cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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).

Original entry on oeis.org

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

Views

Author

Wolfgang Hintze, Apr 16 2015

Keywords

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

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

A257099 From third root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is 1/zeta; sequence gives numerator of b(n).

Original entry on oeis.org

1, -1, -1, -1, -1, 1, -1, -5, -1, 1, -1, 1, -1, 1, 1, -10, -1, 1, -1, 1, 1, 1, -1, 5, -1, 1, -5, 1, -1, -1, -1, -22, 1, 1, 1, 1, -1, 1, 1, 5, -1, -1, -1, 1, 1, 1, -1, 10, -1, 1, 1, 1, -1, 5, 1, 5, 1, 1, -1, -1, -1, 1, 1, -154, 1, -1, -1, 1, 1, -1, -1, 5, -1, 1, 1, 1, 1, -1, -1, 10, -10, 1, -1, -1, 1, 1, 1, 5, -1, -1, 1, 1, 1, 1, 1, 22, -1, 1, 1, 1
Offset: 1

Views

Author

Wolfgang Hintze, Apr 16 2015

Keywords

Comments

Dirichlet g.f. of b(n) = a(n)/A256689(n) is (zeta(x))^(-1/3).
Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/3).
Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...

Links

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 = 3;
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] (* A257099 *)
den = Denominator[t] (* A256689 *)
  • PARI
    for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/3))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

Formula

with k = 3;
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} A257099(j)/A256689(j) ~ n / (Gamma(-1/3) * log(n)^(4/3)) * (1 + 4*(gamma/3 + 1)/(3*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 05 2025

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).

Original entry on oeis.org

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

Views

Author

Wolfgang Hintze, Apr 16 2015

Keywords

Comments

Dirichlet g.f. of b(n) = A257101(n)/A256693(n) is (zeta(x))^(-1/5).
Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/5).
Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...

Links

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 = 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 *)
  • 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
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.