A257101 From fifth root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is 1/zeta; sequence gives numerator of b(n).
1, -1, -1, -2, -1, 1, -1, -6, -2, 1, -1, 2, -1, 1, 1, -21, -1, 2, -1, 2, 1, 1, -1, 6, -2, 1, -6, 2, -1, -1, -1, -399, 1, 1, 1, 4, -1, 1, 1, 6, -1, -1, -1, 2, 2, 1, -1, 21, -2, 2, 1, 2, -1, 6, 1, 6, 1, 1, -1, -2, -1, 1, 2, -1596, 1, -1, -1, 2, 1, -1, -1, 12, -1, 1, 2, 2, 1, -1, -1, 21, -21, 1, -1, -2, 1, 1, 1, 6, -1, -2, 1, 2, 1, 1, 1, 399, -1, 2, 2, 4
Offset: 1
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)
- Vaclav Kotesovec, Graph - the asymptotic ratio (10000 terms)
Crossrefs
Programs
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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]==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] (* A257101 *) den = Denominator[t] (* A256693 *)
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PARI
for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/5))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025
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) = mu(n) for b(m);
a(n) = numerator(b(n)).
Sum_{j=1..n} A257101(j)/A256693(j) ~ n / (Gamma(-1/5) * log(n)^(6/5)) * (1 + 6*(gamma/5 + 1)/(5*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 05 2025
Comments