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A348281 a(n) = Sum_{d|n} d' * mu(d)^2. 0

%I #13 Oct 13 2021 17:21:52

%S 0,1,1,1,1,7,1,1,1,9,1,7,1,11,10,1,1,7,1,9,12,15,1,7,1,17,1,11,1,54,1,

%T 1,16,21,14,7,1,23,18,9,1,68,1,15,10,27,1,7,1,9,22,17,1,7,18,11,24,33,

%U 1,54,1,35,12,1,20,96,1,21,28,90,1,7,1,41,10,23,20,110,1,9,1,45

%N a(n) = Sum_{d|n} d' * mu(d)^2.

%C Sum of the (arithmetic) derivatives of the squarefree divisors of n.

%F a(p^k) = 1 for primes p and k >= 1. For k = 1, we have 1'*mu(1)^2 + p'*mu(p)^2 = 0*1 + 1*1 = 1. For all k >= 2, mu(p^k)^2 = 0. Therefore, a(p^k) = 0*1 + 1*1 + (0 + ... + 0) [k-1 times] = 1.

%e a(10) = 9; a(10) = 1' + 2' + 5' + 10' = 0 + 1 + 1 + 7 = 9.

%o (PARI) ad(n) = vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415

%o a(n) = sumdiv(n, d, ad(d)*moebius(d)^2); \\ _Michel Marcus_, Oct 10 2021

%Y Cf. A003415 (arithmetic derivative), A008683 (mu).

%K nonn

%O 1,6

%A _Wesley Ivan Hurt_, Oct 09 2021

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