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I. Monstrous Moonshine

Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$ for positive integer $d$. Given the Dedekind eta function $\eta(\tau)$, consider the known functions,

\begin{align} j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + \color{red}{196884}q + 21493760q^2 +\cdots\\ j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 = \frac{1}{q} + 104 + \color{red}{4372}q + 96256q^2 +\cdots\\ j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + \color{red}{783}q + 8672q^2 +\cdots\\ j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} = \frac{1}{q} + 24+ \color{red}{276}q + \dots \end{align}

where $196883, 4371, 782$ (the first three red numbers minus $1$) is the smallest degree $>1$ of the irreducible representations of the Monster group, Baby Monster, and Fischer group $Fi_{23}$, respectively. While the fourth as $4\times276=4\times\binom{24}{2} = 1104$ is for the 24-dimensional Leech lattice and Conway group.

II. Definitions

Define the following functions,

$$ \alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1A}(\tau)}}\right)\qquad\qquad\qquad$$

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

III. Conjecture 1

For simplicity, let $\alpha_n = \alpha_n(\tau)$. Then for parameters $s = \frac16,\frac14,\frac13,\frac12,$ we conjecture that those four moonshine functions also have a hypergeometric formula,

$$j_{1A}(\tau) = \frac{432}{\alpha_1\,(1-\alpha_1)}=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\qquad\qquad$$

$$j_{2A}(\tau) = \frac{64}{\alpha_2\,(1-\alpha_2)}=\left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}$$

$$j_{3A}(\tau) = \frac{27}{\alpha_3\,(1-\alpha_3)}=\left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}$$

$$j_{4A}(\tau) = \frac{16}{\alpha_4\,(1-\alpha_4)}=\left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}$$

but avoiding $\tau$ which leads to division by zero.

IV. Conjecture 2

Furthermore, again let $\alpha_n = \alpha_n(\tau)$, then,

$$\left(\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}\right)^2=-1\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}\right)^2=-2\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}\right)^2=-3\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}\right)^2=-4\,\tau^2$$

for $\tau = \sqrt{-d}\,$ and $\tau = \frac{1+\sqrt{-d}}2$.

Note: If $\tau=\frac12\sqrt{-r}$, then the last simplifies to the well-known,

$$\frac{K'(k)}{K(k)} = \sqrt{r}$$

with complete elliptic integral of the first kind $K(k).$ It was the model for conjecturing that similar behavior occurs for the first three. Thus $\alpha_4\big(\tfrac{\tau}2\big)$ is the square of the elliptic modulus $k$,

$$\alpha_4\big(\tfrac{\tau}2\big) = k^2 = \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,e^{\pi i\tau})}{\vartheta_3(0,e^{\pi i\tau})}\right)^4$$

and is also the modular lambda function $\lambda(\tau)$.

V. Question

Q: So are the two conjectures true?

I. Monstrous Moonshine

Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$ for positive integer $d$. Given the Dedekind eta function $\eta(\tau)$, consider the known functions,

\begin{align} j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + \color{red}{196884}q + 21493760q^2 +\cdots\\ j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 = \frac{1}{q} + 104 + \color{red}{4372}q + 96256q^2 +\cdots\\ j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + \color{red}{783}q + 8672q^2 +\cdots\\ j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} = \frac{1}{q} + 24+ \color{red}{276}q + \dots \end{align}

where $196883, 4371, 782$ (the first three red numbers minus $1$) is the smallest degree $>1$ of the irreducible representations of the Monster group, Baby Monster, and Fischer group $Fi_{23}$, respectively. While the fourth as $4\times276=4\times\binom{24}{2} = 1104$ is for the 24-dimensional Leech lattice and Conway group.

II. Definitions

Define the following functions,

$$ \alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1A}(\tau)}}\right)\qquad\qquad\qquad$$

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

III. Conjecture 1

Part A.

For simplicity, let $\alpha_n = \alpha_n(\tau)$. Then for parameters $s = \frac16,\frac14,\frac13,\frac12,$ we conjecture that those four moonshine functions also have a hypergeometric formula,

$$j_{1A}(\tau) = \frac{432}{\alpha_1\,(1-\alpha_1)}=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\qquad\qquad$$

$$j_{2A}(\tau) = \frac{64}{\alpha_2\,(1-\alpha_2)}=\left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}$$

$$j_{3A}(\tau) = \frac{27}{\alpha_3\,(1-\alpha_3)}=\left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}$$

$$j_{4A}(\tau) = \frac{16}{\alpha_4\,(1-\alpha_4)}=\left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}$$

but avoiding $\tau$ which leads to division by zero.

Part B. (Added Sept 17)

As a consequence of the proposed identities for $\alpha_n(1-\alpha)$ discussed in the previous post, we can have some more nice hypergeometric formulas,

$$j_{2A}(\tau) = \left(8\times\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)^6}{A(q)\,B(q)\,C(q)}\right)^2$$

$$j_{3A}(\tau) = \left(3\times\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)^3}{a(q)\,b(q)\,c(q)}\right)^3$$

$$j_{4A}(\tau) = \left(2\times\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)^{3/2}}{\vartheta_2(q)\,\vartheta_3(q)\,\vartheta_4(q)}\right)^4$$

where,

$$A(q)^2 = B(q)^2+C(q)^2$$ $$a(q)^3 = b(q)^3+c(q)^3$$ $$\vartheta_3(q)^4 = \vartheta_2(q)^4+\vartheta_4(q)^4$$

using the set with the odd divisor function $A(q), B(q), C(q)$, the Borwein cubic theta functions $a(q), b(q), c(q)$, and the Jacobi theta functions $\vartheta_n(0,q)=\vartheta(q)$.

For consistency all $q$, including the argument of the Jacobi thetas, is $q = e^{2\pi i \tau}$.

IV. Conjecture 2

Again let $\alpha_n = \alpha_n(\tau)$, then,

$$\left(\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}\right)^2=-1\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}\right)^2=-2\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}\right)^2=-3\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}\right)^2=-4\,\tau^2$$

for $\tau = \sqrt{-d}\,$ and $\tau = \frac{1+\sqrt{-d}}2$.

Note: If $\tau=\frac12\sqrt{-r}$, then the last simplifies to the well-known,

$$\frac{K'(k)}{K(k)} = \sqrt{r}$$

with complete elliptic integral of the first kind $K(k).$ It was the model for conjecturing that similar behavior occurs for the first three. Thus $\alpha_4\big(\tfrac{\tau}2\big)$ is the square of the elliptic modulus $k$,

$$\alpha_4\big(\tfrac{\tau}2\big) = k^2 = \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,e^{\pi i\tau})}{\vartheta_3(0,e^{\pi i\tau})}\right)^4$$

and is also the modular lambda function $\lambda(\tau)$.

V. Question

Q: So are the two conjectures true?

I. Monstrous Moonshine

Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$ for positive integer $d$. Given the Dedekind eta function $\eta(\tau)$, consider the known functions,

\begin{align} j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + \color{red}{196884}q + 21493760q^2 +\cdots\\ j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 = \frac{1}{q} + 104 + \color{red}{4372}q + 96256q^2 +\cdots\\ j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + \color{red}{783}q + 8672q^2 +\cdots\\ j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} = \frac{1}{q} + 24+ \color{red}{276}q + \cdots \end{align}

where $196883, 4371, 782$ (the red numbers minus $1$) is the smallest degree $>1$ of the representations of the Monster group, Baby Monster, and Fischer group $Fi_{23}$, respectively. While $4\times276=4\times\binom{24}{2} = 1104$ is for the 24-dimensional Leech lattice and Conway group.

II. Definitions

Define the following functions,

$$ \alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1A}(\tau)}}\right)\qquad\qquad\qquad$$

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

III. Conjecture 1

For simplicity, let $\alpha_n = \alpha_n(\tau)$. Then for parameters $s = \frac16,\frac14,\frac13,\frac12,$ we conjecture that those four moonshine functions also have a hypergeometric formula,

$$j_{1A}(\tau) = \frac{432}{\alpha_1\,(1-\alpha_1)}=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\qquad\qquad$$

$$j_{2A}(\tau) = \frac{64}{\alpha_2\,(1-\alpha_2)}=\left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}$$

$$j_{3A}(\tau) = \frac{27}{\alpha_3\,(1-\alpha_3)}=\left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}$$

$$j_{4A}(\tau) = \frac{16}{\alpha_4\,(1-\alpha_4)}=\left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}$$

but avoiding $\tau$ which leads to division by zero.

IV. Conjecture 2

Furthermore, again let $\alpha_n = \alpha_n(\tau)$, then,

$$\left(\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}\right)^2=-1\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}\right)^2=-2\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}\right)^2=-3\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}\right)^2=-4\,\tau^2$$

for $\tau = \sqrt{-d}$ and $\tau = \frac{1+\sqrt{-d}}2$.

Note: If $\tau=\frac12\sqrt{-r}$, then the last simplifies to the well-known,

$$\frac{K'(k)}{K(k)} = \sqrt{r}$$

with complete elliptic integral of the first kind $K(k).$ It was the model for conjecturing that similar behavior occurs for the first three. Thus $\alpha_4\big(\tfrac{\tau}2\big)$ is the square of the elliptic modulus $k$,

$$\alpha_4\big(\tfrac{\tau}2\big) = k^2 = \lambda(\tau) = \frac{16}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,e^{\pi i\tau})}{\vartheta_3(0,e^{\pi i\tau})}\right)^4$$

and is the modular lambda function $\lambda(\tau)$.

V. Question

Q: So are the two conjectures true?

I. Monstrous Moonshine

Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$ for positive integer $d$. Given the Dedekind eta function $\eta(\tau)$, consider the known functions,

\begin{align} j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + \color{red}{196884}q + 21493760q^2 +\cdots\\ j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 = \frac{1}{q} + 104 + \color{red}{4372}q + 96256q^2 +\cdots\\ j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + \color{red}{783}q + 8672q^2 +\cdots\\ j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} = \frac{1}{q} + 24+ \color{red}{276}q + \dots \end{align}

where $196883, 4371, 782$ (the first three red numbers minus $1$) is the smallest degree $>1$ of the irreducible representations of the Monster group, Baby Monster, and Fischer group $Fi_{23}$, respectively. While the fourth as $4\times276=4\times\binom{24}{2} = 1104$ is for the 24-dimensional Leech lattice and Conway group.

II. Definitions

Define the following functions,

$$ \alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1A}(\tau)}}\right)\qquad\qquad\qquad$$

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

III. Conjecture 1

For simplicity, let $\alpha_n = \alpha_n(\tau)$. Then for parameters $s = \frac16,\frac14,\frac13,\frac12,$ we conjecture that those four moonshine functions also have a hypergeometric formula,

$$j_{1A}(\tau) = \frac{432}{\alpha_1\,(1-\alpha_1)}=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\qquad\qquad$$

$$j_{2A}(\tau) = \frac{64}{\alpha_2\,(1-\alpha_2)}=\left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}$$

$$j_{3A}(\tau) = \frac{27}{\alpha_3\,(1-\alpha_3)}=\left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}$$

$$j_{4A}(\tau) = \frac{16}{\alpha_4\,(1-\alpha_4)}=\left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}$$

but avoiding $\tau$ which leads to division by zero.

IV. Conjecture 2

Furthermore, again let $\alpha_n = \alpha_n(\tau)$, then,

$$\left(\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}\right)^2=-1\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}\right)^2=-2\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}\right)^2=-3\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}\right)^2=-4\,\tau^2$$

for $\tau = \sqrt{-d}\,$ and $\tau = \frac{1+\sqrt{-d}}2$.

Note: If $\tau=\frac12\sqrt{-r}$, then the last simplifies to the well-known,

$$\frac{K'(k)}{K(k)} = \sqrt{r}$$

with complete elliptic integral of the first kind $K(k).$ It was the model for conjecturing that similar behavior occurs for the first three. Thus $\alpha_4\big(\tfrac{\tau}2\big)$ is the square of the elliptic modulus $k$,

$$\alpha_4\big(\tfrac{\tau}2\big) = k^2 = \lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,e^{\pi i\tau})}{\vartheta_3(0,e^{\pi i\tau})}\right)^4$$

and is also the modular lambda function $\lambda(\tau)$.

V. Question

Q: So are the two conjectures true?

I. Monstrous Moonshine

Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$ for positive integer $d$. Given the Dedekind eta function $\eta(\tau)$, consider the known functions,

\begin{align} j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + \color{red}{196884}q + 21493760q^2 +\cdots\\ j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 = \frac{1}{q} + 104 + \color{red}{4372}q + 96256q^2 +\cdots\\ j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + \color{red}{783}q + 8672q^2 +\cdots\\ j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} = \frac{1}{q} + 24+ \color{red}{276}q + \cdots \end{align}

where $196883, 4371, 782$ (the red numbers minus $1$) is the smallest degree $>1$ of the representations of the Monster group, Baby Monster, and Fischer group $Fi_{23}$, respectively. While $4\times276=4\times\binom{24}{2} = 1104$ is for the 24-dimensional Leech lattice and Conway group.

II. Definitions

Define the following functions,

$$ \alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1A}(\tau)}}\right)\qquad\qquad\qquad$$

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

III. Conjecture 1

For simplicity, let $\alpha_n = \alpha_n(\tau)$. Then for parameters $s = \frac16,\frac14,\frac13,\frac12,$ we conjecture that those four moonshine functions also have a hypergeometric formula,

$$j_{1A}(\tau) = \frac{432}{\alpha_1\,(1-\alpha_1)}=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\qquad\qquad$$

$$j_{2A}(\tau) = \frac{64}{\alpha_2\,(1-\alpha_2)}=\left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}$$

$$j_{3A}(\tau) = \frac{27}{\alpha_3\,(1-\alpha_3)}=\left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}$$

$$j_{4A}(\tau) = \frac{16}{\alpha_4\,(1-\alpha_4)}=\left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}$$

but avoiding $\tau$ which leads to division by zero.

IV. Conjecture 2

Furthermore, again let $\alpha_n = \alpha_n(\tau)$, then,

$$\left(\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}\right)^2=-1\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}\right)^2=-2\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}\right)^2=-3\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}\right)^2=-4\,\tau^2$$

for $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$.

V. Question

Q: So are the two conjectures true?

I. Monstrous Moonshine

Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$ for positive integer $d$. Given the Dedekind eta function $\eta(\tau)$, consider the known functions,

\begin{align} j_{1A}(\tau) &= \left(\frac{E_4(\tau)}{\eta^8(\tau)}\right)^3 = \frac{1}{q} + 744 + \color{red}{196884}q + 21493760q^2 +\cdots\\ j_{2A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 = \frac{1}{q} + 104 + \color{red}{4372}q + 96256q^2 +\cdots\\ j_{3A}(\tau) &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 = \frac{1}{q} + 42 + \color{red}{783}q + 8672q^2 +\cdots\\ j_{4A}(\tau)&=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} = \frac{1}{q} + 24+ \color{red}{276}q + \cdots \end{align}

where $196883, 4371, 782$ (the red numbers minus $1$) is the smallest degree $>1$ of the representations of the Monster group, Baby Monster, and Fischer group $Fi_{23}$, respectively. While $4\times276=4\times\binom{24}{2} = 1104$ is for the 24-dimensional Leech lattice and Conway group.

II. Definitions

Define the following functions,

$$ \alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1A}(\tau)}}\right)\qquad\qquad\qquad$$

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

III. Conjecture 1

For simplicity, let $\alpha_n = \alpha_n(\tau)$. Then for parameters $s = \frac16,\frac14,\frac13,\frac12,$ we conjecture that those four moonshine functions also have a hypergeometric formula,

$$j_{1A}(\tau) = \frac{432}{\alpha_1\,(1-\alpha_1)}=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\qquad\qquad$$

$$j_{2A}(\tau) = \frac{64}{\alpha_2\,(1-\alpha_2)}=\left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}$$

$$j_{3A}(\tau) = \frac{27}{\alpha_3\,(1-\alpha_3)}=\left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}$$

$$j_{4A}(\tau) = \frac{16}{\alpha_4\,(1-\alpha_4)}=\left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}$$

but avoiding $\tau$ which leads to division by zero.

IV. Conjecture 2

Furthermore, again let $\alpha_n = \alpha_n(\tau)$, then,

$$\left(\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}\right)^2=-1\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}\right)^2=-2\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}\right)^2=-3\,\tau^2$$

$$\left(\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}\right)^2=-4\,\tau^2$$

for $\tau = \sqrt{-d}$ and $\tau = \frac{1+\sqrt{-d}}2$.

Note: If $\tau=\frac12\sqrt{-r}$, then the last simplifies to the well-known,

$$\frac{K'(k)}{K(k)} = \sqrt{r}$$

with complete elliptic integral of the first kind $K(k).$ It was the model for conjecturing that similar behavior occurs for the first three. Thus $\alpha_4\big(\tfrac{\tau}2\big)$ is the square of the elliptic modulus $k$,

$$\alpha_4\big(\tfrac{\tau}2\big) = k^2 = \lambda(\tau) = \frac{16}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,e^{\pi i\tau})}{\vartheta_3(0,e^{\pi i\tau})}\right)^4$$

and is the modular lambda function $\lambda(\tau)$.

V. Question

Q: So are the two conjectures true?