A256688 From third root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is zeta function; sequence gives numerator of b(n).

1, 1, 1, 2, 1, 1, 1, 14, 2, 1, 1, 2, 1, 1, 1, 35, 1, 2, 1, 2, 1, 1, 1, 14, 2, 1, 14, 2, 1, 1, 1, 91, 1, 1, 1, 4, 1, 1, 1, 14, 1, 1, 1, 2, 2, 1, 1, 35, 2, 2, 1, 2, 1, 14, 1, 14, 1, 1, 1, 2, 1, 1, 2, 728, 1, 1, 1, 2, 1, 1, 1, 28, 1, 1, 2, 2, 1, 1, 1, 35, 35, 1, 1, 2, 1, 1, 1, 14, 1, 2, 1, 2, 1, 1, 1, 91, 1, 2, 2, 4

Offset: 1

Author

Wolfgang Hintze, Apr 08 2015

Keywords

• nonn
• frac
• mult

Comments

Dirichlet g.f. of A256688(n)/A256689(n) is (zeta(x))^(1/3).

Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...

Examples

			b(1), b(2), ... = 1, 1/3, 1/3, 2/9, 1/3, 1/9, 1/3, 14/81, 2/9, 1/9, 1/3, 2/27, 1/3, 1/9, 1/9, 35/243, ...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

Crossrefs

Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).

Programs

Mathematica

k = 3;
c[1, n_] = b[n];
c[k_, n_] := DivisorSum[n, c[1,#1]*c[k - 1, n/#1] & ]
nn = 100; eqs = Table[c[k, n] == 1, {n, 1, nn}];
sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];
t = Table[b[n], {n, 1, nn}] /. sol[[1]];
num = Numerator[t] (* A256688 *)
den = Denominator[t] (* A256689 *)

PARI

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(1/3))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

Formula

with k = 3;

zeta(x)^(1/k) = Sum_{n>=1} b(n)/n^x;

c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;

Then solve c(k,n) = 1 for b(m);

a(n) = numerator(b(n)).

Sum_{j=1..n} A256688(j)/A256689(j) ~ n / (Gamma(1/3) * log(n)^(2/3)) * (1 + (2*(1 - gamma/3))/(3*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025