set is always a distributive lattice. It is a Boolean algebra if and only if is square-free. A positive integer is square-free if and only if , where denotes the Möbius function. Dirichlet series The absolute value of the Möbius function is the indicator function for the square-free integers – that is, is equal to 1 if is square-free, and 0 if it is not. The Dirichlet series of this indicator function is where is the Riemann zeta function. This follows from the Euler product where the products are taken over the prime numbers. Distribution Let Q(x) denote the number of square-free integers between 1 and x ( shifting index by 1). For large n, 3/4 of the positive integers less than n are not divisible by 4, 8/9 of these numbers are not divisible by 9, and so on. Because these ratios satisfy the multiplicative property (this follows from Chinese remainder theorem), we obtain the approximation: This argument can be made rigorous for getting the estimate (using big O notation) Sketch of a proof: the above characterization gives observing that the last summand is zero for , it results that By exploiting the largest known zero-free region of the Riemann zeta function Arnold Walfisz improved the approximation to for some positive constant . Under the Riemann hypothesis, the error term can be reduced to Recently (2015) the error term has been further reduced to The asymptotic/natural density of square-free numbers is therefore Therefore over 3/5 of the integers are square-free. Likewise, if Q(x,n) denotes the number of n-free integers (e.g. 3-free integers being cube-free integers) between 1 and x, one can show Since a multiple of 4 must have a square factor 4=22, it cannot occur that four consecutive integers are all square-free. On the other hand, there exist infinitely many integers n for which 4n +1, 4n +2, 4n +3 are all square-free. Otherwise, observing that 4n and at least one of 4n +1, 4n +2, 4n +3 among four could be non-square-free for sufficiently large n, half of all positive integers minus finitely many must be non-square-free and therefore for some constant C, contrary to the above asymptotic estimate for . There exist sequences of consecutive non-square-free integers of arbitrary length. Indeed, if n satisfies a simultaneous congruence for distinct primes , then each is divisible by pi 2. On the other hand, the above-mentioned estimate implies