(*Note*: This third method continues from [this post][1].)

Among other things, this allows us to find new level-7 formulas for $1/\pi$ based **both** on the McKay-Thompson series $T_{7A}$ and $T_{7B}$, the former derived from sequence $s_7$ in  [Cooper's paper][2], while the latter uses a new sequence.  

---

**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][3], 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 that 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**

1. Like the previous ones, why does Method 3 work, and how free are its parameters $p,r$? 
2. Can the closed-forms of sequences $t_{7A}$ and $t_{7B}$ be simplified? 
3. 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 $m,n>4$.)


  [1]: https://mathoverflow.net/q/446886/12905
  [2]: https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p
  [3]: https://mathoverflow.net/q/446778/12905