# Anomaly in probability of 3 randomly-selected integers being coprime

The probability that two randomly-selected integers will be coprime is given by $$[\zeta(2)]^{-1}$$, where $$\zeta$$ is the Riemann zeta function. Similarly, for three such integers the probability is $$[\zeta(3)]^{-1}$$ and for four is $$[\zeta(4)]^{-1}$$, coprime in these cases meaning that there is no common divisor of all the integers (and not the stronger condition of being pairwise coprime). So far so logical.

When it comes to evaluating these formula we have, as is well known:

$$[\zeta(2)]^{-1} = \frac{6}{\pi^2}$$

We also have, as can be inferred from this question:

$$[\zeta(4)]^{-1} = \frac{90}{\pi^4}$$

Having seen these results, we might expect that:

$$[\zeta(3)]^{-1} = \frac{N}{\pi^3}$$

with $$N$$ being some integer such that $$6 < N < \pi^3 \approx 31.006$$. However this is not the case: the value of $$[\zeta(3)]^{-1}$$ as given here (formula 2) is approximately $$0.8319$$, the reciprocal of Apery's constant, implying that $$N \approx 25.79$$.

Question: What is the fundamental reason why the cases of $$2$$ and $$4$$ yield these simple formulae in terms of integers and $$\pi$$ but the case of $$3$$ does not, and does the real number $$25.79\dots$$ implicit in the above have any other interesting properties.

Addendum 10/11/2018

I now realise that the first part of the question has effectively been addressed in answers to this question, especially that by Qiaochu Yuan.

• I think it's not even known whether $\zeta(3)$ is a rational multiple of $\pi^3.$ – coffeemath Nov 9 '18 at 23:11