How To Construct a 4D Menger Sponge

The volume is $\dfrac{1}{2} {\pi}^{2}$

which is the volume of the 4-Sphere of radius $1$ , so that the answer is $1+2+2=5$ . In fact, carrying out the same recursive subtractions from n-Cubes of sides $2$ will yield the correct volumes of corresponding n-Spheres of radius $1$ , as in

1D $\;\;$ $2$

2D $\;\;$ $\pi$

3D $\;\;$ $\dfrac{4}{3} \pi$

4D $\;\;$ $\dfrac{1}{2} {\pi}^{2}$

5D $\;\;$ $\dfrac{8}{15} {\pi}^{2}$

6D $\;\;$ $\dfrac{1}{6} {\pi}^{3}$

7D $\;\;$ $\dfrac{16}{105} {\pi}^{3}$

8D $\;\;$ $\dfrac{1}{24} {\pi}^{4}$

9D $\;\;$ $\dfrac{32}{945} {\pi}^{4}$

etc.

For the 4D case, we first find the number of permutations of the set of numbers $(n, n, a, b)$ . There are $4$ cases to consider, which are

$(n, n, n, n)$

$(n, n, n, a)$

$(n, n, a, a)$

$(n, n, a, b)$

where $a=0,1,2,….2n \;$ EXCEPT $n$

Then the number of possible permutations of each are, respectively

$1$

$4 \cdot 2n$

$\dfrac{4\cdot 3}{1\cdot 2} \cdot 2n$

$4 \cdot 3 \cdot \dfrac{1}{2}2n(2n-1)$

Subtracting all of this from ${(2n+1)}^4$ , we end up with

$16{n}^3(n+2)$

so that the infinite product we seek, giving the limit volume, is

${ 2 }^{ 4 }\displaystyle \prod _{ n=1 }^{ \infty }{ \dfrac { 16{ n }^{ 3 }\left( n+2 \right) }{ { \left( 2n+1 \right) }^{ 4 } } }$

which can be rearranged as

${ 2 }^{ 4 }\displaystyle \prod _{ n=1 }^{ \infty }{ \dfrac { \left( 2n \right) ^{ 2 }\left( 4{ n }^{ 2 }+8n \right) }{ { \left( 2n+1 \right) }^{ 4 } } }$

Given the infinite product

$\displaystyle \prod _{ n=1 }^{ \infty }{ \dfrac { \left( n+1 \right) ^{ 2 } }{ { n\left( n+2 \right) } } } =\dfrac{1}{2}$

we then can modify the previous infinite product to as follows

${ 2 }^{ 3 }\displaystyle \prod _{ n=1 }^{ \infty }{ \dfrac { { \left( 2n \right) }^{ 2 }\left( 2n+2 \right) ^{ 2 } }{ { { \left( 2n+1 \right) }^{ 4 } } } }$

But given another well-known infinite product, the Wallis product

$\displaystyle \prod _{ n=1 }^{ \infty }{ \dfrac { { \left( 2n \right) }\left( 2n+2 \right) }{ { { \left( 2n+1 \right) }^{ 2 } } } } =\frac { \pi }{ 4 }$

we have the following volume as the limit

${ 2 }^{ 3 }\dfrac { { \pi }^{ 2 } }{ { 4 }^{ 2 } } =\dfrac { 1 }{ 2 } { \pi }^{ 2 }$

Note: $\;$ The following infinite product will deliver the volume of the $n$ -Sphere of radius $R$

${ \left( 2R \right) }^{ n }\displaystyle \prod _{ k=1 }^{ \infty }{ \dfrac { { \left( 2k \right) }^{ n-1 }\left( 2k+n \right) }{ { \left( 2k+1 \right) }^{ n } } }$

which can be derived from the same method of subtracting volumes from $n$ -Cubes, creating modified $n$ -Menger Sponges. For $n=2$ , the infinite product becomes the Wallis product.