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).

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

Offset: 1

Author

Wolfgang Hintze, Apr 16 2015

Keywords

• sign
• mult
• frac

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

• Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)
• Vaclav Kotesovec, Graph - the asymptotic ratio (10^7 terms)

Crossrefs

Cf. family zeta^(-1/k): A257098/A046644 (k=2), A257099/A256689 (k=3), A257100/A256691 (k=4), A257101/A256693 (k=5).

Cf. family zeta^(+1/k): A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).

Programs

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 *)

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&&dAndrew Howroyd, Aug 08 2018

PARI

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/2))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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)).

Sum_{j=1..n} A257098(j)/A046644(j) ~ -n / (2 * sqrt(Pi) * log(n)^(3/2)) * (1 + 3*(gamma/2 + 1)/(2*log(n))), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, May 05 2025