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A097558
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Sum{k=1 to oo} a(k)/k^r = sqrt(zeta(r) -3/4) +1/2.
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2
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1, 1, 1, 0, 1, -1, 1, 1, 0, -1, 1, 3, 1, -1, -1, -1, 1, 3, 1, 3, -1, -1, 1, -7, 0, -1, 1, 3, 1, 7, 1, 3, -1, -1, -1, -12, 1, -1, -1, -7, 1, 7, 1, 3, 3, -1, 1, 19, 0, 3, -1, 3, 1, -7, -1, -7, -1, -1, 1, -27, 1, -1, 3, -6, -1, 7, 1, 3, -1, 7, 1, 45, 1, -1, 3, 3, -1, 7, 1, 19, -1, -1, 1, -27, -1, -1, -1, -7, 1, -27, -1, 3, -1, -1, -1, -51, 1, 3, 3, -12, 1, 7
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OFFSET
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1,12
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COMMENTS
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The "+ 1/2" in the Dirichlet series generating function was added so the first term of the sequence is an integer. We could have added/subtracted any other integer+1/2 instead and then had the first term equal another integer. "zeta(r)" refers to sum{k=1 to oo} 1/k^r.
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LINKS
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FORMULA
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a(1)=1; for n>=2, a(n) = 1 - sum{k|n, 2<=k<=n-1} a(n/k) a(k).
a(n) depends only on the prime signature of n.
If p is prime, a(p^k) = (-1)^(k+1)*A005043(k-1).
(End)
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MAPLE
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A[1]:= 1:
for n from 2 to 100 do
A[n]:= 1 - add(A[n/k]*A[k], k= numtheory:-divisors(n) minus {1, n})
od:
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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