A256689 From third root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is zeta function; sequence gives denominator of b(n).

1, 3, 3, 9, 3, 9, 3, 81, 9, 9, 3, 27, 3, 9, 9, 243, 3, 27, 3, 27, 9, 9, 3, 243, 9, 9, 81, 27, 3, 27, 3, 729, 9, 9, 9, 81, 3, 9, 9, 243, 3, 27, 3, 27, 27, 9, 3, 729, 9, 27, 9, 27, 3, 243, 9, 243, 9, 9, 3, 81, 3, 9, 27, 6561, 9, 27, 3, 27, 9, 27, 3, 729, 3, 9, 27, 27, 9, 27, 3, 729, 243, 9, 3, 81, 9, 9, 9, 243, 3, 81, 9, 27, 9, 9, 9, 2187, 3, 27, 27, 81

Offset

1,2

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, ...

Links

Wolfgang Hintze, Table of n, a(n) for n = 1..500

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) = denominator(b(n)).

Example

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, ...

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 *)

Crossrefs

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

Sequence in context: A134662 A145444 A165824 * A266533 A151710 A160121

Adjacent sequences: A256686 A256687 A256688 * A256690 A256691 A256692

Keyword

nonn,frac,mult

Author

Wolfgang Hintze, Apr 08 2015

Status

approved