Consider the expansion: $$A006571(n) = q \prod _{k=1}^{\infty } \left(1-q^k\right)^2 \left(1-q^{11 k}\right)^2 \label{1}\tag{1}$$
A006571
and the triple sum:
$$A366450(n)=\sum _{k=1}^n \left(\sum _{y=1}^n \left(\sum _{x=1}^n \frac{g(k)[\gcd (f(x,y),n)=k]}{n}\right)\right) \label{2}\tag{2}$$
A366450
where
- the Weierstrass equation is: $$f(x,y)=x^3-x^2-y^2-y,$$
- $[\;]$ is the Iverson bracket, and
- $g(n)$ is the Dirichlet inverse of Euler totient function A000010, sometimes denoted as: $$g(n) = \varphi^{-1}(n) = \sum_{d \mid n} \mu(d)d.$$
Dividing \eqref{1} and \eqref{2}: $$\text{ratio}(n)=A006571(n)/A366450(n) \label{3}\tag{3}$$
we get a sequence starting:
$$\left\{1,1,1,-\frac{1}{2},1,1,1,0,\frac{2}{3},1,1,-\frac{1}{2},1,1,1,\frac{1}{4},1,\frac{2}{3},\text{Indeterminate},-\frac{1}{2},1,1,1,0,-\frac{4}{5},1,-\frac{5}{9},-\frac{1}{2},\text{Indeterminate},1,1,-\frac{1}{4},1,1,1,-\frac{1}{3},1,\text{Indeterminate},1,0,1,1,1,-\frac{1}{2},\frac{2}{3},1,1,\frac{1}{4},\frac{3}{14},-\frac{4}{5}\right\}$$
Setting $\text{Indeterminate} = 1$, the denominators of \eqref{3} are: $$\{1,1,1,2,1,1,1,1,3,1,1,2,1,1,1,4,1,3,1,2,1,1,1,1,5,1,9,2,1,1,1,4,1,1,1,3,1,1,1,1,1,1,1,2,3,1,1,4,14,5\}$$
For the first few terms we get two partially matching sequences
Both have Dirichlet generating functions given by Amiram Eldar, where the first partial match is: $$\zeta (2 s) \zeta (2 s-1) \prod _{p=\text{prime}} \left(-\frac{1}{p^{2 s}}-\frac{1}{p^{3 s-1}}+\frac{1}{p^s}+1\right)$$ and the second partial match is: $$\zeta (2 s-1) \prod_{p=\text{prime}} \left(\frac{1}{p^{3 s}}-\frac{1}{p^{3 s-1}}+\frac{1}{p^s}+1\right)$$
The greatest common divisor of A056622 and the denominators of the ratios in \eqref{3}: $$\gcd(A056622(n),\text{denominator}(\text{ratio}(n)))$$ matches the 35 first terms before, and the 35 terms after the 36-th term of A056622:
The OEIS made character replacements at $n = 36$ and at $n = 72$ to match the search. After more computation it turns out that the GCD relation differs at: $$n = 36, 72, 76, 100, 116, 128, 144,...$$
Is it possible to find the Dirichlet generating function for either \eqref{3} $$\text{ratio}(n)=A006571(n)/A366450(n) \tag{3}$$ or the denominators of \eqref{3}?
Mathematica 14 for the 50 first terms:
nn = 72;
a[n_] := DivisorSum[n, MoebiusMu[#] # &];
(*a[n_]=DivisorSum[n,MoebiusMu[#] #&];
(*in some mathematica versions you need=instead of:=or was it the \
other way around*)*)
f = (x^3 - x^2 - y^2 - y);
w[n_] :=
SeriesCoefficient[
q (Product[(1 - q^k), {k, 11, n, 11}] Product[
1 - q^k, {k, n}])^2, {q, 0, n}];
A006571 = ParallelTable[w[n], {n, 1, nn}];
A366450 =
ParallelTable[
Sum[Sum[Sum[If[GCD[f, n] == k, 1, 0]*a[k]/n, {x, 1, n}], {y, 1,
n}], {k, 1, n}], {n, 1, nn}];
ratio = A006571/A366450
ReplaceAll[Numerator[%], Indeterminate -> 1]
denominator = Denominator[ratio]
(*matching the 35 first terms and the 35 terms after the 36-th term of A056622*)
Clear[f, p, e, a, n]
f[p_, e_] := If[EvenQ[e], p^(e/2), If[e == 1, 1, p^((e - 3)/2)]];
a[1] = 1;
a[n_] := Times @@ f @@@ FactorInteger[n]; GCD[
Array[a, nn], denominator]
