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 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). 4

%I

%S 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,

%T -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,

%U 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

%N 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).

%C Dirichlet g.f. of b(n) = a(n)/A256689(n) is (zeta (x))^(-1/3).

%C Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/3).

%C Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...

%H Wolfgang Hintze, <a href="/A257099/b257099.txt">Table of n, a(n) for n = 1..500</a>

%F with k = 3;

%F zeta(x)^(-1/k) = Sum_{n>=1} b(n)/n^x;

%F c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;

%F Then solve c(k,n) = mu(n) for b(m);

%F a(n) = numerator(b(n)).

%t k = 3;

%t c[1, n_] = b[n];

%t c[k_, n_] := DivisorSum[n, c[1, #1]*c[k - 1, n/#1] & ]

%t nn = 100; eqs = Table[c[k, n]MoebiusMu[n], {n, 1, nn}];

%t sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];

%t t = Table[b[n], {n, 1, nn}] /. sol[[1]];

%t num = Numerator[t] (* A257099 *)

%t den = Denominator[t] (* A256689 *)

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

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

%K sign

%O 1,8

%A _Wolfgang Hintze_, Apr 16 2015

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Last modified March 18 20:24 EDT 2019. Contains 321294 sequences. (Running on oeis4.)