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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 (list; graph; refs; listen; history; text; internal format)
 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 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) = 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: A066017 A236938 A079834 * A029582 A067095 A070888 Adjacent sequences:  A256685 A256686 A256687 * A256689 A256690 A256691 KEYWORD nonn,frac,mult AUTHOR Wolfgang Hintze, Apr 08 2015 STATUS approved

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Last modified February 16 12:48 EST 2019. Contains 320163 sequences. (Running on oeis4.)