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A257099
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From third root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is 1/zeta; sequence gives numerator of b(n).
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4
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1, -1, -1, -1, -1, 1, -1, -5, -1, 1, -1, 1, -1, 1, 1, -10, -1, 1, -1, 1, 1, 1, -1, 5, -1, 1, -5, 1, -1, -1, -1, -22, 1, 1, 1, 1, -1, 1, 1, 5, -1, -1, -1, 1, 1, 1, -1, 10, -1, 1, 1, 1, -1, 5, 1, 5, 1, 1, -1, -1, -1, 1, 1, -154, 1, -1, -1, 1, 1, -1, -1, 5, -1, 1, 1, 1, 1, -1, -1, 10, -10, 1, -1, -1, 1, 1, 1, 5, -1, -1, 1, 1, 1, 1, 1, 22, -1, 1, 1, 1
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OFFSET
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1,8
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COMMENTS
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Dirichlet g.f. of b(n) = a(n)/A256689(n) is (zeta (x))^(-1/3).
Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/3).
Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...
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LINKS
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FORMULA
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with k = 3;
zeta(x)^(-1/k) = Sum_{n>=1} b(n)/n^x;
c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;
Then solve c(k,n) = mu(n) for b(m);
a(n) = numerator(b(n)).
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MATHEMATICA
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k = 3;
c[1, n_] = b[n];
c[k_, n_] := DivisorSum[n, c[1, #1]*c[k - 1, n/#1] & ]
nn = 100; eqs = Table[c[k, n]==MoebiusMu[n], {n, 1, nn}];
sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];
t = Table[b[n], {n, 1, nn}] /. sol[[1]];
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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