Heuristic derivation of Zudilin’s supercongruences for rational Ramanujan series

Abstract

We derive, using a heuristic method, a p-adic mate of bilateral Ramanujan series. It has (among other consequences) Zudilin’s supercongruences for rational Ramanujan series.

1 Rational Ramanujan series for $\pi ^{-m}$

At the beginning of the twenty first century we discovered new families of Ramanujan-like series, but of greater degree [2], and proved several of them by the WZ (Wilf–Zeilberger) method [3].

We can write the rational Ramanujan-like series as

$$\begin{aligned}{} & {} \sum _{n=0}^{\infty } R(n)=\sum _{n=0}^{\infty } \left( \prod _{i=0}^{2m} \frac{(s_i)_n}{(1)_n} \right) \sum _{k=0}^m a_k n^k z_0^n = \frac{\sqrt{(-1)^m \chi }}{\pi ^m}, \end{aligned}$$

where $z_0$ is a rational such that $z_0 \ne 0$ and $z_0 \ne 1$, the parameters $a_0, a_1,...,a_m$ are positive rationals, and $\chi$ the discriminant of a certain quadratic field (imaginary or real), which is an integer. In case that $|z_0|>1$ we understand the series as its analytic continuation. An example is

$$\begin{aligned}{} & {} \frac{1}{2048} \sum _{n=0}^{\infty } \frac{\left( \frac{1}{2}\right) _n^7 \left( \frac{1}{4}\right) _n \left( \frac{3}{4}\right) _n}{(1)_n^9} (43680n^4+20632n^3+4340n^2+466n+21) \left( \frac{1}{2^{12}} \right) ^n\\{} & {} \quad =\frac{1}{\pi ^4}, \end{aligned}$$

conjectured by Jim Cullen, and recently proved by Kam Cheong Au, using the WZ method [1].

2 Bilateral Ramanujan series

We define the function $f: \mathbb {C} \longrightarrow \mathbb {C}$ in the following way:

$$\begin{aligned} f(x)=e^{-i \pi x} \prod _{s_k} \frac{\cos \pi x - \cos \pi s_k}{1-\cos \pi s_k} \sum _{n \in \mathbb {Z}} R(n+x). \end{aligned}$$

Then there exist coefficients $\alpha _k$ and $\beta _k$ (which we conjecture are rational) such that $f(x)=F(x)$, where

$$\begin{aligned} F(x)=\frac{\sqrt{(-1)^m \chi }}{\pi ^m} \, \left( 1 - \sum _{k=1}^{m} \left( \alpha _k (\cos 2 \pi k x -1) + \beta _k \sin 2 \pi k x \right) \right) , \end{aligned}$$

is the Fourier expansion of f(x).

Proof

The function f(x) is 1-periodic because the product over $s_k$ is 1-periodic as each $s_k=s$, has a companion $s_k=1-s$, and the sum over $\mathbb {Z}$ is clearly 1-periodic as well. In addition f(x) is holomorphic because the zero of $\cos \pi x - \cos \pi s_k$ at $x=-s_k$ cancels the pole of $(s_k)_{n+x}$ at $x=-s_k$, and as f(x) is periodic all the other poles are canceled as well. As f(x) is holomorphic and periodic, it has a Fourier expansion. Finally, we can prove that $f(x)=\mathcal {O}(e^{(2m+1) \pi |\textrm{Im}(x)|})$, and therefore the Fourier expansion terminates at $k=m$.

Example 2.1

$$\begin{aligned}{} & {} \frac{1}{8} \sum _{n \in \mathbb {Z}} \frac{\left( \frac{1}{2}\right) _{n+x}^5}{(1)_{n+x}^5} (20(n+x)^2+8(n+x)+1) \left( \frac{-1}{4}\right) ^{n+x}\\{} & {} \quad \quad =e^{i \pi x} \frac{1-\frac{1}{2} (\cos 2 \pi x -1) + \frac{1}{2} (\cos 4 \pi x -1)}{\pi ^2 \cos ^5\pi x}. \end{aligned}$$

Example 2.2

$$\begin{aligned}{} & {} \frac{1}{384} \sum _{n \in \mathbb {Z}} \frac{\left( \frac{1}{2}\right) _{n+x}\left( \frac{1}{3}\right) _{n+x}\left( \frac{2}{3}\right) _{n+x}\left( \frac{1}{6}\right) _{n+x}\left( \frac{5}{6}\right) _{n+x}}{(1)_{n+x}^5} \left( - \frac{3^6}{4^6} \right) ^{n+x} \\{} & {} \quad \qquad \times \left( 1930(n+x)^2+549(n+x)+45 \right) \\{} & {} \quad \quad = e^{i \pi x} \frac{3-14 (\cos 2\pi x - 1) + 6 (\cos 4\pi x - 1)}{\pi ^2 \cos \pi x \, (4\cos ^2 \pi x-1)(4\cos ^2 \pi x -3)}. \end{aligned}$$

Example 2.3

$$\begin{aligned}{} & {} \frac{1}{32} \sum _{n \in \mathbb {Z}} \frac{\left( \frac{1}{2}\right) _{n+x}^3\left( \frac{1}{4}\right) _{n+x}\left( \frac{3}{4}\right) _{n+x}}{(1)_{n+x}^5} \left( \frac{1}{16} \right) ^{n+x} (120(n+x)^2+34(n+x)+3) \\{} & {} \quad =e^{i \pi x} \, \frac{1-\frac{7}{2} (\cos 2 \pi x -1) + \frac{3}{2} (\cos 4 \pi x - 1) + \left( \frac{1}{2} \sin 2\pi x - \frac{1}{2} \sin 4 \pi x \right) i}{\pi ^2 \cos ^3 \pi x \, (2\cos ^2 \pi x -1)}. \end{aligned}$$

Example 2.4

$$\begin{aligned}{} & {} \frac{1}{6} \sum _{n \in \mathbb {Z}} \frac{(\frac{1}{2})_{n+x}^3(\frac{1}{3})_{n+x}(\frac{2}{3})_{n+x}}{(1)_{n+x}^5} \left[ 28(n+x)^2+18(n+x)+3 \right] \, (-27)^{n+x} \\{} & {} \qquad = e^{i \pi x} \frac{3 + (\cos 2\pi x - 1) + \frac{3}{4} (\cos 4\pi x - 1)}{\pi ^2 \cos ^3 \pi x (4\cos ^2 \pi x - 1)}. \end{aligned}$$

3 Series to the right and to the left

The series to the right hand side is

$$\begin{aligned} \sum _{n=0}^{\infty } R(n+x) = \sum _{n=0}^{\infty } \left( \prod _{i=0}^{2m} \frac{(s_i)_{n+x}}{(1)_{n+x}} \right) \sum _{k=0}^m a_k (n+x)^k \, z_0^{n+x}, \end{aligned}$$

extended by analytic continuation to all $z_0$ different from 0 and 1, and the series on the left hand side is

$$\begin{aligned} \sum _{n=1}^{\infty } R(-n+x)&= \sum _{n=1}^{\infty } \left( \prod _{i=0}^{2m} \frac{(s_i)_{-n+x}}{(1)_{-n+x}} \right) \sum _{k=0}^m a_k (-n+x)^k \, z_0^{-n+x}\\&= x^{2m+1} \sum _{n=1}^{\infty } \left( \prod _{i=0}^{2m} \frac{(1)_{n-x}}{(s_i)_{n-x}} \right) \sum _{k=0}^m a_k (-n+x)^{k-2m-1} \, z_0^{-n+x}\\&=x^{2m+1} z_0^{x} \left( \prod _{i=0}^{2m} \frac{(s_i)_{x}}{(1)_{x}} \right) \sum _{n=1}^{\infty } \left( \prod _{i=0}^{2m} \frac{(1-x)_{n}}{(s_i-x)_{n}} \right) \\&\quad \quad \sum _{k=0}^m a_k (-n+x)^{k-2m-1} \, z_0^{-n}, \end{aligned}$$

extended by analytic continuation to all $z_0$ different from 0 and 1. We see that

$$\begin{aligned}{} & {} \sum _{n=0}^{\infty } R(n) - \sum _{n=0}^{\infty } R(n+x) = \frac{\sqrt{(-1)^m \chi }}{\pi ^m} - e^{i \pi x} \prod _{s_k} \frac{1-\cos \pi s_k}{\cos \pi x - \cos \pi s_k} \frac{\sqrt{(-1)^m \chi }}{\pi ^m} \\{} & {} \quad \left( 1 - \sum _{k=1}^{m} \left( \alpha _k (\cos 2 \pi k x -1) + \beta _k \sin 2 \pi k x \right) \right) \\{} & {} \quad + (A + B x + C x^2 + \cdots )x^{2m+1}, \quad |x|<1, \end{aligned}$$

where $(A + B x + C x^2 + \cdots )x^{2m+1}$ is the development of the series on the left hand side at $x=0$, that is

$$\begin{aligned} z_0^{x} \left( \prod _{i=0}^{2m} \frac{(s_i)_{x}}{(1)_{x}} \right) \sum _{n=1}^{\infty } \left( \prod _{i=0}^{2m} \frac{(1-x)_{n}}{(s_i-x)_{n}} \right) \sum _{k=0}^m a_k (-n+x)^{k-2m-1} \, z_0^{-n} = A + B x + C x^2 + \cdots . \end{aligned}$$

4 Heuristic derivation of a p-adic mate

Let

$$\begin{aligned} S(N) = \sum _{n=0}^ {\infty } R(n) - \sum _{n=0}^{\infty } R(n+N) = \sum _{n=0}^{N-1} R(n). \end{aligned}$$

As in a Ramanujan-like series each $s_k<1/2$ has a companion $1-s_k$, we notice that

$$\begin{aligned} e^{i \pi x} \prod _{s_k} \frac{1-\cos \pi s_k}{\cos \pi x - \cos \pi s_k} = e^{i \pi x} \prod _{s_k=\frac{1}{2}} \frac{1}{\cos \pi x} \prod _{s_k<\frac{1}{2}} \frac{1-\cos ^2 \pi s_k}{\cos ^2 \pi x - \cos ^2 \pi s_k} \end{aligned}$$

tends to 1 as $x \rightarrow N$ because there is an odd number of factors when $s_k=1/2$. Hence for $x \rightarrow N$, we formally have

$$\begin{aligned} S(x)= & {} \sum _{n=0}^{\infty } R(n) - \sum _{n=0}^{\infty } R(n+x) \\= & {} \frac{\sqrt{(-1)^m \chi }}{\pi ^m} \left( \sum _{k=1}^{m} \left( \alpha _k (\cos 2 \pi k x -1) + \beta _k \sin 2 \pi k x \right) \right) \\{} & {} \quad + (A + B x + C x^2 + \cdots )x^{2m+1}. \end{aligned}$$

Let

$$\begin{aligned} G(x)=\frac{\sqrt{(-1)^m \chi }}{\pi ^m} \sum _{k=1}^{m} \left( \alpha _k (\cos 2 \pi k x -1) + \beta _k \sin 2 \pi k x\right) . \end{aligned}$$

For obtaining the p-adic analogues $G_p(x p)$ and $G_p(x)$, we develop G(xp) and G(x) in powers of x. Then, replace the powers of $\pi$ using values of Dirichlet L-functions, and the L-functions with the corresponding p-adic L-functions. Finally, the standard properties of the $L_p$-functions dictate turning even powers of $\pi$ to 0 when $\chi >0$, or odd powers of $\pi$ when $\chi <0$. After making the replacements, we see that

$$\begin{aligned} \lim _{x \rightarrow \nu } \frac{G_p(xp)}{G_p(x)} = p^m. \end{aligned}$$

For $x=\nu$, where $\nu =1,2,3,\dots$, we see that

$$\begin{aligned} z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) \sum _{n=1}^{\infty } \left( \prod _{i=0}^{2m} \frac{(1-\nu )_{n}}{(s_i-\nu )_{n}} \right) \sum _{k=0}^m a_k (-n+\nu )^{k-2m-1} \, z_0^{-n} = A'+B'\nu +C'\nu ^2+\cdots , \end{aligned}$$

where

$$\begin{aligned} A'= z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) A, \quad B'= z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) B, \dots . \end{aligned}$$

On the other hand, we see that

$$\begin{aligned} S(\nu )= & {} (A'+B'\nu +C'\nu ^2+\cdots ) \nu ^{2m+1}\\= & {} z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) (A+B\nu +C\nu ^2+\cdots ) \nu ^{2m+1}. \end{aligned}$$

To get the $p-$adic mate of S(x) we must divide $S_p(\nu p)$ enter $S_p(\nu )$, taking into account that the contribution of G(x) is $(\chi /p) p^m$, and the contribution of the left hand sum is given by

$$\begin{aligned}{} & {} \frac{(A_p+B_p\nu p + C_p \nu ^2 p^2+\cdots ) p^{2m+1} \nu ^{2m+1}}{(A+B\nu + C \nu ^2+\cdots )_p \nu ^{2m+1}} \\{} & {} \qquad = z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) \frac{(A_p+B_p\nu p + C_p \nu ^2 p^2+\cdots ) p^{2m+1} \nu ^{2m+1}}{S_p(\nu )}. \end{aligned}$$

Associating $(\chi /p)$ to $S_p(\nu p)$ and noting that $\Gamma _p(1/2)^{4m}=1$ by the properties of the p-adic $\Gamma$-function, we have

$$\begin{aligned} S(\nu p) = S(\nu ) \left( \frac{\chi }{p}\right) p^m + T(\nu ) (A_p+B_p\nu p + C_p \nu ^2 p^2+\cdots ) p^{2m+1} \nu ^{2m+1}, \end{aligned}$$

where $A_p, B_p,C_p \dots$, are p-adic analogues of $A,B,C\dots$, and

$$\begin{aligned} T(\nu )= z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) . \end{aligned}$$

Observe that taking positive integers values of $\nu$ we can eliminate some of the constants $A_q$, $B_q$,.., and obtain a new kind of supercongruences $\pmod {p^{2m+k}}$. For example, eliminating $A_q$ and $B_q$, we obtain supercongruences $\pmod {p^{2m+3}}$ relating S(p), S(2p) and S(3p).

We can apply a similar technique of bilateral series and p-adic mates to other kind of hypergeometric series, for example to those in [4].

5 Extended Zudilin’s supercongruences

The above p-adic mate has (among other consequences) a generalization for positive integers $\nu$ of Zudilin’s $\nu =1$ supercongruences [7] and [5], namely

$$\begin{aligned} S(\nu p) = S(\nu ) \left( \frac{\chi }{p}\right) p^m \pmod {p^{2m+1}}, \end{aligned}$$

except for very few values of p.

Example 5.1

See the Ramanujan-like series [2, Eq. (1–3)]. Let

$$\begin{aligned} S(N)=\sum _{n=0}^{N-1} \frac{\left( \frac{1}{2}\right) _n^5}{(1)_n^5}\left( \frac{-1}{4}\right) ^n (20n^2+8n+1) \end{aligned}$$

If p is a prime number (except for very few of them), then

$$\begin{aligned} S(\nu p) \equiv S(\nu ) \left( \frac{1}{p}\right) p^2 \pmod {p^5}, \end{aligned}$$

for positive integers $\nu$. Observe that for all prime p we have $(1/p)=1$.

Example 5.2

See the Ramanujan-like series [2, Eq. (4–1)]. Let

$$\begin{aligned} S(N)=\sum _{n=0}^{N-1} \frac{\left( \frac{1}{2}\right) ^7}{(1)_n^7}\left( \frac{1}{64}\right) ^n (168n^3+76n^2+14n+1) \end{aligned}$$

If p is a prime number (except for very few of them), then

$$\begin{aligned} S(\nu p) \equiv S(\nu ) \left( \frac{-4}{p}\right) p^3 \pmod {p^7}, \end{aligned}$$

for positive integers $\nu$.

Example 5.3

See the Ramanujan-like series [2, Eq. (2–4)]. Let

$$\begin{aligned} S(N)=\sum _{n=0}^{N-1} \frac{\left( \frac{1}{2}\right) _n \left( \frac{1}{2}\right) _n \left( \frac{1}{3}\right) _n\left( \frac{2}{3}\right) _n \left( \frac{1}{6}\right) _n \left( \frac{5}{6}\right) _n}{(1)_n^5}\left( \frac{-1}{80^3}\right) ^n (5418n^2+693n+29) \end{aligned}$$

If p is a prime number (except for very few of them), then

$$\begin{aligned} S(\nu p) \equiv S(\nu ) \left( \frac{5}{p}\right) p^2 \pmod {p^5}, \end{aligned}$$

for positive integers $\nu$.

6 Extended Zhao’s supercongruences

By identifying numerical approximations, we conjecture that $A=r L(\chi ,m+1)$, where r is a rational. The p-adic analogue of A is $A_p=r L_p(\chi ,m+1)$. We have the following supercongruences:

$$\begin{aligned} S(\nu p) \equiv \left( \frac{\chi }{p}\right) S(\nu ) p^m + r z_0^{\nu } \left( \prod _{i=0}^{2m} \frac{(s_i)_{\nu }}{(1)_{\nu }} \right) L_p(\chi ,m+1)p^{2m+1} \pmod {p^{2m+2}}. \end{aligned}$$

which generalizes for positive integers $\nu$ the Yue Zhao’s supercongruences for $\nu =1$ (author Y. Zhao at mathoverflow). To check these supercongruences use the following congruences

$$\begin{aligned} L_p(\chi ,m+1){} & {} \equiv L(\chi ,2+m-p) \pmod {p}, \\ \zeta _p(m+1){} & {} \equiv \frac{\textrm{bernoulli}(p-m-1)}{m+1} \pmod {p}. \end{aligned}$$

Observe that $L(1,m+1)=\zeta (m+1)$ and $L_p(1,m+1)=\zeta _p(m+1)$. For Bernoulli numbers associated to $\chi$ see [6].

Example 6.1

See the Ramanujan-like series [2, Eq. (1–3)]. Let

$$\begin{aligned} S(N)=\sum _{n=0}^{N-1} \frac{\left( \frac{1}{2}\right) _n^5}{(1)_n^5}\left( \frac{-1}{4}\right) ^n (20n^2+8n+1), \quad T(\nu )=\left( \frac{-1}{4}\right) ^\nu \frac{\left( \frac{1}{2}\right) _\nu ^5}{(1)_\nu ^5}. \end{aligned}$$

If p is a prime number (except for very few of them), then

$$\begin{aligned} S(\nu p) \equiv S(\nu ) p^2 + 448 T(\nu ) \zeta _p(3) \nu ^5 p^5 \pmod {p^6}, \end{aligned}$$

for positive integers $\nu$.

Example 6.2

See the Ramanujan-like series [2, Eq. (4–1)]. Let

$$\begin{aligned} S(N)=\sum _{n=0}^{N-1} \frac{\left( \frac{1}{2}\right) _n^7}{(1)_n^7}\left( \frac{1}{64}\right) ^n (168n^3+76n^2+14n+1), \quad T(\nu )=\left( \frac{1}{64} \right) ^{\nu } \frac{\left( \frac{1}{2}\right) _\nu ^7}{(1)_\nu ^7}. \end{aligned}$$

If p is a prime number (except for very few of them), then

$$\begin{aligned} S(\nu p) \equiv S(\nu ) \left( \frac{-4}{p}\right) p^3 + 1536 T(\nu ) L_p(-4,4) \nu ^7 p^7 \pmod {p^8}, \end{aligned}$$

for positive integers $\nu$.

Example 6.3

See the Ramanujan-like series [2, Eq. (2–4)]. Let

$$\begin{aligned} S(N)=\sum _{n=0}^{N-1} \frac{\left( \frac{1}{2}\right) _n \left( \frac{1}{2}\right) _n \left( \frac{1}{3}\right) _n\left( \frac{2}{3}\right) _n \left( \frac{1}{6}\right) _n \left( \frac{5}{6}\right) _n}{(1)_n^5}\left( \frac{-1}{80^3}\right) ^n (5418n^2+693n+29), \end{aligned}$$

and

$$\begin{aligned} T(\nu )=\left( \frac{-1}{80^3}\right) ^\nu \frac{\left( \frac{1}{2}\right) _\nu \left( \frac{1}{3}\right) _\nu \left( \frac{2}{3}\right) _\nu \left( \frac{1}{6}\right) _\nu \left( \frac{5}{6}\right) _\nu }{(1)_\nu ^5}. \end{aligned}$$

If p is a prime number (except for very few of them), then

$$\begin{aligned} S(\nu p) \equiv S(\nu ) \left( \frac{5}{p}\right) p^2 + 42000 T(\nu ) L_p(5,3) \nu ^5 p^5 \pmod {p^6}, \end{aligned}$$

for positive integers $\nu$.

7 An application of the extended supercongruences

In next examples, we use the generalized Zudilin’s supercongruences to obtain the rational parameters of the rational Ramanujan series. For that aim (except for a global rational factor) we just need taking a sufficiently large prime p and m values of $\nu$. In addition, we can check that there is a rational r such that Zhao’s supercongruences hold for that prime p and those m values of $\nu$. Hence $A_p=r L_p(\chi , m+1)$, and we conclude that $A=r L(\chi , m+1)$. Finally, observe that if $|z_0|>1$ then the series for A is convergent.

Example 7.1

We want to see that there is a series of the following form:

$$\begin{aligned} \sum _{n=0}^{\infty } \frac{\left( \frac{1}{2} \right) _n^5}{(1)_n^5} \frac{(-1)^n}{4^n} (a_0+a_1 n+ a_2 n^2)=t_0 \frac{\sqrt{\chi }}{\pi ^2}, \quad \chi =1, \end{aligned}$$

where $a_0,a_1,a_2,t_0$ are positive integers. Indeed, using the Wilf–Zeilberger (WZ method) we proved that $a_0=1, a_1=8, a_2=20$. Here

$$\begin{aligned} S(\nu p)-S(\nu )p^2 \equiv 0 \pmod {p^5}, \quad \nu =1,2,3, \dots , \end{aligned}$$

and taking $p=11$, and $\nu =1,2$, we get the linear system

$$\begin{aligned} 103175 a_0 + 126304 a_1 + 81213 a_2&\equiv 0 \pmod {11^5}, \\ 23608 a_0 + 21777 a_1 + 22319 a_2&\equiv 0 \pmod {11^5}. \end{aligned}$$

Let $a_0=t$. From the above equations, we obtain

$$\begin{aligned} -66812987 t - 95491225 a_2&\equiv 0 \pmod {11^4}, \\ -35044211 t - 95491225 a_1&\equiv 0 \pmod {11^4}. \end{aligned}$$

Solving the equations taking into account that the inverse $\pmod {11^4}$ of 95491225 is 12252, we obtain

$$\begin{aligned} a_2=-14621 t \pmod {11^4}&= 20 t, \\ a_1=-14633 t \pmod {11^4}&= 8 t, \end{aligned}$$

Hence the solutions are of the following form:

$$\begin{aligned} a_0=t, \quad a_1=8t, \quad a_2=20t. \end{aligned}$$

Example 7.2

We want to know if there is a series of the following form:

$$\begin{aligned} \sum _{n=0}^{\infty } \frac{\left( \frac{1}{2} \right) _n \left( \frac{1}{3} \right) _n \left( \frac{2}{3} \right) _n \left( \frac{1}{4} \right) _n \left( \frac{3}{4} \right) _n}{(1)_n^5} \frac{(-1)^n}{48^n} (a_0+a_1 n+ a_2 n^2)=t_0 \frac{\sqrt{\chi }}{\pi ^2}, \quad \chi =1, \end{aligned}$$

and where $a_0,a_1,a_2,t_0$ are positive integers. Using the PSLQ algorithm we conjecture that $a_0=5, a_1=63, a_2=252$ and $t_0=48$. Here

$$\begin{aligned} S(\nu p)-S(\nu ) p^2 \equiv 0 \pmod {p^5}, \quad \nu =1,2,3, \dots , \end{aligned}$$

and taking $p=13$, and $\nu =1,2$, we get the linear system

$$\begin{aligned} 155250 a_1 + 1838 a_2 + 327490 a_0&\equiv 0 \pmod {13^5}, \\ 304350 a_1 + 329224 a_2 + 67674 a_0&\equiv 0 \pmod {13^5}. \end{aligned}$$

Let $a_0=5t$. From the above equations, we obtain

$$\begin{aligned} 26628 a_1 + 7535 t&\equiv 0 \pmod {13^4}, \\ 26628 a_2 + 1579 t&\equiv 0 \pmod {13^4}. \end{aligned}$$

As the inverse $\pmod {13^4}$ of 26628 is 9279, we obtain

$$\begin{aligned} a_2=-28309 t \pmod {13^4}&= 252 t, \\ a_1=-28498 t \pmod {13^4}&= 63 t, \end{aligned}$$

Hence the solutions are: $a_0=5t, \quad a_1=63t, \quad a_2=252t$.

Example 7.3

We want to know if there is a series of the following form:

$$\begin{aligned} \sum _{n=0}^{\infty } \frac{\left( \frac{1}{2} \right) _n^7}{(1)_n^7} \left( \frac{1}{64 } \right) ^n (a_0+a_1 n+ a_2 n^2+a_3 n^3)=t_0 \frac{\sqrt{-\chi }}{\pi ^3}, \quad \chi =-4, \end{aligned}$$

where $a_0,a_1,a_2,a_3,t_0$ are positive integers. Using the PSLQ algorithm, we conjecture that $a_0=1, a_1=14, a_2=76,a_3=168$ and $t_0=16$. Here

$$\begin{aligned} S(\nu p) - S(\nu ) \left( \frac{-4}{p} \right) p^3 \equiv 0, \pmod {p^7} \quad \nu =1,2,\dots , \end{aligned}$$

and taking $p=11$, and $\nu =1,2,3$, we get the equations

$$\begin{aligned} 2078533 a_1 + 9963171 a_2 + 11695266 a_3 + 16073136 a_0 \equiv 0 \pmod {11^7}, \end{aligned}$$

$$\begin{aligned} 12453192 a_1 + 988367 a_2 + 3883033 a_3 + 14086913 a_0&\equiv 0 \pmod {11^7}, \\ 17113786 a_1 + 2247378 a_2 + 4011161 a_3 + 7012796a_0&\equiv 0 \pmod {11^7}. \end{aligned}$$

Let $a_0=t$. From the above equations, we obtain

$$\begin{aligned} 7854385 a_1 + 3429250 a_2 + 19159030 t&\equiv 0 \pmod {11^4}, \\ 3851936 a_1 + 8961898 a_2 + 5481146 t&\equiv 0 \pmod {11^4}. \end{aligned}$$

Solving the equations, we obtain

$$\begin{aligned} a_1&= - 11965 t \pmod {11^4} = 14 t, \\ a_2&=- 1255 t \pmod {11^4} = 76 t, \\ a_3&=- 14473 t \pmod {11^4} = 168 t. \end{aligned}$$

Example 7.4

We want to know if there is a series of the following form:

$$\begin{aligned} \sum _{n=0}^{\infty } \frac{\left( \frac{1}{2} \right) _n \left( \frac{1}{3} \right) _n \left( \frac{2}{3} \right) _n \left( \frac{1}{6} \right) _n \left( \frac{5}{6} \right) _n}{(1)_n^5} \left( \frac{-1}{80^3} \right) ^n (a_0+a_1 n+ a_2 n^2)=t_0 \frac{\sqrt{\chi }}{\pi ^2}, \quad \chi =5, \end{aligned}$$

and where $a_0,a_1,a_2,t_0$ are positive integers. Using the PSLQ algorithm we conjecture that $a_0=29, a_1=693, a_2=5418$ and $t_0=128$. Here

$$\begin{aligned} S(\nu p)- S(\nu ) \left( \frac{5}{p} \right) p^2 \equiv 0 \pmod {p^5}, \quad \nu =1,2,3, \dots , \end{aligned}$$

and taking $p=41$, $a_0=29 t$, and $\nu =1,2$, we get the linear system

$$\begin{aligned} 76877806 a_2 + 113924268 a_1 + 43501045 t&\equiv 0 \pmod {41^5}, \\ 88965067 a_2 + 84189111 a_1 + 113390736 t&\equiv 0 \pmod {41^5}. \end{aligned}$$

From the above equations, we obtain

$$\begin{aligned} 38939 a_1 + 32305 t&\equiv 0 \pmod {41^3}, \\ 29982 a_2 + 4321 t&\equiv 0 \pmod {41^3}. \end{aligned}$$

As the inverse of $38939 \pmod {41^3}$ is 55540, we obtain

$$\begin{aligned} a_2=-63503 t \pmod {41^3}&= 5418 t, \\ a_1=-68228 t \pmod {41^3}&= 693 t, \end{aligned}$$

Hence the solutions are: $a_0=29t, \quad a_1=693t, \quad a_2=5418t$.