Okay, so I'm going to call the Error function \(E^m(x)\). And I'll describe it in full detail.
Let's call the function:
\[
F^m(x) = \sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d | k} h_m(d) \frac{(-1)^{m+d}}{2^{k/d}}\\
\]
Then, we will call:
\[
E^{m}(x) = f_m(x)-f_m(0) +(-1)^{m+1} F^m(x)\\
\]
For \(m=1,2\) this function is just \(0\). For \(m=3,4\) it's the same function. For \(m=4,5\) it gets a bit wilder. And as I continue shit starts to get a bit divergey; but that's because I'm doing quick calculations and avoiding large number calculations.
All the coefficients are rational of \(E^m\); and it looks pretty god damn nice, regardless of \(m\). But it also looks pretty random with the placement of these rational coefficients.
Also, it's super super suspect; that the only values which are nonzero are when \(m > 2\); which is just like modular functions--the Eisenstein series is only interesting when \(m > 2\).
The coefficients of \(E^m(x)\) also just seem to be:
\[
a_{mk}/2^k\\
\]
where \(a_{mk}\) is a natural number. So I may be missing an extra term that doesn't appear for \(m=1,2\). Jesus this shit is annoying.
NVM
I figured it out. And I can explain the difference.
Let's write:
\[
f_m^{k}(0) = B_k^m\\
\]
Where:
\[
f_m(x) = \sum_{k=0}^\infty B_k^m x^k\\
\]
Then:
\[
f_m(x) = \sum_{n=0}^\infty \frac{x^n}{\left(1+x^n\right)^m} \frac{1}{2^n}\\
\]
But it also equals:
\[
f_m(x) = \sum_{j=0}^\infty \sum_{n=0}^\infty (-1)^j x^{nj} \frac{x^n}{\left(1+x^n\right)^{m-1}}\frac{1}{2^n}\\
\]
This produces a SIMILAR recursive equation as with \(F_{-k}^m\), but it is not the same. We gather some extra terms
Let's call the function:
\[
F^m(x) = \sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d | k} h_m(d) \frac{(-1)^{m+d}}{2^{k/d}}\\
\]
Then, we will call:
\[
E^{m}(x) = f_m(x)-f_m(0) +(-1)^{m+1} F^m(x)\\
\]
For \(m=1,2\) this function is just \(0\). For \(m=3,4\) it's the same function. For \(m=4,5\) it gets a bit wilder. And as I continue shit starts to get a bit divergey; but that's because I'm doing quick calculations and avoiding large number calculations.
All the coefficients are rational of \(E^m\); and it looks pretty god damn nice, regardless of \(m\). But it also looks pretty random with the placement of these rational coefficients.
Also, it's super super suspect; that the only values which are nonzero are when \(m > 2\); which is just like modular functions--the Eisenstein series is only interesting when \(m > 2\).
The coefficients of \(E^m(x)\) also just seem to be:
\[
a_{mk}/2^k\\
\]
where \(a_{mk}\) is a natural number. So I may be missing an extra term that doesn't appear for \(m=1,2\). Jesus this shit is annoying.
NVM
I figured it out. And I can explain the difference.
Let's write:
\[
f_m^{k}(0) = B_k^m\\
\]
Where:
\[
f_m(x) = \sum_{k=0}^\infty B_k^m x^k\\
\]
Then:
\[
f_m(x) = \sum_{n=0}^\infty \frac{x^n}{\left(1+x^n\right)^m} \frac{1}{2^n}\\
\]
But it also equals:
\[
f_m(x) = \sum_{j=0}^\infty \sum_{n=0}^\infty (-1)^j x^{nj} \frac{x^n}{\left(1+x^n\right)^{m-1}}\frac{1}{2^n}\\
\]
This produces a SIMILAR recursive equation as with \(F_{-k}^m\), but it is not the same. We gather some extra terms
