I am going to do much more reading on this. I thought I'd add, that I posted some of the last stuff as word soup, and I may have misrepresented the function \(J(q)\). As far as I can tell \(J(q) = \frac{-1}{q^2}\). But when we start introducing \(m\) into the picture I'm expecting this to become much more complex.
I'm going to introduce the function:
\[
f_m(\theta; x) = \sum_{n=0}^\infty \frac{x^n}{\left(1+e^{i\theta n}x^n\right)^m} \frac{1}{2^n}\\
\]
We are getting much of the same results here. We can pretty much always assume that \(\theta =0\) without loss of generality. The results are the same upto a change of variables--which is why I think there's something modular happening here.
With this function we have:
\[
\frac{1}{2\pi i} \int_{|z| = 2} f_m(\theta;z) \frac{dz}{z^{k+1}} = 0\,\,\text{for} k \ge 1\\
\]
When \(k=0\) and \(m > 1\) we have:
\[
\frac{1}{2\pi i} \int_{|z| = 2}f_m(\theta;z) \frac{dz}{z^{k+1}} = 2^{-m}\,\,\text{for} k =0 \,\,\wedge m > 1\\
\]
The tricky part happens when \( k < 0\). Something really weird is happening here. It seems to be producing some kind of linear transformation of \(\frac{d^k}{dz^k}\Big{|}_{z=0} f _m(\theta;z)\), but it's not obvious.
This produces something very strange. For \( k > 0\) we have:
\[
b_{kn}^{(m)} = - e^{i\theta(k+1- n)}\sum_{q^n = -1} \text{Res}\left(\frac{z^{n-k-1}}{\left(1+e^{i\theta n}z^n\right)^m},z=q\right)\\
\]
where:
\[
\frac{z^n}{\left(1+e^{i\theta n}z^n\right)^m} = \sum_{k=0}^\infty b_{kn}^{(m)} z^k\\
\]
But when \(k < 0\); there is no obvious formula that appears. I thought we'd get something similar; but there appears to be something tricky going on. I think we're collecting more negatives and stuff. I'm having a little trouble finding a quick formula for the residues; because they are second order or higher residues, and for fucks sakes that's a headache, lol.
I will read your links and get around to brute forcing this stupid residue.
Also, to add to how cool this is:
\[
\sum_{m=1}^\infty f_m(\theta ; x) = \sum_{n=0}^\infty \frac{e^{-i\theta n}}{2^n} = \frac{e^{i\theta}}{e^{i\theta} - \frac{1}{2}}\\
\]
Choosing any coefficients \(a_n\) rather than \(\frac{1}{2^n}\) changes nothing from all the results I've written. I just like using this sequence because it's fast and simple. Getting a prime series will be a consequence of this; choosing a prime indicator function produces your prime series. Actually, choosing \(d_n\) where \(d_n\) is a modular sequence; then:
\[
\sum_{m=1}^\infty f_m(\theta , x) = \mathfrak{M}(\theta)\\
\]
Where \(\mathfrak{M}\) is modular.
This would mean the series:
\[
f_m(\theta; x) = \sum_{n=0}^\infty d_n \frac{x^n}{\left(1+e^{i\theta n}x^n\right)^m}\\
\]
Is intimately related to the modular function:
\[
\mathfrak{M}(\theta) = \sum_{n=0}^\infty d_ne^{i\theta n}\\
\]
So long as we take the modular group mod 2 \pi rather than mod 1; we should probably switch to \(e^{-2\pi i \theta}\) rather than \(e^{i\theta}\); this will clean up some of the algebra...
Okay! So I found the recurrence relationship for the residue. Let's call:
\[
a_{kn}^m = \frac{1}{2\pi i} \int_{|z| = 2} \frac{z^n}{(1+z^n)^m} \frac{dz}{z^{k+1}}\\
\]
Then this is equivalent to:
\[
a_{kn}^{m} = \sum_{j=1}^\infty (-1)^{j+1} a_{(k+jn)n}^{m-1}\\
\]
Because:
\[
\frac{1}{(1+z^n)^m} = \frac{1}{(1+z^n)^{m-1}} \sum_{j=1}^\infty (-1)^{j+1} z^{-j}\\
\]
When \(|z| > 1\).
This is a finite sum, because k+jn is eventually positive, and then \(a_{(k+jn)n}^m = 0\). Then our sum is written as:
\[
F_k^m = \frac{1}{2\pi i} \int_{|z| = 2} f_m(0;z) \frac{dz}{z^{k+1}}\\
\]
Where:
\[
F_k^m = \sum_{n=0}^\infty \frac{a_{kn}^m}{2^n}\\
\]
So, the function:
\[
\begin{align}
F(x) &=\sum_{k=1}^\infty F_{-k}^2 x^k\\
&= \sum_{k=1}^\infty x^k\sum_{n=0}^\infty \frac{a_{kn}^2}{2^n}\\
&= \sum_{k=1}^\infty x^k \sum_{n=0}^\infty \frac{\sum_{j=1}^\infty (-1)^{j+1}a_{(k+jn)n}}{2^n}\\
\end{align}
\]
The value \(a_{kn}\) is non-zero only when \(k = -dn\); upon which it equals \((-1)^d\). So this becomes:
\[
\sum_{j=1}^\infty (-1)^{j+1}a_{(k+jn)n} = (-1)^{d+1}d\\
\]
And Therfore!
\[
F^2(x) = \sum_{k=1}^\infty x^k \sum_{d|k} d\frac{(-1)^{d+1}}{2^{k/d}}\\
\]
..........
ALL WE DO IS ADD A POWER OF \(d\)!!!! I think this will continue and that, my conjecture is:
\[
\sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d|k} d^{m-1}\frac{(-1)^{d+1}}{2^{k/d}}\\
\]
.....
NVM
\[
F_{-k}^3 = \sum_{d|k} \frac{d(d-1)}{2} \frac{(-1)^d}{2^{k/d}}\\
\]
So the answer is:
\[
h_m(d) = \sum_{j_1=1}^d \sum_{j_2 = 1}^{j} ... \sum_{j_{m-1}=1}^{j_{m-2}} 1\\
\]
Which is simply:
\[
h_m(d) = d(d-1)(d-2)\cdots (d-m+2)/(m-1)!\\
\]
which is:
\[
\begin{align}
h_1(d) &= 1\\
h_2(d) &= d\\
h_3(d) &= d(d-1)/2\\
h_4(d) &= d(d-1)(d-2)/6\\
&\vdots\\
\end{align}
\]
Or; in a clean formula:
\[
h_m(d) = \frac{d!}{(d-m+1)!(m-1)!}
\]
And:
\[
F_{-k}^m = \sum_{d | k} h_m(d) \frac{(-1)^{d+m}}{2^{k/d}}\\
\]
Notice we have to add in a negative based on the parity of \(m\)!
And now we get that:
\[
\sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d | k} h_m(d) \frac{(-1)^{d+m}}{2^{k/d}}\\
\]
THIS IS NOT THE COEFFICIENTS OF:
\[
f_m(0;x) = \sum_{k=0}^\infty B_k^m x^k\\
\]
They are close; but something strange is happening here... And I can't map it yet.
If we add in \(\theta\), I can still map this:
\[
F_{-k}^{m}(\theta) = \frac{1}{2\pi i} \int_{|z| = 2} f_m(\theta ; z)z^{k-1}\,dz\\
\]
Then:
\[
F_{-k}^m(\theta) = e^{2 i \theta k}\sum_{d | k} h_m(d) e^{-i\theta k/d}\frac{(-1)^{d+m}}{2^{k/d}}\\
\]
I'm going to introduce the function:
\[
f_m(\theta; x) = \sum_{n=0}^\infty \frac{x^n}{\left(1+e^{i\theta n}x^n\right)^m} \frac{1}{2^n}\\
\]
We are getting much of the same results here. We can pretty much always assume that \(\theta =0\) without loss of generality. The results are the same upto a change of variables--which is why I think there's something modular happening here.
With this function we have:
\[
\frac{1}{2\pi i} \int_{|z| = 2} f_m(\theta;z) \frac{dz}{z^{k+1}} = 0\,\,\text{for} k \ge 1\\
\]
When \(k=0\) and \(m > 1\) we have:
\[
\frac{1}{2\pi i} \int_{|z| = 2}f_m(\theta;z) \frac{dz}{z^{k+1}} = 2^{-m}\,\,\text{for} k =0 \,\,\wedge m > 1\\
\]
The tricky part happens when \( k < 0\). Something really weird is happening here. It seems to be producing some kind of linear transformation of \(\frac{d^k}{dz^k}\Big{|}_{z=0} f _m(\theta;z)\), but it's not obvious.
This produces something very strange. For \( k > 0\) we have:
\[
b_{kn}^{(m)} = - e^{i\theta(k+1- n)}\sum_{q^n = -1} \text{Res}\left(\frac{z^{n-k-1}}{\left(1+e^{i\theta n}z^n\right)^m},z=q\right)\\
\]
where:
\[
\frac{z^n}{\left(1+e^{i\theta n}z^n\right)^m} = \sum_{k=0}^\infty b_{kn}^{(m)} z^k\\
\]
But when \(k < 0\); there is no obvious formula that appears. I thought we'd get something similar; but there appears to be something tricky going on. I think we're collecting more negatives and stuff. I'm having a little trouble finding a quick formula for the residues; because they are second order or higher residues, and for fucks sakes that's a headache, lol.
I will read your links and get around to brute forcing this stupid residue.
Also, to add to how cool this is:
\[
\sum_{m=1}^\infty f_m(\theta ; x) = \sum_{n=0}^\infty \frac{e^{-i\theta n}}{2^n} = \frac{e^{i\theta}}{e^{i\theta} - \frac{1}{2}}\\
\]
Choosing any coefficients \(a_n\) rather than \(\frac{1}{2^n}\) changes nothing from all the results I've written. I just like using this sequence because it's fast and simple. Getting a prime series will be a consequence of this; choosing a prime indicator function produces your prime series. Actually, choosing \(d_n\) where \(d_n\) is a modular sequence; then:
\[
\sum_{m=1}^\infty f_m(\theta , x) = \mathfrak{M}(\theta)\\
\]
Where \(\mathfrak{M}\) is modular.
This would mean the series:
\[
f_m(\theta; x) = \sum_{n=0}^\infty d_n \frac{x^n}{\left(1+e^{i\theta n}x^n\right)^m}\\
\]
Is intimately related to the modular function:
\[
\mathfrak{M}(\theta) = \sum_{n=0}^\infty d_ne^{i\theta n}\\
\]
So long as we take the modular group mod 2 \pi rather than mod 1; we should probably switch to \(e^{-2\pi i \theta}\) rather than \(e^{i\theta}\); this will clean up some of the algebra...
Okay! So I found the recurrence relationship for the residue. Let's call:
\[
a_{kn}^m = \frac{1}{2\pi i} \int_{|z| = 2} \frac{z^n}{(1+z^n)^m} \frac{dz}{z^{k+1}}\\
\]
Then this is equivalent to:
\[
a_{kn}^{m} = \sum_{j=1}^\infty (-1)^{j+1} a_{(k+jn)n}^{m-1}\\
\]
Because:
\[
\frac{1}{(1+z^n)^m} = \frac{1}{(1+z^n)^{m-1}} \sum_{j=1}^\infty (-1)^{j+1} z^{-j}\\
\]
When \(|z| > 1\).
This is a finite sum, because k+jn is eventually positive, and then \(a_{(k+jn)n}^m = 0\). Then our sum is written as:
\[
F_k^m = \frac{1}{2\pi i} \int_{|z| = 2} f_m(0;z) \frac{dz}{z^{k+1}}\\
\]
Where:
\[
F_k^m = \sum_{n=0}^\infty \frac{a_{kn}^m}{2^n}\\
\]
So, the function:
\[
\begin{align}
F(x) &=\sum_{k=1}^\infty F_{-k}^2 x^k\\
&= \sum_{k=1}^\infty x^k\sum_{n=0}^\infty \frac{a_{kn}^2}{2^n}\\
&= \sum_{k=1}^\infty x^k \sum_{n=0}^\infty \frac{\sum_{j=1}^\infty (-1)^{j+1}a_{(k+jn)n}}{2^n}\\
\end{align}
\]
The value \(a_{kn}\) is non-zero only when \(k = -dn\); upon which it equals \((-1)^d\). So this becomes:
\[
\sum_{j=1}^\infty (-1)^{j+1}a_{(k+jn)n} = (-1)^{d+1}d\\
\]
And Therfore!
\[
F^2(x) = \sum_{k=1}^\infty x^k \sum_{d|k} d\frac{(-1)^{d+1}}{2^{k/d}}\\
\]
..........
ALL WE DO IS ADD A POWER OF \(d\)!!!! I think this will continue and that, my conjecture is:
\[
\sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d|k} d^{m-1}\frac{(-1)^{d+1}}{2^{k/d}}\\
\]
.....
NVM
\[
F_{-k}^3 = \sum_{d|k} \frac{d(d-1)}{2} \frac{(-1)^d}{2^{k/d}}\\
\]
So the answer is:
\[
h_m(d) = \sum_{j_1=1}^d \sum_{j_2 = 1}^{j} ... \sum_{j_{m-1}=1}^{j_{m-2}} 1\\
\]
Which is simply:
\[
h_m(d) = d(d-1)(d-2)\cdots (d-m+2)/(m-1)!\\
\]
which is:
\[
\begin{align}
h_1(d) &= 1\\
h_2(d) &= d\\
h_3(d) &= d(d-1)/2\\
h_4(d) &= d(d-1)(d-2)/6\\
&\vdots\\
\end{align}
\]
Or; in a clean formula:
\[
h_m(d) = \frac{d!}{(d-m+1)!(m-1)!}
\]
And:
\[
F_{-k}^m = \sum_{d | k} h_m(d) \frac{(-1)^{d+m}}{2^{k/d}}\\
\]
Notice we have to add in a negative based on the parity of \(m\)!
And now we get that:
\[
\sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d | k} h_m(d) \frac{(-1)^{d+m}}{2^{k/d}}\\
\]
THIS IS NOT THE COEFFICIENTS OF:
\[
f_m(0;x) = \sum_{k=0}^\infty B_k^m x^k\\
\]
They are close; but something strange is happening here... And I can't map it yet.
If we add in \(\theta\), I can still map this:
\[
F_{-k}^{m}(\theta) = \frac{1}{2\pi i} \int_{|z| = 2} f_m(\theta ; z)z^{k-1}\,dz\\
\]
Then:
\[
F_{-k}^m(\theta) = e^{2 i \theta k}\sum_{d | k} h_m(d) e^{-i\theta k/d}\frac{(-1)^{d+m}}{2^{k/d}}\\
\]
