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\title{An Exact Hypergeometric Tail Certificate for\\
Ramanujan Challenge Problem 2.8}
\author{Problem 2.8 submission}
\date{July 2026}

\begin{document}
\maketitle

\begin{abstract}
Let \(G_N=M_0M_1\cdots M_{N-1}\) be the \(4\times4\) transfer
product in Ramanujan Challenge Problem 2.8, and let \(P_{N,j}\) and
\(Q_{N,j}\) be the two official seeded rows evaluated in column \(j\).
We prove
\[
  \lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
  =\frac{\sqrt{10005}}{\pi}
  \qquad (j=1,2,3,4).
\]
Equivalently, \(Q_{N,j}/P_{N,j}\to\pi/\sqrt{10005}\).

The missing connection constant is fixed by an exact rank-three
hypergeometric tail.  A nonterminating \(\F43\) Euler jet is carried
backward by the displayed matrix, while the first denominator is a
terminating adjoint \(\F43\).  Four denominator-cleared Ore identities
and three coefficient identities give an all-\(N\) Pad\'e divisibility
theorem.  Positivity at the negative CM point, a Cauchy estimate, and an
explicit stable-graph contraction prove convergence for all four columns.
Every algebraic identity used below is reproduced by a dependency-free
rational-polynomial checker.  The classical Chudnovsky formula, cited
precisely in Section~2, is the sole imported theorem.
\end{abstract}

\enlargethispage{2\baselineskip}
\tableofcontents

\section{Statement and compact form of the seeds}

Put
\[
 R=151931373056001=53360^3+1,\qquad
 x_0=\frac1R,\qquad
 z=-\frac{x}{1-x}.
\]
Thus the official CM point is
\[
 z_0=-\frac1{R-1}=-\frac1{53360^3}.
\]
Let
\[
 G_N=M_0M_1\cdots M_{N-1},\qquad G_0=I_4,
\]
where the rational family is defined as follows.  Put \(r=x^{-1}\),
\(\omega=u(3u-2)(3u+2)\), and
\begin{align*}
a_1={}&r(144u^5-288u^4+144u^3)
       -99u^5+333u^4-229u^3-114u^2+40u+64,\\
a_2={}&r(432u^4-864u^3+432u^2)
       -243u^4+909u^3-868u^2-80u+272,\\
a_3={}&r(432u^3-864u^2+432u)
       -153u^3+648u^2-860u+360,\\
a_4={}&144r(u-1)^2,\\
b_1={}&-144ru^3+9u^4+63u^3+158u^2+168u+64,\\
b_2={}&216ru^2+36u^3-189u^2-316u-168,\\
b_3={}&108ru+54u^2-189u-158,\\
c_1={}&-288r^2u^3+
r(54u^4+378u^3+948u^2+1008u+384)\\
&\hspace{2.2em}+18u^5+45u^4-251u^3-1086u^2-1384u-576,\\
c_2={}&-432r^2u^2+
r(153u^4-657u^3+1292u^2+2064u+1072)\\
&\hspace{2.2em}-72u^4+702u^3-1069u^2-2508u-1512,\\
c_3={}&-216r^2u+
r(180u^3-891u^2+1450u+1116)\\
&\hspace{2.2em}-108u^3+864u^2-1385u-1422,\\
c_4={}&-4r^2+r(6u^2-33u+\tfrac{536}{9})
       -4u^2+32u-63.
\end{align*}
Define
\begin{equation}\label{eq:deformed-matrix}
\mathcal M(u,x)=
\begin{pmatrix}
a_1/\omega&a_2/\omega&a_3/\omega&a_4/\omega\\
-u^3&-3u^2&-3u&-1\\
xb_1/144&-xb_2/72&-xb_3/36&
 x(-2r-(2u-7))/2\\
x^2c_1/288&x^2c_2/144&x^2c_3/72&x^2c_4/4
\end{pmatrix},
\qquad M_N(x)=\mathcal M(2N+3,x).
\end{equation}
At \(x=x_0\), the only deformed challenge coefficient is restored by
\[
236337691420383=\frac{14R-567}{9}.
\]
Thus every use of Cauchy's theorem below concerns the explicitly displayed
rational \(x\)-family, not an unspecified continuation.

Define four Pascal rows
\[
\begin{aligned}
 b_0&=(1,0,0,0),&
 b_1&=(1,1,0,0),\\
 b_2&=(1,2,1,0),&
 b_3&=(1,3,3,1)
\end{aligned}
\]
and let \(\mathcal P\) be the matrix with rows \(b_0,b_1,b_2,b_3\).
The compact denominator row is
\[
 C(x)=
 \left(
 \frac{18}{x}+\frac{159}{4},\
 \frac{54}{x}+\frac{131}{2},\
 \frac{54}{x}+27,\
 \frac{18}{x}
 \right).
\]
Equivalently,
\[
 C=\frac{18}{x}b_3+\frac54b_0+\frac{23}{2}b_1+27b_2.
\]
Set
\[
 A=13591409,\qquad B=545140134,\qquad S=426880.
\]
The two official initial rows have the exact form
\begin{equation}\label{eq:seed-identities}
 A_1=SC,\qquad
 A_0=AC-\frac54H_0,\qquad
 H_0=Ab_0+Bb_1=(A+B,B,0,0).
\end{equation}
At \(x=x_0\), these identities reproduce the official integer rows
entry by entry:
\[
\begin{aligned}
A_0={}&(37169305760442252761441,\,
111507917281327441564208,\\
&\hspace{4.7em}111507917281327599720129,\,
37169305760442410917362),\\
A_1={}&(1167416361542639692320,\,
3502249084627896132160,\\
&\hspace{4.7em}3502249084627879697280,\,
1167416361542622723840).
\end{aligned}
\]

For \(j=1,\ldots,4\), write
\[
 P_{N,j}=A_0G_N\e_j,\qquad
 Q_{N,j}=A_1G_N\e_j.
\]
We first identify the first-column limit and then invoke the exact cyclic
frame to cover all four columns.

\section{The CM function and its exact value}

Let
\[
 y(z)=\F32\left(
 \begin{matrix}\frac16,\frac12,\frac56\\1,1\end{matrix};z
 \right),
 \qquad
 \theta=z\frac{d}{dz}=(1-x)x\frac{d}{dx},
\]
and define
\begin{equation}\label{eq:phi}
 \Phi(x)=\frac{Ay(z)+B\theta y(z)}{S}.
\end{equation}

We use one external theorem: the classical Chudnovsky identity.  In the
normalization used here it is Theorem~0.1 of Milla's equation-by-equation
derivation \([2]\), whose modular and CM proof is completed in
Theorem~9.7 and Chapter~10:
\[
 \frac1\pi=
 \frac{12}{640320^{3/2}}
 \sum_{k=0}^{\infty}
 \frac{(6k)!}{(3k)!(k!)^3}
 (A+Bk)(-640320^{-3})^k.
\]
The elementary coefficient identity
\[
 \frac{(6k)!}{(3k)!(k!)^3}
 =1728^k
 \frac{(\frac16)_k(\frac12)_k(\frac56)_k}{(k!)^3}
\]
and \(640320=12\cdot53360\) give
\[
 Ay(z_0)+B\theta y(z_0)
 =\frac{640320^{3/2}}{12\pi}
 =\frac{426880\sqrt{10005}}{\pi}.
\]
Consequently,
\begin{equation}\label{eq:CM-value}
 \boxed{\Phi(x_0)=\frac{\sqrt{10005}}{\pi}.}
\end{equation}

\section{The nonterminating adjoint tail}

Put \(n=N+1\) and \(\delta_N=\theta-n\).  Define
\begin{equation}\label{eq:tail}
 F_N(z)=\kappa_Nz^n
 \F43\left(
 \begin{matrix}
 n,n+\frac16,n+\frac12,n+\frac56\\
 2n,2n,2n
 \end{matrix};z\right),
\end{equation}
where
\[
 \kappa_0=\frac5{72},\qquad
 \frac{\kappa_{N+1}}{\kappa_N}
 =-\frac{(6N+7)(6N+11)}
 {576(N+1)^2(2N+3)^2}.
\]
Its Euler jet is
\[
 k_N=\left(F_N,\delta_NF_N,\delta_N^2F_N,\delta_N^3F_N\right)^T.
\]

\begin{proposition}[Exact tail contiguity]\label{prop:tail-contiguity}
 For every \(N\ge0\),
 \begin{equation}\label{eq:tail-contiguity}
  \boxed{M_Nk_{N+1}=k_N.}
 \end{equation}
\end{proposition}

\begin{proof}
Put \(m=(u-1)/2=N+1\).  For the \(r\)-th row of
\(\mathcal M(u,x)\), define
\[
 P_r(t)=\sum_{s=0}^3\mathcal M(u,x)_{r+1,s+1}t^s,\qquad
 \mathcal TQ=(1-x)x\partial_xQ+(t+1)Q.
\]
The shifted tail satisfies
\[
 L_+(t)F_{N+1}=0,
\]
where
\[
L_+(t)=(1-x)t(t+u)^3+
x(t+m+1)(t+m+\tfrac76)(t+m+\tfrac32)(t+m+\tfrac{11}{6}).
\]
There is no division step: direct expansion gives the following four
denominator-cleared polynomial identities:
\begin{align}
u(3u-2)(3u+2)x(\mathcal TP_0-P_1)
 & =144(u-1)^2L_+,\label{eq:ore0}\\
\mathcal TP_1-P_2&=-L_+,\label{eq:ore1}\\
2(\mathcal TP_2-P_3)&=(-2+7x-2ux)L_+,\label{eq:ore2}\\
36(\mathcal TP_3+\ell_3P_3+\ell_2P_2+\ell_1P_1+\ell_0P_0)
 &=q_3L_+,\label{eq:ore3}
\end{align}
with
\begin{align*}
\ell_0={}&(u-1)u(3u-2)(3u+2)x/144,\\
\ell_1={}&[-576+864u-432u^2+72u^3
 +(580-872u+405u^2-36u^3)x]/72,\\
\ell_2={}&[432-432u+108u^2
 +(-436+405u-54u^2)x]/36,\\
\ell_3={}&(-12+6u+11x-2ux)/2,\\
q_3={}&-36+(536-297u+54u^2)x
       +(-567+288u-36u^2)x^2.
\end{align*}

For clarity, the coefficient calculation producing the first row is also
written without a special-function routine.  Set
\[
\begin{aligned}
A(t)&=\frac{144(u-1)^2(t+u)^3}{u(3u-2)(3u+2)},&
B(t)&=P_0(t)-\frac{A(t)}x,\\
\varrho&=-\frac{(3u-2)(3u+2)}{144(u-1)^2u^2},\\
\chi_j&=
\frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)}
     {m(m+\frac16)(m+\frac12)(m+\frac56)}\\
&\quad{}\times
\left(\frac{2m(2m+1)}{(2m+j)(2m+j+1)}\right)^3,\\
\psi_j&=
\frac{(m+j)(m+\frac16+j)(m+\frac12+j)(m+\frac56+j)}
     {(2m+j)^3(j+1)} .
\end{aligned}
\]
The constant and generic coefficient equations are
\begin{equation}\label{eq:tail-coefficients}
\varrho[-A(0)]=1,\qquad
\varrho\left[-A(j)\chi_j+
\frac{(A(j-1)+B(j-1))\chi_{j-1}}{\psi_{j-1}}\right]=1
\quad(j\ge1).
\end{equation}
Clearing the displayed nonzero factors turns
\eqref{eq:ore0}--\eqref{eq:tail-coefficients} into polynomial
equalities with every coefficient zero.  The independent script
\texttt{p28\_standalone\_equations.py} expands precisely these equalities
using only rational addition and multiplication.  Equation
\eqref{eq:tail-coefficients} gives the first row of
\eqref{eq:tail-contiguity}; applying
\eqref{eq:ore0}--\eqref{eq:ore3} successively gives the other three
Euler-jet rows.
\end{proof}

For \(N=0\), write \(y(z)=1+\sum_{k\ge1}c_kz^k\).  The coefficient of
\(z\) is
\[
c_1=\frac{(1/6)(1/2)(5/6)}{1^3}=\frac5{72},
\]
and, after shifting \(k\mapsto k+1\), both sides below have initial
coefficient \(5/72\) and consecutive-coefficient ratio
\[
\frac{(k+\frac16)(k+\frac12)(k+\frac56)}{(k+1)^3}.
\]
Hence coefficient equality, rather than a named ascension rule, gives
\begin{equation}\label{eq:ascension}
 F_0=y-1
 =\frac5{72}z
 \F43\left(
 \begin{matrix}1,\frac76,\frac32,\frac{11}{6}\\2,2,2\end{matrix};z
 \right).
\end{equation}
Because \(\mathcal P\) is the Pascal matrix,
\[
 \mathcal Pk_0=
 (y-1,\theta y,\theta^2y,\theta^3y)^T.
\]

\section{The rank-three error carrier}

Set
\[
 f=(y-1,\theta y,\theta^2y,\theta^3y)^T,\qquad
 \mathcal E_0=\frac54\mathcal P+fC,\qquad
 \mathcal E_N=\mathcal E_0G_N.
\]
The transformed hypergeometric equation is
\begin{equation}\label{eq:transformed-ode}
 72\theta^3y+108x\theta^2y+46x\theta y+5xy=0.
\end{equation}
Indeed the coefficient ratio of \(y\) gives
\[
\left[\theta^3-
z(\theta+\tfrac16)(\theta+\tfrac12)(\theta+\tfrac56)\right]y=0.
\]
Substituting \(z=-x/(1-x)\), using
\(\theta=(1-x)x\partial_x\), and multiplying by \(72(1-x)\)
expands to \eqref{eq:transformed-ode}.
Using the displayed decomposition of \(C\), equations
\eqref{eq:ascension}--\eqref{eq:transformed-ode} give
\[
 Ck_0=-\frac54.
\]
It follows that
\[
 \mathcal E_0k_0=\frac54f+f(Ck_0)=0.
\]
Proposition~\ref{prop:tail-contiguity} therefore implies
\begin{equation}\label{eq:annihilation}
 \boxed{\mathcal E_Nk_N=0\qquad(N\ge0).}
\end{equation}
The same differential equation gives the exact row relation
\begin{equation}\label{eq:row-relation}
 72(\mathcal E_N)_{3,*}
 +108x(\mathcal E_N)_{2,*}
 +46x(\mathcal E_N)_{1,*}
 +5x(\mathcal E_N)_{0,*}=0.
\end{equation}

\section{A discrete valuation lemma}

Let
\[
 H=\diag(x,1,1,1),\qquad J_N=HM_N.
\]
Every entry of \(J_N\) is regular at \(x=0\).  If \(u=2N+3\) and
\[
 a_N=\frac{144(u-1)^2}{u(3u-2)(3u+2)},\qquad
 V_N=(u^3,3u^2,3u,1),
\]
direct substitution in the authoritative matrix gives
\begin{equation}\label{eq:rank-one}
 J_N(0)=
 \begin{pmatrix}a_N\\-1\\-1\\-1\end{pmatrix}V_N.
\end{equation}
Thus \(J_N(0)\) has rank one.

The first nonconstant coefficient of the \(\F43\) in
\eqref{eq:tail} equals
\[
 c_N=\frac{u(3u-2)(3u+2)}{144(u-1)^2}=a_N^{-1}.
\]
Since \(z=-x+O(x^2)\),
\begin{equation}\label{eq:tail-direction}
 Hk_N=x^{N+2}\eta_N
 \left[
 \begin{pmatrix}1\\-c_N\\-c_N\\-c_N\end{pmatrix}
 +O(x)
 \right],\qquad \eta_N\ne0.
\end{equation}
The leading vector in \eqref{eq:tail-direction} is precisely the image
direction in \eqref{eq:rank-one}.
The other expansion required below is equally direct.  Since
\[
F_{N+1}=\kappa_{N+1}z^{N+2}(1+O(z)),\qquad
\delta_{N+1}=\theta-(N+2),
\]
the shifted derivatives kill the leading monomial, and therefore
\begin{equation}\label{eq:next-tail-direction}
k_{N+1}=x^{N+2}\widetilde\eta_N
\left(\e_1+xs_N+O(x^2)\right),
\qquad\widetilde\eta_N\ne0.
\end{equation}
Moreover, \eqref{eq:rank-one} gives
\[
J_N(0)\e_1=u^3(a_N,-1,-1,-1)^T\ne0.
\]

\begin{lemma}[DVR step, including the extra first-column zero]
\label{lem:dvr}
Let \(R_0=\Q[[x]]\), \(H=\diag(x,1,1,1)\), and suppose
\[
 E=x^NLH,\qquad L\in\operatorname{Mat}_4(R_0),\qquad Ek=0.
\]
Assume \(J=HM\in\operatorname{Mat}_4(R_0)\), \(Mk^+=k\), and
\[
\begin{aligned}
 k^+&=x^r\alpha(\e_1+xs+O(x^2)),\\
 Hk&=x^r\beta(v_0+xv_1+O(x^2)),
\end{aligned}
\]
with \(\alpha\beta\ne0\).  If \(J(0)\) has rank one,
\(\operatorname{im}J(0)=\Q v_0\), and \(J(0)\e_1\ne0\), then
\[
 EM=x^{N+1}L^+H
\]
for some \(L^+\in\operatorname{Mat}_4(R_0)\).
\end{lemma}

\begin{proof}
Absorb \(\beta/\alpha\) into \(v_0,v_1\).  From \(Jk^+=Hk\),
\[
 J(\e_1+xs+O(x^2))=v_0+xv_1+O(x^2).
\]
Write \(L=L_0+xL_1+\cdots\).  The equation \(L(Hk)=0\) gives
\[
 L_0v_0=0,\qquad L_0v_1+L_1v_0=0.
\]
Since \(J(0)\) has image \(\Q v_0\), \(L_0J(0)=0\), so \(LJ\) is
entrywise divisible by \(x\).  The coefficient of \(x\) in its first
column is
\[
 L_0(v_1-J(0)s)+L_1v_0=-L_0J(0)s=0.
\]
Hence that column is divisible by \(x^2\).  Therefore
\[
 L^+=x^{-1}LJH^{-1}
\]
is regular and \(EM=x^{N+1}L^+H\).
\end{proof}

\begin{proposition}[All-\(N\) Pad\'e divisibility]\label{prop:divisibility}
For every \(N\ge0\), there is
\(L_N\in\operatorname{Mat}_4(\Q[[x]])\) such that
\begin{equation}\label{eq:divisibility}
 \boxed{\mathcal E_N=x^NL_NH.}
\end{equation}
Thus every row of \(\mathcal E_N\) has componentwise valuations at least
\[
 (N+1,N,N,N).
\]
The last row has the stronger valuations
\[
 (N+2,N+1,N+1,N+1).
\]
\end{proposition}

\begin{proof}
Every component of \(f\) is \(O(x)\), while \(C\) has only a simple pole.
All four components of \(f\) have leading term \(-5x/72\).
Consequently the constant term in the first column of \(fC\) is
\(-5/4\), cancelling the first component of every row of
\((5/4)\mathcal P\).  Hence \(\mathcal E_0=L_0H\).

Apply Lemma~\ref{lem:dvr} inductively, using
\eqref{eq:annihilation}, \eqref{eq:rank-one},
\eqref{eq:tail-direction}, \eqref{eq:next-tail-direction}, the displayed
nonzero first column of \(J_N(0)\), and \(M_Nk_{N+1}=k_N\).
The stronger last-row assertion follows from
\eqref{eq:row-relation}.
\end{proof}

\section{The terminating denominator}

For the first column put
\[
 q_N(x)=CG_N\e_1,\qquad n=N+1,\qquad Q_N(x)=x^nq_N(x),
\]
and define
\[
 \widehat Q_N(z)
 =(1-z)^nQ_N\!\left(-\frac{z}{1-z}\right).
\]
Since \(x=-z/(1-z)\), this is also the first component of
\((-z)^nCG_N\).

\begin{proposition}[Exact terminating denominator]
\label{prop:terminating}
For every \(N\ge0\),
\begin{equation}\label{eq:qhat}
\frac{\widehat Q_N(z)}{\alpha_n}
=\F43\left(
\begin{matrix}
-n,-n-\frac16,-n-\frac12,-n-\frac56\\
1-2n,1-2n,1-2n
\end{matrix};z
\right),
\end{equation}
where
\begin{equation}\label{eq:normalization}
 \alpha_1=18,\qquad
 \frac{\alpha_{n+1}}{\alpha_n}
 =\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}.
\end{equation}
Here and below the hypergeometric expression denotes the unambiguous finite
sum over \(0\le k\le n\); it terminates before any lower Pochhammer symbol
can vanish.
\end{proposition}

\begin{proof}
Write \(p_n=\widehat Q_{n-1}\) for \(n\ge1\).  We use coefficient
induction, because normalization at \(z=0\) alone
would not select a unique solution of the fourth-order equation.  To expose
the matrix-to-scalar bridge, write
\[
\mathcal L_n(t)=t(t+2n-1)^3
-z(t+n)(t+n+\tfrac16)(t+n+\tfrac12)(t+n+\tfrac56)
=\sum_{j=0}^4c_jt^j
\]
and define, for operator polynomials with coefficients on the left,
\[
\Theta Q=z\partial_zQ+tQ.
\]
The four horizontal components are reconstructed from the first by
\[
\pi_0=1,\qquad
\pi_3=\frac{c_4}{c_0}t,\qquad
\pi_2=\frac{c_3}{c_4}\pi_3-\Theta\pi_3,\qquad
\pi_1=\frac{c_2}{c_4}\pi_3-\Theta\pi_2.
\]
The remaining horizontal equation is the direct factorization
\[
\begin{aligned}
\Theta\pi_1+1-\frac{c_1}{c_4}\pi_3
={}&-\frac{72}{n(2n+1)(6n+1)(6n+5)z}\bigl[
t(t-2n)^3\\
&\hspace{4em}
-z(t-n)(t-n-\tfrac16)
(t-n-\tfrac12)(t-n-\tfrac56)\bigr].
\end{aligned}
\]
Let
\[
\mathcal C_n(z)=
\begin{pmatrix}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
-c_0/c_4&-c_1/c_4&-c_2/c_4&-c_3/c_4
\end{pmatrix}.
\]
The propagation of the reconstructed row is the following sixteen-entry
identity:
\begin{equation}\label{eq:full-gauge}
\mathcal C_n[-z\mathcal M(2n+1,-z/(1-z))]
-\theta[-z\mathcal M(2n+1,-z/(1-z))]
-[-z\mathcal M(2n+1,-z/(1-z))]\mathcal C_{n+1}=0.
\end{equation}
Finally, direct contraction with the displayed matrix gives
\begin{equation}\label{eq:matrix-scalar-bridge}
\sum_{r=0}^3\pi_r(t)
\left[-z\mathcal M(2n+1,-z/(1-z))\right]_{r+1,1}
=d_0(t)+zd_1(t).
\end{equation}
These are identities in \(\Q(n,z)(t)\).  The factors depending on the
integer \(n\) are nonzero for \(n\ge1\); no value is obtained by dividing
at \(z=0\), because the verifier cross-multiplies first and the reconstructed
base functions have removable limits there.  The mandatory
sparse-polynomial verifier checks every
reconstruction equation, the closing factorization, and
\eqref{eq:full-gauge}--\eqref{eq:matrix-scalar-bridge} by cross
multiplication.  For any reconstructed horizontal row these identities give
\begin{equation}\label{eq:terminating-step}
p_{n+1}
 =\bigl(d_0(\theta)+zd_1(\theta)\bigr)p_n,
\end{equation}
where
\[
d_0(t)=
\frac{72(2n+1)^2(2n-t)^3}{n(6n+1)(6n+5)}
\]
and
\[
d_1(t)=-\frac{P(n,t)}{n(2n+1)(6n+1)(6n+5)}
\]
with
\begin{align*}
P(n,t)={}&-5n-76n^2+1404n^3+4360n^4+4320n^5+1440n^6\\
&+(5+127n-1760n^2-6536n^3-7632n^4-3024n^5)t\\
&+(-51+659n+3086n^2+4500n^3+2232n^4)t^2\\
&+(-72-432n-864n^2-576n^3)t^3.
\end{align*}
Thus \eqref{eq:terminating-step} is the scalar form of the displayed
matrix recurrence; no differential-equation uniqueness is used below.

Let
\[
h_{n,k}=
\frac{(-n)_k(-n-\frac16)_k(-n-\frac12)_k(-n-\frac56)_k}
{(1-2n)_k^3k!}
\quad(0\le k\le n),
\qquad
\nu_n=\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}.
\]
For \(n\ge1\), clearing the displayed nonzero denominators gives exactly
\begin{align}
d_0(0)&=\nu_n,\label{eq:term-constant}\\
d_0(k)+d_1(k-1)\frac{h_{n,k-1}}{h_{n,k}}
&=\nu_n\frac{h_{n+1,k}}{h_{n,k}}
\qquad(1\le k\le n),\label{eq:term-generic}\\
d_1(n)&=-\nu_n
\frac{(n+\frac76)(n+\frac32)(n+\frac{11}{6})}
{8(2n+1)^3}.\label{eq:term-top}
\end{align}
The base row is not assumed: from the first component of \(C\),
\[
q_0(x)=\frac{18}{x}+\frac{159}{4},\qquad
Q_0(x)=xq_0(x)=18+\frac{159}{4}x,
\]
and therefore
\[
p_1(z)=(1-z)Q_0\!\left(-\frac{z}{1-z}\right)
=18\left(1-\frac{77}{24}z\right).
\]
Direct application of the four displayed \(\pi_r\) operators gives the
four base identities
\begin{equation}\label{eq:base-horizontal-row}
-zC\!\left(-\frac{z}{1-z}\right)
=\left.(\pi_0(\theta)p_1,\pi_1(\theta)p_1,
\pi_2(\theta)p_1,\pi_3(\theta)p_1)\right|_{n=1}.
\end{equation}
The fourth base equation is also direct:
\[
\left[
\theta(\theta-2)^3
-z(\theta-1)(\theta-\tfrac76)
(\theta-\tfrac32)(\theta-\tfrac{11}{6})
\right]p_1=0.
\]
Equivalently, if \(H_1=-zC(-z/(1-z))\), then all four components of
\[
\theta H_1+H_1\mathcal C_1=0
\]
vanish.  The mandatory verifier checks both forms independently.
Moreover,
\[
\deg((d_0(\theta)+zd_1(\theta))p_n)\le n+1.
\]
Equations \eqref{eq:term-constant}--\eqref{eq:term-top} therefore prove,
coefficient by coefficient, that the right side of
\eqref{eq:terminating-step} is
\(\alpha_{n+1}\sum_{k=0}^{n+1}h_{n+1,k}z^k\).
Starting from \eqref{eq:base-horizontal-row},
\eqref{eq:full-gauge} propagates the horizontal form at every step.
This proves \eqref{eq:qhat} and \eqref{eq:normalization} for the actual
row \(CG_N\), not merely for a scalar surrogate.
The mandatory standalone checker independently expands the cleared
identities and rejects any nonzero coefficient.
\end{proof}

\begin{corollary}[Positivity at the CM point]\label{cor:positivity}
At \(z_0=-1/53360^3\),
\[
 \widehat Q_N(z_0)\ge\alpha_n>0.
\]
Moreover,
\[
 \alpha_n\ge18\cdot29^N(N!)^2.
\]
\end{corollary}

\begin{proof}
For \(0\le k\le n\), the coefficient of \(z^k\) in
\eqref{eq:qhat} has sign \((-1)^k\).  Since \(z_0<0\), every summand is
nonnegative.  Also
\[
\frac{576n^2(2n+1)^2}{(6n+1)(6n+5)}-29n^2
=\frac{n^2(1260n^2+1260n+431)}
{(6n+1)(6n+5)}>0.
\]
Iterating \eqref{eq:normalization} proves the lower bound.
\end{proof}

\section{From formal contact to convergence at
\texorpdfstring{\(x_0\)}{x0}}

This step is included to rule out a beyond-all-orders ambiguity.
For \(r=0,1\), let
\[
 E_{N,r}(x)=(\mathcal E_N)_{r,1},\qquad
 \mathcal R_{N,r}(x)=x^nE_{N,r}(x).
\]
Proposition~\ref{prop:divisibility} says that
\(\mathcal R_{N,r}\) has a zero of order at least \(2n\).

Choose \(r_0=1/4\).  On \(|x|=r_0\), \(|z|\le1/3\).  The coefficients of
\(y\) have modulus at most one, so
\[
 |y-1|\le\frac12,\qquad
 |\theta^jy|\le\sum_{k\ge1}k^3(1/3)^k=\frac{33}{8}<5
 \quad(1\le j\le3).
\]
Termwise estimates give \(\|\mathcal E_0\|_\infty<6000\).

For \(m\ge1\), put
\[
 D(m)=\diag(1,m,m^2,m^3),\qquad
 \mathcal B_m=D(m)^{-1}M_mD(m+1)/(m+1)^2.
\]
On \(|x|=1/4\), direct estimates of the authoritative entries give
\[
 |(M_m)_{ij}|\le10^4u^{\,i+2-j},\qquad u=2m+3,
\]
and therefore
\[
 |(\mathcal B_m)_{ij}|
 \le10^4\left(\frac um\right)^{i-1}
          \left(\frac u{m+1}\right)^{3-j}.
\]
For \(m\ge1\), \(u/m\le5\) and \(u/(m+1)\le5/2\).  Because the exponent
\(3-j\) is negative when \(j=4\), we use the exact case split
\[
\left(\frac{u}{m+1}\right)^{3-j}\le
\begin{cases}
(5/2)^{3-j}\le25/4,&j\le3,\\[1mm]
1/2,&j=4,
\end{cases}
\]
where the second line follows from \(u/(m+1)=2+1/(m+1)\ge2\).
Thus every entry is at most \(7{,}812{,}500\), and summing each row gives
\[
 \|\mathcal B_m\|_\infty\le31{,}250{,}000<4\cdot10^8,\qquad
 \|M_0\|_\infty<10^7.
\]
The balancing telescopes:
\[
 G_N=M_0(N!)^2\mathcal B_1\cdots
 \mathcal B_{N-1}D(N)^{-1}.
\]
It follows that, for \(N\ge1\),
\begin{equation}\label{eq:circle-bound}
 \max_{|x|=1/4}|\mathcal R_{N,r}(x)|
 \le6\cdot10^{10}(N!)^2(4\cdot10^8)^{N-1}(1/4)^n.
\end{equation}
We now verify the analytic hypothesis behind the next step.  On
\(|x|\le1/4\),
\[
\left|-\frac{x}{1-x}\right|\le\frac13<1,
\]
so the series defining \(y\) and its first three Euler derivatives are
holomorphic on a neighborhood of the closed disk.  Because \(f=O(x)\),
the simple pole of \(C\) cancels in \(fC\), so \(\mathcal E_0\) is
holomorphic there.  Inspection of \eqref{eq:deformed-matrix} shows that
each \(M_m\) is holomorphic off \(x=0\) in this disk and has at most a
simple pole at \(0\).  Hence
\(\mathcal R_{N,r}=x^n(\mathcal E_N)_{r,1}\), with \(n=N+1\), is
holomorphic on the disk after removing its possible singularity at zero.
Proposition~\ref{prop:divisibility} strengthens its order there to at
least \(2n\).  Therefore
\(\mathcal R_{N,r}(x)/x^{2n}\) has a removable singularity at zero and
is holomorphic on a neighborhood of the closed disk.
Applying the maximum principle to
\(\mathcal R_{N,r}(x)/x^{2n}\) gives
\begin{equation}\label{eq:cauchy}
 |\mathcal R_{N,r}(x_0)|
 \le6\cdot10^{10}(N!)^2(4\cdot10^8)^{N-1}(1/4)^n
 (4x_0)^{2n}.
\end{equation}

Because \(1-z=1/(1-x)\),
\[
 Q_N(x_0)=(1-x_0)^n\widehat Q_N(z_0).
\]
Corollary~\ref{cor:positivity} yields
\[
 Q_N(x_0)\ge
 18\cdot29^N(N!)^2(1-x_0)^n.
\]
Combining this with \eqref{eq:cauchy}, we obtain
\begin{equation}\label{eq:geometric-error}
 \left|\frac{E_{N,r}(x_0)}{q_N(x_0)}\right|
 =\left|\frac{\mathcal R_{N,r}(x_0)}{Q_N(x_0)}\right|
 \le K_0\,\beta(x_0)^N,
\end{equation}
where \(K_0<\infty\) is independent of \(N\) and
\[
 \beta(x_0)=
 \frac{4\cdot10^8}{29}
 \frac{x_0^2}{(1/4)(1-x_0)}
 =\frac{3125}{1307443596565949700399927}
 <4\cdot10^{-19}<1.
\]
Therefore
\begin{equation}\label{eq:error-vanish}
 \frac{E_{N,0}(x_0)}{q_N(x_0)}\longrightarrow0,\qquad
 \frac{E_{N,1}(x_0)}{q_N(x_0)}\longrightarrow0.
\end{equation}

\section{Identification of the first-column limit}

By \eqref{eq:seed-identities} and the definition of \(\Phi\),
\[
 A_0-\Phi A_1
 =-A(\mathcal E_0)_{0,*}-B(\mathcal E_0)_{1,*}.
\]
Multiplying by \(G_N\e_1\), dividing by
\(A_1G_N\e_1=S q_N\), and using
\eqref{eq:error-vanish}, we get
\[
 \lim_{N\to\infty}
 \frac{A_0G_N\e_1}{A_1G_N\e_1}
 =\Phi(x_0).
\]
Equation \eqref{eq:CM-value} therefore proves
\begin{equation}\label{eq:first-column}
 \boxed{
 \lim_{N\to\infty}\frac{P_{N,1}}{Q_{N,1}}
 =\frac{\sqrt{10005}}{\pi}.}
\end{equation}

\section{The other three official columns}

All matrices in this section are evaluated at \(x=x_0=1/R\).
For \(m\ge1\), retain
\[
D_m=\diag(1,m,m^2,m^3),\qquad
\mathcal B_m=D_m^{-1}M_mD_{m+1}/(m+1)^2,
\]
and, for any row \(a\), put
\[
Z_m(a)=\frac{aG_mD_m}{(m!)^2}.
\]
Then the balancing gives the exact recurrence
\begin{equation}\label{eq:balanced-row-recurrence}
Z_{m+1}(a)=Z_m(a)\mathcal B_m.
\end{equation}
Substitution in the displayed matrix shows, entry by entry, that
\(\mathcal B_m-\mathcal S=O(m^{-1})\), where
\[
 \mathcal S=
\begin{pmatrix}
64R-44&96R-54&48R-17&8R\\
-8&-12&-6&-1\\
R^{-1}&-4R^{-1}&-6R^{-1}&-2R^{-1}\\
2R^{-2}&(17R-8)R^{-2}&4(5R-3)R^{-2}&(6R-4)R^{-2}
\end{pmatrix}.
\]
Its characteristic polynomial is \(Q_R(t)/R^2\), where
\[
\begin{aligned}
Q_R(t)={}&R^2t^4-(64R^3-56R^2-4)t^3\\
&+(48R^2-262R+220)t^2-(12R-8)t+1.
\end{aligned}
\]
On \(|t|=1\), the cubic coefficient strictly dominates the sum of the
other four coefficient magnitudes, because
\[
 (64R^3-56R^2-4)-(49R^2-250R+213)
 =64R^3-105R^2+250R-217>0.
\]
To count the roots without an irreducibility or root-finder call, consider
\[
H_s(t)=-(64R^3-56R^2-4)t^3+
s\{R^2t^4+(48R^2-262R+220)t^2-(12R-8)t+1\}.
\]
The strict inequality above gives \(H_s(t)\ne0\) for
\(|t|=1\), \(0\le s\le1\).  Thus the winding number of
\(H_s(e^{i\vartheta})\) about zero cannot change with \(s\); at \(s=0\)
it is \(3\).  Hence \(Q_R=H_1\) has three zeros in \(|t|<1\) and one,
counted with multiplicity, in \(|t|>1\).  Moreover
\[
Q_R(1)=-(64R^3-105R^2+274R-233)<0,\qquad
\lim_{t\to+\infty}Q_R(t)=+\infty.
\]
Consequently the unique exterior zero is a simple real number
\(\rho>1\).

\begin{lemma}[Explicit dominant-product dichotomy]
\label{lem:dominant-product}
Fix \(\tau\) with
\[
\max_{\lambda\ne\rho}|\lambda|<\tau<1,
\]
Then there exist a late index \(m_0\), a linear functional \(\Lambda\) on
rows, nonzero scalars \(L_m\) independent of the row, and a left
\(\rho\)-eigenvector \(w\) of \(\mathcal S\) such that the following
alternatives hold:
\begin{align}
\Lambda(a)=0&\quad\Longrightarrow\quad
\|Z_m(a)\|\le K_a\tau^{m-m_0},\label{eq:exceptional-decay}\\
\Lambda(a)\ne0&\quad\Longrightarrow\quad
Z_m(a)=\Lambda(a)L_m\bigl(w+o_a(1)\bigr).
\label{eq:dominant-asymptotic}
\end{align}
\end{lemma}

\begin{proof}
Choose an invertible \(P\) with
\[
P^{-1}\mathcal SP=
\begin{pmatrix}\rho&0\\0&A\end{pmatrix},
\qquad\operatorname{spr}(A)<1.
\]
Choose \(\operatorname{spr}(A)<\theta<\tau\).  For stable rows define
\[
\|\beta\|_\theta=\sum_{k=0}^{\infty}
\theta^{-k}\|\beta A^k\|_0.
\]
The finite Jordan identity
\[
J_\lambda^k=\sum_{\ell=0}^{s-1}
\binom{k}{\ell}\lambda^{k-\ell}N^\ell
\]
gives, for \(\operatorname{spr}(A)<\eta<\theta\),
\(\|A^k\|_0\le Ck^2\eta^k\); hence the series converges and
\[
\|\beta A\|_\theta
=\theta\sum_{k=1}^{\infty}\theta^{-k}\|\beta A^k\|_0
\le\theta\|\beta\|_\theta.
\]
Use the dual norm for stable columns.

Write
\[
T_m=P^{-1}\mathcal B_mP=
\begin{pmatrix}a_m&b_m\\c_m&E_m\end{pmatrix}.
\]
Then \(a_m\to\rho\), \(b_m,c_m\to0\), and \(E_m\to A\).
Choose
\[
\theta<d_*<\tau<1<a_*<\rho.
\]
For sufficiently large \(m_0\), a positive \(\epsilon\) makes, for
\(m\ge m_0\),
\begin{gather}
|a_m|\ge a_*,\quad\|E_m\|\le d_*,
\quad\|b_m\|,\|c_m\|\le\epsilon,\label{eq:block-bounds}\\
d_*+\epsilon<\tau,\quad a_*-\epsilon>1,\quad
d_*+\epsilon<a_*-\epsilon,\label{eq:block-separation}\\
q:=\frac{d_*}{a_*-\epsilon}
+\frac{(d_*+\epsilon)\epsilon}{(a_*-\epsilon)^2}<1.
\label{eq:graph-contraction-constant}
\end{gather}
For stable columns \(\|h\|\le1\), set
\[
\Psi_m(h)=\frac{E_mh-c_m}{a_m-b_mh}.
\]
Equations \eqref{eq:block-bounds}--\eqref{eq:block-separation} give
\[
\|\Psi_m(h)\|
\le\frac{d_*+\epsilon}{a_*-\epsilon}<1.
\]
For two such columns,
\[
\Psi_m(h)-\Psi_m(k)
=\frac{E_m(h-k)}{a_m-b_mh}
+\frac{(E_mk-c_m)b_m(h-k)}
{(a_m-b_mh)(a_m-b_mk)},
\]
so \eqref{eq:graph-contraction-constant} gives
\[
\|\Psi_m(h)-\Psi_m(k)\|\le q\|h-k\|.
\]

For \(M>m\), set \(h_M^{(M)}=0\) and recurse backward by
\(h_j^{(M)}=\Psi_j(h_{j+1}^{(M)})\).  Then, for \(M'>M\),
\[
\|h_m^{(M')}-h_m^{(M)}\|\le2q^{M-m}.
\]
Thus \(h_m=\lim_{M\to\infty}h_m^{(M)}\) exists and satisfies
\begin{equation}\label{eq:graph-invariance}
h_ma_m+c_m=(h_mb_m+E_m)h_{m+1}.
\end{equation}
The defining recurrence gives the explicit bound
\[
\|h_m\|\le
\frac{\|E_m\|\|h_{m+1}\|+\|c_m\|}
{|a_m|-\|b_m\|}.
\]
Together with \(\|A\|_\theta/\rho<1\), this gives
\[
\limsup_{m\to\infty}\|h_m\|
\le\frac{\theta}{\rho}\limsup_{m\to\infty}\|h_m\|,
\qquad\text{hence}\qquad h_m\to0.
\]

Write
\[
U_m(a)=Z_m(a)P=(\alpha_m,\beta_m),\qquad
\xi_m=\alpha_m-\beta_mh_m,\qquad
d_m=a_m-b_mh_{m+1}.
\]
Using \eqref{eq:graph-invariance} in
\(U_{m+1}=U_mT_m\) gives the exact scalar equation
\[
\xi_{m+1}=d_m\xi_m.
\]
Define the composed seed functional and product
\[
\Lambda(a)=\alpha_{m_0}(a)-\beta_{m_0}(a)h_{m_0},\qquad
L_m=\prod_{\ell=m_0}^{m-1}d_\ell.
\]
Both are now explicit, \(\Lambda\) is linear, and
\(\xi_m=\Lambda(a)L_m\).  Equations
\eqref{eq:block-bounds}--\eqref{eq:block-separation} ensure
\(d_m\ne0\).

If \(\Lambda(a)=0\), then \(\alpha_m=\beta_mh_m\) and
\[
\beta_{m+1}=\beta_m(E_m+h_mb_m),\qquad
\|\beta_{m+1}\|<(d_*+\epsilon)\|\beta_m\|
<\tau\|\beta_m\|,
\]
which proves \eqref{eq:exceptional-decay}.

If \(\Lambda(a)\ne0\), put \(r_m=\beta_m/\xi_m\).  Exact substitution
gives
\[
r_{m+1}
=\frac{b_m+r_m(E_m+h_mb_m)}{d_m}.
\]
Here \(b_m/d_m\to0\) and
\((E_m+h_mb_m)/d_m\to A/\rho\), whose norm is below one.
Enlarge \(m_0\) once more, redefining \(\Lambda\) and \(L_m\) from this
new index, so that for some \(q_1<1\),
\[
\left\|\frac{E_m+h_mb_m}{d_m}\right\|\le q_1
\qquad(m\ge m_0).
\]
Then, for every \(m\ge m_0\),
\[
\|r_m\|\le q_1^{m-m_0}\|r_{m_0}\|
+\sum_{\ell=m_0}^{m-1}q_1^{m-1-\ell}
\left\|\frac{b_\ell}{d_\ell}\right\|\longrightarrow0.
\]
Since \(\alpha_m/\xi_m=1+r_mh_m\to1\),
\[
U_m(a)=\Lambda(a)L_m\bigl((1,0)+o_a(1)\bigr).
\]
Multiplying by \(P^{-1}\) proves
\eqref{eq:dominant-asymptotic} with
\[
w=(1,0)P^{-1},\qquad w\mathcal S=\rho w.
\]
\end{proof}

No GCD or irreducibility decision is needed to show that all four
coordinates of \(w\) are nonzero.  Define the polynomial row \(w(t)\) by
\[
\begin{aligned}
w_1(t)={}&Rt(Rt^2-7)+12R^2t^2+4t^2+44t+10,\\
\frac{w_2(t)}2={}&R^2t((48R-27)t-28)
+(194R-108)t+40R-23,\\
w_3(t)={}&R^2t((48R-17)t-8)
+(198R-68)t+71R-32,\\
w_4(t)={}&2R(8+(17+3R)t+4R^2t^2).
\end{aligned}
\]
Direct multiplication gives the exact polynomial identity
\[
w(t)(tI-\mathcal S)=(Q_R(t),0,0,0).
\]
At \(t=\rho\), this is a left \(\rho\)-eigenvector.  The exterior
eigenspace is one-dimensional, so the vector in
Lemma~\ref{lem:dominant-product} may be rescaled to \(w(\rho)\), with the
inverse rescaling absorbed into \(L_m\).  Since \(R>7\) and \(\rho>1\),
every displayed grouping is positive.  Thus \(w_j(\rho)>0\) for
\(j=1,2,3,4\); below we abbreviate \(w_j=w_j(\rho)\).

Undoing the balancing in \eqref{eq:dominant-asymptotic} gives, whenever
\(\Lambda(a)\ne0\),
\begin{equation}\label{eq:all-column-asymptotic}
aG_m\e_j=(m!)^2m^{-(j-1)}
\Lambda(a)L_m\bigl(w_j+o_a(1)\bigr).
\end{equation}
It remains to verify that the two official rows are not exceptional.
The positivity estimate and \(Q_m=x_0^{m+1}q_m\) give
\begin{equation}\label{eq:q-lower}
\frac{q_m(x_0)}{(m!)^2}
\ge18(R-1)\,[29(R-1)]^m.
\end{equation}
If \(\Lambda(C)=0\), the first coordinate of
\eqref{eq:exceptional-decay} would contradict \eqref{eq:q-lower}.
Therefore
\[
\Lambda(A_1)=S\Lambda(C)\ne0.
\]
If \(\Lambda(A_0)=0\), then
\eqref{eq:exceptional-decay}, \eqref{eq:dominant-asymptotic}, and
\(|d_m|>1\) for large \(m\) would make the first-column quotient tend
to zero, contradicting \eqref{eq:first-column}.  Hence
\(\Lambda(A_0)\ne0\) as well.

Because \(w_j>0\), equation \eqref{eq:all-column-asymptotic} first proves
that every \(Q_{m,j}\) is nonzero for all sufficiently large \(m\), and
only then permits division:
\[
\lim_{m\to\infty}\frac{P_{m,j}}{Q_{m,j}}
=\frac{\Lambda(A_0)}{\Lambda(A_1)}
=\lim_{m\to\infty}\frac{P_{m,1}}{Q_{m,1}}
=\frac{\sqrt{10005}}{\pi}.
\]
We conclude:

\begin{theorem}[Ramanujan Challenge Problem 2.8]\label{thm:main}
For every official column \(j=1,2,3,4\),
\[
 \boxed{
 \lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
 =\frac{\sqrt{10005}}{\pi}.}
\]
Equivalently,
\[
 \boxed{
 \lim_{N\to\infty}\frac{Q_{N,j}}{P_{N,j}}
 =\frac{\pi}{\sqrt{10005}}.}
\]
\end{theorem}

\section{Reproducibility map}

The proof package contains the following certificates.

\begin{center}
\begin{tabular}{
  >{\raggedright\arraybackslash}p{0.41\textwidth}
  p{0.49\textwidth}}
\toprule
File & Exact obligation\\
\midrule
\path{p28_full_closure_certificate.wl}
& Optional independent Wolfram cross-check of the differential gauge and
hypergeometric closure.\\
\path{p28_standalone_equations.py}
& Mandatory dependency-free expansion of the four cleared Ore
factorizations, the tail coefficient equations, the terminating base,
generic, and top identities, ascension, and the \(\F32\) equation.\\
\path{p28_dominant_product_algebra.py}
& Mandatory dependency-free verification of
\(\mathcal B_m=\mathcal S+O(m^{-1})\), the characteristic polynomial,
the root-separation inequalities, the left-eigenvector identity, and
the four positive coordinate rewrites.\\
\path{p28_kernel_contiguity_certificate.sage}
& Optional independent Sage check of the four displayed
factorizations; it performs no Ore division.\\
\path{p28_lattice_hypotheses_certificate.sage}
& Optional exact cross-check of the rank-one factorization, tail
direction, and transformed ODE identities.\\
\path{p28_convergence_constants.py}
& Exact rational verification of the coefficient bounds,
\(\alpha_{n+1}/\alpha_n\ge29n^2\), and \(\beta(x_0)<1\).\\
\path{all_four_columns_certificate.sage}
& Optional Sage cross-check of the balanced limit and exterior-root
algebra; the analytic contraction is proved in Lemma~\ref{lem:dominant-product}.\\
\path{p28_parametric_pade_probe.py}
& Diagnostic finite exact regression of the predicted valuations; it is
not used as proof of an all-\(N\) statement.\\
\bottomrule
\end{tabular}
\end{center}

The two mandatory algebra checkers use only the standard library's
\texttt{fractions.Fraction}, sparse coefficient dictionaries, and explicit
addition, multiplication, differentiation, and determinant expansion.
They invoke no division algorithm, factorizer, root finder, special-function
library, or numerical sample.  Wolfram Language and SageMath are optional
independent cross-checks, not a trust requirement.

\section*{References}
\addcontentsline{toc}{section}{References}

\begin{enumerate}[label={[\arabic*]}]
\item D. V. Chudnovsky and G. V. Chudnovsky,
  ``Approximations and complex multiplication according to Ramanujan,''
  in \emph{Ramanujan Revisited}, Academic Press, 1988, pp.~375--472.
\item L. Milla,
  ``A detailed proof of the Chudnovsky formula with means of basic
  complex analysis,'' arXiv:1809.00533v6, 2021,
  \href{https://arxiv.org/abs/1809.00533}{arXiv:1809.00533}.
\item J. L. Fields,
  ``Rational approximations to generalized hypergeometric functions,''
  \emph{Mathematics of Computation} \textbf{19} (1965), 606--624,
  \href{https://doi.org/10.1090/S0025-5718-1965-0194620-7}
  {doi:10.1090/S0025-5718-1965-0194620-7}.
\item Yu. V. Nesterenko,
  ``Hermite--Pad\'e approximants of generalized hypergeometric
  functions,'' \emph{Russian Acad. Sci. Sb. Math.}
  \textbf{83} (1995), 189--219.
\item The Ramanujan Machine,
  \href{https://www.ramanujanmachine.com/ramanujan-challenge/}
  {Ramanujan Challenge}, Problem 2.8.
\end{enumerate}

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