TITLE: Why are values of Eisenstein $E_2^*$ algebraic integers?
QUESTION [23 upvotes]: I'm looking for a proof that the following term is an algebraic integer whenever $\tau_N=\frac{N+\sqrt{-N}}{2}$ is a quadratic irrationality with class number $1$:
$$A_N:=\sqrt{-N}\cdot\frac{E_2(\tau_N)-\frac{3}{\pi\cdot Im(\tau_N)}}{\eta^4(\tau_N)}$$
Here $\eta$ denotes the Dedekind $\eta$-Function and $E_2$ is the Eisenstein series of weight $2$.
As @HenriCohen said here: How to compute Coefficients in Chudnovsky's Formula?
it follows from theorems of complex multiplication, but I couldn't find such theorems.
I calculated the numerical value of $A_N$ for all discriminants with class number 1. The results are:

$A_3 = 0$
$A_4 = 0$
$A_{7}=3\cdot e^{i\pi/3}$
$A_{8}=4\cdot e^{i\pi/2}$
$A_{11}=8\cdot e^{i\pi/3}$
$A_{12}=6\cdot4^{1/3}\cdot e^{i\pi/2}$
$A_{16}=12\cdot2^{1/2}\cdot e^{i\pi/2}$
$A_{19}=24\cdot e^{i\pi/3}$
$A_{27}=24\cdot9^{1/3}\cdot e^{i\pi/3}$
$A_{28}=54\cdot e^{i\pi/2}$
$A_{43}=144\cdot e^{i\pi/3}$
$A_{67}=456\cdot e^{i\pi/3}$
$A_{163}=8688\cdot e^{i\pi/3}$

Thus we get numerically, that these $A_N$ are algebraic integers, but I don't see how I can prove it. Does anyone know how to do that?
EDIT: Thanks to the answer of Nikos Bagis, only the $A_N$ with odd $N$ remain to be proven. I moved the remaining part of the question here, where a complete answer was given by Michael Griffin.
EDIT: Complete Solution:
The answer of Michael Griffin (see here) can now be found in more details in the appendix of this arXiv-preprint.
Edit: Ramanujan-Sato-Series
Tito Piezas III found some Ramanujan-Sato-Series of level 9 which can be expressed with these numbers $A_N$ (see this question).

REPLY [4 votes]: This is too long to fit in a comment.

Ramanujan established in his monumental paper Modular Equations and Approximations to $\pi$ that the desired expression is an algebraic number if $\tau_n=\sqrt{-n} $ where $n$ is a positive rational number.
Let $$P(q) =1-24\sum_{j=1}^{\infty}\frac{jq^{j} }{1-q^{j}}\tag{1}$$ and then Ramanujan proved that $$P(e^{-2\pi\sqrt{n}}) =\left( \frac{2K}{\pi}\right)^2A_n+\frac{3}{\pi\sqrt{n}}\tag{2}$$ where $A_n$ is an algebraic number dependent on $n$ provided that $n$ is a positive rational number.  Here $K=K(k) $ is the complete elliptic integral of first kind with modulus $k$ and $k$ corresponds to nome $q$ so that $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},q=e^{-\pi\sqrt{n}}\tag{3}$$ and $k$ is an algebraic number if $n$ is a positive rational number.  Ramanujan's proof is presented in one of my blog posts. 
If $q=\exp(\pi i\tau) $ then we have $P(q^2) =E_2(\tau)$. Further we have $$\eta(\tau) =q^{1/12}\prod_{j=1}^{\infty}(1-q^{2j}),q=e^{\pi i\tau}\tag{4}$$ It is well known that the eta function can be expressed in terms of $k, K$ as $$\eta(\tau)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{5}$$ and therefore the equation $(2)$ can be written as $$A_n=\dfrac{E_{2}(\tau_n)-\dfrac{3}{\pi\Im{\tau_n}}}{\eta^4(\tau_n)}$$ which is an algebraic number.
Note that if $\tau_n=\dfrac{n+\sqrt{-n} }{2}$ then we have $E_2(\tau_n)=P(q^2)$ if $n$ is even and $q=e^{\pi i\tau_n} $ and if $n$ is odd then $E_2(\tau_n)=P(-q^2)$. Using Ramanujan's technique one can prove that $$P(e^{-\pi\sqrt{n}})=\left(\frac{2K}{\pi}\right) ^2B_n+\frac{6}{\pi\sqrt{n}}\tag{6}$$ (just replace $n$ in $(2)$ by $n/4$ and $B_n=A_{n/4}$) and $$P(-e^{-\pi\sqrt{n}}) =\left(\frac{2K}{\pi}\right)^2C_n+\frac{6}{\pi\sqrt{n}}\tag{7}$$ where $B_n, C_n$ are algebraic numbers and $n$ is a positive rational number and thus the expression mentioned in the question is an algebraic number. Proving that it is an algebraic integer is unfortunately not possible via Ramanujan methods.