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Let $n \in N$ and $q\geq 2$.

I am trying to calculate the following sum: $$ \sum_{i=0}^{\sqrt n/2}\sum_{j= i \sqrt n }^{(i+1)\sqrt n}\frac{(-1)^q2^q(\frac{n}{2}-j)^q}{(n-j)!j!} $$

Any help will be appreciated.

Thank you.

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    Is there a question?2012-04-17
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    How are you interpreting your bounds on the sums when $\frac{n}{2}$ and $\sqrt{n}$ are not whole numbers?2012-04-17
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    j can go to (n^3/2)/2, so (n-j)! will at some points refer to factorials of negative numbers. Expected?2012-04-17
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    Yes. We can 'stop' the sum with negative factorials. Can think that they are 0.2012-04-17
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    @Wonder: If we reexpress OP's sum as $$\frac{(-2)^q}{n!} \sum_{i=0}^{n/2} \left(\frac{n}{2}-i\right)^q \sum_{j=i\sqrt{n}}^{(i+1)\sqrt{n}} \binom{n}{j}$$ and take $\dbinom{n}{k}=0$ if $n < k$, it all works out.2012-04-17
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    I am very sorry, I had typo. The right sum is $\sum_{i=0}^{\sqrt n/2}\sum_{j= i \sqrt n }^{(i+1)\sqrt n}\frac{(-1)^q2^q(\frac{n}{2}-j)^q}{(n-j)!j!} $2012-04-17
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    Shouldn't you be using the floor or ceiling function in the limits? As brc is pointing out, how would you summ up to $\sqrt 3/2$?2012-04-17
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    If $n$ is NOT a square, then bounds for $j$ are 0 or irrational, and the expression equals to (assuming that the sums are over integers within the bounds): $\frac{(-1)^q}{n!}\sum_{j=0}^{\left[\left(\left[\frac{\sqrt{n}}{2}\right]+1\right)\sqrt{n}\right]}\binom{n}{j}\left(n-2j\right)^q$ (basically, in this case you do need the first sum, and the second sum is for all $j$'s in the interval). If $n$ is a square, then you take some values of $j$ twice: for example, if $n=4$ then first you sum over $j=0,1,2$ and then $j=2,3,4$, i.e. you have two terms for $j=2$. Is this what you really mean here?2012-04-17
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    @J.M Yes, it seems to work out.2012-04-17
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    @ Vadim: thank you. Vadim, could you please elaborate-why we dont need summation over i in the first case. And also, how to calcukate the obtained sum? Thank you very much.2012-04-17

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