# Are these two facts related?

I was told some days ago that the possibility of two randomly picked numbers are relatively prime to each other is $6/(\pi^2)$. And it is well known that the value of Riemann zeta function at 2 is $(\pi^2)/6$. So I guess there is a correspondence between them. Maybe the possibility of $n$ randomly picked numbers are relatively prime to each other(there are two cases here: (1)these $n$ numbers are pairly relatively prime to each other (2)the common divisor of all of these $n$ numbers is 1) equals $1/\zeta (n)$? And I think when we consider $n=1$, the possibility of one randomly picked number is prime is 0, and meanwhile $\zeta(1)=\infty$. So in this case with this sense this proposition still holds. But I think I must be daydreaming... Please let me know if you find the formula above is wrong for some $n$.

• It seems like the argument here shouldn't be hard to generalize: math.stackexchange.com/questions/64498/… – Cocopuffs Nov 24 '12 at 12:41
• @Cocopuffs then we can use this method to calculate zeta(3)? By some method to calculate the probability like using conditional probability? – lee Nov 24 '12 at 12:55
• There's no harm in trying but I wouldn't bet on it. – Cocopuffs Nov 24 '12 at 13:19
• @Cocopuffs Okay.. I find that I didn't change the problem. :-( – lee Nov 24 '12 at 14:40
• I suppose I should clarify: I don't think you'll find anything which isn't already known. Of course this doesn't mean that you shouldn't pursue it, it's just that $\zeta(3)$ has been very well studied. – Cocopuffs Nov 24 '12 at 22:41

Pick two random numbers less than $n$, then

• $\lfloor n/2\rfloor^2$ pairs are both divisible by 2.
• $\lfloor n/3\rfloor^2$ pairs are both divisible by 3.
• $\lfloor n/5\rfloor^2$ pairs are both divisible by 5. ...

The number of relatively prime pairs less than or equal to $n$ is:

$$n^2 - \sum\lfloor \frac np\rfloor^2 + \sum\lfloor \frac n{pq}\rfloor^2- \sum\lfloor \frac n{pqr}\rfloor^2 + ...$$

Sums are taken over the distinct primes $p,q,r,...$ less than n. Let $\mu(x)$ be the Möbius function this is

$$\sum\mu(k)\lfloor n/k\rfloor^2$$

The probability is the limit as $n$ goes to infinity divided by $n^2$, or

$$\sum\frac{\mu(k)}{k^2} .$$

Now, the Dirichlet series that generates the Möbius function is the (multiplicative) inverse of the Riemann zeta function $$\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}.$$ So we get $\frac{1}{\zeta(2)}=\frac6{\pi^2}$.

MathWork says:

This result is related to the fact that the greatest common divisor of $m$ and $n$, $(m,n)=k$, can be interpreted as the number of lattice points in the plane which lie on the straight line connecting the vectors $(0,0)$ and $(m,n)$ (excluding $(m,n)$ itself). In fact, $6/\pi^2$ is the fractional number of lattice points visible from the origin (Castellanos 1988, pp. 155-156).

• My question is can we replace 2 with n? – lee Nov 24 '12 at 13:01
• Opps sorry: Given three integers $(k,m,n)$ chosen at random, the probability that no common factor will divide them all is $\zeta(3)^{-1}$. Also from Mathworks... – draks ... Nov 24 '12 at 13:03