(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 $m,n>4$.)
Best Answer
For Question $3$ about the recurrence relations, using my code from MMA question 285008 for $a_n := T_{7A}(n)$ I used
findseqrecur[4, 4, Array[t7A, 33, 1], 1, "a", k, -1]to get$$ 0 = 14(n+1)(n+2)(2n+3) a_n \\ -3(n+2)(19n^2+76n+80) a_{n+1} \\ + 5(2n+5)(3n^2+15n+19) a_{n+2} \\ - (n+3)^3 a_{n+3}. $$
For $b_n := T_{7B}(n)$ there are several recurrences.
For degree $4$ polynomials I used
findseqrecur[6, 5, Array[t7B, 34, 1], 1, "b", k, -4]to get$$ 0 = -7^5(k-17)(k-2)^3 b_{k-3}\quad \\ -7^3(19k^4-678k+2218k^2-2640k+1113)b_{k-2}\;\; \\ -7(85k^4-8707k^3+9978k^2-7072k+2090)b_{k-1} \\ +(85k^4+8707k^3+9978k^2+7072k+2090)b_{k}\, \\ +(19k^4+678k+2218k^2+2640k+1113)b_{k+1} \\ \quad +(k+17)(k+2)^3 b_{k+2}. $$
For degree $3$ polynomials I used
findseqrecur[8, 4, Array[t7B, 38, 1], 1, "b", k, -5]to get$$ 0 = 7^7(k-3)^3b_{k-4}\; \\ + 7^5(47k^3-300k^2+646k-470)b_{k-3} \;\\ + 2\cdot 7^3(480k^3-1830k^2+2483k-1206)b_{k-2}\quad \\ + 7^2 (1578k^3-2001k^2+1513k-433)b_{k-1} \\ + 7 (1578k^3+2001k^2+1513k+433)b_{k}\;\; \\ \;\; + 2 (480k^3+1830k^2+2483k+1206)b_{k+1} \\ \; + (47k^3+300k^2+646k+470)b_{k+2} \\ \; + (k+3)^3 b_{k+3}. $$
Notice the symmetry of these two recursions.