Define the following,


$$j(\tau) = \Big(\tfrac{E\_4(\tau)}{\eta^8(\tau)}\Big)^3 = {1 \over q} + 744 + \color{blue}{196884} q + 21493760 q^2 + 864299970 q^3 + \cdots \tag{1}$$


$$j\_{2A}(\tau) =\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 = \tfrac{1}{q} + 104 + \color{blue}{4372}q + 96256q^2 + 1240002q^3+\cdots \tag{2}$$


$$j\_{3A}(\tau) =\Big(\big(\tfrac{\eta(\tau)}{\eta(3\tau)}\big)^{6}+3^3 \big(\tfrac{\eta(3\tau)}{\eta(\tau)}\big)^{6}\Big)^2 = \tfrac{1}{q} + 42 + \color{blue}{783}q + 8672q^2 +65367q^3+\dots \tag{3}$$


$$j\_{4A}(\tau)=\Big(\big(\tfrac{\eta(\tau)}{\eta(4\tau)}\big)^{4}+4^2 \big(\tfrac{\eta(4\tau)}{\eta(\tau)}\big)^{4}\Big)^2 = \tfrac{1}{q} + 24+ \color{blue}{276}q + 2048q^2 +11202q^3+\dots\\ \tag{4}$$
$$j\_{7A}(\tau)=\Big(\big(\tfrac{\eta(\tau)}{\eta(7\tau)}\big)^{2}+7 \big(\tfrac{\eta(7\tau)}{\eta(\tau)}\big)^{2}\Big)^2 = \tfrac{1}{q} + 10+ \color{blue}{51}q + 204q^2 +681q^3+\dots\\ \tag{5}$$


where $\eta(\tau)$ is the *Dedekind eta function*.


(Note that $196883, 4371, 782, 276, 51$ are degrees of the irreducible representations of the *Monster*, *Baby Monster*, *Fischer Fi23*, *Conway Co1*, and *Held* groups, respectively.) 


The asymptotic formula (by Rademacher?) for the coefficient of $q^n$ of $(1)$ is given by,


$$a(n) \approx \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\,n^{3/4}}\tag{for 1}$$


For example, let $a(n)$ be the coefficient, and $a'(n)$ given by the formula, then,


$$\begin{array}{cccc}
n&100&200&300\\
a(n)&8.38\,\text{x}\,10^{52}&2.011\,\text{x}\,10^{75}&3.293\,\text{x}\,10^{92}\\
a'(n)&8.40\,\text{x}\,10^{52}&2.016\,\text{x}\,10^{75}&3.299\,\text{x}\,10^{92}\\
\end{array}$$


***Question***: What are the analogous coefficient formulas for $(2), (3), (4), (5)$? 


$\color{blue}{\text{Edit}}$: After perusing the [***OEIS***](https://oeis.org/A097340), it seems that the fourth has, 


$$d(n) \approx \frac{e^{2\pi\sqrt{n}}}{2\,n^{3/4}}\tag{for 4}$$


$$\begin{array}{cccc}
n&100&200&300\\
d(n)&3.04\,\text{x}\,10^{25}&3.64\,\text{x}\,10^{36}&1.26\,\text{x}\,10^{45}\\
d'(n)&3.06\,\text{x}\,10^{25}&3.66\,\text{x}\,10^{36}&1.27\,\text{x}\,10^{45}\\
\end{array}$$


though I have no proof that this is its correct asymptotic formula.



Asymptotic formulas for Monster-related modular functions?
In a recent paper with Ken Ono and John Duncan, we give exact formulas for the coefficients of the Monsterous Moonshine modules. I will give a link to a preprint a few days when I can, but the relevant part relies on prior work of Bringmann and Ono (<http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/115.pdf>) which gives exact formulas for coefficients of modular forms assuming knowledge of their poles at cusps.


Each of your $j\_g(\tau)=\sum a\_g(n)q^n$ is a Hauptmodul for a group of the form $\Gamma\_g=\left\langle \Gamma\_0(N),\begin{pmatrix}0&-1\\N&0\end{pmatrix}\right\rangle,$ where $N=1,2,3,$ or $4$. In this case, the dominant term in the exact formulas for $a\_g(n)$ is $$\frac{2\pi}{\sqrt{N n}}I\_1\left(4\pi \sqrt{n/N}\right)\sim \frac{e^{4\pi \sqrt{n/N}}}{\sqrt{2}N^{1/4}n^{3/4}}.$$ 
Here $I\_k$ is the usual $I$-Bessel function.


