(This continues from level 10.) Given some moonshine functions $j_{N}$. There are nice descending and consistent relations for levels $6m$ with $m= 2,3,5,$
$$j_{12A} = \left(\sqrt{j_{12H}} + \frac{\color{blue}{1}}{\sqrt{j_{12H}}}\right)^2 =\left(\sqrt{j_{12E}} + \frac{\color{blue}4}{\sqrt{j_{12E}}}\right)^2 = \left(\sqrt{j_{12B}} + \frac{\color{blue}3}{\sqrt{j_{12B}}}\right)^2+4$$
$$j_{18B} = \left(\sqrt{j_{18E}} + \frac{\color{blue}{1}}{\sqrt{j_{18E}}}\right)^2 = \left(\sqrt{j_{18A}} + \frac{\color{blue}3}{\sqrt{j_{18A}}}\right)^2 = \left(\sqrt{j_{18C}} + \frac{\color{blue}2}{\sqrt{j_{18C}}}\right)^2+4$$
$$j_{30B} = \left(\sqrt{j_{30F}} + \frac{\color{blue}{1}}{\sqrt{j_{30F}}}\right)^2 = \left(\sqrt{j_{30C}} + \frac{\color{blue}2}{\sqrt{j_{30C}}}\right)^2 = \left(\sqrt{j_{30A}} + \frac{\color{blue}1}{\sqrt{j_{30A}}}\right)^2+4$$
As before, if the "auxiliary" integer sequence for the LHS function is known, then one gets at least four sequences for that level by using the blue numbers. Each can then be used to find formulas for $1/\pi.$
I. Level-12 functions
For brevity, this is only for level-12. Given the Dedekind eta function $\eta(\tau)$, define,
\begin{align} j_{12A}(\tau) &= \left(\frac{\eta^2(2\tau)\,\eta^2(6\tau)}{\eta(\tau)\,\eta(3\tau)\,\eta(4\tau)\,\eta(12\tau)}\right)^{6} \\ j_{12B}(\tau) &= \left(\frac{\eta(\tau)\,\eta(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta(3\tau)\,\eta(12\tau)}\right)^{4} \\ j_{12E}(\tau) &= \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(4\tau)\,\eta(12\tau)}\right)^{2} \\ j_{12H}(\tau) &= \left(\frac{\eta(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta(12\tau)}\right)^{4} \\ j_{12I}(\tau)\, &= \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^{2} \end{align}
Conway and Norton found these five moonshine functions obey (or a version thereof),
$$j_{12A}+2j_{12I} = j_{12B}+j_{12E}+j_{12H}+8$$
II. Sequences
From "The Level 12 Analogue of Ramanujan's Function $k$", it turns out the auxiliary sequence for $j_{12A}(\tau)$ are the Domb numbers. (Edit: In Chan et al's paper, it also appears in level-6). From this sequence, we can derive three more,
\begin{align} s_{12A}(k) &= \sum_{j=0}^k \binom{k}{j}^2 \binom{2k-2j}{k-j} \binom{2j}{j}\\ s_{12H}(j) &=\sum_{k=0}^j (-u)^{j-k}\binom{j+k}{j-k}\,s_{12A}(k)\\ s_{12E}(j) &=\sum_{k=0}^j (-v)^{j-k}\binom{j+k}{j-k}\,s_{12A}(k)\\ s_{12B}(n) &=\sum_{j=0}^n \sum_{k=0}^j(-w)^{n-j}\,\binom{n+j}{n-j}\binom{j}{k}\binom{2j}{j}\binom{2k}{k}^{-1}s_{12A}(k) \end{align}
where $u = \color{blue}1$, $v = \color{blue}4$, $w = \color{blue}3.$ Using the variable $n$ for uniformity, the first few terms are,
\begin{align} s_{12A}(n) &=1, 4, 28, 256, 2716, 31504,\ldots\\ s_{12H}(n) &=1, 3, 17, 139, 1305, 13307,\ldots\\ s_{12E}(n) &=1, 0, -4, 16, -36, 32, 64, 0, -2404,\ldots\\ s_{12B}(n) &=1, -1, 1, 7, -103, 1015, -9087, 78615,\ldots \end{align}
such that all $s_{12}(0) = 1.$ The sequences $(s_{12A},\,s_{12H},\, s_{12E},\, s_{12B})$ have an $m$-term recurrence relation with $m = 3,6,6,7,$ respectively (similar to level-10).
III. Pi formulas
A. These four sequences can generate new formulas for $1/\pi$ of level 12. For example, let $\tau = \frac16\sqrt{-15}$, then,
\begin{align} j_{12A}(\tau) &= 64\\ j_{12D}(\tau) &= (4+\sqrt{15})^2\\ j_{12E}(\tau) &= 4(2+\sqrt{3})^2\\ j_{12B}(\tau) &= 3(2+\sqrt{5})^2\qquad \end{align}
and we get,
\begin{align} \quad\frac1{\pi} &= \sqrt{3}\sum_{n=0}^\infty s_{12A}(n)\,\frac{\;5n+1}{(64)^{n+1/2}}\\ \frac1{\pi} &= \frac{4}{\sqrt{5}}\,\sum_{n=0}^\infty s_{12D}(n)\,\frac{5n+1+\psi_1\quad}{\big((4+\sqrt{15})^2\big)^{n+1/2}}\\ \frac1{\pi} &= \sqrt4\,\sum_{n=0}^\infty s_{12E}(n)\,\frac{5n+1+\psi_2}{\,\big(4(2+\sqrt{3})^2\big)^{n+1/2}}\\ \frac1{\pi} &= \sqrt{3}\,\sum_{n=0}^\infty s_{12B}(n)\,\frac{An+B+\psi_3}{\,\big(3(2+\sqrt{5})^2\big)^{n+1/2}} \end{align}
where $\psi_1 = \frac{3}{8\,(4+\sqrt{15})}$ and $\psi_2 = \frac{3}{4\,(2+\sqrt{3})}.$ (Note: The fourth to be added later.)
B. Furthermore, if within the radius of convergence, it seems that,
$$x=\sum_{n=0}^\infty s_{12A}(n)\,\frac{1}{\;\big(j_{12A}\big)^{n+1/2}} = \sum_{n=0}^\infty s_{12H}(n)\,\frac{1}{\;\big(j_{12H}\big)^{n+1/2}} = \\ \sum_{n=0}^\infty s_{12E}(n)\,\frac{1}{\;\big(j_{12E}\big)^{n+1/2}} = \sum_{n=0}^\infty s_{12B}(n)\,\frac{1}{\;\big(j_{12B}\big)^{n+1/2}}$$
though level-12 is special since $x$ has a closed-form,
$$x=\frac{\eta^4(2\tau)\,\eta^4(6\tau)}{\eta(\tau)\,\eta(3\tau)\,\eta(4\tau)\,\eta(12\tau)}$$
IV. Questions
- I have no proof for the pi formulas above, and other similar level-12 I’ve tested. Are they indeed true? (Cooper's paper cited does not include any pi formulas.)
- As usual, do the sequences' closed-forms for level-12 have simpler versions just like for level-6?
Edit. Turns out the paper "Domb’s numbers and Ramanujan–Sato type series for 1/pi" by H. Chan, S. Chan, and Z. Liu has the first sequence and pi formula, but not the other three sequences.)
