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For $n\ge 2\in\mathbb N$, let $S_n$ be the volume of a $(n-1)$ dimensional solid which satisfies $$\sum_{i=1}^{n}x_i=0, |x_i|\le1\ (i=1,2,\cdots,n).$$

Then, here is my question.

Question : Can we represent $S_n$ by $n$ ?

Remark : This question has been asked previously on math.SE without receiving any answers.

Motivation : I've been interested in this simple question. I've got the followings :

$$S_2=2\sqrt 2, S_3=3\sqrt 3, S_4=\frac{32}{3}.$$

$S_4$ is the volume of a regular octahedron, whose edge length is $\sqrt 8$, which passes through the following six points :

$$(1,1,-1,-1),(1,-1,1,-1),(1,-1,-1,1),(-1,1,1,-1),(-1,1,-1,1),(-1,-1,1,1).$$

However, I don't have any good idea for $n$ in general. Can anyone help?

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    $\begingroup$ Lazy but curious, I still would like to see $S_5$ computed explicitly, perhaps $S_6$ too. $\endgroup$ – Włodzimierz Holsztyński Nov 15 '13 at 16:53
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This question (in a much more general form) is answered in this preprint by Marichal and Mosinghoff. They point out that the answer to your question actually goes back to Polya's PhD thesis.

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Section 2 in the paper http://arxiv.org/abs/math/0503115 expresses this volume in terms of sinc integrals $$\sigma_n=\frac2\pi\int_0^\infty \left(\frac{\sin x}{x}\right)^n\,dx$$ Namely, it is shown that the volume you are interested in is equal to $2^{n-1}\sqrt{n}\sigma_n$.

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Using the formula for the pdf of the Irwin-Hall distribution one gets $$S_n = \frac{\sqrt{n}}{(n-1)!}\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}(-1)^{k}{n \choose k}\left(n-2k\right)^{n-1}$$

It's fairly straightforward to see why, imagine you're drawing random point in your cube, how many will have coordinates that sum to less than $\epsilon$ in absolute value? This gives you a $2\sqrt{n}\epsilon$ thick slice of hypercube around the hyperplane. Take the limit as $\epsilon \rightarrow 0$

The first values are, $1,2\sqrt{2}, 3\sqrt{3}, \frac{32}{3}, \frac{115\sqrt{5}}{12},\frac{88\sqrt{6}}{5},\ldots$

Using the central-limit theorem gives the asymptotic $$S_n \sim \sqrt{\frac{6}{\pi}}2^{n-1}$$

This paper proposes an algorithm for a slight generalisation

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