A046644 -id:A046644 - cp's OEIS frontend

A318512 Denominators (in their lowest terms) of the sequence whose Dirichlet convolution with itself yields squares (A000290), or equally A064549.

1, 1, 2, 1, 2, 1, 2, 1, 8, 1, 2, 1, 2, 1, 4, 1, 2, 4, 2, 1, 4, 1, 2, 1, 8, 1, 16, 1, 2, 2, 2, 1, 4, 1, 4, 4, 2, 1, 4, 1, 2, 2, 2, 1, 16, 1, 2, 1, 8, 4, 4, 1, 2, 8, 4, 1, 4, 1, 2, 2, 2, 1, 16, 1, 4, 2, 2, 1, 4, 2, 2, 2, 2, 1, 16, 1, 4, 2, 2, 1, 128, 1, 2, 2, 4, 1, 4, 1, 2, 8, 4, 1, 4, 1, 4, 1, 2, 4, 16, 4, 2, 2, 2, 1, 8

Offset: 1

Antti Karttunen, Aug 30 2018

These are also denominators (in their lowest terms) for the sequence whose Dirichlet convolution with itself yields A064549, n * Product_{primes p|n} p.
From Antti Karttunen, Sep 02 2018: (Start)
Proof for the above claim:
This sequence is defined as the denominator (given in the lowest terms) of rational valued function r(1) = 1, r(n) = (1/2) * (A000290(n) - Sum_{d|n, d>1, d 1. Define sequence Ay(n) as the denominator of function s(n), with otherwise similar definition, but with A064549 in place of A000290. Let Ay(n) be the denominator of s(n), reduced also into the lowest terms. (Corresponding numerators are A318649 and A318511 respectively. Note that the denominators in both cases must always be of the form 2^k, with k >= 0).
  By applying the distributive property of Dirichlet Convolution [which says that for any completely multiplicative function f, it doesn't matter whether one multiplies the result of convolution afterwards, or whether one multiplies the operands separately before convolution: f(g * g) = (fg) * (fg)], with A000027 in the role of f in both cases, one obtains a pair of equations:
    A318649(n)     A318681(n)     n*A299149(n)
    ----------  =  ----------  =  ------------
    A318512(n)     A299150(n)      A299150(n)
  and
    A318511(n)     A318680(n)     n*A318653(n)
    ----------  =  ----------  =  ------------
       Ay(n)       A299150(n)      A299150(n)
  where the leftmost ratios are reduced into their lowest terms.
  Sequence A318656 gives the 2-adic valuation of ratio A318649(n)/A318512(n), and because there are no even terms neither in A299149 nor in A318653, it also gives the 2-adic valuation of the latter ratio. As A318511/Ay is given in the lowest terms (not both of A318511(n) and Ay(n) can be even at same n), this implies that Ay must indeed be identical to A318512, and furthermore that A318655(n) = A007814(A318649(n)) = A007814(A318511(n)).
(End)

Antti Karttunen, Table of n, a(n) for n = 1..16384
Wikipedia, Dirichlet convolution

Cf. A000290, A064549, A046644, A299150, A318513, A318651, A318653, A318654, A318655, A318656, A318658, A318680, A318681.

Cf. A318649, A318511 (numerators).

f[1] = 1; f[n_] := f[n] = 1/2 (n*Times @@ FactorInteger[n][[All, 1]] - Sum[f[d] f[n/d], {d, Divisors[n][[2 ;; -2]]}]); Table[Denominator[f[n]], {n, 1, 100}] (* Vaclav Kotesovec, May 10 2025 *)

up_to = 65537;
A064549(n) = { my(f=factor(n)); for (i=1, #f~, f[i, 2]++); factorback(f); };
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&&dA064549(n)));
A318512(n) = denominator(v318511_12[n]);

for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-p^2*X)^(1/2))[n]), ", ")) \\ Vaclav Kotesovec, May 09 2025

a(n) = denominator of f(n), where f(1) = 1, f(n) = (1/2) * (A000290(n) - Sum_{d|n, d>1, d 1. [Equally, one could use A064549 in place of A000290.]
a(n) = 2^A318513(n).
a(n) = A046644(n)/A318651(n).
a(2n-1) = A046644(2n-1) = A318658(2n-1), for all n >= 1.

The main definition changed, more formulas added by Antti Karttunen, Aug 31 2018

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

A087003 a(2n) = 0 and a(2n+1) = mu(2n+1); also the sum of Mobius function values computed for terms of 3x+1 trajectory started at n, provided that Collatz conjecture is true.

Original entry on oeis.org

1, 0, -1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 1, 0, -1, 0, 0, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, 0, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 0, 0, 1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 0, 0, 1, 0, -1, 0, 0, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, 0, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, 1, 0, -1

Offset: 1

Labos Elemer, Oct 02 2003

Observe that (these summatory) terms are from {-1,0,1}, so behave like Mobius function values, not like Mertens function values. Moreover, empirically: a(n) deviates from mu(initial-value) = mu(n) only if iv = n is an even squarefree number (i.e., it is from A039956). - This comment, like also the next one, concerns the original Collatz-related definition of this sequence. - Antti Karttunen, Sep 18 2017
From Marc LeBrun, Feb 19 2004: (Start)
Absolute values are the same as those of A091069. First consider the descending parts of Collatz (or 3x+1) trajectories, those that begin with even numbers 2^p k, with k odd. These go 2^p*k, 2^(p-1)*k, ... 2k, k. All but 2k and k are divisible by 4, a (rational) square, hence their mu values are all 0 and so they contribute nothing to the sum.
Then at the end, since mu(2k) = -mu(k), the last two steps cancel each other out. So every descending chain in a trajectory contributes 0. Of course the full trajectory of every even number consists entirely of descending chains, so A087003 is 0 for all even n.
On the other hand, the trajectory of every odd number consists of just that number followed by the trajectory of an even number (which contributes nothing) so A087003 is indeed equal to mu(n) for odd n.
(End)
The sequence is multiplicative; it may be defined as the Dirichlet inverse of the integers modulo 2 (A000035). - Gerard P. Michon, Apr 29 2007
a(n) appears in the second column of A156241 at every second row. - Mats Granvik, Feb 07 2009

Antti Karttunen, Table of n, a(n) for n = 1..10000
G. P. Michon, The Collatz problem.
G. P. Michon, Multiplicative functions.
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370, A008683, A039956, A292273, A099990, A209229.

Cf. A000035 (the Dirichlet inverse), A318657/A318658 (the "Dirichlet Square Root").

c[x_] := (1-Mod[x, 2])*(x/2)+Mod[x, 2]*(3*x+1); c[1]=1; fpl[x_] := Delete[FixedPointList[c, x], -1] lf[x_] := Length[fpl[x]] Table[Apply[Plus, Table[MoebiusMu[Part[fpl[w], j]], {j, 1, lf[w]}]], {w, 1, 256}]
Riffle[MoebiusMu[Range[1,121,2]],0] (* Harvey P. Dale, Jan 24 2025 *)

A006370(n) = if(n%2, 3*n+1, n/2); \\ This function from Michael B. Porter, May 29 2010
A087003(n) = { my(s=1); while(n>1, s += moebius(n); n = A006370(n)); (s); }; \\ Antti Karttunen, Sep 14 2017

a(n)={sumdiv(n, d,  my(e=valuation(d, 2)); if(d==1<Andrew Howroyd, Aug 04 2018

A087003(n) = ((n%2)*moebius(n)); \\ Antti Karttunen, Sep 01 2018

a(n) = A008683(n) + A292273(n). - Antti Karttunen, Sep 14 2017
Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
From Jianing Song, Aug 04 2018: (Start)
Multiplicative with a(2^e) = 0, a(p^e) = (-1 + (-1)^e)/2 for odd primes p.
Dirichlet g.f.: 1/((1 - 2^(-s))*zeta(s)).
(End)
From Antti Karttunen, Sep 01 2018: (Start)
a(n) = A000035(n)*A008683(n).
Dirichlet convolution of A318657/A046644 with itself.
(End)
Sum_{n>=1} a(n)/n^2 = A217739 . Sum_{n>=1} a(n)/n^3 = A233091. Sum_{n>=1} a(n)/n^4 = A300707. - R. J. Mathar, Dec 17 2024

a(2n) = 0, a(2n+1) = mu(2n+1) added to the name as the new primary definition by Antti Karttunen, Sep 18 2017

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

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

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, ...
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

			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(denominator(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) = 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

A299149 Numerators of the positive solution to n = Sum_{d|n} a(d) * a(n/d).

Original entry on oeis.org

1, 1, 3, 3, 5, 3, 7, 5, 27, 5, 11, 9, 13, 7, 15, 35, 17, 27, 19, 15, 21, 11, 23, 15, 75, 13, 135, 21, 29, 15, 31, 63, 33, 17, 35, 81, 37, 19, 39, 25, 41, 21, 43, 33, 135, 23, 47, 105, 147, 75, 51, 39, 53, 135, 55, 35, 57, 29, 59, 45, 61, 31, 189, 231, 65, 33

Offset: 1

Gus Wiseman, Feb 03 2018

Dirichlet convolution of a(n)/A046644(n) with itself yields A000265. - Antti Karttunen, Aug 30 2018

			Sequence begins: 1, 1, 3/2, 3/2, 5/2, 3/2, 7/2, 5/2, 27/8, 5/2, 11/2, 9/4, 13/2, 7/2.

Antti Karttunen, Table of n, a(n) for n = 1..65537 (first 1000 terms Andrew Howroyd)
Vaclav Kotesovec, Graph - the asymptotic ratio (65537 terms)
Wikipedia, Dirichlet convolution.

Cf. A000010, A000265, A003958, A007431, A018804, A023900, A029935, A046643, A046644, A165825, A257098, A298971, A299119, A299150 (denominators), A299151, A317848, A318319, A318321, A318649.

nn=50;
sys=Table[n==Sum[a[d]*a[n/d],{d,Divisors[n]}],{n,nn}];
Numerator[Array[a,nn]/.Solve[sys,Array[a,nn]][[2]]]
odd[n_] := n/2^IntegerExponent[n, 2]; f[p_, e_] := odd[p^e*Binomial[2*e, e]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Apr 30 2023 *)

a(n)={my(v=factor(n)[,2]); numerator(n*prod(i=1, #v, my(e=v[i]); binomial(2*e, e)/4^e))} \\ Andrew Howroyd, Aug 09 2018

\\ 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 09 2018

for(n=1, 100, print1(numerator(direuler(p=2, n, 1/(1-p*X)^(1/2))[n]), ", ")) \\ Vaclav Kotesovec, May 09 2025

a(n) = numerator(n*A317848(n)/A165825(n)) = A000265(n*A317848(n)). - Andrew Howroyd, Aug 09 2018

Sum_{k=1..n} A299149(k)/A299150(k) ~ n^2 / (2*sqrt(Pi*log(n))) * (1 + (1-gamma) / (4*log(n))), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, May 09 2025

Keyword:mult added by Andrew Howroyd, Aug 09 2018

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

cp's OEIS Frontend

A318512 Denominators (in their lowest terms) of the sequence whose Dirichlet convolution with itself yields squares (A000290), or equally A064549.

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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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A087003 a(2n) = 0 and a(2n+1) = mu(2n+1); also the sum of Mobius function values computed for terms of 3x+1 trajectory started at n, provided that Collatz conjecture is true.

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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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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).

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A299149 Numerators of the positive solution to n = Sum_{d|n} a(d) * a(n/d).

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