square-free number

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

$$x=\prod _{i=1}^r{p}_i^{a_i}$$

is the prime factorization of $x$ into $r$ distinct primes, then $a_i\ge 2$ isalways false for square-free $x$.

Note: we assume here that $x$ itself must be greater than 1; hence 1 is not considered square-free. However, one must be alert to the particular context in which “square-free” is used as to whether this is considered the case.

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 $x$ (namely, $p_i^2$.)

1 Asymptotic Analysis

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 (http://planetmath.org/SieveOfEratosthenes2) as follows:

$A(n)$$={\displaystyle \sum _{k\le n}}[k\text{ is squarefree }]$

$={\displaystyle \sum _{k\le n}}{\displaystyle \sum _{d^2|k}}\mu (d)$

$={\displaystyle \sum _{d\le \sqrt{n}}}\mu (d){\displaystyle \sum _{\begin{array}{c}k\le n\\ d^2|n\end{array}}}1$

$={\displaystyle \sum _{d\le \sqrt{n}}}\mu (d)\lfloor {\displaystyle \frac{n}{d^2}}\rfloor$

$=n{\displaystyle \sum _{d\le \sqrt{n}}}{\displaystyle \frac{\mu (d)}{d^2}}+O(\sqrt{n})$

$=n{\displaystyle \sum _{d\ge 1}}{\displaystyle \frac{\mu (d)}{d^2}}+O(\sqrt{n})$

$=n{\displaystyle \frac{1}{\zeta (2)}}+O(\sqrt{n})$

$=n{\displaystyle \frac{6}{{\pi }^2}}+O(\sqrt{n}).$

It was shown that the Riemann Hypothesis implies error term $O(n^{7/22+ϵ})$ in the above [1].