The $j$-ivariant has the following Fourier expansion $$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$ Here is $q=e^{2\pi i \tau}$.

Is there some simple effective bound on the coefficients $a_n$?

Backround.

This question comes from On the “gap” in a theorem of Heegner. Let $D$ be a negative discriminant such that $h(D)=1$. We want to show that $J=j(\sqrt{D})$ generates a cubic extension of $\mathbf Q$. Since we have at our disposal a monic cubic polynomial with rational coefficients, the modular equation $\Phi_2(X,j)$, whose root is $J$, and the other two roots are non-real, it is sufficient to show that $J$ is not an integer.

In this case $j=j\left(\frac{-1+\sqrt D}{2} \right)$ is also an integer. Set

$$t=e^{2\pi i(-1+\sqrt D)/2}.$$

Then

$$J=\frac{1}{t^2}+744+196884t^2+O(t^4)$$

and

$$j^2-1488j+160512-J=42987520t+O(t^2)$$.

On the left there is an integer. However the right side tends to zero as $|D|$ gets large. Stark asserts that $|D|>60$ is enough. Why is it enough?

The $j$-ivariant has the following Fourier expansion $$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$ Here is $q=e^{2\pi i \tau}$.

Is there some simple effective bound on the coefficients $a_n$?

Backround.

This question comes from On the “gap” in a theorem of Heegner. Let $D$ be a negative discriminant such that $h(D)=1$. We want to show that $J=j(\sqrt{D})$ generates a cubic extension of $\mathbf Q$. Since we have at our disposal a monic cubic polynomial with rational coefficients, the modular equation $\Phi_2(X,j)$, whose root is $J$, and the other two roots are non-real, it is sufficient to show that $J$ is not an integer.

In this case $j=j\left(\frac{-1+\sqrt D}{2} \right)$ is also an integer. Set

$$t=e^{2\pi i(-1+\sqrt D)/2}.$$

Then

$$J=\frac{1}{t^2}+744+196884t^2+O(t^4),$$

and

$$j^2-1488j+160512-J=42987520t+O(t^2).$$.

On the left there is an integer. However the right side tends to zero as $|D|$ gets large. Stark asserts that $|D|>60$ is enough for the RHS to be less than 1. Why is it enough?