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%I #15 Oct 13 2021 16:44:39
%S 0,1,1,1,1,3,1,1,2,3,1,3,1,2,2,1,1,5,1,3,3,3,1,3,2,3,2,2,1,6,1,1,3,3,
%T 2,5,1,3,2,3,1,6,1,3,4,2,1,3,2,4,2,3,1,5,2,2,3,3,1,6,1,3,5,1,2,7,1,3,
%U 3,5,1,5,1,3,4,3,2,6,1,3,2,3,1,6,3,2,2,3,1,10,2,2
%N a(n) = Sum_{d|n, d>1} mu(d')^2.
%C Number of squarefree derivatives of the divisors of n.
%F a(p) = 1 for primes p, since we have mu(p')^2 = mu(1)^2 = 1.
%e a(6) = 3; the divisors of 6 are 1,2,3,6 and their (arithmetic) derivatives are 0,1,1,5 respectively. 1,1,5 are squarefree, so a(6) = 3.
%o (PARI) ad(n) = vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415
%o a(n) = sumdiv(n, d, my(x=ad(d)); if (x>0, moebius(ad(d))^2)); \\ _Michel Marcus_, Oct 10 2021
%Y Cf. A003415, A008683 (mu).
%K nonn
%O 1,6
%A _Wesley Ivan Hurt_, Oct 09 2021