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.