# Rational Hypergeometric Ramanujan Identities for $1/\pi^{c}$: Survey and
Generalizations

Henri Cohen and Jesús Guillera

###### Abstract

We give a simple unified proof for _all_ existing rational hypergeometric
ramanujan identities for $1/\pi$, and give a complete survey (without proof)
of several generalizations: rational hypergeometric identities for
$1/\pi^{c}$, Taylor expansions, upside-down formulas, and supercongruences.

##  1 Introduction

In a famous paper [20], S. Ramanujan gave $17$ formulas for $1/\pi$. These
formulas were proved and generalized much later by numerous authors. In the
present paper, which is mainly a survey and does not claim originality, we
have several goals. First, we want to show that the $36$ rational
hypergeometric formulas for $1/\pi$ follow by specialization from a _single_
general formula. Second, we give a list of all known rational hypergeometric
formulas for $1/\pi^{c}$ with $c\geq 2$, many unproved. Third, we will give
Taylor expansions of which the $1/\pi^{c}$ formulas are only the constant
term. Fourth, we give a list of what can be called _upside-down_ formulas.
Finally, we give the _supercongruences_ corresponding to all the $1/\pi^{c}$
formulas. A large part of the formulas of this paper apart from the initial
$1/\pi$ formulas are due to the second author.

This survey is meant to be exhaustive, which means that we would appreciate
feedback from readers who are aware of formulas that are not in our list (and
evidently of errors). Note that we do not list formulas involving algebraic
(as opposed to rational) parameters, nor do we list the second author’s two-
sided formulas where the sums are over $n\in{\mathbb{Z}}$ instead of $n\geq
0$, or other generalizations.

Acknowledgment: We heartily thank Wadim Zudilin for enlightening
conversations.

We recall that the _Pochhammer symbol_ $(x)_{n}$ is defined by
$(x)_{n}=x(x+1)\cdots(x+n-1)=\Gamma(x+n)/\Gamma(x)$, this last formula
allowing us to define it also when $n$ is not an integer.

###### Definition 1.1

  1. (1)

Let $d\geq 2$ be an integer. For all $n\geq 0$ we define

$R_{n}(d)=\prod_{\begin{subarray}{c}1\leq i\leq d\\\
\gcd(i,d)=1\end{subarray}}\dfrac{(i/d)_{n}}{n!}\;.$

  2. (2)

A _rational hypergeometric product_ $H$ is a sequence of the form
$H_{n}=\prod_{d\in I}R_{n}(d)^{v_{d}}$ for some finite index set $I$ and
_positive_ exponents $v_{d}$. We define the _degree_ of $H$ by
$\deg(H)=\sum_{d\in I}\phi(d)$.

  3. (3)

A _rational hypergeometric Ramanujan series_ is a series of the form

$S(H,a,P)=\sum_{n\geq 0}P(n)\dfrac{H_{n}}{a^{n}}\;,$

with $P\in{\mathbb{Z}}[X]$ and $a\in{\mathbb{Q}}^{*}$.

Some comments are in order:

  1. (1)

We only consider coefficients which are hypergeometric _products_ (as opposed
to quotients). We could allow $v_{d}<0$, or more general coefficients, and
indeed there is a vast amount of formulas for such general series, but we will
restrict to those, although we will later give “upside-down” formulas where
_all_ the $v_{d}$ are negative. In the tables that we give below, we will
abbreviate $\prod_{d\in I}R_{n}(d)^{v_{d}}$ as $\prod d^{v_{d}}$.

  2. (2)

In addition, we restrict to such products which are _rational_ in the
hypergeometric motive sense, in other words such that if some irreducible
fraction $i/d$ occurs, then all irreducible fractions in $]0,1[$ with
denominator $d$ occur.

  3. (3)

We only consider $P\in{\mathbb{Z}}[X]$ (or $P\in{\mathbb{Q}}[X]$, which is the
same up to a multiplicative constant). Ramanujan himself gave formulas where
$P$ has coefficients in a quadratic extension of ${\mathbb{Q}}$, but we will
not consider those. Similarly, we restrict to $a\in{\mathbb{Q}}^{*}$.

  4. (4)

It is immediate to see that $H_{n}\sim C/n^{\deg(H)/2}$ for some constant $C$,
so the series $S$ converges for $|a|>1$, also for $a=-1$ if
$\deg(P)<\deg(H)/2$, and for $a=1$ if $\deg(P)<\deg(H)/2-1$. In fact, in all
the examples that we will see we have $\deg(P)=(\deg(H)-1)/2$ (and in
particular $\deg(H)$ is odd).

  5. (5)

The hypergeometric function $\sum_{n\geq 0}H_{n}z^{n}$ satisfies a linear
differential equation of degree $\deg(H)$, in other words there exists a
polynomial $Q(n,z)$ of degree $\deg(H)$ in $n$ such that $\sum_{n\geq
0}Q(n,z)H_{n}z^{n}=0$. We may thus restrict to polynomials $P$ such that
$\deg(P)<\deg(H)$, since formulas with $P$ differing by a multiple of $Q$ are
trivially equivalent. We will give examples of this below.

In fact, the only hypergeometric products that we will consider in this paper
are products of the following $R_{n}(d)$:

$\displaystyle R_{n}(p)$
$\displaystyle=p^{-pn}\dfrac{(pn)!}{n!^{p}}\text{\quad for $p=2,\ 3,\ 5$\;,}$
$\displaystyle R_{n}(4)$
$\displaystyle=2^{-6n}\dfrac{(4n)!}{(2n)!n!^{2}}\;,\text{\quad
and\quad}R_{n}(6)=2^{-4n}3^{-3n}\dfrac{(6n)!}{(3n)!(2n)!n!}\;.$

###### Definition 1.2

A (convergent) hypergeometric Ramanujan series will be called a
$1/\pi^{c}$-formula if its sum $S(H,a,P)$ is equal to an algebraic number
divided by $\pi^{c}$.

A search through the (abundant) literature (together with additional personal
investigations) shows that the only algebraic numbers that occur are of the
form $\sqrt{k}$ for some $k\in{\mathbb{Q}}^{*}$, and we have been able to find
exactly $36$ series whose sum is of the form $\sqrt{k}/\pi$, $10$ whose sum is
of the form $\sqrt{k}/\pi^{2}$, plus a single example for $1/\pi^{3}$ due to
B. Gourevitch and two examples for $1/\pi^{4}$, one due to J. Cullen the other
to Y. Zhao. In addition, there are a number of “divergent” hypergeometric
series for $1/\pi^{c}$ which we will mention.

##  2 Rational Hypergeometric Formulas for $1/\pi$

There are (at least) three methods for proving such formulas. The first, due
to Ramanujan, is the use of elliptic functions and generalizations, the second
is the use of modular functions, and the third is to use WZ-type summation
methods. In the present paper we only use modular functions. In fact, we will
show that _all_ the known rational formulas for $1/\pi$ follow from a single
general formula giving an identity between complex _functions_ , which can
then be specialized to any CM point that we want, and in particular to CM
points giving rational formulas.

###  2.1 A General Identity

An important theorem which can be found for instance in [21] states that any
modular form (or function) $F$ of weight $k$ (say on a congruence subgroup of
$\Gamma$), expressed locally in terms of a modular _function_ $h$ (of weight
$0$), is a solution of a _linear_ differential equation of order $k+1$ with
algebraic coefficients, which can be explicitly constructed; if in addition
$h$ is a _Hauptmodul_ , the coefficients can be chosen to be polynomials.
Equivalently, there exists a sequence $u(n)$ satisfying a polynomial
recurrence relation such that for $\Im(\tau)$ sufficiently large we have
$F(\tau)=\sum_{n\geq 0}u(n)h(\tau)^{n}$. We then have the following easy
result, where from now on we denote by $D$ the differential operator
$D=(1/2\pi i)d/d\tau=qd/dq$:

###### Proposition 2.1

As above, let $F$ be a modular function of weight $k$ on some congruence
subgroup of $\Gamma$, $h$ a modular function (of weight $0$), write
$F(\tau)=\sum_{n\geq 0}u(n)h(\tau)^{n}$ for $\Im(\tau)$ sufficiently large,
and finally set

$G^{*}(\tau)=\dfrac{D(F)/F}{D(h)/h}(\tau)-\dfrac{k}{4\pi\Im(\tau)D(h)/h}\;,$

which is nonholomorphic but modular of weight $0$. We have the following
general formula, which can be considered as a formula for $1/\pi$:

$\sum_{n\geq
0}(n-G^{*}(\tau))u(n)h(\tau)^{n}=\dfrac{k}{4\pi\Im(\tau)}\dfrac{F}{D(h)/h}(\tau)\;.$

Proof. First apply $D$ to the formula expressing $F$ in terms of $h$, so that

$\dfrac{D(F(\tau))}{D(h(\tau))/h(\tau)}=\sum_{n\geq 0}nu(n)h(\tau)^{n}\;.$

Thus, if we set $G=(D(F)/F)/(D(h)/h)$, the left hand side is $GF=G\sum_{n\geq
0}u(n)h^{n}$, so we have the identity

$\sum_{n\geq 0}(n-G(\tau))u(n)h(\tau)^{n}=0\;.$

Now $D(h)/h$ is a modular function of weight $2$, but $D(F)/F$ is only quasi-
modular: $D^{*}(F)/F=D(F)/F-(k/(4\pi\Im(\tau)))$ is truly modular
nonholomorphic of weight $2$, Thus, we set $G^{*}=(D^{*}(F)/F)/(D(h)/h)$, and
this gives both the formula for $G^{*}$ and the desired identity.
$\sqcap$$\sqcup$

Now CM theory tells us that if $\tau$ is a CM point and $h(\tau)$ and
$F(\tau)$ have algebraic Fourier coefficients, then both $h(\tau)$ and
$G^{*}(\tau)$ will be algebraic numbers, and $(F/(D(h)/h))(\tau)$, which has
weight $k-2$, will be an algebraic number times $\Omega_{\tau}^{k-2}$, where
$\Omega_{\tau}$ is a suitable period, for instance
$\Omega_{\tau}=\eta(\tau)^{2}$. Thus, it will itself be algebraic if $k=2$,
otherwise be equal to an algebraic number times a product of values of the
gamma function at rational arguments by the Lerch, Chowla–Selberg formula.

The rest of the work consists simply in specializing the above general
argument to specific modular functions $F$ and $h$ and specific CM points
$\tau$.

Remark. Under this modular interpretation the existence of these $1/\pi$
formulas is due exclusively to the existence of the modularity-preserving
nonholomorphic modification $D^{*}$ of the differential operator $D$ seen
above, which involves $1/\pi$.

###  2.2 First Special Case: Level $1$

We first consider modular forms on the full modular group. It is known at
least since Klein–Fricke that we have the hypergeometric representation

$E_{4}^{1/4}={}_{2}F_{1}(1/12,5/12;1;1/J_{1})\;,$

where $E_{k}=1-B_{k}/(2k)\sum_{n\geq 1}\sigma_{k-1}(n)q^{n}$ and
$J_{1}(\tau)=j(\tau)/1728$. Thanks to the Clausen identity, we deduce that

$E_{4}^{1/2}={}_{3}F_{2}(1/2,1/6,5/6;1,1;1/J_{1})\;.$

This is modular of weight $2$, so we apply the above proposition to
$F=E_{4}^{1/2}$ and $h=1/J_{1}$. We compute that $D(h)/h=-D(j)/j=E_{6}/E_{4}$,
$D(F)/F=(1/6)(E_{2}-E_{6}/E_{4})$, hence
$D^{*}(F)/F=(1/6)(E_{2}^{*}-E_{6}/E_{4})$ where
$E_{2}^{*}=E_{2}-3/(\pi\Im(\tau))$, so $G^{*}=-(1/6)(1-E_{2}^{*}E_{4}/E_{6})$,
so the general identity specializes to

$\sum_{n\geq
0}\left(6n+1-\dfrac{E_{2}^{*}E_{4}}{E_{6}}(\tau)\right)\dfrac{R_{n}(2)R_{n}(6)}{J_{1}(\tau)^{n}}=\dfrac{3}{\pi\Im(\tau)}\dfrac{E_{4}^{3/2}}{E_{6}}(\tau)\;,$

an identity due to the Chudnovsky brothers.

For comparison with higher levels, we set $s_{1}=1/6$, so that for instance
$E_{4}^{1/4}={}_{2}F_{1}(s_{1}/2,(1-s_{1})/2;1;1/J_{1})$.

###  2.3 Special Cases: Levels $2$ and $3$

There is no difference for higher levels compared to level $1$, apart from the
need to give explicitly the modular functions used and the hypergeometric
identities.

As it happens, levels $2$ and $3$ can be treated together. For $N=2$ and $3$
set

$\displaystyle F_{2}(\tau)$
$\displaystyle=\dfrac{NE_{2}(N\tau)-E_{2}(\tau)}{N-1}\;,\quad
F_{4}(\tau)=\dfrac{N^{2}E_{4}(N\tau)-E_{4}(\tau)}{N^{2}-1}\;,$ $\displaystyle
J_{N}(\tau)$
$\displaystyle=\dfrac{F_{2}^{4}}{F_{2}^{4}-F_{4}^{2}}\;,\text{\quad
and\quad}P_{2}(\tau)=\dfrac{NE_{2}(N\tau)+E_{2}(\tau)}{N+1}\;.$

The hypergeometric identity is

$F_{2}^{1/2}={}_{2}F_{1}(s_{N}/2,(1-s_{N})/2;1;1/J_{N})\;,\text{\quad
with\quad}s_{N}=(N+1)/12\;,$

so by Clausen

$F_{2}={}_{3}F_{2}(1/2,s_{N},1-s_{N};1,1;1/J_{N})\;.$

We apply the proposition to $F=F_{2}$ and $h=1/J_{N}$. We compute that
$D(h)/h=F_{4}/F_{2}$, $D(F)/F=s_{N}(P_{2}-F_{4}/F_{2})$, hence
$D^{*}(F)/F=s_{N}(P_{2}^{*}-F_{4}/F_{2})$ with
$P_{2}^{*}=P_{2}-6/((N+1)\pi\Im(\tau))$, so
$G^{*}=-s_{N}(1-P_{2}^{*}F_{2}/F_{4})$, and since $6/(N+1)=1/(2s_{N})$, the
general identity specializes to the two identities

$\displaystyle\sum_{n\geq
0}\left(4n+1-\dfrac{P_{2}^{*}F_{2}}{F_{4}}(\tau)\right)\dfrac{R_{n}(2)R_{n}(4)}{J_{2}(\tau)^{n}}$
$\displaystyle=\dfrac{2}{\pi\Im(\tau)}\dfrac{F_{2}^{2}}{F_{4}}(\tau)\;,$
$\displaystyle\sum_{n\geq
0}\left(3n+1-\dfrac{P_{2}^{*}F_{2}}{F_{4}}(\tau)\right)\dfrac{R_{n}(2)R_{n}(3)}{J_{3}(\tau)^{n}}$
$\displaystyle=\dfrac{3}{2\pi\Im(\tau)}\dfrac{F_{2}^{2}}{F_{4}}(\tau)\;.$

###  2.4 Special Case: Level $4$

Here we set

$\displaystyle F_{2}(\tau)$
$\displaystyle=\dfrac{4E_{2}(4\tau)-E_{2}(\tau)}{3}\;,\quad
G_{2}(\tau)=4E_{2}(4\tau)-4E_{2}(2\tau)+E_{2}(\tau)$ $\displaystyle J_{4}$
$\displaystyle=\dfrac{F_{2}^{2}}{F_{2}^{2}-G_{2}^{2}}\;,\text{\quad
and\quad}P_{2}(\tau)=E_{2}(2\tau)\;.$

The hypergeometric identity is again

$F_{2}^{1/2}={}_{2}F_{1}(s_{N}/2,(1-s_{N})/2;1;1/J_{4})\;,\text{\quad
with\quad}s_{4}=1/2\;,$

so by Clausen $F_{2}={}_{3}F_{2}(1/2,s_{4},1-s_{4};1,1;1/J_{4})$. We apply the
proposition to $F=F_{2}$ and $h=1/J_{4}$. We compute that $D(h)/h=G_{2}$,
$D(F)/F=(P_{2}-G_{2})/3$, hence $D^{*}(F)/F=(P_{2}^{*}-G_{2})/3$ with
$P_{2}^{*}=P_{2}-3/(2\pi\Im(\tau))$, so $G^{*}=-(1-P_{2}^{*}/G_{2})/3$, hence
the general identity specializes to

$\sum_{n\geq
0}\left(3n+1-\dfrac{P_{2}^{*}}{G_{2}}(\tau)\right)\dfrac{R_{n}(2)^{3}}{J_{4}(\tau)^{n}}=\dfrac{3}{2\pi\Im(\tau)}\dfrac{F_{2}}{G_{2}}(\tau)\;.$

##  3 The Basic List of Rational Hypergeometric $1/\pi$ Formulas

From the above four specializations it is now immediate to obtain as many
hypergeometric $1/\pi$ formulas as we like. To obtain such formulas which are
_rational_ in the above sense, we first need $J_{N}(\tau)$ to be rational.
This trivially implies that the coefficients of $1/(\pi\Im(\tau))$ on the
right-hand side of the formulas are square roots of rational numbers. Thus, if
we want the coefficient of $1/\pi$ to be algebraic, we need both $J_{N}(\tau)$
rational and $\Im(\tau)$ algebraic, and a transcendence theorem (well-known
for $N=1$, but proved similarly for $N>1$) implies that $\tau$ is a CM point,
i.e., of the form $(a+\sqrt{D})/b$ with $D<0$ and $a$, $b$ integral. In turn
this implies (less trivially) that the other coefficients involved will be
square roots of a rational number.

The following table summarizes the results obtained in this way: each formula
is of the form $\sum_{n\geq 0}P(n)H_{N}(n)/a^{n}=\sqrt{k}/\pi$, where the
function $H_{N}(n)=(1/2)_{n}(s_{N})_{n}(1-s_{N})_{n}/n!^{3}$ is the
coefficient of $x^{n}$ in ${}_{3}F_{2}(1/2,s_{N},1-s_{N};1,1;x)$, so that
$H_{1}(n)=R_{n}(2)R_{n}(6)$, $H_{2}(n)=R_{n}(2)R_{n}(4)$,
$H_{3}(n)=R_{n}(2)R_{n}(3)$, and $H_{4}(n)=R_{n}(2)^{3}$. For uniqueness, we
always choose $P$ with content $1$ and positive leading coefficient, $k$ is
given in factored form, and the square root of $k$ is always the positive one.
For future reference, we assign a number from 1 to 36 to each formula.

$\sum_{n\geq 0}P(n)\dfrac{H_{N}(n)}{a^{n}}=\dfrac{\sqrt{k}}{\pi}$

# $N$ $H_{N}$ $\tau$ $a=J_{N}(\tau)$ $P$ $k$ 1 $1$ $2\cdot 6$
$(1+\sqrt{-7})/2$ $-2^{-6}5^{3}$ $63x+8$ $3\cdot 5^{3}$ 2 $1$ $2\cdot 6$
$(1+\sqrt{-11})/2$ $-2^{9}3^{-3}$ $154x+15$ $2^{11}$ 3 $1$ $2\cdot 6$
$(1+\sqrt{-19})/2$ $-2^{9}$ $342x+25$ $2^{11}\cdot 3$ 4 $1$ $2\cdot 6$
$(1+\sqrt{-27})/2$ $-2^{9}3^{-2}5^{3}$ $506x+31$ $2^{11}\cdot 3^{-3}\cdot
5^{3}$ 5 $1$ $2\cdot 6$ $(1+\sqrt{-43})/2$ $-2^{12}5^{3}$ $5418x+263$
$2^{14}\cdot 3^{-1}\cdot 5^{3}$ 6 $1$ $2\cdot 6$ $(1+\sqrt{-67})/2$
$-2^{9}5^{3}11^{3}$ $261702x+10177$ $2^{11}\cdot 3\cdot 5^{3}\cdot 11^{3}$ 7
$1$ $2\cdot 6$ $(1+\sqrt{-163})/2$ $-2^{12}5^{3}23^{3}29^{3}$
$545140134x+13591409$ $2^{14}\cdot 3\cdot 5^{3}\cdot 23^{3}\cdot 29^{3}$ 8 $1$
$2\cdot 6$ $\sqrt{-2}$ $3^{-3}5^{3}$ $28x+3$ $5^{3}$ 9 $1$ $2\cdot 6$
$\sqrt{-3}$ $2^{-2}5^{3}$ $11x+1$ $2^{-2}\cdot 3^{-1}\cdot 5^{3}$ 10 $1$
$2\cdot 6$ $\sqrt{-4}$ $2^{-3}11^{3}$ $63x+5$ $2^{-4}\cdot 3\cdot 11^{3}$ 11
$1$ $2\cdot 6$ $\sqrt{-7}$ $2^{-6}5^{3}17^{3}$ $133x+8$ $2^{-2}\cdot
3^{-5}\cdot 5^{3}\cdot 17^{3}$ 12 $2$ $2\cdot 4$ $(1+\sqrt{-5})/2$ $-2^{2}$
$20x+3$ $2^{6}$ 13 $2$ $2\cdot 4$ $(1+\sqrt{-7})/2$ $-2^{-8}3^{4}7^{2}$
$65x+8$ $3^{4}\cdot 7$ 14 $2$ $2\cdot 4$ $(1+\sqrt{-9})/2$ $-2^{4}3$ $28x+3$
$2^{8}\cdot 3^{-1}$ 15 $2$ $2\cdot 4$ $(1+\sqrt{-13})/2$ $-2^{2}3^{4}$
$260x+23$ $2^{6}\cdot 3^{4}$ 16 $2$ $2\cdot 4$ $(1+\sqrt{-25})/2$
$-2^{6}3^{4}5$ $644x+41$ $2^{10}\cdot 3^{4}\cdot 5^{-1}$ 17 $2$ $2\cdot 4$
$(1+\sqrt{-37})/2$ $-2^{2}3^{4}7^{4}$ $21460x+1123$ $2^{6}\cdot 3^{4}\cdot
7^{4}$ 18 $2$ $2\cdot 4$ $\sqrt{-1}$ $2^{-5}3^{4}$ $7x+1$ $2^{-2}\cdot 3^{4}$
19 $2$ $2\cdot 4$ $\sqrt{-6}/2$ $3^{2}$ $8x+1$ $2^{2}\cdot 3$ 20 $2$ $2\cdot
4$ $\sqrt{-10}/2$ $3^{4}$ $10x+1$ $2^{-3}\cdot 3^{4}$ 21 $2$ $2\cdot 4$
$\sqrt{-18}/2$ $7^{4}$ $40x+3$ $3^{-3}\cdot 7^{4}$ 22 $2$ $2\cdot 4$
$\sqrt{-22}/2$ $3^{4}11^{2}$ $280x+19$ $2^{2}\cdot 3^{4}\cdot 11$ 23 $2$
$2\cdot 4$ $\sqrt{-58}/2$ $3^{8}11^{4}$ $26390x+1103$ $2^{-3}\cdot 3^{8}\cdot
11^{4}$ 24 $3$ $2\cdot 3$ $(3+\sqrt{-27})/6$ $-2^{4}3^{-2}$ $5x+1$ $2^{4}\cdot
3^{-1}$ 25 $3$ $2\cdot 3$ $(3+\sqrt{-51})/6$ $-2^{4}$ $51x+7$ $2^{4}\cdot
3^{3}$ 26 $3$ $2\cdot 3$ $(3+\sqrt{-75})/6$ $-2^{4}5$ $9x+1$ $2^{4}\cdot
3\cdot 5^{-1}$ 27 $3$ $2\cdot 3$ $(3+\sqrt{-123})/6$ $-2^{10}$ $615x+53$
$2^{10}\cdot 3^{3}$ 28 $3$ $2\cdot 3$ $(3+\sqrt{-147})/6$ $-2^{4}3^{3}7$
$165x+13$ $2^{4}\cdot 3^{6}\cdot 7^{-1}$ 29 $3$ $2\cdot 3$ $(3+\sqrt{-267})/6$
$-2^{4}5^{6}$ $14151x+827$ $2^{4}\cdot 3^{3}\cdot 5^{6}$ 30 $3$ $2\cdot 3$
$\sqrt{-6}/3$ $2$ $6x+1$ $3^{3}$ 31 $3$ $2\cdot 3$ $\sqrt{-12}/3$
$2^{-1}3^{3}$ $15x+2$ $2^{-4}\cdot 3^{6}$ 32 $3$ $2\cdot 3$ $\sqrt{-15}/3$
$2^{-2}5^{3}$ $33x+4$ $2^{-2}\cdot 3^{3}\cdot 5^{2}$ 33 $4$ $2^{3}$
$(1+\sqrt{-2})/2$ $-1$ $4x+1$ $2^{2}$ 34 $4$ $2^{3}$ $(1+\sqrt{-4})/2$
$-2^{3}$ $6x+1$ $2^{3}$ 35 $4$ $2^{3}$ $\sqrt{-3}/2$ $2^{2}$ $6x+1$ $2^{4}$ 36
$4$ $2^{3}$ $\sqrt{-7}/2$ $2^{6}$ $42x+5$ $2^{8}$

Rational hypergeometric formulas for $1/\pi$

Note that a generalization of the proof of the class number $1$ problem for
imaginary quadratic fields _proves_ that the values for $a$ listed above
(together with the values for the divergent series that we will give below)
are the _only_ rational values of $J_{N}(\tau)$ at CM arguments $\tau$,
outside of the values $0$ and $1$ which cannot be used.

Note that we do not claim that we have found all possible rational $1/\pi$
formulas, but only that, as far as we can tell, no other such formula exists
in the literature, and a rather long search using linear dependence algorithms
did not find any additional ones, except from trivial modifications coming
from the fact that ${}_{2}F_{1}$ is solution of a linear differential equation
of order $2$. For instance, we have the following formulas, which are
trivially equivalent to the last three formulas of the above list:

$\displaystyle\sum_{n\geq 0}(6n^{3}+n^{2})\dfrac{R_{n}(2)^{3}}{(-8)^{n}}$
$\displaystyle=-\dfrac{\sqrt{2}/6}{\pi}\;,\quad\sum_{n\geq
0}(2n^{3}-n^{2})\dfrac{R_{n}(2)^{3}}{4^{n}}=\dfrac{1/3}{\pi}\;,$
$\displaystyle\sum_{n\geq 0}(210n^{3}-5n^{2}+n)\dfrac{R_{n}(2)^{3}}{64^{n}}$
$\displaystyle=\dfrac{4/3}{\pi}\;.$

Note that the same method allows us to find _divergent_ series because
$|a|<1$:

# $N$ $H_{N}$ $\tau$ $a=J_{N}(\tau)$ $P$ $k$ sign 37 $2$ $2\cdot 4$
$(-1+\sqrt{-3})/2$ $-2^{-4}3^{2}$ $5x+1$ $3$ $+$ 38 $2$ $2\cdot 4$
$(1+\sqrt{-7})/4$ $2^{-8}3^{4}$ $35x+8$ $-2^{2}3^{4}$ $-$ 39 $3$ $2\cdot 3$
$(2+\sqrt{-2})/6$ $2\cdot 3^{-3}$ $10x+3$ $-2^{2}5^{2}$ $-$ 40 $3$ $2\cdot 3$
$(1+\sqrt{-11})/6$ $2^{4}3^{-3}$ $11x+3$ $-2^{4}3^{2}$ $+$ 41 $3$ $2\cdot 3$
$(3+\sqrt{-15})/6$ $-2^{-2}$ $15x+4$ $3^{3}$ $+$ 42 $4$ $2^{3}$
$(1+\sqrt{-1})/2$ $-2^{-3}$ $3x+1$ $1$ $+$ 43 $4$ $2^{3}$ $(3+\sqrt{-7})/8$
$2^{-6}$ $21x+8$ $-2^{4}$ $-$ 44 $4$ $2^{3}$ $(1+\sqrt{-3})/4$ $2^{-2}$ $3x+1$
$-2^{2}$ $-$

The values of $k$ given in this table are those coming from the general
formula, but correspond to the values obtained from the analytic continuation
of the hypergeometric series only when $a<0$. On the other hand, we will see
below that they all lead to supercongruences, as well as so-called upside-down
formulas. Here the sign of the square root of $k$ can vary, so is indicated in
the last column with respect to the principal determination.

##  4 Additional Consequences of Proposition 2.1

In the previous section, we have only used Proposition 2.1 for some very
specific pairs $(F,h)$ which lead to rational hypergeometric formulas for
$1/\pi$. It is evidently possible to use it for other pairs: in particular we
could use other subgroups of $\Gamma$, and in particular the groups
$\Gamma_{0}^{*}(N)$ for $N>4$, or still the same subgroups that we have
already considered, but with different $F$ (since the subgroups
$\Gamma_{0}^{*}(N)$ for $N\leq 4$ all have genus $0$, there is not much point
in changing the function $h$, since the resulting formulas could also be
obtained by standard hypergeometric identities).

Using $\Gamma_{0}^{*}(N)$ for $N>4$ will not lead to identities involving
hypergeometric functions, but for instance to more general functions called
_Heun functions_.

Using different functions $F$ does give additional formulas. We simply give
two examples, without proof since they are once again direct applications of
Proposition 2.1. These are examples in level $1$, so we keep $h=J_{1}=j/1728$.

First, we choose $F=E_{4}^{1/4}$, and we find the general formula

$\sum_{n\geq
0}\left(12n+1-\dfrac{E_{2}^{*}E_{4}}{E_{6}}(\tau)\right)\dfrac{(1/12)_{n}(5/12)_{n}/n!^{2}}{J_{1}(\tau)^{n}}=\dfrac{3}{\pi\Im(\tau)}E_{4}(\tau)^{-1/4}\dfrac{E_{4}^{3/2}}{E_{6}}(\tau)\;.$

Specializing to $\tau=\sqrt{-3}$ and $\tau=\sqrt{-4}$, and as mentioned above
using the Chowla–Selberg formula to compute the values of $E_{4}(\tau)$, we
obtain the following identities:

$\displaystyle\sum_{n\geq
0}(22n+1)\dfrac{(1/12)_{n}(5/12)_{n}}{n!^{2}}\dfrac{1}{(125/4)^{n}}$
$\displaystyle=\dfrac{(2^{4/3}5^{5/4}/3)\pi}{\Gamma(1/3)^{3}}=\dfrac{2^{1/3}5^{5/4}3^{-1/2}}{B(1/3,1/3)}$
$\displaystyle\sum_{n\geq
0}(126n+5)\dfrac{(1/12)_{n}(5/12)_{n}}{n!^{2}}\dfrac{1}{(11/2)^{3n}}$
$\displaystyle=\dfrac{2^{1/2}3^{1/4}11^{5/4}\pi^{1/2}}{\Gamma(1/4)^{2}}=\dfrac{2^{1/2}3^{1/4}11^{5/4}}{B(1/4,1/4)}\;,$

where $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$ is the beta function.

Choosing instead $F=E_{6}^{1/6}$ leads to the following general formula:

$\sum_{n\geq
0}\left(12n+1-\dfrac{E_{2}^{*}E_{6}}{E_{4}^{2}}(\tau)\right)\dfrac{(1/12)_{n}(7/12)_{n}/n!^{2}}{(1-J_{1}(\tau))^{n}}=\dfrac{3}{\pi\Im(\tau)}E_{4}(\tau)^{-1/4}\left(\dfrac{E_{4}^{3/2}}{E_{6}}(\tau)\right)^{-7/6}\;,$

and specializing to the same values of $\tau$ gives

$\displaystyle\sum_{n\geq
0}(150n+7)\dfrac{(1/12)_{n}(7/12)_{n}}{n!^{2}}\dfrac{1}{(-121/4)^{n}}$
$\displaystyle=\dfrac{2^{4/3}11^{7/6}\pi}{\Gamma(1/3)^{3}}=\dfrac{2^{1/3}3^{1/2}11^{7/6}}{B(1/3,1/3)}$
$\displaystyle\sum_{n\geq
0}(726n+29)\dfrac{(1/12)_{n}(7/12)_{n}}{n!^{2}}\dfrac{1}{(-1323/8)^{n}}$
$\displaystyle=\dfrac{2^{1/2}3^{5/2}7^{7/6}\pi^{1/2}}{\Gamma(1/4)^{2}}=\dfrac{2^{1/2}3^{5/2}7^{7/6}}{B(1/4,1/4)}\;.$

##  5 Generalization I: Rational Hypergeometric Formulas for $1/\pi^{c}$

_Finding_ rational hypergeometric identities for $1/\pi^{c}$ is easily done
using linear dependence algorithms based on the LLL algorithm. _Proving_ them
is more difficult: among the methods used is the WZ method, but it is not the
only one. In fact some of the identities (and the three known identities for
$c\geq 3$) are still conjectural. _Explaining_ them, as we have done for
$1/\pi$ formulas has only started to be done in a recent paper by Dembelé et
al. [6]: the $1/\pi^{2}$ formulas are linked to Asai $L$-functions attached to
Hilbert modular forms for real quadratic fields. However, this apparently
still does not prove all of them.

In the following table, we list all known formulas, as before coding $H$ as
$\prod_{d\in I}d^{v_{d}}$, meaning that $H_{n}=\prod_{d\in
I}R_{n}(d)^{v_{d}}$, and the formula is of the form

$\sum_{n\geq 0}P(n)\dfrac{H_{n}}{a^{n}}=\dfrac{\sqrt{k}}{\pi^{c}}\;.$

# $c$ $H$ $a$ $P$ $k$ 1 $2$ $2^{5}$ $-2^{2}$ $20x^{2}+8x+1$ $2^{6}$ 2 $2$
$2^{5}$ $-2^{10}$ $820x^{2}+180x+13$ $2^{14}$ 3 $2$ $2^{3}\cdot 3$
$2^{6}3^{-3}$ $74x^{2}+27x+3$ $2^{8}\cdot 3^{2}$ 4 $2$ $2^{3}\cdot 4$ $2^{4}$
$120x^{2}+34x+3$ $2^{10}$ 5 $2$ $2\cdot 3\cdot 4$ $-2^{4}3$ $252x^{2}+63x+5$
$2^{8}\cdot 3^{2}$ 6 $2$ $2\cdot 3\cdot 6$ $-2^{12}3^{-6}$ $1930x^{2}+549x+45$
$2^{14}\cdot 3^{2}$ 7 $2$ $2\cdot 3\cdot 6$ $-2^{12}5^{3}$ $5418x^{2}+693x+29$
$2^{14}\cdot 5$ 8 $2$ $2\cdot 3\cdot 6$ $3^{-6}5^{6}$ $532x^{2}+126x+9$
$2^{-4}\cdot 3^{2}\cdot 5^{6}$ 9 $2$ $2\cdot 4\cdot 6$ $-2^{10}$
$1640x^{2}+278x+15$ $2^{16}\cdot 3^{-1}$ 10 $2$ $2\cdot 8$ $7^{4}$
$1920x^{2}+304x+15$ $2^{6}\cdot 7^{3}$ 11 $3$ $2^{7}$ $2^{6}$
$168x^{3}+76x^{2}+14x+1$ $2^{10}$ 12 $4$ $2^{5}\cdot 3\cdot 4$ $-2^{8}3^{-3}$
$4528x^{4}+3180x^{3}+972x^{2}+147x+9$ $2^{16}\cdot 3^{2}$ 13 $4$ $2^{7}4$
$2^{12}$ $43680x^{4}+20632x^{3}+4340x^{2}+466x+21$ $2^{22}$

Rational hypergeometric formulas for $1/\pi^{c}$

Most formulas for $1/\pi^{2}$ were found by the second author, the formula for
$1/\pi^{3}$ was found by B. Gourevitch and the two formulas for $1/\pi^{4}$
were found by Y. Zhao and J. Cullen respectively.

We can also find in the literature the following divergent formulas for
$1/\pi^{c}$:

# $c$ $H$ $a$ $P$ $k$ sign 14 $2$ $2^{5}$ $-2^{-2}$ $10x^{2}+6x+1$ $2^{4}$ $+$
15 $2$ $2^{5}$ $-2^{-10}$ $205x^{2}+160x+32$ $2^{8}$ $+$ 16 $2$ $2^{3}3$
$-3^{-3}$ $28x^{2}+18x+3$ $2^{2}3^{2}$ $+$ 17 $2$ $2\cdot 5$ $-2^{8}5^{-5}$
$483x^{2}+245x+30$ $2^{8}5^{2}$ $+$ 18 $2$ $2\cdot 3\cdot 4$ $-2^{4}3^{-3}$
$172x^{2}+75x+9$ $2^{8}3^{2}$ $+$ 19 $3$ $2^{7}$ $2^{-6}$
$21x^{3}+22x^{2}+8x+1$ $-2^{2}3^{2}$ $-$ 20 $3$ $2^{5}3$ $2^{2}3^{-3}$
$92x^{3}+84x^{2}+27x+3$ $-2^{8}3^{2}$ $-$ 21 $4$ $2^{5}5$ $-2^{10}5^{-5}$
$5532x^{4}+5600x^{3}+2275x^{2}+425x+30$ $2^{16}5^{2}$ $+$

##  6 Generalization II: Taylor Expansions of $1/\pi^{c}$ Formulas

###  6.1 Taylor Expansions of $1/\pi$ Formulas

Following ideas of the second author [11], [14], [15], we are going to
generalize the above formulas by considering them as constant terms of Taylor
expansions. Note that for any $y$ we can define
$(x)_{n+y}=\Gamma(x+n+y)/\Gamma(x)$, hence since $n!=(1)_{n}$:

$R_{n+x}(d)=\prod_{\begin{subarray}{c}1\leq i\leq d\\\
\gcd(i,d)=1\end{subarray}}\dfrac{(i/d)_{n+x}}{(1)_{n+x}}\;.$

Thus $H_{n+x}$ makes sense, so we could define the generalized sum as

$\sum_{n\geq 0}P(n+x)\dfrac{H_{n+x}}{a^{n+x}}\;.$

However, when $a<0$ the factor $a^{x}$ introduces parasitic imaginary terms,
so we prefer to define

$S(H,a,P;x)=\sum_{n\geq
0}P(n+x)\dfrac{H_{n+x}}{\operatorname{sign}(a)^{n}|a|^{n+x}}=\sum_{n\geq
0}P(n+x)\dfrac{H_{n+x}}{a^{n}|a|^{x}}\;,$

which is equal to $\operatorname{sign}(a)^{x}$ times the previous one. Note
that this is not the only possible normalization. We could also shift all the
Pochhammer indices by $x$ instead of shifing $n$. In all cases, this would
give the above series multiplied by a quotient of products of gamma functions
involving $x$, so the transformation from one to the other is immediate.

It is clear that
$S(H,a,P;x+1)=\operatorname{sign}(a)(S(H,a,P;x)-P(x)H_{x}/|a|^{x})$, so we may
assume if necessary that $x\in[0,1[$.

In view of the existing literature, we can ask at least two questions: first,
give (at least the initial terms of) the power series expansion of
$S(H,a,P;x)$ around $x=0$. Second, give the value of $S(H,a,P;1/2)$.

One observes that if the value of the sum is $a_{0}\sqrt{-D}/\pi$ with
$a_{0}\in{\mathbb{Q}}^{*}$ and $D$ a negative fundamental discriminant, the
expansion is always of the form

$S(H,a,P;x)=a_{0}|D|\left(\dfrac{\sqrt{-D}}{\pi}+0x-a_{2}|D|L(D,1)x^{2}-a_{3}D^{2}L(D,2)x^{3}+O(x^{4})\right)\;,$

with the $a_{i}$ rational, and where we write $L(D,m)$ for $\sum_{n\geq
1}\mbox{$\left(\frac{D}{n}\right)$}/n^{m}$ (of course, since $D<0$ we have
$a_{2}|D|L(D,1)=a^{\prime}_{2}\sqrt{-D}\pi$ for some rational
$a^{\prime}_{2}$). Note that, as mentioned above, it is in principle possible
to compute the coefficient of $x^{4}$, but by laziness we have done so only
for cases (33), (35), and (36), see below.

The following table uses the same numbering of the formulas as that given
above; the column $C_{3}$ is related to supercongruences and will be explained
below:

$\displaystyle S(H,a,P;x)$
$\displaystyle=a_{0}|D|\left(\dfrac{\sqrt{-D}}{\pi}+0x-a_{2}|D|L(D,1)x^{2}-a_{3}D^{2}L(D,2)x^{3}+O(x^{4})\right)\;,$
$\displaystyle S_{p}(H,a,P)$ $\displaystyle\equiv
P(0)\mbox{$\left(\dfrac{D}{p}\right)$}p+C_{3}L(D,3-p)p^{3}\allowbreak\
({\rm{mod}}\,\,p^{4})\;.$

# $D$ $a_{0}$ $a_{2}$ $a_{3}$ $\pi^{2}S(H,a,P;1/2)/(a_{0}|D|\sqrt{-D})$
$C_{3}$ 1 $-15$ $1/3$ $3/4$ $1/2$ $\log(3^{3}/5)$ $20$ 2 $-8$ $2$ $7/2$ $4$
$\log(2)$ $15$ 3 $-24$ $2/3$ $15/4$ $2$ $\log(2^{5}/3^{3})$ $5/2$ 4 $-120$
$2/27$ $23/8$ $1/3$ $\log(3^{3}.5/2^{7})$ $5/12$ 5 $-15$ $128/9$ $39/4$ $12$
$\log(2^{2}3^{9}/5^{7})$ $5/64$ 6 $-1320$ $2/3$ $63/16$ $1/26$
$\log(2^{13}11^{5}/(3^{3}5^{11}))$ $5/104$ 7 $-40020$ $16/3$ $159/128$
$1/11560$ $\log(3^{21}5^{13}29^{5}/(2^{38}23^{11}))$ $5/36992$ 8 $-20$ $1/8$
$1$ $1/2$ $2\operatorname{asin}(3/5)$ $-15/2$ 9 $-15$ $1/18$ $2$ $3/2$
$2\operatorname{asin}(7/5^{2})$ $-5/8$ 10 $-132$ $1/96$ $3/2$ $1/8$
$2\operatorname{asin}(41/(3^{3}11))$ $-5/4$ 11 $-255$ $1/162$ $1$ $1/12$
$2\operatorname{asin}(4207/(5^{4}17^{2}))$ $-5/81$ 12 $-4$ $1$ $3$ $4$
$2\log(2)$ $6$ 13 $-7$ $9/7$ $5/2$ $5/2$ $2\log((88+13\sqrt{7})/3^{4})$ $20/3$
14 $-3$ $16/9$ $21/2$ $20$ $(3/2)\log(3^{3}/2^{4})$ $15/8$ 15 $-4$ $9$ $11$
$20$ $2\log(3^{2}/2^{3})$ $10/3$ 16 $-20$ $36/25$ $23/4$ $4$
$\log(2^{18}/(3^{4}5^{5}))$ $1/6$ 17 $-4$ $441$ $35$ $100$
$2\log(2.3^{10}/7^{6})$ $50/147$ 18 $-4$ $9/16$ $2$ $5/2$
$2\operatorname{asin}(7/3^{2})$ $-10/3$ 19 $-3$ $2/3$ $6$ $10$ $\pi/3$ $-15/8$
20 $-8$ $9/64$ $4$ $4$ $2\operatorname{asin}(17/3^{4})$ $-1/3$ 21 $-3$ $49/27$
$24$ $60$ $2\operatorname{asin}(239/(2.7^{4}))$ $-45/392$ 22 $-11$ $18/11$
$10$ $10$ $2\operatorname{asin}(353/(2/3^{8}))$ $-5/24$ 23 $-8$ $9801/64$ $28$
$60$ $2\operatorname{asin}(8668855388657/(3^{8}11^{12}))$ $-5/1089$ 24 $-3$
$4/9$ $5/2$ $10/3$ $\log(3^{3}/2^{2})$ $5/2$ 25 $-3$ $4$ $13/2$ $10$
$3\log(4/3)$ $15/2$ 26 $-15$ $4/75$ $7/4$ $1$ $\log(3^{9}/(2^{2}5^{5}))$ $1/4$
27 $-3$ $32$ $37/2$ $40$ $3\log(2^{8}/3^{5})$ $15/4$ 28 $-7$ $108/49$ $15/2$
$10$ $\log(7^{7}/(2^{10}3^{6}))$ $5/18$ 29 $-3$ $500$ $85/2$ $130$
$3\log(5^{6}/(2^{6}3^{5}))$ $39/50$ 30 $-3$ $1$ $2$ $5/2$
$(2/3)(\pi-\operatorname{asin}(1633/3^{9}))$ $-15/4$ 31 $-4$ $27/32$ $4$ $5$
$2\operatorname{asin}(329/3^{6})$ $-20/9$ 32 $-3$ $5/2$ $8$ $13$
$2\operatorname{asin}(239/3^{6})$ $-78/25$ 33 $-4$ $1/4$ $1$ $1$
$8L(-4,2)/\pi$ $2$ 34 $-8$ $1/8$ $3/2$ $1$ $4L(-4,2)/\pi$ $1$ 35 $-4$ $1/2$
$2$ $2$ $\pi/2$ $-2$ 36 $-4$ $2$ $6$ $8$ $\pi/6$ $-2$

###  6.2 Observations for $1/\pi$

Concerning the Taylor expansions around $x=0$, note that the coefficient
$a_{1}$ of $x^{1}$ always vanishes, and that the numerator of the first seven
$a_{2}$ are equal to $|D(\tau)|-4$, where $D(\tau)=-7$, $-11$, $-19$, $-27$,
$-43$, $-67$, and $-163$ is the discriminant of the corresponding $\tau$ (not
to be confused with the $D$ occurring in the result).

In addition, we also notice a common pattern for the value at $x=1/2$:

  1. (1)

In the value of $S(H,a,P;1/2)$ for $a<0$: with only a few exceptions listed
below, we have $S(H,a,P;1/2)=c_{0}|D|\sqrt{-D}\log(c_{1})/\pi^{2}$, where
$c_{0}$ and $c_{1}$ are rational. The exceptions are as follows:

     * •

In cases (33) and (34) we have $S(H,a,P;1/2)=c_{0}|D|\sqrt{-D}L(-4,2)/\pi^{3}$
with $c_{0}$ rational.

     * •

In case (13) $c_{1}$ is not rational but in the quadratic field
${\mathbb{Q}}(\sqrt{7})$.

  2. (2)

In the value of $S(H,a,P;1/2)$ for $a>0$: in all cases we have
$S(H,a,P;1/2)=c_{0}|D|\sqrt{-D}\operatorname{asin}(c_{1})/\pi^{2}$ where
$c_{0}$ and $c_{1}$ are rational (note that $c|D|\sqrt{-D}/\pi$ is of course
of this form, for instance by choosing $c_{1}=1$).

It is also possible to guess the coefficient of $x^{4}$: for instance in cases
(33), (35), and (36), set

$C_{1}=\sum_{n\geq 1}(-1)^{n}\dfrac{H_{2n}}{(2n+1)^{2}}\text{\qquad
and\qquad}C_{2}=\sum_{n\geq 1}(-1)^{n}\dfrac{H_{n}}{(2n+1)^{2}}\;,$

where here $H_{n}=\sum_{1\leq j\leq n}1/j$ is the $n$th harmonic sum. Then

$\displaystyle S(H,a,P;x)$
$\displaystyle=a_{0}|D|\left(\dfrac{\sqrt{-D}}{\pi}+0x-a_{2}|D|L(D,1)x^{2}\right.$
$\displaystyle\left.\phantom{=}-a_{3}D^{2}L(D,2)x^{3}+a_{4}x^{4}+O(x^{5})\right)\;,$

with $D=-4$ and

$\displaystyle a_{4}$
$\displaystyle=(8/3)(50C_{1}-11C_{2}-22L(-4,2)\log(2))\;,$ $\displaystyle
a_{4}$ $\displaystyle=(128/3)(10C_{1}-C_{2}-2L(-4,2)\log(2))\;,$
$\displaystyle a_{4}$ $\displaystyle=128(22C_{1}-C_{2}-2L(-4,2)\log(2))$

for cases (33), (35), and (36) respectively.

###  6.3 Taylor Expansions of $1/\pi^{c}$ Formulas for $c\geq 2$

For $c=2$ one observes that if the value of the sum is $a_{0}\sqrt{D}/\pi^{2}$
with $a_{0}\in{\mathbb{Q}}^{*}$ and $D$ a fundamental discriminant, the
expansion is always of the form

$\displaystyle S(H,a,P;x)$
$\displaystyle=a_{0}D\left(\dfrac{\sqrt{D}}{\pi^{2}}+0x-a_{2}D\sqrt{D}x^{2}+0x^{3}\right.$
$\displaystyle\phantom{=}\left.+a_{4}D^{3}L(D,2)x^{4}-a_{5}a_{0}D^{4}L(D,3)x^{5}+O(x^{6})\right)\;,$

with the $a_{i}$ rational (of course, when $D>0$ we have
$a_{4}D^{3}L(D,2)=a^{\prime}_{4}\pi^{2}D\sqrt{D}$ for some rational
$a^{\prime}_{4}$).

$\displaystyle S(H,a,P;x)$
$\displaystyle=a_{0}D\left(\dfrac{\sqrt{D}}{\pi^{2}}+0x-a_{2}D\sqrt{D}x^{2}+0x^{3}\right.$
$\displaystyle\phantom{=}\left.+a_{4}D^{3}L(D,2)x^{4}-a_{5}a_{0}D^{4}L(D,3)x^{5}+O(x^{6})\right)\;,$
$\displaystyle S_{p}(H,a,P)$ $\displaystyle\equiv
P(0)\mbox{$\left(\dfrac{D}{p}\right)$}p^{2}+C_{5}L(D,4-p)p^{5}\allowbreak\
({\rm{mod}}\,\,p^{6})\;.$

# $D$ $a_{0}$ $a_{2}$ $a_{4}$ $a_{5}$ $\pi^{3}S(1/2)/(a_{0}D\sqrt{D})$ $C_{5}$
1 $1$ $8$ $1/2$ $25/4$ $7$ $14\zeta(3)/\pi^{2}$ $-7/2$ 2 $1$ $128$ $5/2$
$305/4$ $7$ $2\zeta(3)/\pi^{2}$ $-7/2$ 3 $1$ $48$ $1/3$ $4$ $7/9$ $2\pi/3$
$21$ 4 $1$ $32$ $1$ $20$ $7$ $\pi/3$ $21/2$ 5 $1$ $48$ $3/2$ $157/4$ $91/9$
$4\log(2^{5}/3^{3})$ $-91/9$ 6 $1$ $384$ $5/6$ $85/4$ $7/9$ $2\log(2)$ $-315$
7 $5$ $128/5$ $3/2$ $887/32$ $21/8$ $\log(2^{74}5^{5}/3^{54})$ $-35/216$ 8 $1$
$375/4$ $4/3$ $40$ $6944/1125$ $2\operatorname{asin}(164833/5^{8})$ $1953/50$
9 $12$ $32/9$ $7/24$ $757/1152$ $3/16$ $\log(3^{9}/2^{14})$ $-15/2$ 10 $28$
$1$ $1/7$ $31/336$ $1/21$ $2\operatorname{asin}(2241857/2^{25})$ $35/8$

The last three (one for $1/\pi^{3}$ and two for $1/\pi^{4}$) obey completely
similar expansions, but we give them one by one:

11:

$\displaystyle S(H,a,P;x)$
$\displaystyle=32\left(\dfrac{1}{\pi^{3}}+0x-\dfrac{1}{\pi}x^{2}+0x^{3}+(16/3)L(-4,1)x^{4}+0x^{5}\right.$
$\displaystyle\left.\phantom{=}-(8224/45)L(-4,3)x^{6}+32^{2}\cdot
48L(-4,4)x^{7}+O(x^{8})\right)\;,$ $\displaystyle S(H,a,P;1/2)$
$\displaystyle=\dfrac{8}{\pi^{3}}\;,$ $\displaystyle S_{p}(H,a,P)$
$\displaystyle\equiv\mbox{$\left(\dfrac{-4}{p}\right)$}p^{3}-6L(-4,5-p)p^{7}\allowbreak\
({\rm{mod}}\,\,p^{8})\;.$

12:

$\displaystyle S(H,a,P;x)$
$\displaystyle=768\left(\dfrac{1}{\pi^{4}}+0x-\dfrac{1/2}{\pi^{2}}x^{2}+0x^{3}+(3/8)x^{4}+0x^{5}-(147/8)\zeta(2)x^{6}\right.$
$\displaystyle\left.\phantom{=}+0x^{7}+(471187/1344)\zeta(4)x^{8}-3968\zeta(5)x^{9}+O(x^{10})\right)\;,$
$\displaystyle S(H,a,P;1/2)$ $\displaystyle=\dfrac{9216\zeta(3)}{\pi^{7}}\;,$
$\displaystyle S_{p}(H,a,P)$ $\displaystyle\equiv
9p^{4}-(837/2)\zeta(6-p)p^{9}\allowbreak\ ({\rm{mod}}\,\,p^{10})\;.$

13:

$\displaystyle S(H,a,P;x)$
$\displaystyle=2048\left(\dfrac{1}{\pi^{4}}+0x-\dfrac{2}{\pi^{2}}x^{2}+0x^{3}+(11/3)x^{4}+0x^{5}-(908/15)\zeta(2)x^{6}\right.$
$\displaystyle\left.\phantom{=}+0x^{7}+(53932/7)\zeta(4)x^{8}-95232\zeta(5)x^{9}+O(x^{10})\right)\;,$
$\displaystyle S(H,a,P;1/2)$ $\displaystyle=\dfrac{2048/15}{\pi^{4}}\;,$
$\displaystyle S_{p}(H,a,P)$ $\displaystyle\equiv
21p^{4}+(279/4)\zeta(6-p)p^{9}\allowbreak\ ({\rm{mod}}\,\,p^{10})\;.$

The observations for $1/\pi^{c}$ are essentially identical to the case of
$1/\pi$, in particular the coefficients of $x^{2j-1}$ for $1\leq j\leq c$
vanish.

##  7 Generalization III: Upside-Down Series

For completeness, we list the upside-down series (i.e., with $H_{n}$ in the
denominator) given in [17]. We do not know if all have been proved, but
probably not all among those with $c>1$.

The general recipe is as follows: if $\sum_{n\geq
0}P(n)H_{n}/a^{n}=\sqrt{k}/\pi^{c}$ is a _divergent_ or semi-convergent series
(i.e., with $|a|<1$ or $a=-1$) then

$\sum_{n\geq 1}\dfrac{P(-n)}{n^{2c+1}H_{n}(1/a)^{n}}=A\cdot L(D,c+1)\;,$

where $D$ is the fundamental discriminant corresponding to $(-1)^{c}k$ and
$A\in{\mathbb{Q}}^{*}$. Thus, the only new value is that of $A$. However, for
the reader’s convenience, we give the list explicitly, the numbering
corresponding to that of the initial series (which is different for $c=1$ and
$c\geq 2$).

$\sum_{n\geq 1}\dfrac{Q(n)}{n^{2c+1}H_{n}b^{n}}=A\cdot L(D,c+1)\;.$

# $c$ $H_{N}$ $b=1/a$ $Q$ $D$ $A$ 37 $1$ $2\cdot 4$ $-2^{4}3^{-2}$ $5x-1$ $-3$
$-45/2$ 38 $1$ $2\cdot 4$ $2^{8}3^{-4}$ $35x-8$ $1$ $72$ 39 $1$ $2\cdot 3$
$2^{-1}\cdot 3^{3}$ $10x-3$ $1$ $3$ 40 $1$ $2\cdot 3$ $2^{-4}3^{3}$ $11x-3$
$1$ $48$ 41 $1$ $2\cdot 3$ $-2^{2}$ $15x-4$ $-3$ $-27$ 42 $1$ $2^{3}$ $-2^{3}$
$3x-1$ $-4$ $-2$ 43 $1$ $2^{3}$ $2^{6}$ $21x-8$ $1$ $1$ 44 $1$ $2^{3}$ $2^{2}$
$3x-1$ $1$ $3$ 33 $1$ $2^{3}$ $-1$ $4x-1$ $-4$ $-16$ 14 $2$ $2^{5}$ $-2^{2}$
$10x^{2}-6x+1$ $1$ $-28$ 15 $2$ $2^{5}$ $-2^{10}$ $205x^{2}-160x+32$ $1$ $-2$
16 $2$ $2^{3}3$ $-3^{3}$ $28x^{2}-18x+3$ $1$ $-14$ 17 $2$ $2\cdot 5$
$-2^{-8}5^{5}$ $483x^{2}-245x+30$ $1$ $-896$ 18 $2$ $2\cdot 3\cdot 4$
$-2^{-4}3^{3}$ $172x^{2}-75x+9$ $1$ $-1792$ 19 $3$ $2^{7}$ $2^{6}$
$21x^{3}-22x^{2}+8x-1$ $1$ $45/4$ 20 $3$ $2^{5}3$ $2^{-2}3^{3}$
$92x^{3}-84x^{2}+27x-3$ $1$ $720$ 21 $4$ $2^{5}5$ $-2^{-10}5^{5}$
$5532x^{4}-5600x^{3}+2275x^{2}-425x+30$ $1$ $-380928$

Upside-down series

In particular, note that the last formula gives a series for $\zeta(5)$.

We finish by giving an example of a nonrational $1/\pi$ formula, but there
exist almost a hundred involving only quadratic irrationals, listed in [1]:

$\sum_{n\geq
0}\dfrac{R_{n}(2)^{3}}{(2+\sqrt{3})^{4n}}(12n+(3-\sqrt{3}))=\dfrac{(2+\sqrt{3})(4/3)^{1/4}}{\pi}\;.$

##  8 Generalization IV: Supercongruences

It has been noted long ago by several authors that to _every_ $1/\pi^{c}$
formula (including divergent ones) corresponds a congruence modulo $p$ to a
higher power than could be expected, what is now called a _supercongruence_.

The main observation is as follows: if there exists a $1/\pi^{c}$-formula of
the form $\sum_{n\geq 0}P(n)H_{n}/a^{n}=\sqrt{k}/\pi^{c}$, then for all primes
$p$ such that $v_{p}(a)=v_{p}(k)=0$ and not dividing any $d$ occuring in $H$
we should have the following precise supercongruence:

$\sum_{n=0}^{p-1}P(n)\dfrac{H_{n}}{a^{n}}\equiv
P(0)\mbox{$\left(\dfrac{(-1)^{c}4k}{p}\right)$}p^{c}\allowbreak\
({\rm{mod}}\,\,p^{2c+1})\;.$

For instance, we have the following supercongruences:

$\displaystyle\sum_{n=0}^{p-1}(154n+15)\dfrac{R_{2}(n)R_{6}(n)}{(-8/3)^{3n}}$
$\displaystyle\equiv 15\mbox{$\left(\dfrac{-8}{p}\right)$}p\allowbreak\
({\rm{mod}}\,\,p^{3})$
$\displaystyle\sum_{n=0}^{p-1}(5418n^{2}+693n+29)\dfrac{R_{2}(n)R_{3}(n)R_{6}(n)}{(-80)^{3n}}$
$\displaystyle\equiv 29\mbox{$\left(\dfrac{20}{p}\right)$}p^{2}\allowbreak\
({\rm{mod}}\,\,p^{5})$

The same phenomenon is valid for the _divergent_ series for $1/\pi^{c}$ that
we have given. For instance we have

$\sum_{n=0}^{p-1}(35n+8)\dfrac{R_{2}(n)R_{4}(n)}{(3/4)^{4n}}\equiv
8p\allowbreak\ ({\rm{mod}}\,\,p^{3})$

However it has been noticed by several authors that these supercongruences can
be refined to a higher power of $p$ (more precisely to a congruence modulo
$p^{2c+2}$ instead of $p^{2c+1}$): the recipe, made precise by the second
author, is simply to replace the $L(D,c+1)$ occurring in the coefficient of
$x^{2c+1}$ of the Taylor expansions by $L(D,c+2-p)$ times a suitable rational
number. In other words,

$\displaystyle S_{p}(H,a,P)$
$\displaystyle:=\sum_{n=0}^{p-1}P(n)\dfrac{H_{n}}{a^{n}}$ $\displaystyle\equiv
P(0)\mbox{$\left(\dfrac{(-1)^{c}4k}{p}\right)$}p^{c}+C_{2c+1}p^{2c+1}L(D,c+2-p)\allowbreak\
({\rm{mod}}\,\,p^{2c+2})\;.$

The coefficients $C_{2c+1}$ have been given for all the convergent series in
the above tables. The remaining coefficients for the divergent series are as
follows:

For the divergent $1/\pi$ formulas,

$C_{3}=(15/4,0,0,0,12,2,0,0)$

for formulas (37) to (44).

For the divergent $1/\pi^{c}$ formulas for $c\geq 2$,

$C_{2c+1}=(-7/2,-64,-21/2,-210,-63,0,0,-1395)$

for formulas (14) to (21).

The coefficients $0$ of course mean that the congruence is valid modulo
$p^{2c+2}$ with no correction terms.

We can observe that $C_{3}$ is almost always divisible by $5$, $C_{5}$ is
almost always divisible by $7$, and $C_{9}$ is always divisible by
$279=3^{2}\cdot 31$. We have no explanation for this phenomenon.

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Henri Cohen, Université de Bordeaux, LFANT, IMB, U.M.R. 5251 du C.N.R.S, 351
Cours de la Libération, 33405 Talence Cedex, FRANCE.

Jesús Guillera, Universidad de Zaragoza, Departamento de Matemáticas, 50009
Zaragoza, SPAIN.