Given $q = e^{2\pi i \tau}$ and the Eisenstein series $E\_{2k}(\tau)$, i.e.,


$$E\_2(\tau) = 1-24\sum\_{n=1}^\infty \frac{n q^n}{1-q^n}$$


$$E\_4(\tau) = 1+240\sum\_{n=1}^\infty \frac{n^3 q^n}{1-q^n}$$


and so on. Define the function,


$$F\_{2k}(\tau) = \frac{E\_{2k}(\tau)}{\left(E\_2(\tau)-\frac{3}{\pi\; \Im(\tau)}\right)^k}$$


for $k \geq 2$, where $\tau = \frac{1+\sqrt{-d}}{2}$, $\Im(\tau)$ is the imaginary part of $\tau$, and $d$ has class number $h(-d) = m$. For example, we have,


$$F\_4\left(\tfrac{1+\sqrt{-163}}{2}\right) = \frac{5\cdot23\cdot29\cdot163}{2^2\cdot3\cdot181^2}$$


$$F\_6\left(\tfrac{1+\sqrt{-163}}{2}\right) = \frac{7\cdot11\cdot19\cdot127\cdot163^2}{2^9\cdot181^3}$$


$$F\_8(\tau) = F\_4^2(\tau)$$


and so on.


**Question**: In general, is it true that for $k \geq 2$ the function $F\_{2k}$, ***like the j-function, is an algebraic number of degree m = h(-d)***? (I've tested it with *d* with higher class numbers, and it seems to be true.)



Eisenstein series and 163?
I inadvertently came across the partial answer to my own question. It turns out Ramanujan had already explored something similar. Let $q = e^{2\pi i \tau}$, $\tau=\tfrac{1+\sqrt{-n}}{2}$, and,


$$P\_n = 1-24\sum\_{k=1}^\infty \frac{k q^k}{1-q^k}$$


$$Q\_n = 1+240\sum\_{k=1}^\infty \frac{k^3 q^k}{1-q^k}$$


$$R\_n = 1-504\sum\_{k=1}^\infty \frac{k^5 q^k}{1-q^k}$$


$$a\_n = \frac{b\_n}{6} \left(1-\frac{Q\_n}{R\_n}\left(P\_n-\frac{6}{\pi\sqrt{n}}\right)\right)$$


$$b\_n = \sqrt{n(1728-j(\tau))}$$


and *[j-function](http://mathworld.wolfram.com/j-Function.html)* $j(\tau)$, then,


$$\frac{1}{F\_4(\tau)} = \frac{1}{Q\_n}\left(P\_n-\frac{6}{\pi\sqrt{n}}\right)^2 = \left(1-\frac{6\,a\_n}{b\_n}\right)^2\left(\frac{-1728+j(\tau)}{j(\tau)}\right)$$


which is ***Theorem 7.1*** (p.13) of Berndt's "*[Ramanujan's contributions to Eisenstein Series](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.115.2404&rep=rep1&type=pdf)*". 


Thus, it remains to establish that the square factor of the *RHS* is an algebraic number with the same degree as $j(\tau)$. (It seems it suffices to prove it for $a\_n$.)


P.S. Note that for $n>3$,


$$\sum\_{k=0}^\infty (-1)^k\frac{(6k)!}{(3k)!(k!)^3}\frac{a\_n+b\_n k}{(-j(\tau))^{k+1/2}} = \frac{1}{\pi}$$


in particular, for $n=163$, we have the Chudnovsky's formula,


$$\sum\_{k=0}^\infty (-1)^k\frac{(6k)!}{(3k)!(k!)^3}\frac{163096908+6541681608 k}{(640320^3)^{k+1/2}} = \frac{1}{\pi}$$


