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

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

This paper proposes an algorithm for a slight generalisation

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

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

It's fairly straightforward to see why, imagine you're drawing random point in your cube, how many 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 $\epsilon \rightarrow 0$

This paper proposes an algorithm for a slight generalisation

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

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

This paper proposes an algorithm for a slight generalisation

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

It's fairly straightforward to see why, imagine you're drawing random point in your cube, how many fall within a distance $\epsilon$ of the hyperplane? Now consider that the thickness of the slice is $2\sqrt{n}\epsilon$. Take the limit as $\epsilon \rightarrow 0$

This paper proposes an algorithm for a slight generalisation

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

It's fairly straightforward to see why, imagine you're drawing random point in your cube, how many 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 $\epsilon \rightarrow 0$

This paper proposes an algorithm for a slight generalisation