1,3

Race between the n-th square-free number and round(n*zeta(2)).

Daniel Forgues, Table of n, a(n) for n=1..60794

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

a(n) = A005117(n) - n * Sum[{i, 1, inf}, {1/(i^2)}] = O(sqrt(n))

a(n) = A005117(n) - n * (pi^2)/6 = O(sqrt(n))

Cf. A005117 Square-free numbers.

Cf. A013929 Not square-free numbers.

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

Cf. A158819 Number of square-free numbers <= n minus round(n/zeta(2)).

Sequence in context: A089367 A130192 A104564 this_sequence A156384 A064656 A056608

Adjacent sequences: A160761 A160762 A160763 this_sequence A160765 A160766 A160767

sign

Daniel Forgues (squid(AT)zensearch.com), May 26 2009