Transformations of Ramanujan's 1/pi formulas $\sum_{n=0}^{\infty} s(n)\frac{An+ B}{C^n}$ and Monster moonshine functions

Question

Someone with many papers on Ramanujan's work asked me how I managed to find the closed-forms for the binomial sums of level $10$ in a recent MO post. (A colleague of his wasn't able to find them.) I found about six general transformation methods, one of which is relevant to his question.

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I. Method 1

Given the binomial coefficient $\binom{n}{k}$, some free parameters $p, r,$ and a sequence $s_1(n)$. Define a second sequence as,

$$s_2(m) = \sum_{n=0}^m (-r)^{m-pn} \binom{m+pn}{m-pn} s_1(n)$$

Then we have the transformation,

$$\sum_{n=0}^{\infty} s_1(n)\,\frac{An+ B}{C^n}=\frac{C^{1/p}}{\alpha-r}\sum_{m=0}^{\infty} s_2(m)\,\frac{A/p\,m+ B+D_1}{\alpha^m}$$

where we find $\alpha$ and $D_1$ as,

$$C^{1/p}=\Big(\sqrt{\alpha}+\frac{r}{\sqrt{\alpha}}\Big)^2$$ $$D_1 = \frac{r\,(A/p-2B)}{\alpha+r}$$

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II. Level 2

As an example of the method, consider Ramanujan's formula,

$$\frac1{\pi}=\frac{2\sqrt2}{99^2} \sum_{k=0}^\infty \binom{2k}{k}^2\binom{4k}{2k} \frac{26390k+1103}{396^{4k}}$$

so we have $A =26390, B=1103, C = 396^4.$ But we also have the known relation,

$$j_{2A}=\Big(\sqrt{j_{2B}}+\frac{64}{\sqrt{j_{2B}}}\Big)^2= 396^4$$

where $j_M$ is one of the "moonshine functions" (which span a linear space of 163 dimensions). We could choose other $p,r$, but for this case, the appropriate choice is $p=1$ and $r=64$, and giving us the 12th power of the fundamental unit $U_{29}$,

$$\alpha = j_{2B} = 64\left(\frac{5+\sqrt{29}}2\right)^{12}$$

With $\alpha,p,r$ known, then $D_1$ follows, and we get the transformed formula,

$$\frac{1}{\pi} = \frac{16\sqrt{2}}{\sqrt{\alpha}}\sum_{k=0}^\infty s_2(k)\, \frac{99^2\sqrt{29}\left(k+\frac12\right)-24184}{\alpha^k}$$

using the second sequence,

$$s_2(k) = \sum_{j=0}^k(-64)^{k-j}\binom{k+j}{k-j}\binom{2j}{j}^2\binom{4j}{2j}$$

See also Ramanujan-Sato series for level 2.

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III. Level 10

A known formula is,

$$\frac{1}{\pi} = \frac{5}{76\sqrt{95}}\sum_{k=0}^\infty \sum_{j=0}^k \binom{k}{j}^4\frac{408k+47}{76^{2k}}$$

However, there are the relations,

$$j_{10A} = \left(\sqrt{j_{10B}} + \frac{4}{\sqrt{j_{10B}}}\right)^2 = \left(\sqrt{j_{10D}} + \frac{-1}{\sqrt{j_{10D}}}\right)^2 = 76^2$$

From these, we can get $j_{10B}$ and $j_{10D}$. And just like before, we chose $p = 1$, then $r_B = 4$ and $r_D = -1$. In addition to the 1st sequence, and with minor change of variables, this gives us the 2nd and 3rd sequences,

\begin{align} s_{10A}(k) &=\sum_{j=0}^k \binom{k}{j}^4 = 1,2,18,164,1810,\dots \\ s_{10B}(k) &=\sum_{j=0}^k(-4)^{k-j}\binom{k+j}{k-j}\sum_{m=0}^j \binom{j}{m}^4 = 1,−2,10,−68,514,−4100,\dots\\ s_{10D}(k) &= \sum_{j=0}^k\binom{k+j}{k-j} \sum_{m=0}^j \binom{j}{m}^4 = 1,3,25,267,3249,\dots \end{align}

However, there is still $j_{10C}.$ (All four are found in this post.) But since its relation with $j_{10A}$ is only a near-square, deriving its sequence $s_{10C}(k)$ is trickier but doable (which can be discussed in another post).

Using Method 1, this level 10 formula can be transformed just like Ramanujan's level 2 formula. So with four sequences and subscripts $\small{A, B, C, D}$, this implies there are at least four basic families of 1/pi for level $10$.

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IV. Questions

1. I found Method 1 empirically. I tested it with various "seed" sequences (even the Fibonacci sequence and others) with free parameters $p,r$ chosen carefully and, as long as it is within the radius of convergence, it seems valid. So when it works, then why does it work?
2. For prudence, when testing the method, I only chose small integers for parameters $p,r$. But how arbitrary can they be?

Answer 1

(This is an addendum to add more details but not to clutter the main post.)

As more evidence, the method gives alternative closed-forms for the level-$6$ binomial sums. To recall, we have,

$$j_{6A} = \left(\sqrt{j_{6B}} + \frac{-\color{blue}1}{\sqrt{j_{6B}}}\right)^2 = \left(\sqrt{j_{6C}} + \frac{\color{blue}8}{\sqrt{j_{6C}}}\right)^2$$

Using the method to generate a new sequence from the "seed", we chose $p=1,$ then $r=-\color{blue}1$ and $r=\color{blue}8.$ The 1st sequence is the "seed",

\begin{align} s_{6A}(k) &=\binom{2k}{k}\sum_{j=0}^k \binom{k}{j}^3\quad\quad\\ &=1, 4, 60, 1120, 24220,\ldots\quad \end{align}

and we find an alternative form for the 2nd sequence (which are Apéry's numbers $b_n$),

\begin{align} s_{6B}(k) &=\sum_{j=0}^k \binom{k}{j}^2\binom{k+j}{j}^2\\ &=\sum_{j=0}^k(\color{blue}1)^{k-j}\,\binom{k+j}{k-j}\binom{2j}j\sum_{m=0}^k \binom{j}{m}^3\\ &=1, 5, 73, 1445, 33001,\ldots\\ \end{align}

and for the 3rd as,

\begin{align} s_{6C}(k) &= (-1)^k \sum_{j=0}^k \binom{k}{j}^2 \binom{2(k-j)}{k-j} \binom{2j}{j}\\ &=\sum_{j=0}^k(-\color{blue}8)^{k-j}\,\binom{k+j}{k-j}\binom{2j}j\sum_{m=0}^k \binom{j}{m}^3\\ &=1, -4, 28, -256, 2716,\ldots \end{align}

The 4th using $j_{6D}$ is trickier since its relation with $j_{6A}$ is only a near-square but again, still doable. But this makes me wonder if the level-$10$ analogues have simpler single-summation versions.