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a(n) = Sum_{1 <= i <= j <= n} mu(i*j)*floor((n/i)/j).
0

%I #16 Dec 10 2023 18:13:50

%S 1,1,1,1,1,2,2,2,2,3,3,4,4,5,6,6,6,7,7,8,9,10,10,11,11,12,12,13,13,13,

%T 13,13,14,15,16,17,17,18,19,20,20,20,20,21,22,23,23,24,24,25,26,27,27,

%U 28,29,30,31,32,32,32,32,33,34,34,35,35,35,36,37,37,37,38,38,39,40,41,42,42,42,43,43,44,44,44,45,46,47,48,48,48

%N a(n) = Sum_{1 <= i <= j <= n} mu(i*j)*floor((n/i)/j).

%C 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.

%F a(n) ~ n/2 (n->infinity).

%o (PARI) a(n) = sum(j=1, n, sum(i=1, j, moebius(i*j)*floor(n/i/j)))

%K nonn

%O 1,6

%A _Benoit Cloitre_, Mar 01 2018