Finding the dimension of the sphere cube

Question

If you take an $2r\times 2r\times 2r$ cube, and divide it to 27 equal cubes, and then remove all the "axis" cubes (all the cubes which are straight left, straight right, straight up, etc. from the middle cube) then divide each cube into 125 equal cubes and remove all the axis cubes, and repeat this process for every odd number to the power of three, you will get a nice fractal cube.

If we try to work out the area of this cube we get (after some short algebra):$$A=8r^3\prod_{n=1}^{\infty}\frac{(2n)^2(2n+3)}{(2n+1)^3}=8r^3\frac{2\times2\times\not5\times4\times4\times\not7\times6\times6\times\dots}{3\times3\times3\times\not5\times5\times5\times\not7\times7\times7\times\dots}$$ Now, we extract the Wallis product and get$$A=8r^3\frac{\pi}6$$Which is the area of a sphere with radius $r$(!!!)

Now I want to know what is the fractal dimension of this sphere cube.

Answer 1

I don't know much about Hausdorff dimension, but a quick scan of the Wikipedia account suggests that if you performed your construction up to the stage where you divide into $p^3$ cubes for some odd $p$ but then continued using that value rather than increasing it, the result would have dimension $log(p^3 -3p +2)/log(p)$. This expression tends to 3 as p increases. I suspect your construction will have higher dimension than any of these, and so has dimension 3.

Answer 2

Assume w.l.o.g. that $r = 1/2$.

Then the remaining cubes at each successive level $n = 1, 2, \ldots$ have side lengths $$\frac{1}{3}, \frac{1}{3\cdot5}, \frac{1}{3\cdot5\cdot7}, \ldots, \prod_{k=1}^{n}\frac{1}{2k+1} =: \epsilon(n) $$ and their counts are $$20, 20\cdot112, 20\cdot112\cdot324, \ldots, \prod_{k=1}^n (2k+1)^3 - 3\cdot(2k+1) + 2 =: N(n)$$ Then the box-counting dimension is $$\dim_{\text{box}} = \lim_{n\to\infty} \frac{\log(N(n))}{\log(1/\epsilon(n))} = 3$$ where the limit is calculated using SageMath (which itself uses Maxima):

limit(log(product((2*k+1)^3-3*(2*k+1)+2,k,1,n))/log(product(2*k+1,k,1,n)),n=infinity)