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a(n) = mu(n)*Sum_{k=1..n} (n/k) where mu(n) is Möbius (or Moebius) function and (x/y) is Kronecker's symbol.
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%I #11 Apr 01 2015 10:22:55

%S 1,-1,0,0,0,2,-4,0,0,2,-2,0,0,4,6,0,0,0,-2,0,0,2,-6,0,0,6,0,0,0,-4,

%T -12,0,0,4,4,0,0,6,12,0,0,-4,0,0,0,4,-10,0,0,0,4,0,0,0,18,0,0,2,-6,0,

%U 0,8,0,0,0,-8,2,0,0,-4,-14,0,0,10,0,0,0,-4,-22,0,0,4,-4,0,0,10,14,0,0,0,2,0,0,8,14,0,0,0,0,0,0,-4,-22,0,0

%N a(n) = mu(n)*Sum_{k=1..n} (n/k) where mu(n) is Möbius (or Moebius) function and (x/y) is Kronecker's symbol.

%C Conjecture: the partial sum of this sequence changes sign infinitely often.

%C All terms are congruent to 0 (mod 2) for n>2.

%F a(n) = A008683(n)*A071961(n).

%t f[n_] := MoebiusMu[n] * Sum[ KroneckerSymbol[n, k], {k, n}]; Array[f, 100]

%o (PARI) a(n) = moebius(n)*sum(k=1, n, kronecker(n, k)); \\ _Michel Marcus_, Mar 24 2015

%Y Cf. A008683, A071961.

%K sign

%O 1,6

%A _Robert G. Wilson v_, Mar 23 2015