Now, the above formula is mildly annoying because if \$$k\$$ is an integer

$\Gamma(k) = (k-1)!$

(where, by the way, the "-1" offers absolutely no advantage: it's a completely stupid convention), while

$\Gamma(k + \frac{1}{2}) = {(2k)! \over 4^k k!} \sqrt{\pi} = \frac{(2k-1)!!}{2^k} \sqrt{\pi}$

so

$\text{moment}(d,n) = \frac{2^{d+n-2} \Gamma(\frac{d}{2}) \Gamma(\frac{1}{2} (d+n-1))}{\sqrt{\pi} \, \Gamma(d+ \frac{n}{2} - 1)}$

seems to involve 4 cases, depending on the parity of both \$$d\$$ and \$$n\$$.

It makes plenty of sense to only focus on the _even_ moments, \$$n\$$ even, say \$$n = 2m\$$, and then

$\text{moment}(d,2m) = \frac{2^{d+2m-2} \Gamma(\frac{d}{2}) \Gamma(m+ \frac{1}{2} (d-1))}{\sqrt{\pi} \, \Gamma(d+ m - 1)}$

but we still have two cases, \$$d\$$ even and \$$d\$$ odd.

I'm more interested in the case \$$d\$$ even because Greg Egan has done computer calculations suggesting that in these cases, _only finitely many primes_ show up as denominators of the moments \$$\text{moment}(d,2m) \$$ written in lowest terms:

> I tallied all the denominators for even n=2,...,100,000:

> d=6: {1,49234},{3,766}

> d=8: {1,49017},{2,13},{5,970}

> d=10: {1,47149},{5,1893},{7,911},{35,47}

> d=12: {1,46505},{2,10},{3,1211},{7,1823},{9,358},{14,2},{21,67},{63,24}

> For odd d, the denominators quickly get large and various.

I think the first number in these ordered pairs is the denominator while the second is how many times it shows up in \$$\text{moment}(d,n) \$$ for even \$$n\$$ between 2 and 100,000.