Erdos-Kac for squarefree numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:59:03Z http://mathoverflow.net/feeds/question/84754 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84754/erdos-kac-for-squarefree-numbers Erdos-Kac for squarefree numbers Zev 2012-01-02T16:57:39Z 2012-01-04T18:46:12Z <p>In its usual form, the Erdos-Kac Theorem states that if $f(n) : \mathbb{N} \rightarrow \mathbb{R}$ is a strongly additive function with $|f(p)| \le 1$ for all primes $p$, then $$\frac{\#\{n \le x : \frac{f(n) - A(x)}{\sqrt{B(x)}} \le z\}}{x}$$ converges weakly to the Gaussian distribution $G(z)$, where $$A(x) = \sum_{p \le x}\frac{f(p)}{p} \text{ and } B(x) =\sum_{p \le x}\frac{f(p)^2}{p}$$ as long as $B(x) \rightarrow \infty$ as $x \rightarrow \infty$.</p> <p>With some work and using results from "Sieving and the Erdos-Kac theorem" by Granville and Soundararajan, I am able to show that $$\frac{\#\{n \le x, n \text{ squarefree } : \frac{f(n) - A(x)}{\sqrt{B(x)}} \le z\}}{\#\{n \le x : n \text{ squarefree }\}}$$ converges weakly to $G(z)$ under the same hypotheses.</p> <p>This seems like it should be a standard result but I haven't been able to find a reference for it. Does anyone know of a reference for this? </p>