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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.)

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Just a remark: I suspect knowing your conjecture for $k =2$ and $k = 3$ would imply it for all $k$, since one can express $E_{2k}$ in terms of $E_4$ and $E_6$. – David Loeffler Sep 18 '12 at 14:20
up vote 2 down vote accepted

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 $j(\tau)$, 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".

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,

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

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Nice. Just a quick question: why do P, Q and R have a subscript n? – Vincent Dec 29 '15 at 21:57
@Vincent: Since their argument is $q=e^{2\pi i \tau}$ and $\tau = \frac{1+\sqrt{\color{blue}{-n}}}{2}$. – Tito Piezas III Dec 30 '15 at 3:22
aah, right, thank you! – Vincent Dec 30 '15 at 11:02

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