Strengthened supercongruences for Ramanujan-type formulas for $1/\pi^k$

Question

The question below is again a follow-up of an old question.

Motivation: Zhi-Wei Sun listed a number of supercongruences attached to Ramanujan-type $1/\pi$ formulas in the arXiv paper which can be found here. A typical example of these congruences can be found on p. 18 of this paper:

$$\begin{split}\sum_{k=0}^{p-1}\frac{(1/2)_k(1/3)_k(2/3)_k}{(1)_k^3}\left(-\frac{1}{500^2}\right)^k(14151k+827)\equiv\\ 827\left(\frac{-3}{p}\right)p+\frac{13}{150}p^3B_{p-2}\left(\frac{1}{3}\right)\pmod{p^4},p>5\end{split}$$

which is stronger than the formula (24) in W. Zudilin's paper. $B_n(x)$ is a Bernoulli polynomial and $p$ is a prime number.

Bernoulli polynomials and Euler polynomials occur quite frequently in other supercongruences conjectured by Sun. It is well known that Bernoulli numbers and Euler number also occur in the values of $\zeta$ and $L_{-4}$ at negative integers, which implies that one can replace Bernoulli polynomials and Euler polynomials by the non-zero values of degree 1 $L$-functions at negative integers.

Experiment: When one compared the supercongruences raised by Sun with the factorization of $L$-functions of hypergeometric motives attached to Ramanujan $1/\pi$ formulas(see M. Watkins' table on p.29), one will see Euler numbers(or Bernoulli numbers) in the supercongruences where the L-functions attached to hypergeometric motives have $L_{-4}$(or $\zeta$) as a factor.

Example: We denote the truncated sum $$\sum_{k=0}^{p-1}\frac{(1/2)_k(1/4)_k(3/4)_k}{(1)_k^3}\left(\frac{1}{99}\right)^{4k}(26390k+1103)$$ by $S_p$. Since the $L$-function attached the hypergeometric motive has $L_{-8}$ as a factor, we denote $$T_p=\left(S_p-1103\left(\frac{-2}{p}\right)p\right)/(p^3L_{-8}(3-p)).$$

Sun's strengthened supercongruences implies $$T_p\equiv\frac{x}{y}\pmod{p}$$ for every prime $p>5$ and integers $x,y$. The integers $x,y$ in the system of linear congruences can be determined by lattice reduction. We then determine $x=-5,y=1089$.

It seems that the strengthened supercongruences can be found in every Ramanujan-type formula $1/\pi^k$ with the same method, e.g.,$$\begin{split}\sum_{k=0}^{p-1}\frac{(1/2)_k(1/3)_k(2/3)_k(1/6)_k(5/6)_k}{(1)_k^5}\left(-\frac{1}{80^3}\right)^k(5418k^2+693k+29)\equiv\\ 29\left(\frac{5}{p}\right)p^2-\frac{35}{216}p^5L_5(4-p)\pmod{p^6},p>5,\end{split}$$ where $L_5$ in the congruence is the Dirichlet $L$-function attached to the primitive real character with period $5$. We also note that $L_5$ is a factor of the $L$-function attached to the hypergeometric motive of the formula (29) in Zudilin's paper.

Questions: Why do the values of degree 1 $L$-factor at negative integers occur in the strengthened supercongruences?

Answer 1

For the two examples in your question I conjecture $p$-adic expansions that begin as: $$ S_p = 1103 \left(\frac{-2}{p}\right)p - \frac{5}{1089} L_{-8,p}(2) p^3 + \cdots, $$ and $$ S_p = 29 \left(\frac{5}{p}\right) p^2 - \frac{35}{216} L_{5,p}(3) p^5 + \cdots, $$ where $L_{-8, p}(k)$ is the $p$-adic analogue of $L_{-8}(k)$, and $L_{5, p}(k)$ the $p$-adic analogue of $L_{5}(k)$.

As $$L_{-8, p}(2) \equiv L_{-8}(3-p) \pmod{p},$$ and $$L_{5,p}(3) \equiv L_5(4-p) \pmod{p},$$ the above $p$-adic expansions imply the supercongruences in the question.

For details and support of the conjectured $p$-adic expansions, see http://arxiv.org/abs/1910.01961)