OFFSET
1,6
COMMENTS
We have Sum_{k=1..n} mu(k)*floor(n/k) = 1 and lim_{n -> infinity} Sum_{1 <= i <= j <= n} (mu(i*j)/i)/j = 1/2.
FORMULA
a(n) ~ n/2 (n->infinity).
PROG
(PARI) a(n) = sum(j=1, n, sum(i=1, j, moebius(i*j)*floor(n/i/j)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Mar 01 2018
STATUS
approved