# Other kind of supercongruences for rational Ramanujan-like series

We can write rational Ramanujan-like series with rational parameters in the following form:

$$\sum_{n=0}^{\infty} \left( \prod_{i=0}^{2m} \frac{(s_i)_{n}}{(1)_{n}} \right) z_0^{n} \sum_{k=0}^m a_k n^k = \frac{\sqrt{(-1)^m \chi}}{\pi^m},$$

where $$\chi$$ is the character (an integer).

Zudilin’s supercongruences are related to them, namely

$$\sum_{n=0}^{p-1} \left( \prod_{i=0}^{2m} \frac{(s_i)_{n}}{(1)_{n}} \right) z_0^{n} \sum_{k=0}^m a_k n^k \equiv \left( \frac{\chi}{p} \right) p^m \pmod{p^{2m+1}},$$ for primes $$p$$ which are not factors of $$z_0$$.

We conjecture the following alternative supercongruences:

$$\sum_{n=0}^{\frac{p-1}{2}} t(n) \left( \prod_{i=0}^{2m} \frac{(s_i-\frac{p}{2})_{n}}{(1-\frac{p}{2})_{n}} \right) z_0^{n} \sum_{k=0}^m a_k \left(n-\frac{p}{2}\right)^k \equiv 0 \pmod{p^{2m+1}},$$

for primes $$p$$ which are not factors of $$z_0$$, where $$0 \leq n < \frac{p-1}{2} \Rightarrow t(n)=1, \quad t\left(\frac{p-1}{2}\right)=1-\left(\frac{\varepsilon}{p}\right)p^j,$$

being $$2j+1$$ the number of $$s_i$$ that are equal to $$1/2$$. The character $$\varepsilon$$ is related to $$\chi$$, and the values of $$\varepsilon$$ are given in the tables of arXiv:1908.05123.

When $$\varepsilon=1$$ and $$j=0$$ these new supercongruences take a simpler form, namely

$$\sum_{n=0}^{\frac{p-3}{2}} \left( \prod_{i=0}^{2m} \frac{(s_i-\frac{p}{2})_{n}}{(1-\frac{p}{2})_{n}} \right) z_0^{n} \sum_{k=0}^m a_k \left(n-\frac{p}{2}\right)^k \equiv 0 \pmod{p^{2m+1}}.$$

We show two examples:

1-) For the Chudnovskys’ series we have the supercongruences

$$\begin{multline} \sum_{n=0}^{\frac{p-3}{2}} \frac{\left(\frac12-\frac{p}{2} \right)_n \left(\frac16-\frac{p}{2} \right)_n\left(\frac56-\frac{p}{2} \right)_n}{\left( 1-\frac{p}{2} \right)_n^3} \left( \frac{-1}{53\,360^3} \right)^n \\ \left(545\,140\, 134\left(n-\frac{p}{2}\right)+13\,591\,409 \right) \equiv 0 \pmod{p^3}, \end{multline}$$

which are much simpler than Zudilin’s supercongruences

$$\begin{multline} \sum_{n=0}^{p-1} \frac{\left(\frac12 \right)_n \left(\frac16 \right)_n\left(\frac56 \right)_n}{\left( 1 \right)_n^3}\left(545\,140\,134n+13\,591\,409 \right) \left( \frac{-1}{53\,360^3} \right)^n \\ {}\equiv \left( \frac{-40\,020}{p} \right)p^2 \pmod{p^3}. \end{multline}$$

2-) For the following Ramanujan-like series for $$1/\pi^2$$:

$$\sum_{n=0}^{\infty} \frac{\left( \frac12 \right)_n \left( \frac13 \right)_n \left( \frac23 \right)_n \left( \frac16 \right)_n \left(\frac56 \right)_n}{(1)_n^5} \frac{(-1)^n}{80^{3n}} (5418n^2+693n+29) = \frac{128 \sqrt 5}{\pi^2},$$

we have the supercongruences

$$\sum_{n=0}^{\frac{p-3}{2}} \frac{\left( \frac12-\frac{p}{2} \right)_n \left( \frac13-\frac{p}{2} \right)_n \left( \frac23-\frac{p}{2} \right)_n \left( \frac16-\frac{p}{2} \right)_n \left(\frac56-\frac{p}{2} \right)_n}{(1-\frac{p}{2})_n^5} \frac{(-1)^n}{80^{3n}} \\ \left(5418(n-\frac{p}{2})^2+693(n-\frac{p}{2})+29\right) \equiv 0 \pmod{p^5}.$$

The same values of $$\varepsilon$$ appear in the paper arXiv:1906.07384 in Asai $$L$$-functions of Hilbert modular forms. I find it intriguing. What could be an explanation?