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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).
10
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
1,4
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
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)).
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: A368811 A342458 A358722 * A372326 A029582 A067095
KEYWORD
nonn,frac,mult
AUTHOR
Wolfgang Hintze, Apr 08 2015
STATUS
approved