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A253179
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.
0
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, -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, 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
OFFSET
1,6
COMMENTS
Conjecture: the partial sum of this sequence changes sign infinitely often.
All terms are congruent to 0 (mod 2) for n>2.
FORMULA
a(n) = A008683(n)*A071961(n).
MATHEMATICA
f[n_] := MoebiusMu[n] * Sum[ KroneckerSymbol[n, k], {k, n}]; Array[f, 100]
PROG
(PARI) a(n) = moebius(n)*sum(k=1, n, kronecker(n, k)); \\ Michel Marcus, Mar 24 2015
CROSSREFS
Sequence in context: A371648 A347679 A028586 * A300723 A263788 A355335
KEYWORD
sign
AUTHOR
Robert G. Wilson v, Mar 23 2015
STATUS
approved