| Version 6 |
Version 5 |
| A square-free number is a natural number that contains no powers greater than 1 in its prime factorization. In other words, if $x$ is our number, and |
A square-free number is a natural number that contains no powers greater than 1 in its prime factorization. In other words, if $x$ is our number, and |
| $$ x = \prod_{i=1}^r p_i^{a_i} $$ |
$$ x = \prod_{i=1}^r p_i^{a_i} $$ |
| is the prime factorization of $x$ into $r$ distinct primes, then $a_i \ge 2$ is |
is the prime factorization of $x$ into $r$ distinct primes, then $a_i \ge 2$ is |
| always false for square-free $x$. |
always false for square-free $x$. |
| The name derives from the fact that if any $a_i$ were to be greater than or equal to two, we could be sure that at least one square divides the $x$ (namely, $p_i^2$.) |
The name derives from the fact that if any $a_i$ were to be greater than or equal to two, we could be sure that at least one square divides the $x$ (namely, $p_i^2$.) |
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The asymptotic density of square-free numbers is $\frac{6}{\pi^2}$ which can be proved by application of a square-free variation of the sieve of Eratosthenes as follows:
|
The asymptotic density of square-free numbers is $\frac{6}{\pi^2}$ which can be proved by application of a square-free variation of a sieve of Eratosthenes as follows:
|
| \begin{align*} |
\begin{align*} |
| A(n)&=\sum_{k\leq n} [n \text{ is a squarefree}]\\ |
A(n)&=\sum_{k\leq n} [n \text{ is a squarefree}]\\ |
| &=\sum_{k\leq n} \sum_{d^2 | k} \mu(d)\\ |
&=\sum_{k\leq n} \sum_{d^2 | k} \mu(d)\\ |
| &=\sum_{d \leq \sqrt{n}} \mu(d) \sum_{\substack{k \leq n\\d^2 | n}} 1 \\ |
&=\sum_{d \leq \sqrt{n}} \mu(d) \sum_{\substack{k \leq n\\d^2 | n}} 1 \\ |
| &=\sum_{d \leq \sqrt{n}} \mu(d) \left\lfloor{\frac{n}{d^2}}\right\rfloor\\ |
&=\sum_{d \leq \sqrt{n}} \mu(d) \left\lfloor{\frac{n}{d^2}}\right\rfloor\\ |
| &=n \sum_{d \leq \sqrt{n}} \frac{\mu(d)}{d^2}+O(\sqrt{n})\\ |
&=n \sum_{d \leq \sqrt{n}} \frac{\mu(d)}{d^2}+O(\sqrt{n})\\ |
| &=n \sum_{d} \frac{\mu(d)}{d^2}+O(\sqrt{n})\\ |
&=n \sum_{d} \frac{\mu(d)}{d^2}+O(\sqrt{n})\\ |
| &=n \frac{1}{\zeta(2)} + O(\sqrt{n}) = n \frac{6}{\pi^2}+O(\sqrt{n}). |
&=n \frac{1}{\zeta(2)} + O(\sqrt{n}) = n \frac{6}{\pi^2}+O(\sqrt{n}). |
| \end{align*} |
\end{align*} |
| It was shown that Riemann Hypothesis implies error term $O(n^{7/22+\epsilon})$ in the above. |
It was shown that Riemann Hypothesis implies error term $O(n^{7/22+\epsilon})$ in the above. |
