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A256691 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives denominator of b(n).

%I A256691 #32 May 05 2025 09:34:04
%S A256691 1,4,4,32,4,16,4,128,32,16,4,128,4,16,16,2048,4,128,4,128,16,16,4,512,
%T A256691 32,16,128,128,4,64,4,8192,16,16,16,1024,4,16,16,512,4,64,4,128,128,
%U A256691 16,4,8192,32,128,16,128,4,512,16,512,16,16,4,512,4,16,128,65536,16,64,4,128,16,64,4,4096,4,16,128,128,16,64,4,8192,2048,16,4,512,16,16,16,512,4,512,16,128,16,16,16,32768,4,128,128,1024
%N A256691 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives denominator of b(n).
%C A256691 Dirichlet g.f. of A256690(n)/A256691(n) is (zeta(x))^(1/4).
%C A256691 Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...
%H A256691 Vaclav Kotesovec, <a href="/A256691/b256691.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..500 from Wolfgang Hintze)
%F A256691 with k = 4;
%F A256691 zeta(x)^(1/k) = Sum_{n>=1} b(n)/n^x;
%F A256691 c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;
%F A256691 Then solve c(k,n) = 1 for b(m);
%F A256691 a(n) = denominator(b(n)).
%F A256691 Sum_{j=1..n} A256690(j)/A256691(j) ~ n / (Gamma(1/4) * log(n)^(3/4)) * (1 + (3*(1 - gamma/4))/(4*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - _Vaclav Kotesovec_, May 04 2025
%e A256691 b(1), b(2), ... = 1, 1/4, 1/4, 5/32, 1/4, 1/16, 1/4, 15/128, 5/32, 1/16, 1/4, 5/128, 1/4, 1/16, 1/16, 195/2048, ...
%t A256691 k = 4;
%t A256691 c[1, n_] = b[n];
%t A256691 c[k_, n_] := DivisorSum[n, c[1,#1]*c[k - 1, n/#1] & ]
%t A256691 nn = 100; eqs = Table[c[k, n] == 1, {n, 1, nn}];
%t A256691 sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];
%t A256691 t = Table[b[n], {n, 1, nn}] /. sol[[1]];
%t A256691 num = Numerator[t] (* A256690 *)
%t A256691 den = Denominator[t] (* A256691 *)
%o A256691 (PARI) for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-X)^(1/4))[n]), ", ")) \\ _Vaclav Kotesovec_, May 04 2025
%Y A256691 Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).
%K A256691 nonn,frac,mult
%O A256691 1,2
%A A256691 _Wolfgang Hintze_, Apr 08 2015