Question

We know that the probability $P(k)$ of $k$ randomly chosen integers $(k \ge 2)$ from the set of natural number are coprime is

$$ P(k) = \frac{1}{\zeta(k)}. $$

I am looking at a special case of this problem. Let $S_n$ be the set of all natural numbers which do not have a prime factor greater than $n$-th prime (i.e $S_n$ is the set of natural numbers that can be formed using only the first $n$ prime numbers). What is the probability $P(k, S_n)$ that $k$ randomly chosen integers $(k \ge 2)$ from the set $S_n$ are coprime?

I do not know the answer but I think it could be in a parametric form involving $n$ such that in the trivial case when $n\to \infty$, $P(k, S_{\infty}) = P(k) = 1/\zeta(k)$.

Edit: Explained the meaning of "all natural numbers."

Answer 1

First of all, note that there is no canonical notion of equidistribution on a countable set like the integers.

When asking for the probability that $k$ 'randomly chosen' integers are coprime, it is more-or-less intuitively clear what is meant by 'randomly chosen'. This is basically because the density of integers divisible by a given prime $p$ is (up to 'differences from rounding') the same in any interval $\{1, \dots, n\}$, namely $1/p$.

In your question this is not the case: for example in $S_4 \cap \{1, \dots, 10\}$, 5 of 10 numbers (= 50 percent) are even, while in $S_4 \cap \{1, \dots, 10^6\}$, 1070 of 1273 integers are even (which is about 84 percent) and in $S_4 \cap \{1, \dots, 10^{30}\}$, already 445064 of 462692 are even, which is about 96 percent.

So in order to make your question well-defined you need to be explicit about what you mean by 'randomly chosen'.

Answer 2

A reasonable thing to do would be to look at the number of pairs in $S_k \cap \{1 \cdots N\}$ which are relatively prime. For fixed $k$ this would go to zero and indeed the probability would go to $1$ that a randomly selected element is not relatively prime to anything else in the set (other than $1$ of course).

Looking further at the set $S_4 \cap \{2..10^{30}\}$ we can also say that over $78\%$ of the members are less than $10^{28}$ (So in the bottom $1\%$ sizewise) and that over $73\%$ are not relatively prime to anything else (i.e. divide by $2,3,5$ and $7$). The number of relatively prime pairs is $0.013\%.$

Perhaps looking at integers $m$ with all prime factors less than $m^{2/3}$ or something like that would be interesting but I do not have a feel for what the right question would be.