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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<n, u[d]*u[n/d], 0)))/2); u}

apply(numerator, DirSqrt(vector(100, n, moebius(n)))) \\ Andrew Howroyd, Aug 08 2018

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).

Sequence in context: A196755 A199510 A146306 * A322356 A119788 A059592

Adjacent sequences:  A257095 A257096 A257097 * A257099 A257100 A257101

KEYWORD

sign,mult

AUTHOR

Wolfgang Hintze, Apr 16 2015

STATUS

approved

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Last modified January 17 18:48 EST 2019. Contains 319251 sequences. (Running on oeis4.)