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 Fi _{23}*,

*Conway Co*, and

_{1}*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**, 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.