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A256693 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives denominator of b(n).

%I A256693 #25 May 05 2025 09:34:43
%S A256693 1,5,5,25,5,25,5,125,25,25,5,125,5,25,25,625,5,125,5,125,25,25,5,625,
%T A256693 25,25,125,125,5,125,5,15625,25,25,25,625,5,25,25,625,5,125,5,125,125,
%U A256693 25,5,3125,25,125,25,125,5,625,25,625,25,25,5,625,5,25,125,78125,25,125,5,125,25,125,5,3125,5,25,125,125,25,125,5,3125,625,25,5,625,25,25,25,625,5,625,25,125,25,25,25,78125,5,125,125,625
%N A256693 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives denominator of b(n).
%C A256693 Dirichlet g.f. of A256692(n)/A256693(n) is (zeta(x))^(1/5).
%C A256693 Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...
%C A256693 In general, for m > 0, if Dirichlet g.f. is zeta(s)^m, then Sum_{j=1..n} a(j) ~ n*log(n)^(m-1)/Gamma(m) * (1 + (m-1)*(m*gamma - 1)/log(n)), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - _Vaclav Kotesovec_, May 04 2025
%H A256693 Vaclav Kotesovec, <a href="/A256693/b256693.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..500 from Wolfgang Hintze)
%F A256693 with k = 5;
%F A256693 zeta(x)^(1/k) = Sum_{n>=1} b(n)/n^x;
%F A256693 c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;
%F A256693 Then solve c(k,n) = 1 for b(m);
%F A256693 a(n) = denominator(b(n)).
%F A256693 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 A256693 b(1), b(2), ... =
%e A256693 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 A256693 k = 5;
%t A256693 c[1, n_] = b[n];
%t A256693 c[k_, n_] := DivisorSum[n, c[1, #1]*c[k - 1, n/#1] & ]
%t A256693 nn = 100; eqs = Table[c[k, n] == 1, {n, 1, nn}];
%t A256693 sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];
%t A256693 t = Table[b[n], {n, 1, nn}] /. sol[[1]];
%t A256693 num = Numerator[t] (* A256692 *)
%t A256693 den = Denominator[t] (* A256693 *)
%o A256693 (PARI) for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-X)^(1/5))[n]), ", ")) \\ _Vaclav Kotesovec_, May 04 2025
%Y A256693 Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).
%K A256693 nonn,frac,mult
%O A256693 1,2
%A A256693 _Wolfgang Hintze_, Apr 08 2015