Square free integer

The square-free integers up to 120, which remain after removing the multiples of the squares of the minor cousins than Ã120.

An integer n is square-free if there is no prime number p such that p² divides n. This means that the prime factors of n are all distinct, so

n=p1p2 pk= i한 한 {1, ,k!pCousinpi.{displaystyle n=p_{1}p_{2}cdots p_{k}=prod _{iin {1,cdotsk} atop p{text{ cousin}}}}}{p_{i}. !

Thus, 10=2 5 is free of squares, but 20=2² 5 is not, because it is divisible by a square. The first few square-free integers are:

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39,... (A005117 Succession in OEIS)

Alternatively, if the number a, when expressed as a product of prime factors, all of them have exponent 1, a is said to be an integer free from squares.

Dirichlet generating function

Yeah. q(n#1, where n is an integer that contains no square in its factoring and q(n? n contains one or more squares in its factoring, the function q(n) is defined as q(n)=日本語μ μ (n)日本語{displaystyle q(n)= gift(n)}, being μ(n) the function of Möbius. So, Dirichlet's generating function for square-free integers is

␡ ␡ n=1∞ ∞ q(n)ns=␡ ␡ n=1∞ ∞ 日本語μ μ (n)日本語ns=γ γ (s)γ γ (2s){displaystyle sum _{n=1}^{infty }{frac {q(n)}}{n^{s}}}}=sum _{n=1}{infty }{frac {associatedmu(n)}{n^{s}}}}={frac {zeta(s)}{zeta (2s)}}}}}}}}}}}}}}}}{

where ζ(s) is the Riemann zeta function. This can easily be seen from the Euler product

γ γ (s)γ γ (2s)= p(1− − p− − 2s)(1− − p− − s)= p(1+p− − s).{displaystyle {frac {zeta(s)}{zeta (2s)}}}=prod _{p}{frac {(1-p^{-2s})}{(1-p^{-s})}}}=prod _{p}(1+p^{-s}). !

Distribution of square-free numbers

If Q(x) indicates the number of free numbers of squares less than or equal to x, then

Q(x)=6xπ π 2+O(x){displaystyle Q(x)={frac {6x}{pi ^{2}}}{sqrt {x}}})}

(see π).
The density of the free numbers of squares is, therefore,

limx→ → ∞ ∞ Q(x)x=6π π 2{displaystyle lim _{xto infty }{frac {Q(x)}{x}}}}{frac {6}{pi ^{2}}}}}}}}}}