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 1-10 of 12 results. Next

A257090 a(n) = log_2 (A256691(n)).

Original entry on oeis.org

0, 2, 2, 5, 2, 4, 2, 7, 5, 4, 2, 7, 2, 4, 4, 11, 2, 7, 2, 7, 4, 4, 2, 9, 5, 4, 7, 7, 2, 6, 2, 13, 4, 4, 4, 10, 2, 4, 4, 9, 2, 6, 2, 7, 7, 4, 2, 13, 5, 7, 4, 7, 2, 9, 4, 9, 4, 4, 2, 9, 2, 4, 7, 16, 4, 6, 2, 7, 4, 6, 2, 12, 2, 4, 7, 7, 4, 6, 2, 13, 11, 4, 2, 9, 4, 4, 4, 9, 2, 9, 4, 7, 4, 4, 4, 15, 2, 7, 7, 10
Offset: 1

Views

Author

Wolfgang Hintze, Apr 16 2015

Keywords

Comments

a(n) is the logarithm to the base 2 of the denominator of the Dirichlet series of zeta(s)^(1/4). For details, see A256691.

Links

Crossrefs

Cf. A046645 (k = 2, log_2), A257089 (k = 3, log_3), A257090 (k = 4, log_2), A257091 (k = 5, log_5).

Formula

2^a(n) = A256691(n).

A256689 From third root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is zeta function; sequence gives denominator of b(n).

Original entry on oeis.org

1, 3, 3, 9, 3, 9, 3, 81, 9, 9, 3, 27, 3, 9, 9, 243, 3, 27, 3, 27, 9, 9, 3, 243, 9, 9, 81, 27, 3, 27, 3, 729, 9, 9, 9, 81, 3, 9, 9, 243, 3, 27, 3, 27, 27, 9, 3, 729, 9, 27, 9, 27, 3, 243, 9, 243, 9, 9, 3, 81, 3, 9, 27, 6561, 9, 27, 3, 27, 9, 27, 3, 729, 3, 9, 27, 27, 9, 27, 3, 729, 243, 9, 3, 81, 9, 9, 9, 243, 3, 81, 9, 27, 9, 9, 9, 2187, 3, 27, 27, 81
Offset: 1

Views

Author

Wolfgang Hintze, Apr 08 2015

Keywords

Comments

Dirichlet g.f. of A256688(n)/A256689(n) is (zeta(x))^(1/3).
Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...

Examples

			b(1), b(2), ... = 1, 1/3, 1/3, 2/9, 1/3, 1/9, 1/3, 14/81, 2/9, 1/9, 1/3, 2/27, 1/3, 1/9, 1/9, 35/243, ...

Links

Crossrefs

Cf. 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] == 1, {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] (* A256688 *)
den = Denominator[t] (* A256689 *)
  • PARI
    for(n=1, 100, print1(denominator(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) = 1 for b(m);
a(n) = denominator(b(n)).
Sum_{j=1..n} A256688(j)/A256689(j) ~ n / (Gamma(1/3) * log(n)^(2/3)) * (1 + (2*(1 - gamma/3))/(3*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025

A256688 From third root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is zeta function; sequence gives numerator of b(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 14, 2, 1, 1, 2, 1, 1, 1, 35, 1, 2, 1, 2, 1, 1, 1, 14, 2, 1, 14, 2, 1, 1, 1, 91, 1, 1, 1, 4, 1, 1, 1, 14, 1, 1, 1, 2, 2, 1, 1, 35, 2, 2, 1, 2, 1, 14, 1, 14, 1, 1, 1, 2, 1, 1, 2, 728, 1, 1, 1, 2, 1, 1, 1, 28, 1, 1, 2, 2, 1, 1, 1, 35, 35, 1, 1, 2, 1, 1, 1, 14, 1, 2, 1, 2, 1, 1, 1, 91, 1, 2, 2, 4
Offset: 1

Views

Author

Wolfgang Hintze, Apr 08 2015

Keywords

Comments

Dirichlet g.f. of A256688(n)/A256689(n) is (zeta(x))^(1/3).
Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...

Examples

			b(1), b(2), ... = 1, 1/3, 1/3, 2/9, 1/3, 1/9, 1/3, 14/81, 2/9, 1/9, 1/3, 2/27, 1/3, 1/9, 1/9, 35/243, ...

Links

Crossrefs

Cf. 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] == 1, {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] (* A256688 *)
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) = 1 for b(m);
a(n) = numerator(b(n)).
Sum_{j=1..n} A256688(j)/A256689(j) ~ n / (Gamma(1/3) * log(n)^(2/3)) * (1 + (2*(1 - gamma/3))/(3*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025

A256693 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives denominator of b(n).

Original entry on oeis.org

1, 5, 5, 25, 5, 25, 5, 125, 25, 25, 5, 125, 5, 25, 25, 625, 5, 125, 5, 125, 25, 25, 5, 625, 25, 25, 125, 125, 5, 125, 5, 15625, 25, 25, 25, 625, 5, 25, 25, 625, 5, 125, 5, 125, 125, 25, 5, 3125, 25, 125, 25, 125, 5, 625, 25, 625, 25, 25, 5, 625, 5, 25, 125, 78125, 25, 125, 5, 125, 25, 125, 5, 3125, 5, 25, 125, 125, 25, 125, 5, 3125, 625, 25, 5, 625, 25, 25, 25, 625, 5, 625, 25, 125, 25, 25, 25, 78125, 5, 125, 125, 625
Offset: 1

Views

Author

Wolfgang Hintze, Apr 08 2015

Keywords

Comments

Dirichlet g.f. of A256692(n)/A256693(n) is (zeta(x))^(1/5).
Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...
In general, for m > 0, if Dirichlet g.f. is zeta(s)^m, then Sum_{j=1..n} a(j) ~ n*log(n)^(m-1)/Gamma(m) * (1 + (m-1)*(m*gamma - 1)/log(n)), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025

Examples

			b(1), b(2), ... =
1, 1/5, 1/5, 3/25, 1/5, 1/25, 1/5, 11/125, 3/25, 1/25, 1/5, 3/125, 1/5, 1/25, 1/25, 44/625, 1/5, 3/125, 1/5, 3/125, 1/25, 1/25, 1/5, 11/625

Links

Crossrefs

Cf. 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] == 1, {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] (* A256692 *)
den = Denominator[t] (* A256693 *)
  • PARI
    for(n=1, 100, print1(denominator(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) = 1 for b(m);
a(n) = denominator(b(n)).
Sum_{j=1..n} A256692(j)/A256693(j) ~ n / (Gamma(1/5) * log(n)^(4/5)) * (1 + (4*(1 - gamma/5))/(5*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025

A256690 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives numerator of b(n).

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 1, 15, 5, 1, 1, 5, 1, 1, 1, 195, 1, 5, 1, 5, 1, 1, 1, 15, 5, 1, 15, 5, 1, 1, 1, 663, 1, 1, 1, 25, 1, 1, 1, 15, 1, 1, 1, 5, 5, 1, 1, 195, 5, 5, 1, 5, 1, 15, 1, 15, 1, 1, 1, 5, 1, 1, 5, 4641, 1, 1, 1, 5, 1, 1, 1, 75, 1, 1, 5, 5, 1, 1, 1, 195, 195, 1, 1, 5, 1, 1, 1, 15, 1, 5, 1, 5, 1, 1, 1, 663, 1, 5, 5, 25
Offset: 1

Views

Author

Wolfgang Hintze, Apr 09 2015

Keywords

Comments

Dirichlet g.f. of A256690(n)/A256691(n) is (zeta(x))^(1/4).
Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...

Examples

			b(1), b(2), ... = 1, 1/4, 1/4, 5/32, 1/4, 1/16, 1/4, 15/128, 5/32, 1/16, 1/4, 5/128, 1/4, 1/16, 1/16, 195/2048, ...

Links

Crossrefs

Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).

Programs

  • Mathematica
    k = 4;
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] == 1, {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] (* A256690 *)
den = Denominator[t] (* A256691 *)
  • PARI
    for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(1/4))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

Formula

with k = 4;
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) = 1 for b(m);
a(n) = numerator(b(n)).
Sum_{j=1..n} A256690(j)/A256691(j) ~ n / (Gamma(1/4) * log(n)^(3/4)) * (1 + (3*(1 - gamma/4))/(4*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025

A256692 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives numerator of b(n).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 11, 3, 1, 1, 3, 1, 1, 1, 44, 1, 3, 1, 3, 1, 1, 1, 11, 3, 1, 11, 3, 1, 1, 1, 924, 1, 1, 1, 9, 1, 1, 1, 11, 1, 1, 1, 3, 3, 1, 1, 44, 3, 3, 1, 3, 1, 11, 1, 11, 1, 1, 1, 3, 1, 1, 3, 4004, 1, 1, 1, 3, 1, 1, 1, 33, 1, 1, 3, 3, 1, 1, 1, 44, 44, 1, 1, 3, 1, 1, 1, 11, 1, 3, 1, 3, 1, 1, 1, 924, 1, 3, 3, 9
Offset: 1

Views

Author

Wolfgang Hintze, Apr 08 2015

Keywords

Comments

Dirichlet g.f. of A256692(n)/A256693(n) is (zeta(x))^(1/5).
Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...

Examples

			b(1), b(2), ... =
1, 1/5, 1/5, 3/25, 1/5, 1/25, 1/5, 11/125, 3/25, 1/25, 1/5, 3/125, 1/5, 1/25, 1/25, 44/625, 1/5, 3/125, 1/5, 3/125, 1/25, 1/25, 1/5, 11/625

Links

Crossrefs

Cf. 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] == 1, {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] (* A256692 *)
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) = 1 for b(m);
a(n) = numerator(b(n)).
Sum_{j=1..n} A256692(j)/A256693(j) ~ n / (Gamma(1/5) * log(n)^(4/5)) * (1 + (4*(1 - gamma/5))/(5*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025

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

A257100 From fourth root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is 1/zeta; sequence gives numerator of b(n).

Original entry on oeis.org

1, -1, -1, -3, -1, 1, -1, -7, -3, 1, -1, 3, -1, 1, 1, -77, -1, 3, -1, 3, 1, 1, -1, 7, -3, 1, -7, 3, -1, -1, -1, -231, 1, 1, 1, 9, -1, 1, 1, 7, -1, -1, -1, 3, 3, 1, -1, 77, -3, 3, 1, 3, -1, 7, 1, 7, 1, 1, -1, -3, -1, 1, 3, -1463, 1, -1, -1, 3, 1, -1, -1, 21, -1, 1, 3, 3, 1, -1, -1, 77, -77, 1, -1, -3, 1, 1, 1, 7, -1, -3, 1, 3, 1, 1, 1, 231, -1, 3, 3, 9
Offset: 1

Views

Author

Wolfgang Hintze, Apr 16 2015

Keywords

Comments

Dirichlet g.f. of b(n) = a(n)/A256691(n) is (zeta(x))^(-1/4).
Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/4).
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 = 4;
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] (* A257100 *)
den = Denominator[t] (* A256691 *)
  • PARI
    for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/4))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

Formula

with k = 4;
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} A257100(j)/A256691(j) ~ n / (Gamma(-1/4) * log(n)^(5/4)) * (1 + 5*(gamma/4 + 1)/(4*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
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