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I would like to calculate, or bound from above, the following sum $$ \sum_{i=0}^n(n-2i)^p{p \choose i}, $$ here $p\geq 2$.

Any references are very welcome.

Thank you.

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Well, it is not bigger than $(2 n)^p$ – Igor Rivin Apr 13 2012 at 16:35
I would like to get something sharper :-) The idea to compute it: split sum into 'pieces' like $n/2-c\sqrt n\leq i\leq n/2+c \sqrt n$. – David Apr 13 2012 at 16:40
I assume that you mean ${p \choose i}=0$ when $n>p$ correct? – Daniel Parry Apr 13 2012 at 18:58
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 The paper info.tuwien.ac.at/panholzer/Papers/P13.pdf has some results that seem quite similar, so perhaps the methods used there can be adapted to Michael's problem. – Richard Stanley Apr 28 2012 at 21:30
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 But why crosspost under different names? Or am I missing something? – Felix Goldberg Sep 18 at 13:06
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1 Answer

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It is supposed to be a comment, but button "comment'' does not appear on my screen. Please turn this 'answer' into a 'comment'.

Yes, $n$ is fixed and $p$ goes to $\infty$. In the P. Laplace, Théorie analytique des probabilités, Paris, 1812, can be found the formula: $$\sum_{j=0}^{\lfloor n/2\rfloor}(-1)^j{n \choose j}(n-2j)^n=2^{n-1}n!$$

Can we now say something about $\sum_{i=0}^n(2i-n)^p{p \choose i}$?

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7 
 I think the usual protocol here is to edit your own question, not to comment on it. Anyway, if you fix $n>0$ and let $p \rightarrow \infty$ then each factor $p \choose i$ is $O(p^n)$ while $(2i-n)^p$ depends exponentially on $n$, so the dominant terms are those with $i=n$ and $i=0$, giving an asymptotic of $\bigl( {p \choose n} + (-1)^p \bigr) n^p + O(p^{n-1} (n-2)^p)$. – Noam D. Elkies Apr 14 2012 at 4:28
Thank you. But what will happend if both $n$ and $p$ goes to $\infty$, thinking that $n$ goes to $\infty$ faster then $p$ does? – David Apr 18 2012 at 3:29

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