A257099 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).

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

Offset

1,8

Comments

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, ...

Links

Wolfgang Hintze, Table of n, a(n) for n = 1..500

Formula

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)).

Mathematica

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]];
num = Numerator[t] (* A257099 *)
den = Denominator[t] (* A256689 *)

Crossrefs

Cf. family zeta^(-1/k): A257098/A046644 (k=2), A257099/A256689 (k=3), A257100/A256691 (k=4), A257101/A256693 (k=5).

Cf. family zeta^(+1/k): A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).

Sequence in context: A098087 A354987 A361793 * A348948 A291578 A165485

Adjacent sequences: A257096 A257097 A257098 * A257100 A257101 A257102

Keyword

sign

Author

Wolfgang Hintze, Apr 16 2015

Status

approved