A256690 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives numerator of b(n).

1, 1, 1, 5, 1, 1, 1, 15, 5, 1, 1, 5, 1, 1, 1, 195, 1, 5, 1, 5, 1, 1, 1, 15, 5, 1, 15, 5, 1, 1, 1, 663, 1, 1, 1, 25, 1, 1, 1, 15, 1, 1, 1, 5, 5, 1, 1, 195, 5, 5, 1, 5, 1, 15, 1, 15, 1, 1, 1, 5, 1, 1, 5, 4641, 1, 1, 1, 5, 1, 1, 1, 75, 1, 1, 5, 5, 1, 1, 1, 195, 195, 1, 1, 5, 1, 1, 1, 15, 1, 5, 1, 5, 1, 1, 1, 663, 1, 5, 5, 25

Offset

1,4

Comments

Dirichlet g.f. of A256690(n)/A256691(n) is (zeta (x))^(1/4).

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 = 4;

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)).

Example

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, ...

Mathematica

k = 4;
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] (* A256690 *)
den = Denominator[t] (* A256691 *)

Crossrefs

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

Sequence in context: A353958 A285486 A230368 * A181985 A304320 A130511

Adjacent sequences: A256687 A256688 A256689 * A256691 A256692 A256693

Keyword

nonn,frac,mult

Author

Wolfgang Hintze, Apr 09 2015

Status

approved