom nk=p_a(n)2^{n-a}.$ So you really to to write $x^a$ in the basis $p_0,p_1,\dots,p_a$ to find your result. So $x^3=x(x-1)(x-2)+3x(x-1)+x$ gives $$\sum k^3\binom nk=(n(n-1)(n-2)+6n(n-1)+4n)2^{n-3}$$ – Thomas Andrews Sep 20 '23 at 08:54
  • @QiaochuYuan I think that answers the main question, though I am still curious on how the hypergeometric function relates to the Stirling numbers of the second kind answer – Sherlock9 Sep 20 '23 at 09:12
  • @Sherlock9: Your conjecture is sound. I just wanted to add a corresponding answer, but regrettably this post has been closed. – Markus Scheuer Sep 20 '23 at 18:58
  • @epi163sqrt Sorry, I didn't realize "private feedback" meant closing the question as a duplicate. Just edited the question and I hope to have it reopened soon. – Sherlock9 Sep 20 '23 at 19:02
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    Please do not change the question just to avoid closure. If you have a new question, ask it as one. – Nij Sep 20 '23 at 19:51
  • @Nij Sorry. Will do. – Sherlock9 Sep 20 '23 at 22:01

1 Answers1

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$$S_n=\sum_{k=0}^n k^n \binom{n}{k}$$ is sequence $A072034$ in $OEIS$

Vaclav Kotesovec proposed a superb approximation $$S_n \sim \frac{1}{\sqrt{1+W\left(\frac{1}{e}\right)}}\left(\frac{n}{e\, W\left(\frac{1}{e}\right)}\right)^n$$ where $W(.)$ is Lambert function.

Claude Leibovici
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