A256693 -id:A256693 - cp's OEIS frontend

A257091 a(n) = log_5 (A256693(n)).

0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 3, 1, 6, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 4, 1, 2, 3, 7, 2, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 2, 3, 1, 5, 4, 2, 1, 4, 2, 2, 2, 4, 1, 4, 2, 3, 2, 2, 2, 7, 1, 3, 3, 4

Offset: 1

Wolfgang Hintze, Apr 16 2015

nonn

a(n) is the logarithm to the base 5 of the denominator of the Dirichlet series of zeta(s)^(1/5). For details, see A256693.

Robert Israel and Wolfgang Hintze, Table of n, a(n) for n = 1..10000 (up to 500 from Wolfgang Hintze)
MathOverflow, The number of prime factors of a natural number.

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

F:= proc(n) local e,m;
add(add(floor(e/5^m),m=0..floor(log[5](e))),e=map(t-> t[2],ifactors(n)[2]));
end proc:
seq(F(i),i=1..100);

F[n_] := Sum[Sum[Floor[e/5^m], {m, 0, Floor[Log[5, e]]}], {e, FactorInteger[n][[All, 2]]}];
F[1] = 0;
Array[F, 100] (* Jean-François Alcover, Jun 18 2020, after Maple *)

5^a(n) = A256693(n).
For n<=10000, if n = Product_i p_i^(e_i) is the prime factorization of n, a(n) = A001222(n) + Sum_i floor(e_i/5). - Robert Israel, May 13 2016
If n = Product_i p_i^(e_i) is the prime factorization of n, a(n) = Sum_{j >= 0} Sum_i floor(e_i/5^j). - Robert Israel, May 16 2016

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

Wolfgang Hintze, Apr 08 2015

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, ...

			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, ...

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

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

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 *)

for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-X)^(1/3))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

Wolfgang Hintze, Apr 08 2015

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, ...

			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, ...

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

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

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 *)

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(1/3))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

A256691 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives denominator of b(n).

Original entry on oeis.org

1, 4, 4, 32, 4, 16, 4, 128, 32, 16, 4, 128, 4, 16, 16, 2048, 4, 128, 4, 128, 16, 16, 4, 512, 32, 16, 128, 128, 4, 64, 4, 8192, 16, 16, 16, 1024, 4, 16, 16, 512, 4, 64, 4, 128, 128, 16, 4, 8192, 32, 128, 16, 128, 4, 512, 16, 512, 16, 16, 4, 512, 4, 16, 128, 65536, 16, 64, 4, 128, 16, 64, 4, 4096, 4, 16, 128, 128, 16, 64, 4, 8192, 2048, 16, 4, 512, 16, 16, 16, 512, 4, 512, 16, 128, 16, 16, 16, 32768, 4, 128, 128, 1024

Offset: 1

Wolfgang Hintze, Apr 08 2015

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, ...

			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, ...

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

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

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 *)

for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-X)^(1/4))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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) = denominator(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

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

Wolfgang Hintze, Apr 09 2015

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, ...

			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, ...

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

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

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 *)

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(1/4))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

Wolfgang Hintze, Apr 08 2015

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, ...

			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

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

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

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 *)

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(1/5))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

Wolfgang Hintze, Apr 16 2015

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

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)

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).

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 *)

\\ 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

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

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

Wolfgang Hintze, Apr 16 2015

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, ...

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)

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).

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 *)

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/3))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

Wolfgang Hintze, Apr 16 2015

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, ...

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)

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).

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 *)

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/4))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

Wolfgang Hintze, Apr 16 2015

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, ...

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)

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).

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 *)

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-X)^(-1/5))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025

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

cp's OEIS Frontend

A257091 a(n) = log_5 (A256693(n)).

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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).

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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).

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A256691 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives denominator of b(n).

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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).

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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).

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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).

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