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We can use the Lindberg condition to show the distribution of number of prime divisors of an integer approaches Gaussian.

  1. Is there a similar probabilistic formulation for square free numbers? That is, is it reasonable to say the probability of an uniformly random integer being square free is $\frac6{\pi^2}$ with suitable probability distribution?

  2. Are there non-trivial and deceptive situations where such probability distribution interpretations break down?

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The standard way to formalize this thought is via the concept of "density". This is because you of course cannot uniformly randomly select $n\in\mathbb{N}$, but you can instead examine (for a subset $A\subseteq\mathbb{N}$, using the notation $[n] = \{1,\dots,n\}$):

$$d(A) = \lim_{n\to\infty}\frac{|[n]\cap A|}{|[n]|}$$

I believe this can analogously be written as: $$d(A) = \lim_{n\to\infty}\Pr_{x\leftarrow\mathcal{U}([n])}[x\in A]$$ Highlighting the probabilistic aspect of it.

It is known that the natural density of square free numbers is $6/\pi^2$, precisely as you expect. Natural density has many properties similar to what one would expect for "probabilistic" statements about "uniformly random integers", but has some downfalls (it is not defined for all subsets $A\subseteq\mathbb{N}$). There are other notions of densities which can be examined in those situations, but I am no expert on the benefits/shortcomings of them.

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  • $\begingroup$ The link you presented on distribution does not talk about probability distribution in Erdos Kac sense. Perhaps all square free numbers conspire against non square free numbers? I am looking for valid arguments for probability distributions. $\endgroup$
    – VS.
    Commented Feb 4, 2020 at 12:30

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