Is there a closed form for $$ \sum_{k=1}^n (-1)^k\frac{k^p}{k!(n-k)!},\quad n=0,1,2\ldots,\,p=0,1,2\ldots. $$ I tried to identify the sum with Stirling numbers...
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I notice that the expression equals $$\frac{1}{n!}\sum_{k=0}^n (-1)^k \binom{n}{k} k^p,$$ which looks a lot like ${p \brace n}$, except it has $k^p$ instead of $(n-k)^p$. – A.E Sep 10 '14 at 16:58
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@A.E Do I have to expand $(n-k)^p$? Thanks. – Randy Sep 10 '14 at 17:01
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1Take $k\to n-k$, profit. – Pedro Sep 10 '14 at 17:06
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@A.E What is $n\brace p$. Maybe a binomial coefficient variant? – Тyma Gaidash Dec 18 '22 at 02:04
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@TymaGaidash see https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind – A.E Mar 13 '23 at 02:06
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From my comment and Pedro Tamaroff's suggestion, let $k \leftarrow n - k$.
Then we get $$\dfrac{1}{n!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{n-k} (n-k)^p = \dfrac{1}{n!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (n-k)^p = (-1)^n{ p \brace n} $$
This result only holds if $p > 0$. If $p = 0$, $$\dfrac{1}{n!} \sum_{k=1}^n (-1)^{k} \binom{n}{k} = -\dfrac{1}{n!} $$
A.E
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