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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: A294746 A326180 A064085 * A228637 A352431 A152795 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 August 10 06:08 EDT 2022. Contains 356029 sequences. (Running on oeis4.)