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 A256692 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives numerator of b(n). 10
 1, 1, 1, 3, 1, 1, 1, 11, 3, 1, 1, 3, 1, 1, 1, 44, 1, 3, 1, 3, 1, 1, 1, 11, 3, 1, 11, 3, 1, 1, 1, 924, 1, 1, 1, 9, 1, 1, 1, 11, 1, 1, 1, 3, 3, 1, 1, 44, 3, 3, 1, 3, 1, 11, 1, 11, 1, 1, 1, 3, 1, 1, 3, 4004, 1, 1, 1, 3, 1, 1, 1, 33, 1, 1, 3, 3, 1, 1, 1, 44, 44, 1, 1, 3, 1, 1, 1, 11, 1, 3, 1, 3, 1, 1, 1, 924, 1, 3, 3, 9 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Dirichlet g.f. of A256692(n)/A256693(n) is (zeta (x))^(1/5). 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 = 5; 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, 1/5, 1/5, 3/25, 1/5, 1/25, 1/5, 11/125, 3/25, 1/25, 1/5, 3/125, 1/5, 1/25, 1/25, 44/625, 1/5, 3/125, 1/5, 3/125, 1/25, 1/25, 1/5, 11/625 MATHEMATICA k = 5; 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] (* A256692 *) den = Denominator[t] (* A256693 *) CROSSREFS Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5). Sequence in context: A096066 A294746 A064085 * A228637 A152795 A121585 Adjacent sequences:  A256689 A256690 A256691 * A256693 A256694 A256695 KEYWORD nonn,frac,mult AUTHOR Wolfgang Hintze, Apr 08 2015 STATUS approved

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Last modified March 18 19:58 EDT 2019. Contains 321293 sequences. (Running on oeis4.)