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A256692 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives numerator of b(n).

%I A256692 #29 May 05 2025 09:34:23
%S A256692 1,1,1,3,1,1,1,11,3,1,1,3,1,1,1,44,1,3,1,3,1,1,1,11,3,1,11,3,1,1,1,
%T A256692 924,1,1,1,9,1,1,1,11,1,1,1,3,3,1,1,44,3,3,1,3,1,11,1,11,1,1,1,3,1,1,
%U A256692 3,4004,1,1,1,3,1,1,1,33,1,1,3,3,1,1,1,44,44,1,1,3,1,1,1,11,1,3,1,3,1,1,1,924,1,3,3,9
%N A256692 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives numerator of b(n).
%C A256692 Dirichlet g.f. of A256692(n)/A256693(n) is (zeta(x))^(1/5).
%C A256692 Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...
%H A256692 Vaclav Kotesovec, <a href="/A256692/b256692.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..500 from Wolfgang Hintze)
%F A256692 with k = 5;
%F A256692 zeta(x)^(1/k) = Sum_{n>=1} b(n)/n^x;
%F A256692 c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;
%F A256692 Then solve c(k,n) = 1 for b(m);
%F A256692 a(n) = numerator(b(n)).
%F A256692 Sum_{j=1..n} A256692(j)/A256693(j) ~ n / (Gamma(1/5) * log(n)^(4/5)) * (1 + (4*(1 - gamma/5))/(5*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - _Vaclav Kotesovec_, May 04 2025
%e A256692 b(1), b(2), ... =
%e A256692 1, 1/5, 1/5, 3/25, 1/5, 1/25, 1/5, 11/125, 3/25, 1/25, 1/5, 3/125, 1/5, 1/25, 1/25, 44/625, 1/5, 3/125, 1/5, 3/125, 1/25, 1/25, 1/5, 11/625
%t A256692 k = 5;
%t A256692 c[1, n_] = b[n];
%t A256692 c[k_, n_] := DivisorSum[n, c[1, #1]*c[k - 1, n/#1] & ]
%t A256692 nn = 100; eqs = Table[c[k, n] == 1, {n, 1, nn}];
%t A256692 sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];
%t A256692 t = Table[b[n], {n, 1, nn}] /. sol[[1]];
%t A256692 num = Numerator[t] (* A256692 *)
%t A256692 den = Denominator[t] (* A256693 *)
%o A256692 (PARI) for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(1/5))[n]), ", ")) \\ _Vaclav Kotesovec_, May 04 2025
%Y A256692 Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).
%K A256692 nonn,frac,mult
%O A256692 1,4
%A A256692 _Wolfgang Hintze_, Apr 08 2015