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A257100 From fourth root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is 1/zeta; sequence gives numerator of b(n).

%I A257100 #32 May 05 2025 09:50:26
%S A257100 1,-1,-1,-3,-1,1,-1,-7,-3,1,-1,3,-1,1,1,-77,-1,3,-1,3,1,1,-1,7,-3,1,
%T A257100 -7,3,-1,-1,-1,-231,1,1,1,9,-1,1,1,7,-1,-1,-1,3,3,1,-1,77,-3,3,1,3,-1,
%U A257100 7,1,7,1,1,-1,-3,-1,1,3,-1463,1,-1,-1,3,1,-1,-1,21,-1,1,3,3,1,-1,-1,77,-77,1,-1,-3,1,1,1,7,-1,-3,1,3,1,1,1,231,-1,3,3,9
%N A257100 From fourth root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is 1/zeta; sequence gives numerator of b(n).
%C A257100 Dirichlet g.f. of b(n) = a(n)/A256691(n) is (zeta(x))^(-1/4).
%C A257100 Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/4).
%C A257100 Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...
%H A257100 Vaclav Kotesovec, <a href="/A257100/b257100.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..500 from Wolfgang Hintze)
%F A257100 with k = 4;
%F A257100 zeta(x)^(-1/k) = Sum_{n>=1} b(n)/n^x;
%F A257100 c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;
%F A257100 Then solve c(k,n) = mu(n) for b(m);
%F A257100 a(n) = numerator(b(n)).
%F A257100 Sum_{j=1..n} A257100(j)/A256691(j) ~ n / (Gamma(-1/4) * log(n)^(5/4)) * (1 + 5*(gamma/4 + 1)/(4*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - _Vaclav Kotesovec_, May 05 2025
%t A257100 k = 4;
%t A257100 c[1, n_] = b[n];
%t A257100 c[k_, n_] := DivisorSum[n, c[1, #1]*c[k - 1, n/#1] & ]
%t A257100 nn = 100; eqs = Table[c[k, n]==MoebiusMu[n], {n, 1, nn}];
%t A257100 sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];
%t A257100 t = Table[b[n], {n, 1, nn}] /. sol[[1]];
%t A257100 num = Numerator[t] (* A257100 *)
%t A257100 den = Denominator[t] (* A256691 *)
%o A257100 (PARI) for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/4))[n]), ", ")) \\ _Vaclav Kotesovec_, May 04 2025
%Y A257100 Cf. family zeta^(-1/k): A257098/A046644 (k=2), A257099/A256689 (k=3), A257100/A256691 (k=4), A257101/A256693 (k=5).
%Y A257100 Cf. family zeta^(+1/k): A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).
%K A257100 sign,frac
%O A257100 1,4
%A A257100 _Wolfgang Hintze_, Apr 16 2015