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 A257098 From square root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose square is 1/zeta; sequence gives numerator of b(n). 12
 1, -1, -1, -1, -1, 1, -1, -1, -1, 1, -1, 1, -1, 1, 1, -5, -1, 1, -1, 1, 1, 1, -1, 1, -1, 1, -1, 1, -1, -1, -1, -7, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, 1, 1, 1, -1, 5, -1, 1, 1, 1, -1, 1, 1, 1, 1, 1, -1, -1, -1, 1, 1, -21, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, 1, 1, -1, -1, 5, -5, 1, -1, -1, 1, 1, 1, 1, -1, -1, 1, 1, 1, 1, 1, 7, -1, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,16 COMMENTS Dirichlet g.f. of b(n) = A257098(n)/A046644(n) is (zeta (x))^(-1/2). Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/2). Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ... The sequence of rationals a(n)/A046644(n) is the Moebius transform of A046643/A046644 which is multiplicative. This sequence is then also multiplicative. - Andrew Howroyd, Aug 08 2018 LINKS Wolfgang Hintze, Table of n, a(n) for n = 1..500 FORMULA with k = 2; 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) = mu(n) for b(m); a(n) = numerator(b(n)). MATHEMATICA k = 2; 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]==MoebiusMu[n], {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] (* A257098 *) den = Denominator[t] (* A046644 *) PROG (PARI) \\ DirSqrt(v) finds u such that v = v[1]*dirmul(u, u). DirSqrt(v)={my(n=#v, u=vector(n)); u[1]=1; for(n=2, n, u[n]=(v[n]/v[1] - sumdiv(n, d, if(d>1&&d

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Last modified October 21 07:03 EDT 2019. Contains 328292 sequences. (Running on oeis4.)