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 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). 11
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified August 10 03:38 EDT 2022. Contains 356029 sequences. (Running on oeis4.)