(Note: This third method continues from this post.)
There are level-$7$ pi formulas based on the McKay-Thompson series $T_{7A}$ and Cooper's $s_7$ sequence in this paper. This third method, among other things, will enable us to use $T_{7B}$.
I. Method 3
Given the binomial coefficient $\binom{n}{k}$, some free parameters $p, r,$ and a sequence $s_1(n)$. Define a second sequence as,
$$s_2(m) = \sum_{n=0}^m r^{m-pn}\binom{m}{pn} s_1(n)$$
Then we have the transformation,
$$\sum_{n=0}^{\infty} s_1(n)\,\frac{An+B}{C^n}=\left(\frac{C^{1/p}}{C^{1/p}+r}\right)^2\,\sum_{m=0}^{\infty} s_2(m)\,\frac{A/p\,m+ B-D_3}{(C^{1/p}+r)^m}$$
where,
$$D_3 = \frac{r\,(A/p-B)}{C^{1/p}}$$
II. Examples
Given the Dedekind eta function $\eta(\tau)$. First define the functions,
\begin{align} j_{7A}(\tau) &= \left(\sqrt{j_{7B}(\tau)}+\frac{7}{\sqrt{j_{7B}(\tau)}}\right)^2\\ j_{7B}(\tau) &= \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4 \end{align}
Let $\tau = \frac{7+\sqrt{-427}}{14},$ note that $427 = 7\times61$, and we get,
\begin{align} j_{7A}(\tau) &= -22^3+1 = -(39\sqrt7)^2\\ j_{7B}(\tau) &= -7\left(\frac{39+5\sqrt{61}}{2}\right)^2 \end{align}
where the latter involves the fundamental unit $U_{61}$. We have Cooper's formula,
$$\frac{1}{\pi} = \frac{\sqrt7}{22^3}\sum_{j=0}^\infty s_7(j)\, \frac{11895j+1286}{(-22^3)^j}$$
However, we wish to find a sequence that uses the whole $j_{7A}(\tau) = -22^3+1$ as this will lead to a second sequence that uses $j_{7B}(\tau)$. Thus $r=1$, and applying Method 3, we get,
$$\frac{1}{\pi} = \frac{\sqrt7}{(-22^3+1)^2}\sum_{k=0}^\infty t_{7A}(k)\, \frac{22^3(11895k+1286)-(-22^3+39)}{(-22^3+1)^{k}}$$
Then using Method 1, we get,
$$\frac{1}{\pi} = \frac{1}{(-22^3+1)\sqrt{-\,j_{7B}}}\sum_{h=0}^\infty t_{7B}(h)\, \frac{1272437 - 207636\sqrt{61}(1+2h)}{(j_{7B})^{h}}$$
where $j_{7B} = -7\left(\frac{39+5\sqrt{61}}{2}\right)^2$ as above.
III. Sequences
Starting with Cooper's sequence,
\begin{align}s_7(j) &= \sum_{m=0}^j \binom{j}{m}^2\binom{2m}{j}\binom{j+m}{m}\\ &= 1, 4, 48, 760, 13840, 273504\dots \end{align}
we derive,
\begin{align}t_{7A}(k) &= \sum_{j=0}^k\binom{k}{j}\sum_{m=0}^j \binom{j}{m}^2\binom{2m}{j}\binom{j+m}{m}\quad\\ &= 1, 5, 57, 917, 17185, 350805\dots\quad \end{align}
\begin{align} t_{7B}(h) &= \sum_{k=0}^h(-7)^{h-k}\binom{h+k}{h-k}\sum_{j=0}^k\binom{k}{j}\sum_{m=0}^j \binom{j}{m}^2\binom{2m}{j}\binom{j+m}{m}\\ &= 1, -2, 1, 49, -602, 5257, -39095\dots \end{align}
The advantage of Cooper's sequence $s_{7}$ is it only has a 3-term recurrence relation. The recurrence status of $t_{7A}$ and $t_{7B}$ is unknown. However, we recover the nice relation,
$$\sum_{n=0}^\infty t_{7A}(n)\,\frac{1}{\;\big(j_{7A}\big)^{n+1/2}} = \sum_{n=0}^\infty t_{7B}(n)\,\frac{1}{\;\big(j_{7B}\big)^{n+1/2}}$$
with closed-forms for the sequences, so it is now found in levels $L = 1,2,3,4,6,7,8,10,$ (but not yet in $L=5,9$).
IV. Questions
- Like the previous ones, why does Method 3 work, and how free are its parameters $p,r$?
- Can the closed-forms of sequences $t_{7A}$ and $t_{7B}$ be simplified?
- Lastly, what are their recurrence relations? (I've tested them, got nowhere, and I think it is an $m$-term relation with coefficients as polynomials of deg-$n$ where $m,n>4$.)
