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 $n$.
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Pick two random numbers less than $n$, then
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:
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