# ρ — Adic Analogues of Ramanujan Type Formulas for 1/π

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Department of Mathematics and Statistics, University of Calgary, Calgary AB, T2N 1N4, Canada

Department of Mathematics, University of Washington, Seattle, WA 98195, USA

Department of Mathematics, Cornell University, Ithaca, NY 14853, USA

Department of Mathematics, Iowa State University, Ames, IA 50011, USA

Lehrstuhl D für Mathematik, RWTH Aachen University, 52056 Aachen, Germany

Department of Mathematics, Oregon State University, Corvallis, OR 97331, USA

Author to whom correspondence should be addressed.

Received: 18 February 2013
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Revised: 26 February 2013
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Accepted: 1 March 2013
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Published: 13 March 2013

Following Ramanujan's work on modular equations and approximations of $\pi $ , there are formulas for $1/\pi $ of the form Following Ramanujan's work on modular equations and approximations of $\pi $ , there are formulas for $1/\pi $ of the form $\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{2}\right)}_{k}{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{3}}(ak+1){\left({\lambda}_{d}\right)}^{k}=\frac{\delta}{\pi}$ for $d=2,3,4,6,$ where ${\u0142}_{d}$ are singular values that correspond to elliptic curves with complex multiplication, and $a,\delta $ are explicit algebraic numbers. In this paper we prove a $p-$ adic version of this formula in terms of the so-called Ramanujan type congruence. In addition, we obtain a new supercongruence result for elliptic curves with complex multiplication.

Ramanujan [1] gave a list of infinite series identities of the form
for $d=2,3,4,6,$ where ${\lambda}_{d}$ are singular values that correspond to elliptic curves with complex multiplication, and $a,\delta $ are explicit algebraic numbers. Below is an example of one identity from Ramanujan’s list,

$$\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{2}\right)}_{k}{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{3}}(ak+1){\left({\lambda}_{d}\right)}^{k}=\frac{\delta}{\pi}$$

$$\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{2}\right)}_{k}^{3}}{k{!}^{3}}(6k+1)\frac{1}{{4}^{k}}=\frac{4}{\pi}$$

Here, ${\left(a\right)}_{k}$ denotes the rising factorial ${\left(a\right)}_{k}=a(a+1)\cdots (a+k-1).$ In fact, the following similar identity was given earlier by Bauer [2],

$$\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{2}\right)}_{k}^{3}}{k{!}^{3}}(4k+1){(-1)}^{k}=\frac{2}{\pi}$$

Proofs of these formulas were first given by J. Borwein and P. Borwein [3] and D. Chudnovsky and G. Chudnovsky [4]. Both approaches rely on the arithmetic of elliptic integrals of the first and second kind, including the Legendre relation at singular values. Finding new formulas for $1/{\pi}^{k}$ using various techniques has been an active research area. We refer interested readers to two survey papers, one by Baruah, Berndt, and Chan [5] and another by Zudilin [6], as well as a list of conjectures due to Z.-W. Sun [7]. One of the motivations for studying Ramanujan formulas for $1/\pi $ is to efficiently compute the decimal digits of $\pi .$

In 1997, van Hamme discovered several surprising $p-$adic analogues of Ramanujan formulas for $1/\pi $ [8]. For each prime $p>3$ he conjectured
where $\left(\frac{\xb7}{p}\right)$ is the Legendre symbol.

$$\begin{array}{ccc}\hfill \sum _{k=0}^{p-1}\frac{{\left(\frac{1}{2}\right)}_{k}^{3}}{k{!}^{3}}(4k+1){(-1)}^{k}& \equiv & \left(\frac{-1}{p}\right)p\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{3})\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \sum _{k=0}^{\frac{p-1}{2}}\frac{{\left(\frac{1}{2}\right)}_{k}^{3}}{k{!}^{3}}(6k+1)\frac{1}{{4}^{k}}& \equiv & \left(\frac{-1}{p}\right)p\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{4})\hfill \end{array}$$

The congruence (1.1) was first proved by Mortenson using hypergeometric evaluation identities [9]. This method is dependent upon the availability of such corresponding identities. Later, Equation (1.1) was also established by Zudilin using the techniques of Wilf and Zeilberger [10], which, unfortunately, are not easy to use in general. Motivated by work of Mortenson [9] as well as by McCarthy and Osburn [11], the third author proved congruence (1.2) [12]. Recently, Z.-W. Sun has given a refinement of congruence (1.1) modulo ${p}^{4}$ by adding a factor involving Euler numbers [13].

It is interesting to note that both Equations (1.1) and (1.2) hold modulo ${p}^{3}$ with either $p-1$ or $(p-1)/2$ for the limit of summation. However, this is not true modulo ${p}^{4}$ for either Equation (1.2) or Sun’s refinement of Equation (1.1).

In this paper, we prove a general result on Ramanujan type congruences modulo ${p}^{2}$ under further assumptions. In Section 2, we introduce necessary notation and state our main result. Our method relies on the arithmetic and geometry of elliptic curves, which includes Picard–Fuchs equations, formal expansions of the invariant differentials of elliptic curves, the Chowla–Selberg formula for periods of elliptic curves with complex multiplication, as well as Atkin and Swinnerton–Dyer congruences and results due to Katz. In Section 3, we review a method to prove Ramanujan type formulas for $1/\pi $ utilizing Picard–Fuchs equations associated to families of elliptic curves, originating in the work of Chowla and Selberg [14]. In Section 4 we discuss arithmetic of certain families of elliptic curves. We conclude in Section 5 with the proof of our result.

For r a nonnegative integer and ${\alpha}_{i},{\beta}_{i}\in \mathbb{C}$, the hypergeometric series ${}_{r+1}{F}_{r}$ is defined by
which converges for $\left|x\right|<1$. We write
to denote the truncation of the series after the ${x}^{n}$ term.

$${}_{r+1}{F}_{r}\left[\begin{array}{ccc}{\alpha}_{1}& \dots & {\alpha}_{r+1}\\ {\beta}_{1}& \dots & {\beta}_{r}\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]=\sum _{k=0}^{\infty}\frac{{\left({\alpha}_{1}\right)}_{k}{\left({\alpha}_{2}\right)}_{k}\dots {\left({\alpha}_{r+1}\right)}_{k}}{{\left({\beta}_{1}\right)}_{k}\dots {\left({\beta}_{r}\right)}_{k}}\xb7\frac{{x}^{k}}{k!}$$

$${}_{r+1}{F}_{r}{\left[\begin{array}{ccc}{\alpha}_{1}& \dots & {\alpha}_{r+1}\\ {\beta}_{1}& \dots & {\beta}_{r}\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]}_{n}=\sum _{k=0}^{n}\frac{{\left({\alpha}_{1}\right)}_{k}{\left({\alpha}_{2}\right)}_{k}\dots {\left({\alpha}_{r+1}\right)}_{k}}{{\left({\beta}_{1}\right)}_{k}\dots {\left({\beta}_{r}\right)}_{k}}\xb7\frac{{x}^{k}}{k!}$$

For $d\in \{2,3,4,6\}$ let ${\tilde{E}}_{d}\left(t\right)$ denote the following families of elliptic curves parameterized by t,

$$\begin{array}{cc}\hfill {\tilde{E}}_{2}\left(t\right):& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{y}^{2}=x(x-1)(x-t)\hfill \\ \hfill {\tilde{E}}_{3}\left(t\right):& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{y}^{2}+xy+\frac{t}{27}y={x}^{3}\hfill \\ \hfill {\tilde{E}}_{4}\left(t\right):& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{y}^{2}=x({x}^{2}+x+\frac{t}{4})\hfill \\ \hfill {\tilde{E}}_{6}\left(t\right):& \phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{y}^{2}+xy={x}^{3}-\frac{t}{432}\hfill \end{array}$$

There are in fact many ways to choose such models. Recall that an elliptic curve over a number field is said to have complex multiplication (CM) if its endomorphism ring over $\overline{\mathbb{Q}}$ is an order of an imaginary quadratic field.

For t such that ${\tilde{E}}_{d}\left(t\right)$ has CM, let ${\lambda}_{d}=-4t(t-1)$, and write ${E}_{d}\left({\lambda}_{d}\right)={\tilde{E}}_{d}\left(\frac{1-\sqrt{1-{\lambda}_{d}}}{2}\right)$. We note that $t=\frac{1-\sqrt{1-{\lambda}_{d}}}{2}$ is determined up to a choice of square root, but we will see this does not affect our conclusion. Assume $|{\lambda}_{d}|<1$ for any embedding. Then there is a Ramanujan type formula
for some unique algebraic numbers $a,\delta $ depending on d and ${\lambda}_{d}$. To be more explicit, a can be computed from a so-called singular value function [3]. Specific choices of CM values of ${\lambda}_{d}$ as well as the corresponding constants $a,\delta $ can be derived from various data given by the Borweins [3].

$$\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{2}\right)}_{k}{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{3}}{\left({\lambda}_{d}\right)}^{k}(ak+1)=\frac{\delta}{\pi}$$

$$\sum _{k=0}^{p-1}\frac{{\left(\frac{1}{2}\right)}_{k}{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{3}}{\left({\lambda}_{d}\right)}^{k}(ak+1)\equiv \text{sgn}\xb7\left(\frac{1-{\lambda}_{d}}{p}\right)\xb7p\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2})$$

Remark 1. In fact, Zudilin conjectured the above to be true modulo ${p}^{3}$ when $d=2,3,4,6$.

Remark 2. Our conclusion in Theorem 1 also holds for totally real singular moduli with $|{\lambda}_{d}|\ge 1$. In either case, the values of a are predicted by the Chowla–Selberg formula [14]. See work of Guillera and Zudilin for existing “divergent” Ramanujan type supercongruences [15].

Remark 3. When $d=2$, there is an underlying $K3$ surface ${X}_{\lambda}$ described by the equation
When $\lambda \ne -1$, this manifold is related to the one-parameter family of elliptic curves of the form
via the so-called Shioda–Inose structure [16]. The arithmetic relation between ${X}_{\lambda}$ and ${E}_{\lambda}$ is obtained by Ahlgren, Ono, and Penniston [17]. The $j-$invariant of ${E}_{\lambda}$ is
In fact, ${E}_{\lambda}$ is isomorphic to ${E}_{2}\left(\lambda \right)$ as above. In the next section, we will use the curve ${E}_{\lambda}$ to demonstrate some basic ingredients.

$${X}_{\lambda}:{z}^{2}=x(x-1)y(y-1)(x-\lambda y)$$

$${E}_{\lambda}:{y}^{2}=(x-1)\left({x}^{2}-\frac{1}{1-\lambda}\right)$$

$$j\left({E}_{\lambda}\right)=64\frac{{(4-\lambda )}^{3}}{{\lambda}^{2}}$$

Moreover, we obtain the following supercongruence. The case for $d=2$ was proved by Kibelbek, Long, Moss, Sheller and Yuan [18].

$$\sum _{k=0}^{p-1}\frac{{\left(\frac{1}{2}\right)}_{k}{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{3}}{\left({\lambda}_{d}\right)}^{k}\equiv L\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2})$$

In this section we discuss one method for obtaining Ramanujan type formulas for expansions of $1/\pi $. These formulas were even known prior to Ramanujan by mathematicians including Bauer [2]. Systematic ways to obtain such formulas have been studied both by the Borwein brothers [3] and the Chudnovsky brothers [4]. Formulas of the form Equation (2.1) are obtained through either the classical Legendre relation between periods and quasi-periods at singular values or the Wronskian of the Picard–Fuchs equation associated to the corresponding families of elliptic curves. In the next section we demonstrate an algebraic perspective that dates back to Chowla and Selberg [14] and has recently been recast by Zudilin [6]. The family of elliptic curves we will analyze is ${E}_{\lambda}$ given by Equation (2.2) above.

Here, we collect some well-known hypergeometric series identities that are useful for our discussion. The first two formulas have been described by Andrews, Askey, and Roy [19]. For parameters $a,b,c$ and value of x such that both sides are well-defined,
where we note that Equation (3.1) is due to Pfaff. It follows that
as
Thus, we state the following lemma.

$$\begin{array}{ccc}\hfill {}_{2}{F}_{1}\left[\begin{array}{cc}a& b\\ & c\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]& =& {(1-x)}^{-a}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}a& c-b\\ & c\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{x}{x-1}\right]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {}_{2}{F}_{1}\left[\begin{array}{cc}a& b\\ & a-b+1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]& =& {(1-x)}^{-a}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{a}{2}& \frac{1+a}{2}-b\\ & a-b+1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{-4x}{{(1-x)}^{2}}\right]\hfill \end{array}$$

$${}_{2}{F}_{1}\left[\begin{array}{cc}1-a& a\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]={\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1-a}{2}& \frac{a}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}-4x(x-1)\right]$$

$$\begin{array}{ccc}\hfill {}_{2}{F}_{1}\left[\begin{array}{cc}1-a& a\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]& =& {(1-x)}^{a-1}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}1-a& 1-a\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{x}{x-1}\right]\phantom{\rule{1.em}{0ex}}\text{by}\phantom{\rule{4.pt}{0ex}}\text{Equation}\phantom{\rule{3.33333pt}{0ex}}\left(3.1\right)\hfill \\ & =& {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1-a}{2}& \frac{a}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}-4x(x-1)\right]\phantom{\rule{1.em}{0ex}}\text{by}\phantom{\rule{4.pt}{0ex}}\text{Equation}\phantom{\rule{3.33333pt}{0ex}}\left(3.2\right).\hfill \end{array}$$

$${}_{2}{F}_{1}\left[\begin{array}{cc}1-a& a\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{1\pm \sqrt{1-x}}{2}\right]={\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1-a}{2}& \frac{a}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]$$

Clausen’s formula [19] states that
Combining Clausen’s formula with Lemma 3 yields the following
In the following section, these identities will be used to formulate solutions to the Picard–Fuchs equations of ${E}_{d}\left({\lambda}_{d}\right)$.

$${}_{2}{F}_{1}{\left[\begin{array}{cc}a& b\\ & a+b+\frac{1}{2}\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]}^{2}={\phantom{\rule{0.166667em}{0ex}}}_{3}{F}_{2}\left[\begin{array}{ccc}2a& 2b& a+b\\ & a+b+\frac{1}{2}& 2a+2b\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]$$

$${}_{2}{F}_{1}{\left[\begin{array}{cc}1-a& a\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{1\pm \sqrt{1-x}}{2}\right]}^{2}={\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1-a}{2}& \frac{a}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]}^{2}={\phantom{\rule{0.166667em}{0ex}}}_{3}{F}_{2}\left[\begin{array}{ccc}\frac{1}{2}& 1-a& a\\ & 1& 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}x\right]$$

We now compute the Picard–Fuchs equation satisfied by the periods of ${E}_{\lambda}:{y}^{2}=(x-1)({x}^{2}-\frac{1}{1-\lambda})$. The differential of the first kind, ${\omega}_{\lambda}$, up to a scalar is given by
while a differential of the second kind, ${\eta}_{\lambda},$ is given by

$${\omega}_{\lambda}=\frac{dx}{y}=\frac{dx}{\sqrt{(x-1)({x}^{2}-\frac{1}{1-\lambda})}}$$

$${\eta}_{\lambda}=(x-1)\frac{dx}{y}=\frac{(x-1)dx}{\sqrt{(x-1)({x}^{2}-\frac{1}{1-\lambda})}}=\sqrt{\frac{x-1}{{x}^{2}-\frac{1}{1-\lambda}}}dx$$

Note that $y={(x-1)}^{\frac{1}{2}}{({x}^{2}-(1/(1-\lambda )))}^{\frac{1}{2}}$ and, to ease notation, let
We may view ${\omega}_{\lambda}=:\omega $ as a function of the parameter λ. Denote ${\omega}^{\prime}:=\frac{d}{d\lambda}\omega $, and ${\omega}^{\prime \prime}:=\frac{{d}^{2}}{d{\lambda}^{2}}\omega $. One may compute that
As $d{g}_{\lambda}$ is an exact form, the integral of $d{g}_{\lambda}$ is 0 and we obtain that the periods of the elliptic curve ${E}_{\lambda}$ satisfy the Picard–Fuchs equation

$$\begin{array}{ccc}q\hfill & :=\hfill & 3/8{x}^{4}-{x}^{3}/2-(2-\lambda )/\left(8(1-\lambda )\right){x}^{2}+(2-\lambda )/\left(4(1-\lambda )\right)x-1/\left(8(1-\lambda )\right)\hfill \\ {g}_{\lambda}\hfill & :=\hfill & q/\left({y}^{3}\right)\hfill \end{array}$$

$$d{g}_{\lambda}=\lambda {(1-\lambda )}^{2}{\omega}^{\prime \prime}+{(1-\lambda )}^{2}{\omega}^{\prime}-3/16\omega $$

$$P{F}_{\lambda}:\frac{{d}^{2}}{d{\lambda}^{2}}f+\frac{1}{\lambda}\frac{d}{d\lambda}f+\frac{3}{16\lambda {(1-\lambda )}^{2}}f=0$$

$$f\left(\lambda \right):={(1-\lambda )}^{1/4}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{4}& \frac{1}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\lambda \right]={\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{4}& \frac{3}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{\lambda}{\lambda -1}\right]\phantom{\rule{1.em}{0ex}}by\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}Equation\phantom{\rule{3.33333pt}{0ex}}\left(3.1\right).$$

Proof. Write
so that $f\left(\lambda \right)={(1-\lambda )}^{1/4}a\left(\lambda \right)$. Then the coefficients ${a}_{k}$ satisfy
Using the product rule, we see that

$$a\left(\lambda \right):={\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{4}& \frac{1}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\lambda \right]=\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{4}\right)}_{k}^{2}}{k{!}^{2}}{\lambda}^{k}=\sum _{k=0}^{\infty}{a}_{k}{\lambda}^{k}$$

$${a}_{k+1}{(k+1)}^{2}={a}_{k}{(k+1/4)}^{2}$$

$$\begin{array}{c}\lambda {(1-\lambda )}^{2}\frac{{d}^{2}}{d{\lambda}^{2}}f+{(1-\lambda )}^{2}\frac{d}{d\lambda}f+\frac{3}{16}f=\\ \text{\hspace{1em}\hspace{1em}}={(1-\lambda )}^{5/4}\left(\frac{-1}{16}a+\left(1-\frac{3}{2}\lambda \right){a}^{\prime}+(\lambda -{\lambda}^{2}){a}^{\prime \prime}\right)\hfill \\ \text{\hspace{1em}\hspace{1em}}={(1-\lambda )}^{5/4}\sum _{k=0}^{\infty}\left(-\frac{1}{16}{a}_{k}+(k+1){a}_{k+1}-\frac{3}{2}k{a}_{k}-k(k-1){a}_{k}+k(k+1){a}_{k+1}\right){\lambda}^{k}\hfill \\ \text{\hspace{1em}\hspace{1em}}={(1-\lambda )}^{5/4}\sum _{k=0}^{\infty}\left({(k+1)}^{2}{a}_{k+1}-{(k+1/4)}^{2}{a}_{k}\right){\lambda}^{k}\hfill \\ \text{\hspace{1em}\hspace{1em}}=0\hfill \end{array}$$

☐

It is well-known that Picard–Fuchs equations arising from algebraic varieties are Fuchsian equations with only regular singularities. Also, attached to each differential equation with isolated singularities is the so-called monodromy representation of the fundamental group of the base curve with singularities removed [20]. The image of this representation, well-defined up to conjugation, is called the monodromy group of the differential equation. Shortly, we shall elaborate upon the monodromy group for $P{F}_{\lambda}$.

We gather some results regarding modular forms. Modular forms are functions, F, defined on the Poincaré upper half plane, $\mathcal{H}$, that satisfy the weight k modular invariance condition
for all $z\in \mathcal{H}$ and $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in some subgroup $\Gamma \subseteq S{L}_{2}\left(\mathbb{Z}\right)$ of finite index. As well, they have moderate growth conditions at the cusps of Γ, which are equivalence classes of $\mathbb{Q}\cup \left\{\infty \right\}$ under the action of Γ. Here, for simplicity, we assume that k is an integer. Holomorphic integral weight modular forms for Γ form a finitely generated $\mathbb{C}-$algebra
graded by the weight k. A modular form is said to be weakly holomorphic if it is holomorphic on $\mathcal{H}$ and is either holomorphic or meromorphic at the cusps. Two of the most well-known weakly holomorphic weight 0 modular forms are the modular $j-$function, for $\Gamma =S{L}_{2}\left(\mathbb{Z}\right)$, and the modular lambda function for the principal level$-2$ congruence subgroup $\Gamma \left(2\right)$, defined here by
The function $\eta \left(z\right)$ is the classical eta-function
The modular $j-$function and lambda function are related by the identity

$$F\left(\frac{az+b}{cz+d}\right)={(cz+d)}^{k}F\left(z\right)$$

$$M(\Gamma )=\u2a01{M}_{k}(\Gamma )$$

$$L\left(z\right)=16\frac{\eta {\left(2z\right)}^{4}\eta {(z/2)}^{2}}{\eta {\left(z\right)}^{6}}$$

$$\eta \left(z\right)={q}^{1/24}\prod _{n\ge 1}(1-{q}^{n}),\phantom{\rule{1.em}{0ex}}q={e}^{2\pi iz}$$

$$j={2}^{8}\frac{{({L}^{2}-L+1)}^{3}}{{L}^{2}{(L-1)}^{2}}=64\frac{{(2L(2L-2)+4)}^{3}}{{\left(2L(2L-2)\right)}^{2}}$$

Every meromorphic modular form F has a $q-$expansion at infinity
where again $q={e}^{2\pi iz}$ and $\mu \in \mathbb{N}$ is called the cusp width of Γ at infinity. Recall the weight 2 quasi-modular Eisenstein series
Then
is a weight 2 non-holomorphic modular form. We state an essential fact which can be found in work of Zagier [21].

$$F\left(z\right)=\sum _{n=-{n}_{0}}^{\infty}{a}_{n}{q}^{n/\mu}$$

$${\mathsf{E}}_{2}\left(z\right)=1-24\sum _{n\ge 1}\frac{n{q}^{n}}{1-{q}^{n}}$$

$${\mathsf{E}}_{2}^{*}\left(z\right)={\mathsf{E}}_{2}\left(z\right)-\frac{3}{\pi \text{Im}\left(z\right)}$$

$$F\left(\tau \right)\xb7{\Omega}_{K}^{-k}\in \overline{\mathbb{Q}}$$

In a manner similar to Zagier, we define the derivatives ${\vartheta}_{k}$ and ${\partial}_{k}$ [21]. For a weight k modular form F with algebraic coefficients, let
interpreting ${F}^{\prime}\left(z\right)$ as $\frac{dF}{dq}=\frac{1}{2\pi i}\frac{\partial F}{\partial z}$. Then ${\vartheta}_{k}F$ is weight $k+2$ modular form with algebraic coefficients and
By Proposition 5, the values
for all $\tau \in K\cap \mathcal{H}.$ For future reference we state this as a lemma.

$${\vartheta}_{k}F\left(z\right):={F}^{\prime}\left(z\right)-\frac{k}{12}{\mathsf{E}}_{2}\left(z\right)F\left(z\right),\phantom{\rule{1.em}{0ex}}{\partial}_{k}F\left(z\right):={F}^{\prime}\left(z\right)-\frac{k}{4\pi \text{Im}\left(z\right)}F\left(z\right)$$

$${\vartheta}_{k}F\left(z\right)={\partial}_{k}F\left(z\right)-\frac{k}{12}{\mathsf{E}}_{2}^{*}\left(z\right)F\left(z\right)$$

$${\vartheta}_{k}F\left(\tau \right){\Omega}_{K}^{-k-2}\in \overline{\mathbb{Q}},\text{and}{\partial}_{k}F\left(\tau \right){\Omega}_{K}^{-k-2}\in \overline{\mathbb{Q}}$$

Remark 4. In our setting, we assume ${E}_{\lambda}$ has complex multiplication. Then R, the endomorphism ring of ${E}_{\lambda}$, is an order of an imaginary quadratic field K. In this case, there are two cycles ${C}_{1},{C}_{2}$ in ${H}_{1}({E}_{\lambda},\mathbb{Z})$ such that

$$\frac{{\int}_{{C}_{1}}{\omega}_{\lambda}}{{\int}_{{C}_{2}}{\omega}_{\lambda}}=\tau \in K\cap \mathcal{H}$$

It is known that hypergeometric series can be related to theta series [3]; for instance
where ${\theta}_{3}\left(z\right)={\sum}_{n\in \mathbb{Z}}{q}^{{n}^{2}/2}$ is a weight $1/2$ modular form, and the modular Lambda function $L\left(z\right)$ is as in Equation (3.7). In fact this is a special case of a general result of Stiller [22] regarding solutions of suitable degree 2 ordinary Fuchsian differential equations, which are weight 1 modular forms of the corresponding monodromy groups.

$${}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}L\left(z\right)\right]={\theta}_{3}^{2}\left(z\right)$$

Proof. Recall that the $j-$invariant of the elliptic curve ${E}_{\lambda}$ is
By Equation (3.8), we see that λ can be viewed as a degree 2 polynomial in L, i.e.,
Hence $\lambda \left(z\right)$ is a modular function over $\Gamma \left(2\right)$ with algebraic coefficients.

$$j\left({E}_{\lambda}\right)=64\frac{{(4-\lambda )}^{3}}{{\lambda}^{2}}$$

$$\lambda \left(z\right)=-2L\left(z\right)(2L\left(z\right)-2)=-4L{\left(z\right)}^{2}+4L\left(z\right)$$

Inserting $\lambda :=-2L(2L-2)$ as a function of $q={e}^{2\pi iz}$ and using Equations (3.3) and (3.9), we can make the following statement.

$${}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{4}& \frac{1}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\lambda \left(z\right)\right]={\theta}_{3}^{2}\left(z\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4.pt}{0ex}}and\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{(1-\lambda \left(z\right))}^{1/4}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{4}& \frac{1}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\lambda \left(z\right)\right]$$

We now return to the method for obtaining Ramanujan type formulas for expansions of $1/\pi $. Let E be an elliptic curve defined over $\overline{\mathbb{Q}}$ with complex multiplication, R the endomorphism ring of E over $\mathbb{C}$ which is assumed to be an order of the imaginary quadratic field $K=\mathbb{Q}\left(\sqrt{-d}\right)$ of discriminant $-d$, and ω its invariant differential defined over $\overline{\mathbb{Q}}$. Then Chowla and Selberg [14] (see also the work of Gross [23]) proved that for any $C\in {H}_{1}(E,\mathbb{Z})$, the transcendental part of the period integral ${\int}_{C}\omega $ is as follows:
where n is the order of unit group in $\mathbb{Q}\left(\sqrt{-d}\right)$, ε is a primitive Dirichlet character modulo d for the quadratic field $\mathbb{Q}\left(\sqrt{-d}\right)$, and h is the class number of $\mathbb{Q}\left(\sqrt{-d}\right)$. The well-known Gross–Koblitz formula is the $p-$adic analogue of the Chowla–Selberg formula. Moreover, consider ω as an element of ${H}_{DR}^{1}(E,\mathbb{C})$, the dual of ${H}_{1}(E,\mathbb{Z}){\otimes}_{\mathbb{Z}}\mathbb{C}$. The $2-$dimensional space ${H}_{DR}^{1}(E,\mathbb{C})$ is endowed with the action of the endomorphism ring R. In particular, ω is an eigenfunction of R. There is another eigenfunction ν of R in ${H}_{DR}^{1}\left(E\right)$ that is not parallel to ω. Then ${\int}_{C}\nu $ is an algebraic multiple of $2\pi i/{b}_{K}$. Putting this together, one obtains the relation
where ∼ denotes equality up to multiplication by an algebraic number.

$${b}_{K}:=\sqrt{\pi}\prod _{0<a<d}\Gamma {(a/d)}^{n\epsilon \left(a\right)/4h}$$

$${\int}_{C}\omega \xb7{\int}_{C}\nu \sim \pi $$

We will now to return to our discussion of ${E}_{\lambda}$ by relating the corresponding hypergeometric series to the Chowla–Selberg formulas. Consider ${E}_{\lambda}$ now as an elliptic curve defined over $\mathbb{C}\left(\lambda \right)$. Recall that
is the holomorphic differential $1-$form on ${E}_{\lambda}$, which is unique up to scalars. Let ${C}_{\lambda}$ be a family of $1-$cycles on ${E}_{\lambda}$ that moves complex analytically as λ varies. Then
describes the variation of the periods. Since applying the Picard–Fuchs differential equation to ${\omega}_{\lambda}$ yields an exact form $d{g}_{\lambda}$,
by Green’s theorem. As $p\left(\lambda \right)$ is a holomorphic solution to $P{F}_{\lambda}$ near 0,
for some constant term A, and f as defined in Lemma 4,
Furthermore, f can be viewed as a weight 1 modular form with respect to the variable $z\in \mathcal{H}$ by Lemma 8. Note that since λ has algebraic coefficients with respect to ${q}^{1/2}={e}^{\pi iz}$, as remarked in the proof of Lemma 7, so does ${f}_{\lambda}\left(z\right)$. To write down the corresponding ${\nu}_{\lambda}$, which is also an eigenfunction of R, we use the fact that ${H}_{DR}^{1}({E}_{\lambda}/\mathbb{C}\left[\left[\lambda \right]\right])$ admits the Gauss–Manin connection, so that
Thus ${\nu}_{\lambda}$ can be written as the form $\frac{d{f}_{\lambda}}{d\lambda}-c{f}_{\lambda}$ for some constant c. In terms of modular forms, Lemma 6 ensures that both

$${\omega}_{\lambda}=\frac{dx}{y}=\frac{dx}{\sqrt{(x-1)({x}^{2}-\frac{1}{1-\lambda})}}$$

$$p\left(\lambda \right):={\int}_{{C}_{\lambda}}{\omega}_{\lambda}$$

$$P{F}_{\lambda}\circ p\left(\lambda \right)={\int}_{{C}_{\lambda}}d{g}_{\lambda}=0$$

$$p\left(\lambda \right)=A\xb7f\left(\lambda \right)$$

$${f}_{\lambda}\left(z\right):=f\left(\lambda \left(z\right)\right)={(1-\lambda )}^{1/4}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1}{4}& \frac{1}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\lambda \left(z\right)\right]$$

$${\partial}_{\lambda}{\omega}_{\lambda}\in {H}_{DR}^{1}({E}_{\lambda}/\mathbb{C}\left[\left[\lambda \right]\right])$$

$$\mu :={\partial}_{1}\left({f}_{\lambda}\right)\left(\tau \right){\Omega}_{K}^{-3}\in \overline{\mathbb{Q}},\text{and}w:={f}_{\lambda}\left(\tau \right){\Omega}_{K}^{-1}\in \overline{\mathbb{Q}}$$

$${f}_{\lambda}\left(\tau \right)\left(\frac{d{f}_{\lambda}}{d\lambda}-{c}^{\prime}{f}_{\lambda}\right)\left(\tau \right)\sim 1/\pi $$

Proof. We have that
and $\frac{dq}{d\lambda}={\left({\partial}_{0}\lambda \right)}^{-1}$ is a modular form of weight $-2$. By Proposition 5 there is some $\xi \in \overline{\mathbb{Q}}$ such that $\frac{dq}{d\lambda}\left(\tau \right){\Omega}_{K}^{2}=\xi $. Put ${c}^{\prime}:=\mu \xi /w$; then

$$\frac{d{f}_{\lambda}}{d\lambda}=\frac{d{f}_{\lambda}}{dq}\frac{dq}{d\lambda}\left(\tau \right)=\left({\partial}_{1}{f}_{\lambda}\left(\tau \right)+\frac{1}{4\pi \text{Im}\left(\tau \right)}{f}_{\lambda}\left(\tau \right)\right)\frac{dq}{d\lambda}$$

$${f}_{\lambda}\left(\tau \right)\left(\frac{d{f}_{\lambda}}{d\lambda}-{c}^{\prime}{f}_{\lambda}\right)\left(\tau \right)={\Omega}_{K}w\left({\Omega}_{K}\mu \xi +\frac{1}{4\pi \text{Im}\left(\tau \right)}{\Omega}_{K}^{-1}w\xi -\frac{\mu \xi}{w}{\Omega}_{K}w\right)=\frac{{w}^{2}\xi}{4\pi \text{Im}\left(\tau \right)}\sim 1/\pi $$

☐

Combining Equations (3.11) and (3.12) with the results of Chowla–Selberg, we obtain the following corollary.

$$p\left(\lambda \right)\sim \pi {f}_{\lambda}$$

Remark 5. Recall, ${\nu}_{\lambda}$ can be written as the form $\frac{d{f}_{\lambda}}{d\lambda}-c{f}_{\lambda}$ for some constant c. In Corollary 10, $c=\mu \xi /w$, with $\mu ,\xi ,w$ as in Lemma 9.

Remark 6. If τ is an imaginary quadratic number, then $\lambda \left(\tau \right)$ is algebraic, so is ${(1-\lambda \left(\tau \right))}^{1/4}$. That is, one can draw a similar conclusion for the series ${\tilde{f}}_{\lambda}\left(\tau \right)=\phantom{\rule{0.166667em}{0ex}}{}_{2}{F}_{1}\left[\begin{array}{cc}1/4& 1/4\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\lambda \left(\tau \right)\right]$. Hence, there is an algebraic constant C such that

$${\tilde{f}}_{\lambda}\left(\tau \right)\xb7\left(\frac{d{\tilde{f}}_{\lambda}}{d\lambda}-C{\tilde{f}}_{\lambda}\left(\tau \right)\right)\sim \frac{1}{\pi}$$

$$\sum _{k=0}^{\infty}\frac{{\left(\frac{1}{2}\right)}_{k}^{3}}{k{!}^{3}}(ak+1){\left(\lambda \right)}^{k}\sim \frac{1}{\pi}$$

Remark 7. Using the same technique, one can obtain Ramanujan type formulas for $1/\pi $ corresponding to other series with $d=3,4,6$ as mentioned in the introduction. In fact, the Picard–Fuchs equation of each family of elliptic curves ${\tilde{E}}_{d}\left(t\right)$ has local solutions ${}_{2}{F}_{1}\left[\begin{array}{cc}1/d& (d-1)/d\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]$ near $t=0$ with $d=2,3,4,6.$ The case of $d=2$ is well-known. The other cases can be verified by using the results of Stienstra and Beukers [24]. In the literature, one can find families of elliptic curves ${\tilde{E}}_{d}\left(t\right)$ in weighted projective space whose Picard–Fuchs equation has ${}_{2}{F}_{1}\left[\begin{array}{cc}1/d& (d-1)/d\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]$ as local holomorphic solutions at 0 [25]. (Note that these equations are in terms of ${t}^{1/d}$ and not t for $d=2,3,4,6.$) The corresponding monodromy groups of the Picard–Fuchs equations are four genus zero hyperbolic triangular groups:
respectively [26]. Here ${\Gamma}_{0}^{*}\left(n\right)$ is the group generated by the congruence subgroup ${\Gamma}_{0}\left(n\right)$ consisting of elements in $S{L}_{2}\left(\mathbb{Z}\right)$ that are upper triangular modulo n, and the Fricke involution ${W}_{n}=\left(\begin{array}{cc}0& -1\\ n& 0\end{array}\right)$. Similar to Lemma 8, inserting t with suitable modular functions of the corresponding genus zero groups, one obtains explicit weight 1 modular forms.

$${\Gamma}_{0}^{*}\left(4\right)\cong \Gamma \left(2\right),\phantom{\rule{4pt}{0ex}}{\Gamma}_{0}^{*}\left(3\right),\phantom{\rule{4pt}{0ex}}{\Gamma}_{0}^{*}\left(2\right),\phantom{\rule{4pt}{0ex}}S{L}_{2}\left(\mathbb{Z}\right)$$

Motivated by the Hypergeometric Transformation Formula Stated in Lemma 3, we fix a choice of square root $\sqrt{1-{\lambda}_{d}}$ for $d=2,3,4,6$ and let ${E}_{d}\left({\lambda}_{d}\right)={\tilde{E}}_{d}\left(\frac{1-\sqrt{1-{\lambda}_{d}}}{2}\right),$ as in Section 2. We will show that the choice of square root does not matter in this setting.

$$\frac{dx}{2y+A}=H\left(z\right)dz=\sum _{n\ge 1}{H}_{d,n-1}\left(t\right){z}^{n-1}dz$$

$${H}_{d,p-1}\left(t\right)\equiv {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{d}& \frac{d-1}{d}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{p-1}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)$$

Proof. By Equation (1.2) in the work of Ditters [27], ${H}_{d,n-1}\left(t\right)$ is the coefficient of ${x}^{n-1}$ in the polynomial

$$\sum _{\begin{array}{c}s+j=n-1\\ j\le s\end{array}}\left(\genfrac{}{}{0pt}{}{s}{j}\right){B}^{j}{A}^{s-j}$$

- For $d=2$: $A=0$, $B=x(x-1)(x-t)$.$$\begin{array}{ccc}\hfill {H}_{2,p-1}\left(t\right)& =& {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}\left[\begin{array}{cc}\frac{1-p}{2}& \frac{1-p}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]\hfill \\ & \equiv & {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{(p-1)}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)\hfill \end{array}$$
- For $d=3$: $A=x+\frac{t}{27}$, $B={x}^{3}$. ${H}_{3,p-1}\left(t\right)$ is the coefficient of ${x}^{p-1}$ in$$\sum _{\begin{array}{c}s+j=p-1\\ j\le s\end{array}}\left(\genfrac{}{}{0pt}{}{s}{j}\right){B}^{j}{A}^{s-j}$$By a similar argument, we obtain$$\begin{array}{ccc}\hfill {H}_{3,p-1}\left(t\right)& =& \sum _{j=0}^{\frac{p-1}{3}}\left(\genfrac{}{}{0pt}{}{p-1-j}{j}\right)\left(\genfrac{}{}{0pt}{}{p-1-2j}{p-1-3j}\right){\left(\frac{t}{27}\right)}^{j}\hfill \\ & =& \phantom{\rule{0.166667em}{0ex}}{}_{3}{F}_{2}\left[\begin{array}{ccc}\frac{1-p}{3}& \frac{2-p}{3}& \frac{3-p}{3}\\ & 1& 1-p\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]\hfill \\ & \equiv & {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{3}& \frac{2}{3}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{(p-1)}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)\hfill \end{array}$$
- For $d=4$: $A=0$, $B=x({x}^{2}+x+\frac{t}{4})$.$$\begin{array}{ccc}\hfill {H}_{4,p-1}& =& \sum _{i=0}^{\frac{p-1}{4}}\left(\genfrac{}{}{0pt}{}{\frac{p-1}{2}}{j,j,\frac{p-1}{2}-2j}\right){\left(\frac{t}{4}\right)}^{j}\hfill \\ & =& {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1-p}{4}& \frac{3-p}{4}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{(p-1)}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)\hfill \end{array}$$
- For $d=6$: $A=x$, $B={x}^{3}-\frac{t}{432}$.$$\begin{array}{ccc}\hfill {H}_{6,p-1}\left(t\right)& \equiv & \sum _{j=0}^{\frac{p-1}{6}}\left(\genfrac{}{}{0pt}{}{p-1-3j}{3j}\right)\left(\genfrac{}{}{0pt}{}{3j}{j}\right){\left(\frac{-t}{432}\right)}^{j}\hfill \\ & \equiv & \phantom{\rule{0.166667em}{0ex}}{}_{6}{F}_{5}\left[\begin{array}{cccccc}\frac{1-p}{6}& \frac{2-p}{6}& \frac{3-p}{6}& \frac{4-p}{6}& \frac{5-p}{6}& \frac{6-p}{6}\\ & \frac{1-p}{3}& \frac{2-p}{3}& \frac{3-p}{3}& 1& \frac{1}{2}\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]\hfill \\ & \equiv & {\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{6}& \frac{5}{6}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{(p-1)}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)\hfill \end{array}$$

☐

We observe that the congruences above hold for any $t\in {\mathbb{F}}_{p}$, and for the four values of d in consideration and for $t=\frac{1-\sqrt{1-{\lambda}_{d}}}{2},$ the coefficient ${H}_{d,p-1}\left(t\right)$ can be stated explicitly.

$$\begin{array}{c}{H}_{d,p-1}\left(\frac{1-\sqrt{1-{\lambda}_{d}}}{2}\right)\hfill \\ \hfill \equiv \left\{\begin{array}{cc}{}_{2}{F}_{1}{\left[\begin{array}{c}\frac{1}{2d}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\frac{1}{2}-\frac{1}{2d}\\ 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}{\lambda}_{d}\right]}_{p-1}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p),\hfill & if\phantom{\rule{4.pt}{0ex}}p\equiv 1,d-1\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}2d),\hfill \\ {(1-{\lambda}_{d})}^{1/2}\phantom{\rule{0.166667em}{0ex}}{}_{2}{F}_{1}{\left[\begin{array}{c}\frac{1}{2}+\frac{1}{2d}\phantom{\rule{1.em}{0ex}}1-\frac{1}{2d}\\ \phantom{\rule{1.em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}{\lambda}_{d}\right]}_{p-1}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p),\hfill & otherwise.\hfill \end{array}\right.\end{array}$$

Proof. Let ${P}_{n}\left(x\right):=\phantom{\rule{0.166667em}{0ex}}{}_{2}{F}_{1}\left[\begin{array}{cc}-n& n+1\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{1-x}{2}\right]$ be the nth Legendre polynomial, following the notation given by Andrews, Askey, and Roy [19]. ${P}_{n}\left(x\right)$ is an odd (resp. even) function of x if n is odd (resp. even). By Lemma 12,
where $n=\frac{p-1}{d}$ if $p\equiv 1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}d)$, and $n=\frac{p+1-d}{d}$ if $p\equiv -1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}d)$. If n is even, i.e., when $p\equiv 1$, or $p\equiv d-1\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}2d)$, ${P}_{n}\left(\sqrt{1-{\lambda}_{d}}\right)$ is a polynomial in $(1-{\lambda}_{d})$, which is congruent modulo p to $\phantom{\rule{0.166667em}{0ex}}{}_{2}{F}_{1}{\left[\begin{array}{cc}1/2d& 1/2-1/2d\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}{\lambda}_{d}\right]}_{p-1}$ from Lemma 3. When n is odd, ${P}_{n}\left(\sqrt{1-{\lambda}_{d}}\right)$ is $\sqrt{1-{\lambda}_{d}}$ times a polynomial in $(1-{\lambda}_{d})$. Then the claim follows from Lemma 3 and the following transformation of Euler [19],

$${H}_{d,p-1}\left(\frac{1-\sqrt{1-{\lambda}_{d}}}{2}\right)\equiv {P}_{n}\left(\sqrt{1-{\lambda}_{d}}\right)\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)$$

$${}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{2}-\frac{a}{2}& \frac{a}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}{\lambda}_{d}\right]}_{p-1}={(1-{\lambda}_{d})}^{1/2}{\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{2}+\frac{a}{2}& 1-\frac{a}{2}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}{\lambda}_{d}\right]}_{p-1}$$

☐

Similarly, squares of ${H}_{d,p-1}\left(t\right),$ in this setting, can be described precisely using Equation (3.5).

$${H}_{d,p-1}{\left(\frac{1-\sqrt{1-{\lambda}_{d}}}{2}\right)}^{2}\equiv {\phantom{\rule{0.166667em}{0ex}}}_{3}{F}_{2}{\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{d}& \frac{d-1}{d}\\ & 1& 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}{\lambda}_{d}\right]}_{p-1}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)$$

For $d=2$, it is straightforward to check that the $j-$invariant of ${E}_{2}\left({\lambda}_{2}\right)$ is in $\mathbb{Q}\left({\lambda}_{2}\right)$. Consequently, ${E}_{2}\left({\lambda}_{2}\right)$ has a model defined over $\mathbb{Q}\left({\lambda}_{2}\right)$. To be more explicit, rewrite ${E}_{2}\left({\lambda}_{2}\right)$ in the form
Applying the quartic twist $(x,y)\mapsto (\sqrt{1-{\lambda}_{2}}\phantom{\rule{0.166667em}{0ex}}X,\sqrt[4]{1-{\lambda}_{2}}\phantom{\rule{0.166667em}{0ex}}Y)$, gives the equation
defined over $\mathbb{Q}\left({\lambda}_{2}\right)$, which is isomorphic to ${E}_{{\lambda}_{2}}:{Y}^{2}=(X-1)({X}^{2}-\frac{1}{1-{\lambda}_{2}})$ over $\mathbb{Q}\left({\lambda}_{2}\right)$.

$${y}^{2}={x}^{3}-27(16-{\lambda}_{2})x-432({\lambda}_{2}+8)\sqrt{1-{\lambda}_{2}}$$

$${\widehat{E}}_{2}\left({\lambda}_{2}\right):\phantom{\rule{4pt}{0ex}}{Y}^{2}=(1-{\lambda}_{2}){X}^{3}-27(16-{\lambda}_{2})X-432({\lambda}_{2}+8)$$

Similarly for $d=6$, ${E}_{6}\left({\lambda}_{6}\right)$ is isomorphic over $\mathbb{Q}\left(\sqrt{1-{\lambda}_{6}}\right)$ to
Applying $(x,y)\mapsto (\sqrt{1-{\lambda}_{4}}\phantom{\rule{0.166667em}{0ex}}X,\sqrt[4]{1-{\lambda}_{4}}\phantom{\rule{0.166667em}{0ex}}Y)$, yields the twisted curve
In the cases $d=2,6$,

$${y}^{2}={x}^{3}-27x+54\sqrt{1-{\lambda}_{6}}$$

$${\widehat{E}}_{6}\left({\lambda}_{6}\right):{Y}^{2}=(1-{\lambda}_{6}){X}^{3}-27X+54$$

$$\frac{-x}{y}=\frac{-X}{Y}\sqrt[4]{1-{\lambda}_{d}}$$

When $d=3$, ${E}_{d}\left({\lambda}_{d}\right)$ has no model over $\mathbb{Q}\left({\lambda}_{d}\right)$. However, rewrite ${E}_{3}\left({\lambda}_{3}\right)$ as
and apply a quartic twist $(x,y)\mapsto ({\alpha}^{2}X,{\alpha}^{3}Y)$ to ${E}_{3}\left({\lambda}_{3}\right)$ with $\alpha =\sqrt[4]{1-{\lambda}_{3}}$ to get
The curve is $3-$isogenous over $\mathbb{Q}(\sqrt{1-{\lambda}_{3}},\sqrt{3})$ to
(see work of Top [28]), which is isomorphic to its conjugate under a different embedding of $\sqrt{1-{\lambda}_{3}}$.

$${y}^{2}={x}^{3}+{\left(\frac{x}{2}+\frac{1-\sqrt{1-{\lambda}_{3}}}{108}\right)}^{2}$$

$${Y}^{2}={X}^{3}+\frac{1}{4{\alpha}^{2}}{\left(X+\frac{1-\sqrt{1-{\lambda}_{3}}}{54{\alpha}^{2}}\right)}^{2}$$

$${\widehat{E}}_{3}\left({\lambda}_{3}\right):\phantom{\rule{4pt}{0ex}}{Y}^{2}={X}^{3}-\frac{27}{4{c}^{2}}{\left(X+\frac{-1-\sqrt{1-{\lambda}_{3}}}{2{\alpha}^{2}}\right)}^{2}$$

Similarly, when $d=4$, ${E}_{d}\left({\lambda}_{d}\right)$ is $2-$isogenous to ${E}_{d}^{\prime}\left({\lambda}_{d}\right)={\tilde{E}}_{d}\left(\frac{1+\sqrt{1-{\lambda}_{d}}}{2}\right)$. To see this, we apply a similar quartic twist to get
which is $2-$isogenous over $\mathbb{Q}(\sqrt{1-{\lambda}_{4}},\sqrt{2})$ to
The curve is isomorphic over $\mathbb{Q}(\sqrt{1-{\lambda}_{4}},\sqrt{2})$ to ${\widehat{E}}_{4}\left({\lambda}_{4}\right)$ under a different embedding of $\sqrt{1-{\lambda}_{4}}$.

$${\widehat{E}}_{4}\left({\lambda}_{4}\right):{Y}^{2}=X\left({X}^{2}+\frac{1}{{\alpha}^{2}}X+\frac{1-\sqrt{1-{\lambda}_{4}}}{8{\alpha}^{4}}\right)$$

$${\widehat{E}}_{4}^{\prime}\left({\lambda}_{4}\right):{Y}^{2}=X\left({X}^{2}-\frac{2}{{\alpha}^{2}}X+\frac{1+\sqrt{1-{\lambda}_{4}}}{2{\alpha}^{4}}\right)$$

In the cases $d=3,4$,

$$\frac{-x}{y}=\frac{-X}{Y}\frac{1}{\sqrt[4]{1-{\lambda}_{d}}}$$

For any number field F, we use ${G}_{F}$ to denote $\text{Gal}(\overline{F}/F)$. Let ℓ be an arbitrary prime number and ${\widehat{\rho}}_{d,\ell}$ the Galois representations of ${G}_{K}$ attached to Tate modules of ${\widehat{E}}_{d}\left({\lambda}_{d}\right)$. We show that for each $d=2,3,4,6$ there is an associated Galois representation $\tilde{\rho}$ that is odd.

When $d=2,6$, $K=\mathbb{Q}\left({\lambda}_{d}\right)$, which is assumed to be totally real. Recall that for any conjugacy class c of complex conjugation in ${G}_{\mathbb{Q}\left({\lambda}_{d}\right)}$, ${\widehat{\rho}}_{d,\ell}\left(c\right)$ is not a scalar matrix, as ${\widehat{E}}_{d}\left({\lambda}_{d}\right)$ has CM, and the Galois representation ${\widehat{\rho}}_{d,\ell}$ is odd. In this setting, we write $\tilde{\rho}={\widehat{\rho}}_{d,\ell}$.

When $d=3,4$, $K=\mathbb{Q}\left(\sqrt{1-{\lambda}_{d}}\right)$. If $K=\mathbb{Q}\left({\lambda}_{d}\right)$, as in the setting above, we consider $\tilde{\rho}={\widehat{\rho}}_{d,\ell}$. Suppose $K\ne \mathbb{Q}\left({\lambda}_{d}\right)$. Let σ be any element in ${G}_{K}$ which sends $\sqrt{1-{\lambda}_{d}}$ to $-\sqrt{1-{\lambda}_{d}}$. The above discussion implies that
is isomorphic to ${\psi}^{\sigma}$, its conjugation by σ, where ${s}_{3}=3,{s}_{4}=2$. As ψ is obtained as a restriction, ψ is indeed isomorphic to its conjugate by any element in $\text{Gal}(\mathbb{Q}(\sqrt{1-{\lambda}_{d}},\sqrt{{s}_{d}})/\mathbb{Q}\left({\lambda}_{d}\right))$. This implies that if ψ is absolutely irreducible, then it can be lifted to a $2-$dimensional odd representation $\phi $ of ${G}_{\mathbb{Q}\left({\lambda}_{d}\right)}$, which is well-defined up to a character of $\text{Gal}(\mathbb{Q}(\sqrt{1-{\lambda}_{d}},\sqrt{{s}_{d}})/\mathbb{Q}\left({\lambda}_{d}\right))$. Otherwise, $\psi ={\chi}_{1}\oplus {\chi}_{2}$ for some $1-$dimensional characters ${\chi}_{i}$ of ${G}_{\mathbb{Q}(\sqrt{1-{\lambda}_{d}},\sqrt{{s}_{d}})}$ for $i=1,2$. As ψ is isomorphic to ${\psi}^{\sigma}$, either ${\chi}_{1}^{\sigma}\cong {\chi}_{1}$, which implies ${\chi}_{1}$ can be lifted to a $1-$dimensional character of ${G}_{\mathbb{Q}({\lambda}_{d},\sqrt{1-{\lambda}_{d}})}$, contradicting the fact that ${\chi}_{1}$ originated from a CM elliptic curve, or ${\chi}_{2}\cong {\chi}_{1}^{\sigma}$, which implies that
is absolutely irreducible. The induced character $\phi $ can be extended to a $2-$dimensional Galois representation $\tilde{\rho}$ of ${G}_{\mathbb{Q}\left({\lambda}_{d}\right)}$, well-defined up to a finite character of $\text{Gal}(\mathbb{Q}(\sqrt{1-{\lambda}_{d}},\sqrt{{s}_{d}})/\mathbb{Q}\left({\lambda}_{d}\right))$.

$$\psi :={\widehat{\rho}}_{d,\ell}{|}_{{G}_{\mathbb{Q}(\sqrt{1-{\lambda}_{d}},\sqrt{{s}_{d}})}}$$

$$\phi ={\text{Ind}}_{{G}_{\mathbb{Q}(\sqrt{1-{\lambda}_{d}},\sqrt{{s}_{d}})}}^{{G}_{\mathbb{Q}({\lambda}_{d},\sqrt{{s}_{d}})}}{\chi}_{1}$$

In either case, when $K\ne \mathbb{Q}\left({\lambda}_{d}\right)$ or $K=\mathbb{Q}\left({\lambda}_{d}\right)$, the representation $\tilde{\rho}$ is odd, which can be seen from the fact that it corresponds to an automorphic representation of $G{L}_{2}\left(\mathbb{Q}\left({\lambda}_{d}\right)\right)$ due to our CM assumption. For any prime p such that ${\lambda}_{d}$ embeds in ${\mathbb{Q}}_{p}$, fixing an embedding of ${\lambda}_{d}$ into ${\mathbb{Q}}_{p}$ corresponds to fixing a place v of ${\mathcal{O}}_{\mathbb{Q}\left({\lambda}_{d}\right)}$ above p with relative degree 1. Let ${\text{Fr}}_{v}$ be the corresponding geometric Frobenius conjugacy class in ${G}_{\mathbb{Q}\left({\lambda}_{d}\right)}$. Then by the oddness of $\tilde{\rho}$, we have the following.

Atkin and Swinnerton-Dyer [29] proved the following. Let $E:{y}^{2}={x}^{3}-Bx-C$ be a nonsingular elliptic curve over a totally real field. Let $p>3$ be any prime such that $B,C$ can be embedded in ${\mathbb{Z}}_{p}$ and E has good reduction modulo $p.$ Let ξ be any local uniformizer of E at infinity over ${\mathbb{Z}}_{p}$ that is a formal power series of $\frac{-x}{y}$ with coefficients in ${\mathbb{Z}}_{p}$ that starts with $\frac{-x}{y}$. Then the holomorphic differential
has coefficients in ${\mathbb{Z}}_{p}.$ Moreover, for the coefficient $a\left(n\right)$ of the differential, the following congruence holds for all $n\ge 1$,
where we define $a\left(n\right)=0$ when n is not a positive integer. Note that the assumption of E being defined over a totally real field is necessary to conclude that the last coefficient is p. Otherwise, it is only determined up to a ± sign. Atkin and Swinnerton-Dyer’s result inspired the work of Cartier [30] and Katz [31]. Here we shall use Katz’s approach with a slight generalization using the approach outlined by Stienstra and Beukers [24] so that the base ring is not restricted to ${\mathbb{Z}}_{p}$.

$$-\frac{dx}{2y}=(1+{\sum}_{n\ge 1}a\left(n\right){\xi}^{n})d\xi $$

$$a(np-1)-[p+1-\#(E/{\mathbb{F}}_{p})]\xb7a(n-1)+p\xb7a(n/p-1)\equiv 0\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{1+{\mathrm{o}rd}_{p}n})$$

Throughout this section, we assume that E is an elliptic curve defined over a number field $\mathbb{Q}\left(\sqrt{1-{\lambda}_{d}}\right)$ by a cubic equation in terms of x and y with identity $\underline{O}$. We further assume that E has CM by an order of an imaginary quadratic field $\mathbb{Q}\left(\tau \right)$. The endomorphism ring of $E/\mathbb{C}$ is denoted by R.

Choose some prime p such that ${\lambda}_{d}$ can be embedded into ${\mathbb{Z}}_{p}$ and such that p is unramified in $\mathbb{Q}\left(\sqrt{1-{\lambda}_{d}}\right)$. We extend the $p-$adic valuation of ${\mathbb{Q}}_{p}$ to ${\mathbb{Q}}_{p}\left(\sqrt{1-{\lambda}_{d}}\right)$ so that $\left|p\right|=1/p$ and let $A=\{x\in {\mathbb{Q}}_{p}\left(\sqrt{1-{\lambda}_{d}}\right):|x|\le 1\}$. The ring A is a local ring with maximal ideal $\left(p\right)=pA$. This information can be organized in the following diagram.
Note that A is the ring of Witt vectors of $k={\mathbb{F}}_{p}\left(\sqrt{1-{\lambda}_{d}}\right)$. Let σ be the nontrivial automorphism on $A=W\left({\mathbb{F}}_{p}\left(\sqrt{1-{\lambda}_{d}}\right)\right)$ lifted from the Frobenius automorphism $z\mapsto {z}^{p}$ on ${\mathbb{F}}_{p}$. It satisfies the relation $\sigma \left(a\right)\equiv {a}^{p}\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$ for all $a\in A$. If $A={\mathbb{Z}}_{p}$, σ is simply the identity map on ${\mathbb{Z}}_{p}$. Otherwise, σ is conjugation, mapping $\sqrt{1-{\lambda}_{d}}$ to $-\sqrt{1-{\lambda}_{d}}$. We further assume that E has a model over A and has good reduction modulo $pA$. From Katz we know the de Rham cohomology space ${H}_{DR}^{1}(E/A)$ is isomorphic to
Further details are given by Katz [31]. Both eigenfunctions ω and ν of the endomorphism ring R, as discussed earlier, in ${H}_{DR}^{1}(E,\mathbb{C})$ can be embedded in ${H}_{DR}^{1}(E/A)$ under our assumption. We choose a local uniformizer ξ of E over A near $\underline{O}$ that is a formal power series of $-\frac{x}{y}$ starting with $-\frac{x}{y}$. Using expansion at infinity with ξ to get ${H}_{DR}^{1}(\widehat{E}/A)$, consisting of formal differentials of E of the form $\sum c\left(n\right){\xi}^{n}d\xi $, where $c\left(n\right)\in A$. By formal integration, these differentials will be represented as ${\sum}_{n\ge 1}\frac{c(n-1)}{n}{\xi}^{n}$ in ${H}_{DR}^{1}(\widehat{E}/A)$. Below, we assume
with coefficients in A. Under this notation, any degree–p Frobenius lifting Φ of $\xi \mapsto {\xi}^{p}$ from ${\mathbb{F}}_{p}$ to A acts semi-linearly, which sends the class represented by
to the class represented by

$$\frac{\{f\in {\mathbb{Q}}_{p}\left[\sqrt{1-\lambda}\right]\left[[x,y]\right]:f\left(\underline{O}\right)=0,df\mathrm{integralin}A\}}{\{f\in A\left[[x,y]\right]:f\left(\underline{O}\right)=0\}}$$

$$\omega =\sum _{n\ge 1}\frac{a(n-1)}{n}{\xi}^{n}d\xi ,\phantom{\rule{0.277778em}{0ex}}\text{and}\phantom{\rule{0.277778em}{0ex}}\nu =\sum _{n\ge 1}\frac{b(n-1)}{n}{\xi}^{n}d\xi $$

$$\sum _{n\ge 1}\frac{c\left(n\right)}{n}{\xi}^{n}$$

$$\sum _{n\ge 1}\frac{c{\left(n\right)}^{\sigma}}{n}\Phi {\left(\xi \right)}^{n}$$

Let ${H}_{DR}^{1}(\widehat{E}/A,\left(p\right))$ be the cohomology group defined from a subcomplex of the de Rham complex of E with respect to the divided power ideal $\left(p\right)$. To be more precise, ${H}_{DR}^{1}(\widehat{E}/A,\left(p\right))$ is isomorphic to
In this space, the class representing ω is annihilated by ${\Phi}^{2}-{b}_{1}\Phi +{b}_{2}$ where ${b}_{1},{b}_{2}\in A$. By comparison between étale and crystalline cohomologies, ${T}^{2}-{b}_{1}T+{b}_{2}$ is related to the local Hasse–Weil zeta function of E over K and, consequently, ${b}_{2}^{2}=\pm {p}^{2}$. It follows that
For simplicity, we assume that $a\left(0\right)$ and $b\left(0\right)$ are both 1 after normalization.

$$\frac{\{f\in {\mathbb{Q}}_{p}\left[\sqrt{1-\lambda}\right]\left[\left[\xi \right]\right]:f\left(\underline{O}\right)=0,df\text{intergralin}A\}}{\{f\in pA\left[\left[\xi \right]\right]:f\left(\underline{O}\right)=0\}}$$

$$a(n{p}^{r}-1)-{b}_{1}a{(n{p}^{r-1}-1)}^{\sigma}+{b}_{2}a{(n{p}^{r-2}-1)}^{{\sigma}^{2}}\equiv 0\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{r}A)$$

$$a(p-1)b(p-1)=\text{sgn}\xb7{b}_{2}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2}A)$$

Proof. Recall that ${b}_{2}^{2}=\pm {p}^{2}$. When p is ordinary, i.e., $a(p-1)\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$, and Φ commutes with R, then ω and ν are eigenfunctions of Φ with eigenvalues π and $\overline{\pi}$, which are the two roots of ${T}^{2}-{b}_{1}T+{b}_{2}=0$. In particular, $\overline{\pi}$ is the unit root in A and $\pi ={b}_{2}/\overline{\pi}.$ (See work of Katz [32] for further details.) From comparing the pth coefficients of $\sum \frac{a{\left(n\right)}^{\sigma}}{n}\Phi \left(n\right)=\pi \xb7\frac{a\left(n\right)}{n}{t}^{n}$, we conclude that
Similarly, we have
and thus
Multiplication by p on both sides gives $a(p-1)b(p-1)\equiv {b}_{2}\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2}A)$.

$$1\equiv \pi \frac{a(p-1)}{p}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}\pi A)$$

$$1\equiv \overline{\pi}\frac{b(p-1)}{p}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}\pi A)$$

$$1\equiv \frac{\pi \overline{\pi}a(p-1)b(p-1)}{{p}^{2}}=\frac{{b}_{2}}{p}\xb7\frac{a(p-1)b(p-1)}{p}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$$

In the supersingular case, ${b}_{1}=0$. We first assume that there is a lifting $\varphi $ of $t\mapsto {t}^{p}$ which stabilizes the A–submodule of ${H}_{DR}^{1}(E/A,\left(p\right))$ generated by ω and ν. Assume that the matrix of Φ under the basis consisting of ω and ν is $M=\left(\begin{array}{cc}u& v\\ w& -u\end{array}\right)$. In particular, $-{u}^{2}-vw={b}_{2}$. We now determine M modulo $pA$. Assume that $u\not\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$, then $v,w$ are nonzero modulo $pA$ as well. Otherwise, the determinant cannot be p. As both u and w are units, one can choose a different basis such that the matrix becomes diagonal modulo $\pi A.$ With determinant p, one of the eigenvalues is a unit, contradicting the fact that the trace is 0 modulo $pA$. Thus in fact $u\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$ and $uz-vw\equiv -vw\equiv {b}_{2}\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2}A)$. In this case,
This implies w is a unit and $v\equiv 0\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$ and the desired result follows, namely
In fact, such a lift $\varphi $ exists by adding any p-torsion of the elliptic curve to ${\mathbb{Q}}_{p}\left(\sqrt{1-{\lambda}_{d}}\right)$, which allows us to solve ${\varphi}^{2}=\Phi $. Here Φ is the degree–${p}^{2}$ Frobenius lifting that commutes with R under our assumption. Both ω and ν are eigenfunctions of Φ with the same eigenvalue $-{b}_{2}$. The corresponding field extension is ramified. However, using the corresponding maximal ideal, the above argument shows that
lies in the maximal ideal. Hence, our claim follows.

$$1\equiv v\frac{b(p-1)}{p}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)\phantom{\rule{1.em}{0ex}}\text{and}\phantom{\rule{1.em}{0ex}}1\equiv w\frac{a(p-1)}{p}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$$

$$a(p-1)b(p-1)\equiv \frac{{p}^{2}}{vw}\equiv -{b}_{2}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2}A)$$

$$\frac{a(p-1)b(p-1)-{b}_{2}}{p}$$

☐

The following lemma is an immediate consequence of the fact that the Euler numbers $\phi \left(d\right)\le 2$, for $d\in \{2,3,4,6\}$.

$$\frac{{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{2}}\equiv 0\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}p)$$

$$\frac{{\left(\frac{1}{d}\right)}_{k}{\left(\frac{d-1}{d}\right)}_{k}}{k{!}^{2}}\equiv 0\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2})$$

Consequently, we have the following truncated version of Clausen’s formula.

$${}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{d}& \frac{d-1}{d}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{p-1}^{2}\equiv {\phantom{\rule{0.166667em}{0ex}}}_{3}{F}_{2}{\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{d}& \frac{d-1}{d}\\ & 1& 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}-4t(t-1)\right]}_{p-1}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2})$$

Proof. The Clausen formula can be also stated as
Note that both hand sides are polynomials of degree no larger than $2p-2$. In fact, modulo ${p}^{2}$, the polynomial on the right has only degree $p-1$. It suffices to prove the kth coefficient on the left hand side is also 0 modulo ${p}^{2}$ when $k\ge p$, which follows from Lemma 17. ☐

$${}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{d}& \frac{d-1}{d}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}^{2}={\phantom{\rule{0.166667em}{0ex}}}_{3}{F}_{2}\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{d}& \frac{d-1}{d}\\ & 1& 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}-4t(t-1)\right]$$

We now prove Theorems 1 and 2.

Proof of Theorems 1 and 2. Let $\xi =\frac{-x}{y}$ be the local uniformizer of the elliptic curve. As discussed in Section 3.4, we assume $\nu =c{\partial}_{{\lambda}_{d}}\omega +\omega $ for some constant $c\in A$ under our assumptions. So $a(p-1)={H}_{d,p-1}\left(t\right)$ with $t=\frac{1-\sqrt{1-{\lambda}_{d}}}{2}$ and $b(p-1)$ can be computed accordingly. Note that the leading coefficients of both ω and ν are 1.

Let $d\in \{2,4\}$. Then $\frac{-x}{y}$, as a function on the elliptic curve, has its zeros and poles supported at the 2-torsion points of the elliptic curve. Consequently, in the ordinary case the multiplication by π map, $\left[\pi \right]$, is a degree-p rational function of t, where π is the non-unit root of ${T}^{2}-{b}_{1}T+{b}_{2}$ as discussed in Proposition 16. As π is in the ring of integers of the CM quadratic field that contains the endomorphism ring R, π commutes with R. The formal power series $\sum \frac{a\left(n\right)}{n}{\xi}^{n}$ is the formal logarithm of the formal group arising from the elliptic curve, which is of height 1 when modulo $pA$ due to the ordinary assumption. (The unfamiliar reader is encouraged to consult work of Stienstra and Beukers [24] for a reference on this terminology.) Therefore,
for some ${\Delta}_{d}$ that can be computed explicitly. For further details, consult papers of Kibelbek et al. [18] and Costeret al. [33]. If we replace ξ by $\tilde{\xi}:={\Delta}_{d}^{-1/4}\xi $, then $\left[\pi \right]\left(\tilde{\xi}\right)\equiv {\tilde{t}}^{p}\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$, which allows us to apply Proposition 16. As computed by Kibelbeket al., this change of variable corresponds to the quartic twists for both cases that we discussed in 4.3. Recall that in Section 4.2, after applying the corresponding quartic twist, we find that the constant term is p from the oddness conclusion stated as Lemma 15. Similarly, for $d=3$, the zeros and poles of $\frac{-x}{y}$ locate at the 3-torsion points of the elliptic curve. From the theory of elliptic functions, one can show that the multiplication by π map on t, likewise a degree–p rational map of t, after a similar quartic twisting, can be used as the Frobenius map in Proposition 16. For $d=6$, it follows from $S{L}_{2}\left(\mathbb{Z}\right)$, the monodromy group for ${E}_{6}\left(t\right)$ is an index-6 supergroup of the monodromy group for ${E}_{2}\left(t\right)$, which is $\Gamma \left(2\right)$ (see Remark 7). In each case, we need to replace ξ by $\tilde{\xi}:={\Delta}_{d}^{-1/4}\xi $ to use Proposition 16 and ${\Delta}_{d}^{(p-1)/2}\equiv \left(\frac{1-{\lambda}_{d}}{p}\right)\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}\pi A)$. In conclusion, we have
for the ordinary case. Now we relate $a(p-1)b(p-1){\Delta}_{d}^{(p-1)/2}$ to our theorems.

$$\left[\pi \right]\left(\xi \right)={\Delta}_{d}^{(p-1)/4}{\xi}^{p}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}pA)$$

$$a(p-1)b(p-1){\Delta}_{d}^{(p-1)/2}\equiv p\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{p}^{2}A)$$

We first prove Theorem 2. The case of $d=2$ was proved by Kibelbek et al. [18]; we handle the cases $d=3,4,6$ here. When p is supersingular, the claim follows from Lemma 18. Let p be ordinary. Recall that $a\left(n\right)={H}_{d,n-1}\left(t\right)$ as in Lemma 12. One can use the same formula to compute ${H}_{d,(d-1)p-1}\left(t\right)$. In particular, ${H}_{d,(d-1)-1}\left(t\right)=1$ for $d=3,4,6$. Let $k=1$ or $d-1$. By comparing the ${\tilde{\xi}}^{kp}$th coefficients of
we have
for some $\gamma \in A$, where ${\Delta}_{d}^{(kp-k)/4}=\epsilon (1-k\gamma p)\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{\pi}^{2}A)$ with ε a forth root of unity and ${\epsilon}^{2}=\left(\frac{1-{\lambda}_{d}}{p}\right)$. Meanwhile, for both $k=1$ and $k=d-1$, ${H}_{d,kp-1}\left(t\right)$ are hypergeometric series in t that terminate before the pth term. By the formula given in Lemma 12, the previous two lemmas, and the p-adic analysis given by the third author [12],
for some $\delta \in A$ when $k=1,d-1$. Thus $\gamma \equiv \delta \phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}\pi A)$, which implies
The claim of Theorem 2 follows from Lemma 18. The ordinary case of Theorem 1 also follows.

$$\sum \frac{{\left(a\left(n\right){\Delta}_{d}^{n/4}\right)}^{\sigma}}{n}\Phi {\left(\tilde{\xi}\right)}^{n}=\pi \sum \frac{a\left(n\right){\Delta}_{d}^{n/4}}{n}{\tilde{\xi}}^{n}$$

$$1\equiv \frac{\pi}{p}\xb7{H}_{d,kp-1}\left(t\right){\Delta}_{d}^{(kp-k)/4}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{\pi}^{2}A)$$

$${H}_{p,kp-1}\left(t\right)={\phantom{\rule{0.166667em}{0ex}}}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{d}& \frac{d-1}{d}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}t\right]}_{(p-1)}+k\delta p$$

$${}_{2}{F}_{1}{\left[\begin{array}{cc}\frac{1}{d}& \frac{d-1}{d}\\ & 1\end{array}\phantom{\rule{0.277778em}{0ex}};\phantom{\rule{0.277778em}{0ex}}\frac{1-\sqrt{1-{\lambda}_{d}}}{2}\right]}_{(p-1)}=\frac{p}{\pi \epsilon}\phantom{\rule{10.0pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}{\pi}^{2}A)$$

To conclude the supersingular case for Theorem 1, we note that the multiplication by $-p$ map on $t=\frac{-x}{y}$ results in a degree–${p}^{2}$ rational map on t, which can be used for Proposition 16 after the same quartic twist as the ordinary case. ☐

This project was initiated during a Banff International Research Station (BIRS) workshop called Women in Numbers II (WIN2) in 2011. The authors would like to thank BIRS and the WIN2 organizers for the opportunity to collaborate. Many of the computations were made using the mathematical software Sage, thanks to the Sage Group. The first author would like to thank Karl Dilcher for his interest in the paper. The third author is supported by NSF grant DMS-1001332. She would also like to thank Jonas Kibelbek and Ravi Ramakrishna for many helpful discussions on related topics and Wadim Zudilin for his interest and comments on the paper. Part of the work was done while the third author was visiting Cornell university as an AWM Michler scholar. She is in debt to both AWM and Cornell university. A special thanks to the fourth author’s research group which allowed the third author to visit her at Aachen in June 2012.

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