(*Note*: This third method continues from [this post](https://mathoverflow.net/q/446886/12905).)


There are level-$7$ pi formulas based on the McKay-Thompson series $T\_{7A}$ and Cooper's $s\_7$ sequence in this [paper](https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p). 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](https://mathoverflow.net/q/446778/12905), 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**


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 $m,n>4$.)



Finding a new level-$7$ pi formula using the relation $j_{7A}(\tau) = \left(\sqrt{j_{7B}(\tau)}+\frac{7}{\sqrt{j_{7B}(\tau)}}\right)^2$
For Question $3$ about the recurrence relations, using my code from
[MMA question 285008](https://mathematica.stackexchange.com/q/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.


