(Note: The second method described here continues this post.)
About nine years ago, I made an MO post "Ramanujan's pi formulas with a twist". An answer was informative, but not completely satisfactory. That question focused on levels 1 and 2, but it turns out the phenomenon seem to appear in all levels, and this post will focus on level 10.
In "Level 10 Analogues of Ramanujan's series for 1/pi", Cooper mentions in page 2 three examples, namely,
\begin{align} \frac{1}{\pi} &= \frac{\sqrt{95}}{1444}\sum_{k=0}^\infty t_A(k)\,\frac{408k+47}{\left(76^2\right)^{k}}\\ \frac{1}{\pi} &= \frac{\sqrt{19}}{54910}\sum_{k=0}^\infty t_B(k)\,\frac{34656k+3983}{\left(76^2+4\right)^{k}}\\ \frac{1}{\pi} &= \frac{\sqrt{38}}{27360}\sum_{k=0}^\infty t_C(k)\,\frac{12274k+1427}{\left(76^2-16\right)^{k}} \end{align}
for some sequences $t(k)$. This starts with $j_{10A}(\tau) = 76^2$, but it seems we can apply a general method that uses a denominator $\color{blue}{\beta+4r}$ for some free parameter $r.$
I. Method 2
Given the binomial coefficient $\binom{n}{k}$, some free parameters $p, r,$ and a sequence $s_1(n)$. Define a second sequence as,
$$s_2(m) = \binom{2m}{m}\sum_{n=0}^m r^{m-pn}\binom{m}{pn} \binom{2pn}{pn}^{-1} s_1(n)$$
Then we have the transformation,
$$\sum_{n=0}^{\infty} s_1(n)\,\frac{An+ B}{C^n}=\left(\frac{C^{1/p}}{C^{1/p}+4r}\right)^{3/2}\sum_{m=0}^{\infty} s_2(m)\,\frac{A/p\,m+ B-D_2}{(C^{1/p}+4r)^m}$$
where,
$$D_2 = \frac{2r\,(A/p-2B)}{C^{1/p}}$$
This seems a very general method that transforms one formula to another, with the Ramanujan-type pi series as a special case.
II. Examples
For Cooper's three formulas, the first step is the "seed" sequence,
$$t_A(k)=\sum_{j=0}^k \binom{k}{j}^4 =1, 2, 18, 164, 1810,\ldots$$
For the second and third sequences, we let $p=1$ so what remains is to choose $r$. For eq.$(2)$ it is obviously $\color{blue}{r=1}$ and, after minor change of variables, we get,
$$t_B(k)=\binom{2k}{k}\sum_{j=0}^k (\color{blue}{1})^{k-j} \binom{k}{j} \binom{2j}{j}^{-1} \sum_{m=0}^j \binom{j}{m}^4 =1, 4, 36, 424, 5716,\ldots$$
For eq.$(3)$ it is $\color{blue}{r = -4}$ thus,
$$t_C(k)=\binom{2k}{k}\sum_{j=0}^k (\color{blue}{-4})^{k-j} \binom{k}{j} \binom{2j}{j}^{-1} \sum_{m=0}^j \binom{j}{m}^4 =1, -6, 66, -876, 12786,\ldots$$
These are precisely the sequences in page 10 found by Zudilin. Then applying the Method 2 transformation on $(1)$, and after simplification, we get the forms $(2),(3).$ But apparently one can use other $p,r$ so it seems for level 10 there are indefinitely many families, not just the three above.
III. Other p,r
For example, starting with the Chudnovsky-type formula which is level 1,
$$\frac1{\pi} = \frac{12}{960^{3/2}}\sum_{n=0}^\infty \binom{2n}{n}\binom{3n}{n}\binom{6n}{3n} \frac{16254n+789}{(-960^3)^n}$$
To get rid of the cube exponent, let $p=3.$ We can chose $r=-3$ (or some other small integer), then using the transformation, we get,
$$\frac1{\pi} = \frac{\pm 1}{162\sqrt3}\sum_{k=0}^\infty s_{3C}(k)\,\frac{602k+85}{(-960-12)^{k}}$$
where,
$$s_{3C}(k) =\binom{2k}{k}\sum_{j=0}^k(-3)^{k-3j}\binom{k}{3j}\binom{2j}{j}\binom{3j}{j} = 1, -6, 54, -420, 630,\ldots$$
Of course, as the old MO post discusses, one can use the transformation on the level-1 Chudnovsky, and the level-2 Ramanujan's formula (though I didn't know the general transformation back then). And so on for other levels.
IV. Questions
- Just like Method 1 in this post, I found Method 2 empirically, so what is the basic reason they work?
- Can we actually use any $p,r$ as long as the series remains within the radius of convergence? This implies that for level 10 (and others), there are indefinitely many families.
