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%I #12 Feb 23 2023 01:30:27
%S 1,-1,-1,-2,-1,1,-1,-6,-2,1,-1,2,-1,1,1,-21,-1,2,-1,2,1,1,-1,6,-2,1,
%T -6,2,-1,-1,-1,-399,1,1,1,4,-1,1,1,6,-1,-1,-1,2,2,1,-1,21,-2,2,1,2,-1,
%U 6,1,6,1,1,-1,-2,-1,1,2,-1596,1,-1,-1,2,1,-1,-1,12,-1,1,2,2,1,-1,-1,21,-21,1,-1,-2,1,1,1,6,-1,-2,1,2,1,1,1,399,-1,2,2,4
%N From fifth root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is 1/zeta; sequence gives numerator of b(n).
%C Dirichlet g.f. of b(n) = A257101(n)/A256693(n) is (zeta (x))^(-1/5).
%C Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/5).
%C Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...
%H Wolfgang Hintze, <a href="/A257101/b257101.txt">Table of n, a(n) for n = 1..500</a>
%F with k = 5;
%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 = 5;
%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] (* A257101 *)
%t den = Denominator[t] (* A256693 *)
%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 1,4
%A _Wolfgang Hintze_, Apr 16 2015
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