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A112400
a(n) = Sum_{p|n, p prime} mu(b(p,n)), where mu(k) = A008683(k) (the Moebius function) and p^b(p,n) is the highest power of the prime p dividing n.
1
0, 1, 1, -1, 1, 2, 1, -1, -1, 2, 1, 0, 1, 2, 2, 0, 1, 0, 1, 0, 2, 2, 1, 0, -1, 2, -1, 0, 1, 3, 1, -1, 2, 2, 2, -2, 1, 2, 2, 0, 1, 3, 1, 0, 0, 2, 1, 1, -1, 0, 2, 0, 1, 0, 2, 0, 2, 2, 1, 1, 1, 2, 0, 1, 2, 3, 1, 0, 2, 3, 1, -2, 1, 2, 0, 0, 2, 3, 1, 1, 0, 2, 1, 1, 2, 2, 2, 0, 1, 1, 2, 0, 2, 2, 2, 0, 1, 0, 0, -2, 1, 3, 1, 0, 3, 2, 1, -2, 1, 3, 2, 1, 1, 3, 2, 0, 0, 2, 2, 1, -1, 2
OFFSET
1,6
COMMENTS
The justification for a(1) being 0 is that the sum is empty.
EXAMPLE
12 = 2^3 * 3^1. So a(12) = mu(3) + mu(1) = -1 + 1 = 0.
PROG
(PARI) a(n)=local(v, i, s); v=factor(n); s=0; for(i=1, matsize(v)[1], s+=moebius(v[i, 2])); s \\ (Herrgesell)
(PARI) A112400(n) = vecsum(apply(e -> moebius(e), factorint(n)[, 2])); \\ Antti Karttunen, Jul 07 2017
CROSSREFS
Cf. A008683.
Sequence in context: A092673 A243842 A367405 * A316523 A219185 A365658
KEYWORD
sign
AUTHOR
Leroy Quet, Dec 06 2005
EXTENSIONS
More terms from Lambert Herrgesell (zero815(AT)googlemail.com), Dec 09 2005
STATUS
approved