User:Titus III/sandbox

Temporary 1
Relations


 * $$j_{6A} = \left(\sqrt{j_{6B}} - \frac{\color{blue}1}{\sqrt{j_{6B}}}\right)^2 =\left(\sqrt{j_{6C}} + \frac{\color{blue}8}{\sqrt{j_{6C}}}\right)^2 = \left(\sqrt{j_{6D}} + \frac{\color{blue}9}{\sqrt{j_{6D}}}\right)^2-4$$


 * $$j_{10A} = \left(\sqrt{j_{10D}} - \frac{\color{blue}1}{\sqrt{j_{10D}}}\right)^2 = \left(\sqrt{j_{10B}} + \frac{\color{blue}4}{\sqrt{j_{10B}}}\right)^2 = \left(\sqrt{j_{10C}} + \frac{\color{blue}5}{\sqrt{j_{10C}}}\right)^2-4$$

The formula is,


 * $$\frac{1}{\pi} = \frac{5}{76\sqrt{95}}\,\sum_{k=0}^\infty \sum_{j=0}^k \binom{k}{j}^4\,\frac{408k+47}{76^{2k}}$$

Matthieu group (M11) 11T6 with discriminant $$661^8$$ as real quadratic field with period length 11.


 * $$x^{11} + 20x^9 - 26x^8 + 72x^7 - 137x^6 + 199x^5 - 212x^4 + 98x^3 - 11x^2 + x - 1 = 0$$

Relations
Let $$q = e^{2\pi i\tau} = \exp(2\pi i\tau).$$ All the $$A,\,B,\,C$$ in the identities below obey,


 * $$\big(\sqrt2\,q^{1/8}A\big)^8-B^8+C^8 = 0$$

a consequence of the same relationships obeyed by the Weber modular functions as will be seen later. Getting their ratios,


 * $$\alpha = \sqrt2\,q^{1/8}\,\frac{A}{B} =\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}$$


 * $$\beta = \frac{C}{B} = \frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}$$

then the first relationship immediately implies,


 * $$\alpha^8+\beta^8 = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8 + \left(\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\right)^8 = 1 $$

Furthermore,


 * $$\alpha = \sqrt2\,q^{1/8}\frac{f(-q,-q^3)}{f(-q^2,-q^2)} = \sqrt2\,q^{1/8}\prod_{n=1}\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$

Set 1
Define,


 * $$\begin{align}A_1(q)

&= \frac{f(-q,-q^3)}{f(-q)} \;= q^{-1/12}\frac{\eta(4\tau)}{\eta(2\tau)} \;=\; \frac{1}{\chi(-q^2)}\\[6pt] &= \sum_{n=0}^\infty \frac{q^{n^2+n}} {(q^2;q^2)_n} = \prod_{n=1}^\infty \left(1+q^{2n}\right)\; = \; \prod_{n=1}^\infty\frac{x_{41}}{(1-q^{4n-2})(1-q^{4n-2})}\\[6pt]\end{align}$$


 * $$\begin{align}B_1(q)

&= \frac{f(-q^2,-q^2)}{f(-q)} = q^{1/24}\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} = \chi(q)\\[6pt] &= \sum_{n=0}^\infty \frac{q^{n^2}} {(q^2;q^2)_n} = \prod_{n=1}^\infty \left(1+q^{2n-1}\right) = \prod_{n=1}^\infty\frac{x_{41}}{(1-q^{4n-1})(1-q^{4n-3})}\\[6pt]\end{align}$$


 * $$\begin{align}C_1(q)

&= \;\frac{f(-q,-q)}{f(-q)} \;\;=\;\; q^{1/24}\frac{\eta(\tau)}{\eta(2\tau)} \;\;=\;\; \chi(-q)\\[6pt] &= \sum_{n=0}^\infty \frac{(-1)^n q^{n^2}} {(q^2;q^2)_n} = \prod_{n=1}^\infty \left(1-q^{2n-1}\right) = \prod_{n=1}^\infty\frac{1}{(1+q^n)}\\[6pt]\end{align}$$

where $$x_{41} = \frac{(1-q^{2n})}{(1-q^{4n})}.$$ Then,


 * $$\big(\sqrt2\,q^{1/8}A\big)^8-B^8+C^8 = 0$$

and,


 * $$\alpha = \sqrt2\,q^{1/8}\,\frac{A}{B} = \sqrt2\,q^{1/8}\prod_{n=1}\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$

relations followed by all the sets.

Set 2
Define,


 * $$\begin{align}A_2(q)

&= \frac{f(-q,-q^3)}{f(-q^2)} = q^{-1/24}\frac{\eta(\tau)\,\eta(4\tau)}{\eta^2(2\tau)} = \frac1{\chi(q)}\\[6pt] &= \sum_{n=0}^\infty \frac{(-1)^n q^{n^2+n}} {(q^2;q^2)_n (-q;q^2)_{n+1}} = \prod_{n=1}^\infty \frac{1}{ \left(1+q^{2n-1}\right)} = \prod_{n=1}^\infty \frac{x_{42}}{\big(1-q^{4n-2}\big)\big(1-q^{4n-2}\big)} \\[6pt]\end{align}$$


 * $$\begin{align}B_2(q)

&= \frac{f(-q^2,-q^2)}{f(-q^2)} = q^{1/12}\frac{\eta(2\tau)}{\eta(4\tau)} \;\;=\;\; \chi(-q^2)\\[6pt] &=\, \sum_{n=0}^\infty \frac{(-1)^n \, q^{2n^2}} {(q^4;q^4)_n} \;\; = \;\; \prod_{n=1}^\infty \frac{1}{\left(1+q^{2n}\right)} \;\; = \;\; \prod_{n=1}^\infty \frac{x_{42}}{\big(1-q^{4n-1}\big)\big(1-q^{4n-3}\big)}\\[6pt]\end{align}$$


 * $$\begin{align}C_2(q)

&=\, \left(\frac{f(-q,-q)}{f(-q)}\right)^2 =\, \left(q^{1/24}\frac{\eta(\tau)}{\eta(2\tau)}\right)^2 = \big(\chi(-q)\big)^2\\[6pt] &= \left(\sum_{n=0}^\infty \frac{(-1)^n q^{n^2}} {(q^2;q^2)_n}\right)^2 \;=\; \prod_{n=1}^\infty \left(1-q^{2n-1}\right)^2 \;=\; \prod_{n=1}^\infty\frac{1}{\left(1+q^n\right)^2} \\[6pt]\end{align}$$

where $$x_{42} = \frac{(1-q^{n})}{(1-q^{4n})}.$$.

From their eta quotients and infinite products, it is clear that Set 1 and Set 2 involve reciprocals, namely $$A_1 B_2 = A_2 B_1 = 1.\,$$ And since all the $$A,\,B,\,C$$ obey,


 * $$\big(\sqrt2\,q^{1/8}A\big)^8-B^8+C^8 = 0$$

then Set 1 and Set 2 yield a pair of nice identities purely in terms of $$\chi(\pm q^n)$$,


 * $$\left(\frac{\sqrt2\,q^{1/8}}{\chi(-q^2)}\right)^8-\big(\chi(q)\big)^8+\big(\chi(-q)\big)^8 = 0$$


 * $$\left(\frac{\sqrt2\,q^{1/8}}{\chi(q)}\right)^8-\big(\chi(-q^2)\big)^8+\big(\chi(-q)\big)^{16} = 0$$

Set 3
Define,


 * $$\begin{align}A_3(q)

&= \frac{f(-q^2,-q^2)}{f(-q^2,-q^2)} = \frac{\varphi(-q^2)}{\varphi(-q^2)} \,=\, 1\\[6pt] &= \sum_{n=0}^\infty \frac{q^{n^2+n} (q^2;q^2)_{n+1}} {(-q^3;q^3)_{n+1} (q;q)_n} = \prod_{n=1}^\infty \frac{(1-q^{4n-2})(1-q^{4n-2})}{(1-q^{4n-2})(1-q^{4n-2})} = 1\\[6pt]\end{align}$$


 * $$\begin{align}B_3(q)

&= \frac{f(-q^2,-q^2)}{f(-q,-q^3)} = q^{1/8}\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\\[6pt] &=\; \sum_{n=0}^\infty \frac{q^{n^2} (q;q^2)_n} {\,(q^4;q^4)_n} \; = \; \prod_{n=1}^\infty \frac{(1-q^{4n-2})(1-q^{4n-2})}{(1-q^{4n-1})(1-q^{4n-3})}\\[6pt]\end{align}$$


 * $$\begin{align}C_3(q)

&= \frac{f(-q^2,-q^2)}{f(q,\,q^3)} =\, q^{1/8}\frac{\eta(\tau)}{\eta(4\tau)} = \frac{f(-q,-q)}{f(-q,-q^3)}\\[6pt] &= \sum_{n=0}^\infty \frac{(-1)^n q^{n^2}(-q;q^2)_n}{(q^4;q^4)_n} = \prod_{n=1}^\infty\frac{(1-q^n)}{(1-q^{4n})}\\[6pt] \end{align}$$

As was mentioned, all the $$A,\,B,\,C$$ obey,


 * $$\big(\sqrt2\,q^{1/8}A\big)^8-B^8+C^8 = 0$$

Removing the common numerator $$f(-q^2,-q^2)$$ of this set, one gets the known identity,


 * $$\left(\frac{\sqrt2\,q^{1/8}}{f(-q^2,-q^2)}\right)^8-\left(\frac{1}{f(-q,-q^3)}\right)^8+\left(\frac{1}{f(q,\,q^3)}\right)^8 = 0$$

or simply,


 * $$\;\left(\frac{\sqrt2\,q^{1/8}}{\varphi(-q^2)}\right)^8 - \,\left(\frac{1}{\psi(-q)}\right)^8 \;+\; \left(\frac{1}{\psi(q)}\right)^8 = 0$$

which justifies the numerator in $$A_3(q) = \frac{f(-q^2,-q^2)}{f(-q^2,-q^2)} = 1.$$ Also, note that $$q^{-1/8}B_3(q)$$ is the reciprocal of the continued fraction for mod 4, and $$C_3(q)$$ has been discussed in mod 2.

Set 4
Define,


 * $$\begin{align}A_4(q)

&= \frac{f(-q,-q^3)}{f(-q,-q)} =\frac{f(q,\,q^3)}{f(-q^2,-q^2)} = q^{-1/8}\frac{\eta(4\tau)}{\eta(\tau)}\\[6pt] &= \sum_{n=0}^\infty \frac{q^{(n^2+n)/2}\, (-q;q)_n} {(q;q)_n} = \frac12\sum_{n=0}^\infty \frac{q^{(n^2+n)/2}\, (-1;q)_{n+1}} {(q;q)_n} = \sum_{n=0}^\infty \frac{q^{n^2+n}\, (-q;q^2)_n} {(q;q)_{2n+1}} = \sum_{n=0}^\infty \frac{q^{n^2+n}\,(-q ;q)_{2n}} {(q;q^2)_{n+1} (q^4,q^4)_n}\\[6pt] &= \prod_{n=1}^\infty \frac{(1-q^{4n})}{(1-q^{n})} = \prod_{n=1}^\infty\frac{(1+q^{2n-1})}{(1-q^{4n-2})(1-q^{4n-2})}\\[6pt] \end{align}$$


 * $$\begin{align}B_4(q)

&= \frac{f(-q^2,-q^2)}{f(-q,-q)} =\frac{f(q,\,q^3)}{f(-q,-q^3)} = \frac{\eta^3(2\tau)}{\eta^2(\tau)\,\eta(4\tau)}\\[6pt] &= \sum_{n=0}^\infty \frac{q^{(n^2+n)/2}\, (-q;q)_{n+1}} {(q;q)_n} = \sum_{n=0}^\infty \frac{q^{(n^2+n)/2}\, (-1;q)_{n}} {(q;q)_{n}} \;\;=\;\; \sum_{n=0}^\infty \frac{q^{n^2}\, (-1;q^2)_n} {(q;q)_{2n}} \;=\; \sum_{n=0}^\infty \frac{q^{n^2}\,(-1 ;q)_{2n}} {(q^2;q^2)_{n} (q^2,q^4)_n}\\[6pt] &= \prod_{n=1}^\infty \frac{(1+q^{n})^2}{(1+q^{2n})} \;=\; \prod_{n=1}^\infty\frac{(1+q^{2n-1})}{(1-q^{4n-1})(1-q^{4n-3})}\\[6pt] \end{align}$$


 * $$\begin{align}C_4(q)

&= \,\frac{f(-q,-q)}{f(-q,-q)} \,=\, \frac{\,\varphi(-q)}{\,\varphi(-q)} \,=\, 1\\[6pt] &= \sum_{n=0}^\infty \frac {(-1)^n q^{n^2+n}}{(q^2;q^2)_{n+1}} = \prod_{n=1}^\infty \frac{(1-q^{2n})}{(1-q^{2n})} = 1 \\[6pt] \end{align}$$

Again recall that the $$A,\,B,\,C$$ obey,


 * $$\big(\sqrt2\,q^{1/8}A\big)^8-B^8+C^8 = 0$$

This time, removing the common denominator $$f(-q,-q)=\varphi(-q)$$ of this set, one gets the known identity,


 * $$\big(\sqrt2\,q^{1/8}\,f(-q,-q^3)\big)^8-\left(f(-q^2,-q^2)\right)^8+\big(f(-q,-q)\big)^8 = 0$$

or simply,


 * $$\big(\sqrt2\,q^{1/8}\,\psi(-q)\big)^8-\left(\varphi(-q^2)\right)^8+\big(\varphi(-q)\big)^8 = 0$$

which justifies the denominator in $$C_4(q) = \frac{f(-q,-q)}{f(-q,-q)} = 1.$$

Set 5
Sets 5 and 6 involve numerators of form $$f(q,-q^3)$$ and $$f(-q,\,q^3)$$, hence are complementary.

Define,


 * $$A_5(q) = \frac{f(q,-q^3)}{f(-q^2,-q^2)} = \sum_{n=0}^\infty \frac{q^{n^2+n} (-q^2;q^4)_{n}} {(q,q)_{2n+1} (-q;q^2)_n} \;=\; x_{45}\,\prod_{n=1}^\infty \frac{(1+q^{8n-1})(1+q^{8n-7})}{(1-q^{4n-2})(1-q^{4n-2})}$$


 * $$B_5(q) = \frac{f(q,-q^3)}{f(-q,-q^3)} \,=\, \sum_{n=0}^\infty \frac{q^{n^2} (-1;q^4)_{n} (-q,q^2)_{n}} {(q^2;q^2)_{2n}} = x_{45}\,\prod_{n=1}^\infty \frac{(1+q^{8n-1})(1+q^{8n-7})} {(1-q^{4n-1})(1-q^{4n-3})}$$


 * $$C_5(q) \,=\, \frac{\,f(q,-q^3)}{\,f(q,\,q^3)} \;=\; \sum_{n=0}^\infty\; (?) \;=\; x_{45}\,\prod_{n=1}^\infty \frac{(1+q^{8n-1})(1+q^{8n-7})}{(1+q^{4n-1})(1+q^{4n-3})}$$

where $$x_{45} = \chi(-q^8)\,\frac{f(-q^3,-q^5)}{f(-q^4,-q^4)}.$$

Set 6
Define,


 * $$A_6(q) = \frac{f(-q,\,q^3)}{f(-q^2,-q^2)} = \sum_{n=0}^\infty \frac{q^{n^2+n} (-q^2;q^4)_{n}} {(q,q)_{2n} (-q;q^2)_{n+1}} \;\;=\;\; x_{46}\,\prod_{n=1}^\infty \frac{(1+q^{8n-3})(1+q^{8n-5})}{(1-q^{4n-2})(1-q^{4n-2})}$$


 * $$B_6(q) = \frac{f(-q,\,q^3)}{f(-q,-q^3)} \,=\, \sum_{n=0}^\infty \frac{q^{n^2+2n} (-1;q^4)_{n} (-q,q^2)_{n}} {(q^2;q^2)_{2n}} = x_{46}\,\prod_{n=1}^\infty \frac{(1+q^{8n-3})(1+q^{8n-5})} {(1-q^{4n-1})(1-q^{4n-3})}$$


 * $$C_6(q) \,=\, \frac{\,f(-q,\,q^3)}{\,f(q,\,q^3)} \;=\; \sum_{n=0}^\infty\; (?) \;=\; x_{46}\,\prod_{n=1}^\infty \frac{(1+q^{8n-3})(1+q^{8n-5})}{(1+q^{4n-1})(1+q^{4n-3})}$$

where $$x_{46} = \chi(-q^8)\,\frac{f(-q,\,-q^7)}{f(-q^4,-q^4)}.$$

There are no sums known yet for $$C_5(q)$$ and $$C_6(q)$$, but the last two sets also follow the basic relationships mentioned earlier.

Connection to other functions
These four sets are connected to both the Weber modular functions $$\mathfrak{f}_n(\tau)$$ and the Jacobi theta functions $$\theta_n(q)$$. This can be illustrated by using Set 1. Let,


 * $$\begin{align}

& A = \sqrt2\,q^{1/12}A_1(q) = \mathfrak{f}_2(2\tau)\\ & B = \; q^{-1/24}\,B_1(q) \;=\, \mathfrak{f}\,(2\tau)\\ & C = \; q^{-1/24}\,C_1(q) \,=\, \mathfrak{f}_1(2\tau)\\ \end{align}$$

to get rid of $$q^n$$ so they become radicals. From the well-known property of the Weber modular functions which obey the Fermat curve of degree 8,


 * $$\mathfrak{f}_2(\tau)^8- \mathfrak{f}(\tau)^8+ \mathfrak{f}_1(\tau)^8 = 0$$

then,


 * $$A^8-B^8+C^8 = 0$$

or as sums,


 * $$16q\left(\sum_{n=0}^\infty \frac{q^{n^2+n}} {(q^2;q^2)_n}\right)^8 -\left(\sum_{n=0}^\infty \frac{q^{n^2}} {(q^2;q^2)_n}\right)^8 + \left(\sum_{n=0}^\infty \frac{(-1)^n q^{n^2}} {(q^2;q^2)_n}\right)^8 = 0$$

Compare to the relationship obeyed by the null Jacobi theta functions,


 * $$\quad\left(\sum_{n=-\infty}^\infty q^{n^2+n+1/4}\right)^4 -\left(\sum_{n=-\infty}^\infty q^{n^2}\right)^4 \, + \, \left(\sum_{n=-\infty}^\infty (-1)^n q^{n^2}\right)^4 \,=\, 0$$

which are $$\theta_2(q),\,\theta_3(q),\,\theta_4(q),$$ respectively, with a noticeable difference that the denominator $$(q^2;q^2)_n$$ disappears in the latter.

Mod 2 & 3 Identities
It seems better to combine the mod 2 and mod 3 identities since there are relatively few of them, they usually are just higher mods in disguise, and a ratio of mods 2 and 3 leads to a Rogers-Ramanujan type continued fraction whose form as a Dedekind eta quotient has importance for higher mods.

Pair 1
Define,


 * $$\frac1{A(q)} = \frac{f(-q,-q)}{f(-q,-q^2)} \,=\, \chi(-q) = \sum_{n=0}^\infty \frac {(-1)^n q^{n^2}}{(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{1}{(1+q^{n})} = \prod_{n=1}^\infty(1-q^{2n-1}) = \prod_{n=1}^\infty\frac{(1-q^{n})}{(1-q^{2n})}\,=\, q^{1/24}\frac{\eta(\tau)}{\eta(2\tau)}$$


 * $$\,A(q) \;=\; \frac{f(-q,-q^2)}{f(-q,-q)} = \frac{1}{\chi(-q)} = \sum_{n=0}^\infty \frac {q^{(n^2+n)/2}}{(q;q)_n} = \prod_{n=1}^\infty\big(1+q^{n}\big) = \prod_{n=1}^\infty\frac1{(1-q^{2n-1})} = \prod_{n=1}^\infty\frac{(1-q^{2n})}{(1-q^{n})} = q^{-1/24}\frac{\eta(2\tau)}{\eta(\tau)}$$

Multiplying the two, we get the product of two sums which can be expressed as a sum (in two ways) that naturally equal 1|unity,


 * $$\left(\sum_{n=0}^\infty \frac {(-1)^n q^{n^2}}{(q^2;q^2)_n}\right) \left(\sum_{n=0}^\infty \frac {q^{(n^2+n)/2}}{(q;q)_n}\right) = \sum_{n=0}^\infty \frac {(-1)^n q^{n^2}}{(q;q^2)_{n+1}} = \sum_{n=0}^\infty \frac {(-1)^n q^{n^2+n}}{(q^2;q^2)_{n+1}} = 1$$

Note that,


 * $$\frac1{A(q)} = \frac{f(-q,-q)}{f(-q)} = \frac{f(-q,-q^2)}{f(-q^2)} = \frac{f(-q,-q^3)}{f(-q^4)} = \frac{f(-q,-q^5)}{\varphi(q^3)} = q^{1/24}\frac{\eta(\tau)}{\eta(2\tau)}$$

so will appear in mods 2, 3, 4, 6. And, as will later be seen, $$A(q)$$ has a nice Rogers-Ramanujan type continued fraction.

Pair 2
Define,


 * $$B(q) \,=\, \frac{f(q^0, q^2)}{2\,\psi(-q)} = \frac{\psi(-q)}{f(-q,-q)} = \frac{1}{2}\sum_{n=0}^\infty \frac {q^{n^2-n}(-q;q^2)_n}{(q;q)_{2n}} =\, \prod_{n=1}^\infty \frac{(1-q^{4n})}{(1-q^{n})} = q^{-1/8}\frac{\eta(4\tau)}{\eta(\tau)}$$


 * $$\frac{1}{B(q)} = \frac{2\,\psi(-q)}{f(q^0, q^2)} = \frac{f(-q,-q)}{\psi(-q)} = \sum_{n=0}^\infty \frac {(-1)^n q^{n^2}(-q;q^2)_n}{(q^4;q^4)_n} = \prod_{n=1}^\infty \frac{(1-q^{n})}{(1-q^{4n})} = q^{1/8}\frac{\eta(\tau)}{\eta(4\tau)}$$

Again getting the product,


 * $$\left(\sum_{n=0}^\infty \frac {(-1)^n q^{n^2}(-q;q^2)_n}{(q^4;q^4)_n}\right) \left(\frac12\sum_{n=0}^\infty \frac {q^{n^2-n}(-q;q^2)_n}{(q;q)_{2n}}\right) = \frac12\sum_{n=0}^\infty \frac {q^{(n^2-n)/2}}{(-q;q)_n}= 1$$

though multiplying both sides by 2 implies the sum is actually 2.

Pair 3
As above, define,


 * $$B(q) = \frac{f(q^0, q^2)}{2\,\psi(-q)} = \left(\frac{\varphi(-q^2)}{f(-q)}\right)^3 \left(\frac{\psi(-q)}{\varphi(-q^2)}\right)^2 = q^{-1/8}\frac{\eta(4\tau)}{\eta(\tau)}$$

and its counterpart,


 * $$C(q) \, = \frac{\,f(q,\,q)}{\,\psi(-q)} \,= \left(\frac{\varphi(-q^2)}{f(-q)}\right)^3 = \left(\sum_{n=0}^\infty \frac {q^{n^2}}{(q^2;q^2)_n}\right)^3 = \prod_{n=1}^\infty\left(1+q^{2n-1}\right)^3 = \left(\frac{q^{1/24}\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)}\right)^3 $$

then their ratio shows a relation between mods 2 and 4,


 * $$\frac{B(q)}{C(q)} = \frac12\,\frac{f(q^0,q^2)}{f(q,\,q)} = \left(\frac{\psi(-q)}{\varphi(-q^2)}\right)^2 = \left(\frac{f(-q,\,-q^3)}{f(-q^2,-q^2)}\right)^2$$

Later, it will be shown that a similar relation can be found between mods 3 and 6.

Pair 1
Define,


 * $$\,A_1(q) \,=\, \frac{f(-q,-q^2)}{f(-q,-q)} = \frac{1}{\chi(-q)} = \sum_{n=0}^\infty \frac {q^{(n^2+n)/2}}{(q;q)_n} = \prod_{n=1}^\infty\frac{x_{31}}{(1-q^{3n-1})(1-q^{3n-2})} = q^{-1/24}\frac{\eta(2\tau)}{\eta(\tau)}$$


 * $$\frac1{A_1(q)} = \frac{\,f(-q,-q)}{f(-q,-q^2)} =\, \chi(-q) \, = \sum_{n=0}^\infty \frac {(-1)^n q^{n^2}}{(q^2;q^2)_n} = \, \prod_{n=1}^\infty\frac{(1-q^{3n-1})(1-q^{3n-2})}{x_{31}} \,=\, q^{1/24}\frac{\eta(\tau)}{\eta(2\tau)}$$

where $$x_{31} = \frac{(1-q^{2n})}{(1-q^{3n}}.$$ Recall this has already been discussed in mod 2 as $$A(q).\,$$ In fact, $$A(q)=A_1(q)$$ is a Rogers-Ramanujan type continued fraction discussed in the next section.

Pair 2
Define,


 * $$\begin{align}A_2(q)

&= \frac{f(-q,-q^2)}{\varphi(-q^2)} = \frac{f(-q,-q^3)}{f(-q^2)}\\ &= \sum_{n=0}^\infty \frac{(-1)^n q^{2n^2+n}} {(q^2;q^2)_n (-q;q^2)_{n+1}} = \prod_{n=1}^\infty\frac{x_{32}}{(1-q^{4n-2})(1-q^{4n-2})} = q^{-1/24}\frac{\eta(\tau)\,\eta(4\tau)}{\eta^2(2\tau)}\\[6pt]\end{align}$$


 * $$\begin{align}B_2(q)

&= \frac{f(-q,-q^2)}{\psi(-q)} = \frac{f(-q^2,-q^2)}{f(-q^2)}\\ &=\; \sum_{n=0}^\infty \frac{(-1)^n q^{2n^2}}{(q^4;q^4)_n} \;=\; \prod_{n=1}^\infty\frac{x_{32}}{(1-q^{4n-1})(1-q^{4n-3})} \; = \; q^{1/12}\frac{\eta(2\tau)}{\eta(4\tau)}\\[6pt]\end{align}$$

where $$x_{32} = \frac{(1-q^{n})}{(1-q^{4n})}.$$

This will be discussed more in mod 4.

Pair 3
Define,


 * $$\begin{align}A_3(q)

&= \frac12 \frac{f(q^0,q^3)}{f(-q)} = \frac{f(-q,-q^5)}{\varphi(-q)}\\ &= \sum_{n=0}^\infty \frac{q^{n^2+n} (-q;q)_n} {(q;q)_n (q;q^2)_{n+1}} = \prod_{n=1}^\infty\frac{x_{33}}{(1-q^{6n-3})(1-q^{6n-3})} = q^{-1/3}\frac{\eta^2(6\tau)}{\eta(\tau)\,\eta(3\tau)}\\[6pt]\end{align}$$


 * $$\begin{align}B_3(q)

&= \frac{f(q,\,q^2)}{f(-q)} = \frac{f(-q^3,-q^3)}{\varphi(-q)}\\ &= \sum_{n=0}^\infty \frac{q^{n^2} (-q;q)_n} {(q;q)_n (q;q^2)_{n+1}} = \prod_{n=1}^\infty\frac{x_{33}}{(1-q^{6n-1})(1-q^{6n-5})} = \frac{\eta(2\tau)\,\eta^2(3\tau)}{\eta^2(\tau)\,\eta(6\tau)}\\[6pt]\end{align}$$

where $$x_{33} = \frac{(1-q^{3n})}{(1-q^{n})}.$$

This will be discussed more in mod 6.

Pair 4
Define,


 * $$\begin{align}A_4(q)

&= \frac12 \frac{f(q^0,q^3)}{\psi(q)} = \frac{f(-q,-q^5)}{f(-q^2)}\\ &= \sum_{n=0}^\infty \frac{q^{2n^2+2n} (q;q^2)_n} {(-q;q)_{2n+1} (q^2;q^2)_{n}} = \prod_{n=1}^\infty(1-q^{6n-1})(1-q^{6n-5})\,x_{34} = q^{-1/4}\frac{\eta(\tau)\,\eta^2(6\tau)}{\eta^2(2\tau)\,\eta(3\tau)}\\[6pt]\end{align}$$


 * $$\begin{align}B_4(q)

&= \frac{f(q,\,q^2)}{\psi(q)} = \frac{f(-q^3,-q^3)}{f(-q^2)}\\ &= \sum_{n=0}^\infty \frac{q^{2n^2} (q;q^2)_n} {(-q;q)_{2n} (q^2;q^2)_{n}} \; = \; \prod_{n=1}^\infty \, (1-q^{6n-3})(1-q^{6n-3})\,x_{34} \, = \, q^{1/12}\frac{\eta^2(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\\[6pt]\end{align}$$

where $$x_{34} = \frac{(1-q^{6n})}{(1-q^{2n})}.$$

The functions in Pair 3 and Pair 4 have the same ratio and imply a relation between mods 3 and 6,


 * $$\frac{A_3(q)}{B_3(q)} = \frac{A_4(q)}{B_4(q)} = \frac12 \frac{f(q^0,q^3)}{f(q,\,q^2)} = \frac{f(-q,-q^5)}{f(-q^3,-q^3)}$$

similar to the relation between mods 2 and 4.

Continued fraction
The lowest mod n counterpart to the well-known Rogers-Ramanujan continued fraction is a ratio of a mod 3 and mod 2,


 * $$M(q) = q^{1/24}\frac{f(-q,-q^2)}{f(-q,-q)} = q^{1/24}\prod_{n=1}^\infty \frac{(1-q^{3n})(1-q^{3n-1})(1-q^{3n-2})} {(1-q^{2n})(1-q^{2n-1})(1-q^{2n-1})}$$

The next is a ratio of two mod 4 identities,


 * $$N(q) \,=\, q^{1/8}\frac{f(-q,-q^3)}{f(-q^2,-q^2)} = q^{1/8}\prod_{n=1}^\infty \frac{(1-q^{4n})(1-q^{4n-1})(1-q^{4n-3})} {(1-q^{4n})(1-q^{4n-2})(1-q^{4n-2})}$$

while the Rogers-Ramanujan is a ratio of two mod 5 identities,


 * $$R(q) \,=\, q^{1/5}\frac{f(-q,-q^4)}{f(-q^2,-q^3)} = q^{1/5}\prod_{n=1}^\infty \frac{(1-q^{5n})(1-q^{5n-1})(1-q^{5n-4})} {(1-q^{5n})(1-q^{5n-2})(1-q^{5n-3})}$$

Of course, the second and third have a common factor which just cancels out. The first does not, but also simplifies in its own way. In more detail,


 * $$\begin{align}M(q)

&= q^{1/24}\frac{f(-q,-q^2)}{f(-q,-q)} = q^{1/24}\frac{f(-q)}{\varphi(-q)} \\[6pt] &= q^{1/24}\prod_{n=1}^\infty (1+q^{n}) = q^{1/24}\prod_{n=1}^\infty \frac1{(1-q^{2n-1})}\\[6pt] &= q^{1/24}\sum_{n=0}^\infty \frac {q^{(n^2+n)/2}}{(q;q)_n} = q^{1/24}\prod_{n=1}^\infty \frac{(1-q^{2n})}{(1-q^{n})} = \frac{\eta(2\tau)}{\eta(\tau)}\\[6pt] &= \cfrac{q^{1/24}}{1+\cfrac{-q}{1+\cfrac{q^3-q^2}{1+\cfrac{q^5-q^3}{1+\cfrac{q^7-q^4}{1+\ddots}}}}} \end{align}$$

Definitions
Given the q-Pochhammer symbol,


 * $$(a;q)_n = \prod_{k=0}^{n-1}(1-aq^k)$$
 * $$(a;q)_\infty = \prod_{k=0}^{\infty}(1-aq^k)$$

as well as the Ramanujan theta function,


 * $$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2} $$

In his Notebooks, Ramanujan also defined four one-parameter versions he commonly used as,


 * $$\begin{align}

\varphi(q) &= f(q,\, q)\\ f(-q) &= f(-q, -q^2)\\ \psi(q) &= f(q,\, q^3)\\ \chi(q) &=\frac{f(-q^2,-q^2)}{f(-q,-q^2)}\\ \end{align}$$

There are several simple relations between these four auxiliary functions, one of which is the elegant Fermat curve of degree 8,


 * $$\big[f(-q)\,\chi(-q)\big]^8+\big[\sqrt2\,q^{1/8}\psi(-q)\big]^8 = \big[\varphi(-q^2)\big]^8$$

Let $$q = e^{2\pi i \tau}$$. For simplicity, it will be assumed $$\tau$$ is an imaginary quadratic number so that certain functions below will evaluate as radicals or algebraic numbers.

Mod 2

 * $$A_2(q) = \frac{f(-q,-q)}{f(-q)} \,=\, \frac{\varphi(-q)}{f(-q)} = \sum_{n=0}^\infty \frac {(-1)^n q^{n^2}}{(q^2;q^2)_n} = \prod_{n=1}^\infty(1-q^{2n-1}) \,=\, \prod_{n=1}^\infty\frac{(1-q^n)}{(1-q^{2n})} \,=\, q^{1/24}\frac{\eta(\tau)}{\eta(2\tau)}$$

Mod 3

 * $$A_3(q) = \frac{f(-q,-q^2)}{\varphi(-q)} = \frac{f(-q)}{\varphi(-q)} = \sum_{n=0}^\infty \frac {q^{(n^2+n)/2}}{(q;q)_n} = \prod_{n=1}^\infty\frac{(1-q^{2n})}{(1-q^{3n})(1-q^{3n-1})(1-q^{3n-2})} = q^{-1/24}\frac{\eta(2\tau)}{\eta(\tau)}$$

Note that $$A_2(q) = \frac{1}{A_3(q)} = \chi(-q),$$ hence $$A_2(q)$$ and $$A_3(q)$$ are just reciprocals.

Mod 4

 * $$A(q) = \frac{f(-q,-q^3)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{x_4}{(1-q^{4n-2})(1-q^{4n-2})} = q^{-1/12}\frac{\eta(4\tau)}{\eta(2\tau)}$$


 * $$B(q) = \frac{f(-q^2,-q^2)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{x_4}{(1-q^{4n-1})(1-q^{4n-3})} = q^{1/24}\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}$$

where $$x_4 = \frac{(1-q^{2n})}{(1-q^{4n})}.$$

Let $$ a = \sqrt2\,q^{1/12}A(q)$$ and $$b= q^{-1/24}B(q)$$ to remove the factor $$q^n$$, then $$a, b$$ are just eta quotients and are radicals. Furthermore,


 * $$C_4(q) = \frac{a}{b} = \frac{\sqrt2\,q^{1/12}A(q)}{\;\;q^{-1/24}B(q)} = \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}$$

then $$C_4(q)$$ is also a radical and a known continued fraction analogous to the Rogers-Ramanujan version. The formula for the j-function using $$C_4(q)$$ employs polynomial invariants of the octahedron and the integer $$24$$ in $$q^{1/24}$$ reflects the order of the octahedral group.

Mod 5
Define,


 * $$A(q) = H(q) = \frac{f(-q,-q^4)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}$$


 * $$B(q) = G(q) = \frac{f(-q^2,-q^3)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}$$

Let $$ a = q^{11/60}H(q)$$ and $$b= q^{-1/60}B(q),$$ then $$a, b$$ are radicals. Furthermore,


 * $$R(q) = \frac{q^{11/60} H(q)}{q^{-1/60}G(q)} = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-2})}{(1-q^{5n-2})(1-q^{5n-3})}$$

then $$R(q)$$ is also a radical and is the well-known Rogers-Ramanujan continued fraction. The formula for the j-function using $$R(q)$$ employs polynomial invariants of the icosahedron and the integer $$60$$ in $$q^{1/60}$$reflects the order of the icosahedral group.

Mod 6

 * $$A(q) = \frac{f(-q,-q^5)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}(q,q^2)_n} {(-q,q)_{2n+1} (q^2;q^2)_n} = \prod_{n=1}^\infty\frac{x_6}{(1-q^{6n-3})(1-q^{6n-3})} = q^{-1/4}\frac{\eta(\tau)\eta^2(6\tau)}{\eta^2(2\tau)\eta(3\tau)}$$


 * $$B(q) = \frac{f(-q^3,-q^3)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2}(q,q^2)_n} {(-q,q)_{2n} (q^2;q^2)_n} \;=\; \prod_{n=1}^\infty\frac{x_6}{(1-q^{6n-1})(1-q^{6n-5})} = q^{1/12}\frac{\eta^2(3\tau)}{\eta(2\tau)\eta(6\tau)}$$

where $$x_6 = \frac{(1-q^{n})(1-q^{3n})}{(1-q^{2n})^2}.$$

Let $$ a = q^{1/4}A(q)$$ and $$b= q^{-1/12}B(q)$$ to remove the factor $$q^n$$, then $$a, b$$ are just eta quotients and are radicals. Furthermore,


 * $$C_6(q) = \frac{a}{b} = \frac{\;q^{1/4}A(q)}{q^{-1/12}B(q)} = q^{1/3} \prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})} = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$$

then $$C_6(q)$$ is also a radical, sometimes known as the cubic continued fraction, and analogous to the Rogers-Ramanujan version. The formula for the j-function using $$C_6(q)$$ employs polynomial invariants of the tetrahedron and the integer $$12$$ in $$q^{1/12}$$ reflects the order of the tetrahedral group.

It will be seen later that mods 4, 5, 6 can be connected to the higher mods 8, 10, 12. The first three use $$\tfrac{f(-q,-q^3)}{f(-q)}, \tfrac{f(-q,-q^4)}{f(-q)}, \tfrac{f(-q,-q^5)}{f(-q^2)}$$ and $$q^{1/24}, q^{1/60}, q^{1/12}$$, completing a relation to the three Platonic groups and symmetries. However, this relation can be continued to mods 7 and 14 using $$\tfrac{f(-q,-q^6)}{f(-q^2)},\;\tfrac{f(-q^2,-q^{12})}{f(-q)}$$, but what is involved is the Klein quartic, its order $$168$$, and $$q^{n/168}.$$

Mod 7
Also known as the Rogers-Selberg identities,


 * $$A(q) = \frac{f(-q,-q^6)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}} {(q^2;q^2)_n\,(-q;q)_{2n+1}} = \prod_{n=1}^\infty (1-q^{7n-1})(1-q^{7n-6})\,x_7 \,=\, \frac {(q;q^7)_\infty (q^6;q^7)_\infty (q^7;q^7)_\infty} {(q^2;q^2)_\infty}$$


 * $$B(q) = \frac{f(-q^2,-q^5)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}} {(q^2;q^2)_n\,(-q;q)_{2n}} \,=\, \prod_{n=1}^\infty (1-q^{7n-2})(1-q^{7n-5})\,x_7 \,=\, \frac {(q^2;q^7)_\infty (q^5;q^7)_\infty (q^7;q^7)_\infty} {(q^2;q^2)_\infty}$$


 * $$C(q) = \frac{f(-q^3,-q^4)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2}} {(q^2;q^2)_n\,(-q;q)_{2n}} \,=\, \prod_{n=1}^\infty (1-q^{7n-3})(1-q^{7n-4})\,x_7 \,=\, \frac {(q^3;q^7)_\infty (q^4;q^7)_\infty (q^7;q^7)_\infty} {(q^2;q^2)_\infty}$$

where $$x_7 = \frac{(1-q^{7n})}{(1-q^{2n})}.$$

Let $$a = -q^{61/168}A(q),\,$$ $$b = q^{13/168}B(q),\,$$ and $$c = q^{-11/168}C(q),$$ then $$a,b,c$$ are modular functions and, like their mod 5 versions, are radicals. The integer $$168$$ reflects the order of the Klein quartic and the $$a,b,c$$ obey the beautiful relationship,


 * $$a^3b+b^3c+c^3a = 0$$

which in fact is the Klein quartic algebraic curve.

Mod 8
Also known as the Göllnitz-Gordon identities,


 * $$A(q) = \frac{f(-q,-q^7)}{\psi(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+2n}\,(-q;q^2)_n} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{8n-3})(1-q^{8n-4})(1-q^{8n-5})}$$


 * $$B(q) = \frac{f(-q^3,-q^5)}{\psi(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}\,(-q;q^2)_n} {(q^2;q^2)_n} \;\; = \;\; \prod_{n=1}^\infty\frac{1}{(1-q^{8n-1})(1-q^{8n-4})(1-q^{8n-7})}$$

Let $$a = q^{7/16}A(q)\,$$ and $$b = q^{-1/16}B(q),$$ then $$a,b$$ are radicals. Let


 * $$C_8(q) = \frac{q^{7/16}A(q)}{q^{-1/16}B(q)} = q^{1/2}\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}$$

then it is also a known continued fraction. In fact, $$C_8(q)$$ is connected to the mod 4 version,


 * $$C_4(q) = \sqrt2\,q^{1/8}\prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$

by the quadratic relation,


 * $$\frac{1}{C_8(q)}-C_8(q) = \left(\frac{\eta^3(4\tau)}{\eta(2\tau)\eta^2(8\tau)}\right)^2 =\left(\frac{\sqrt2}{C_4(q^2)}\right)^2 $$

analogous to the mod 5 version,


 * $$\frac{1}{R(q)}-R(q) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$$

Mod 9
Also known as the Bailey mod 9 identities,


 * $$A(q) =\, \frac{f(-q,-q^8)}{f(-q^3)} = \sum_{n=0}^\infty \frac {q^{3n^2+3n}\;(q;q)_{3n+1}} {(q^3;q^3)_n\,(q^3;q^3)_{2n+1}} \quad = \quad \frac {(q;q^9)_\infty (q^8;q^9)_\infty (q^9;q^9)_\infty} {(q^3;q^3)_\infty}$$


 * $$B(q) = \frac{f(-q^2,-q^7)}{f(-q^3)} = \sum_{n=0}^\infty \frac {q^{3n^2+3n}\,(q;q)_{3n}(1-q^{3n+2})} {(q^3;q^3)_n\,(q^3;q^3)_{2n+1}} = \frac {(q^2;q^9)_\infty (q^7;q^9)_\infty (q^9;q^9)_\infty} {(q^3;q^3)_\infty}$$


 * $$C(q) = \frac{f(-q^4,-q^5)}{f(-q^3)} \,=\, \sum_{n=0}^\infty \frac {q^{3n^2}\;(q;q)_{3n}} {(q^3;q^3)_n\,(q^3;q^3)_{2n}} \quad = \quad \frac {(q^4;q^9)_\infty (q^5;q^9)_\infty (q^9;q^9)_\infty} {(q^3;q^3)_\infty}$$

Let $$a = q^{5/9}A(q),\,$$ $$b = q^{2/9}B(q),\,$$ and $$c = -q^{-1/9}C(q),$$ then $$a,b,c$$ are radicals. They obey the relationship,


 * $$a^2b+b^2c+c^2a = 0$$

which should be the algebraic curve of some known mathematical object.

Mod 10

 * $$A(q) \,= \frac{f(-q,-q^9)}{\varphi(-q)} \;=\; \sum_{n=0}^\infty \frac {q^{n(n+3)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}} = \prod_{n=1}^\infty\frac{x_{10}}{(1-q^{10n-3})(1-q^{10n-7})}$$


 * $$B(q) = \frac{f(-q^3,-q^7)}{\varphi(-q)} = \sum_{n=0}^\infty \frac {q^{n(n+1)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}} = \prod_{n=1}^\infty\frac{x_{10}}{(1-q^{10n-1})(1-q^{10n-9})}$$

where $$x_{10} = \frac{(1-q^{10n})^2}{(1-q^{n})(1-q^{5n})}.$$

Let $$a = q^{4/5}A(q)\,$$ and $$b = q^{1/5}B(q),$$ then $$a,b$$ are radicals. Define,


 * $$C_{10}(q) = \frac{q^{4/5}A(q)}{q^{1/5}B(q)} = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})}$$

While no continued fraction is yet known for this, it is connected to the mod 5 version,


 * $$R(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$

simply as,


 * $$C_{10}(q) = R(q)R(q^2)$$

where $$R(q)$$ is the Rogers-Ramanujan continued fraction.

Mod 12

 * $$A(q) = \frac{f(-q,-q^{11})}{f(-q^4)} = \sum_{n=0}^\infty \frac {q^{4n^2+4n}\,(q;q^2)_{2n+1}} {(q^4;q^4)_{2n+1}} = \prod_{n=1}^\infty(1-q^{12n-1})(1-q^{12n-1})\,x_{12}$$


 * $$B(q) = \frac{f(-q^5,-q^{7})}{f(-q^4)} \,=\, \sum_{n=0}^\infty \frac {q^{4n^2}\,(q;q^2)_{2n}} {(q^4;q^4)_{2n}} \quad=\; \prod_{n=1}^\infty(1-q^{12n-5})(1-q^{12n-7})\,x_{12}$$

where $$x_{12} = \frac{(1-q^{12n})}{(1-q^{4n})}.$$

Let $$a = q^{7/8}A(q)\,$$ and $$b = q^{-1/8}B(q),$$ then $$a,b$$ are radicals. Define,


 * $$C_{12}(q) = \frac{q^{7/8}A(q)}{q^{-1/8}B(q)} = q\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}$$

Surprisingly, there is already a continued fraction known for this and studied by Naika, and is connected to the mod 6 version,


 * $$C_6(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}$$

by the quadratic relation,


 * $$\frac1{C_{12}(q)}+C_{12}(q) = \frac1{C_6(q)\,C_6(q^2)}.$$

Mod 14
Also known as the Rogers Mod 14 Identities.


 * $$A(q) = \frac{f(-q^2,-q^{12})}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2+2n}} {(q;q)_n\,(q;q^2)_{n+1}} = \frac {(q^{2};q^{14})_\infty (q^{12};q^{14})_\infty (q^{14};q^{14})_\infty} {(q;q)_\infty}$$


 * $$B(q) = \frac{f(-q^4,-q^{10})}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n\,(q;q^2)_{n+1}} = \frac {(q^{4};q^{14})_\infty (q^{10};q^{14})_\infty (q^{14};q^{14})_\infty} {(q;q)_\infty}$$


 * $$C(q) = \frac{f(-q^6,-q^8)}{f(-q)} \;=\, \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n\,(q;q^2)_{n}} \;\; = \; \frac {(q^{6};q^{14})_\infty (q^{8};q^{14})_\infty (q^{14};q^{14})_\infty} {(q;q)_\infty}$$

Let $$a = -q^{122/168}A(q),\,$$ $$b = q^{26/168}B(q),\,$$ and $$c = q^{-22/168}C(q)$$ then, like their mod 7 versions, the $$a,b,c$$ obey the algebraic curve of the Klein quartic,


 * $$a^3b+b^3c+c^3a = 0.$$

Ramanujan theta function

 * $$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2} $$

with a one-parameter version as,


 * $$f(-q) = f(-q,-q^2)$$

Levels 2 & 3

 * $$\begin{align}M(q)

&= q^{1/24}\frac{f(-q,-q^2)}{f(-q,-q)}\\[6pt] &= q^{1/24}\prod_{n=1}^\infty \frac{(1-q^{2n})}{(1-q^{n})} = q^{1/24}\prod_{n=1}^\infty (1+q^{n})\\[6pt] &= \frac{\eta(2\tau)}{\eta(\tau)}\\ &= \cfrac{q^{1/24}}{1+\cfrac{-q}{1+\cfrac{q^3-q^2}{1+\cfrac{q^5-q^3}{1+\cfrac{q^7-q^4}{1+\ddots}}}}} \end{align}$$

Level 4

 * $$\begin{align}N(q)

&= \sqrt2\,q^{1/8}\frac{f(-q,-q^3)}{f(-q^2,-q^2)}\\[6pt] &= \sqrt2\,q^{1/8}\prod_{n=1}^\infty \frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\[6pt] &= \sqrt2\,q^{1/8}\prod_{n=1}^\infty \frac{(1-q^{2n-1})}{\;(1-q^{4n-2})^2} = \sqrt2\,q^{1/8}\prod_{n=1}^\infty \frac{(1+q^{2n})^2}{(1+q^{n})\;}\\[6pt] &= \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\[6pt] & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}} \end{align}$$

Level 5

 * $$\begin{align}R(q)

&= q^{1/5}\frac{f(-q,-q^4)}{f(-q^2,-q^3)}\\[6pt] &= q^{1/5}\prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\[6pt] &= \cfrac{q^{1/5}} {1+\cfrac{q} {1+\cfrac{q^2} {1+\cfrac{q^3} {1+\ddots}}}} \end{align}$$

Level 6

 * $$\begin{align}S(q)

&= q^{1/3}\frac{f(-q,-q^5)}{f(-q^3,-q^3)}\\[6pt] &= q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\[6pt] &= q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{2n-1})}{\; (1-q^{6n-3})^3} = q^{1/3}\prod_{n=1}^\infty \frac{(1+q^{3n})^3}{(1+q^{n})\;}\\[6pt] &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\[6pt] &= \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}} \end{align}$$

Level 8

 * $$\begin{align}T(q)

&= q^{1/2}\frac{f(-q,-q^7)}{f(-q^3,-q^5)}\\[6pt] &= q^{1/2}\prod_{n=1}^\infty \frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\[6pt] &= \cfrac{q^{1/2}}{1+\cfrac{q+q^2}{1+\cfrac{q^4}{1+\cfrac{q^3+q^6}{1+\cfrac{q^8}{1+\ddots}}}}} = \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\cfrac{q^8}{1+q^9+\ddots}}}}} \end{align}$$

Series

 * $$\begin{align}j_{6A}(\tau) &= j_{6B}(\tau)+\frac{1}{j_{6B}(\tau)}+2 = j_{6C}(\tau)+\frac{16}{j_{6C}(\tau)}+8 = j_{6D}(\tau)+\frac{25}{j_{6D}(\tau)}+10\\

&=\tfrac{1}{q} + 10 + 79q + 352q^2 +1431q^3+\dots \end{align}$$


 * $$\begin{align}j_{6B}(\tau) &= \Big(\tfrac{\eta(2\tau)\eta(3\tau)}{\eta(\tau)\eta(6\tau)}\Big)^{12}=\tfrac{1}{q} + 12 + 78q + 364q^2 + 1365q^3+\dots

\end{align}$$


 * $$\begin{align}j_{6C}(\tau) &= \Big(\tfrac{\eta(\tau)\eta(3\tau)}{\eta(2\tau)\eta(6\tau)}\Big)^{6}=\tfrac{1}{q} -6 + 15q -32q^2 + 87q^3-192q^4+\dots

\end{align}$$


 * $$\begin{align}j_{6D}(\tau) &= \Big(\tfrac{\eta(\tau)\eta(2\tau)}{\eta(3\tau)\eta(6\tau)}\Big)^{4}=\tfrac{1}{q} -4 - 2q + 28q^2 - 27q^3 - 52q^4+\dots\end{align}$$

Geography
1.35417°N, 12.40806°W

0.9275°N, 13.67361°W

17.95167°N, 37.935°W

1°21’15” N, 12°24’29”

0°55’39” N, 13°40’25” E