1,7

Race between the number of square-free numbers and round(n/zeta(2)).

Daniel Forgues, Table of n, a(n) for n=1..100000

A. Granville, ABC means we can count squarefrees

Since zeta(2) = Sum[{i, 1, inf}, {1/(i^2)}] = (pi^2)/6, we get:

a(n) = A013928(n+1) - n/Sum[{i, 1, inf}, {1/(i^2)}] = O(sqrt(n))

a(n) = A013928(n+1) - 6*n/(pi^2) = O(sqrt(n))

Cf. A008966 1 if n is square-free, else 0.

Cf. A013928 Number of square-free numbers < n.

Cf. A100112 If n is the k-th square-free number then k else 0.

Cf. A057627 Number of non-square-free numbers not exceeding n.

Cf. A005117 Square-free numbers.

Cf. A013929 Not square-free numbers.

Sequence in context: A086597 A031214 A056059 this_sequence A031279 A124778 A037831

Adjacent sequences: A158816 A158817 A158818 this_sequence A158820 A158821 A158822

nonn

Daniel Forgues (squid(AT)zensearch.com), Mar 27 2009