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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 authoritative matrix, while the first denominator is a
terminating adjoint \(\F43\).  Their common differential gauge gives an
all-\(N\) Pad\'e divisibility theorem.  Positivity of the terminating
denominator at the negative CM point, together with a balanced-transfer
Cauchy estimate, turns that formal divisibility into a direct fixed-point
convergence proof.  The symbolic contiguity and adjoint identities are
included as reproducible Wolfram Language and SageMath certificates.
\end{abstract}

\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 \(M_N=M(N,x)\) is the authoritative transfer matrix in the analytic
deformation
\[
 236337691420383\ \longmapsto\ \frac{14/x-567}{9}.
\]
At \(x=x_0\), this is the exact identity
\(236337691420383=(14R-567)/9\).  Thus every later use of Cauchy's theorem
concerns this explicitly defined rational \(x\)-family.
The complete entries of \(M(N,x)\) appear verbatim in the accompanying
CAS certificates.

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.

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}

The classical Chudnovsky identity is
\[
 \frac1\pi=
 \frac{12}{640320^{3/2}}
 \sum_{k=0}^{\infty}
 \frac{(6k)!}{(3k)!(k!)^3}
 (A+Bk)(-640320^{-3})^k.
\]
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}
The first row is verified coefficientwise from the ratio of consecutive
\(\F43\) coefficients.  For the other rows, let \(t=\delta_{N+1}\).
The shifted tail satisfies
\[
 \left[
 (1-x)t(t+u)^3+
 x(t+n+1)(t+n+\tfrac76)(t+n+\tfrac32)(t+n+\tfrac{11}{6})
 \right]F_{N+1}=0,
\]
where \(u=2N+3\).  Each of the remaining three row differences is divided
by this degree-four Ore polynomial; its remainder is identically zero in
\(\Q(N,x)[t]\).  The exact coefficient identity, the three Ore divisions,
and the normalization ratio are checked in
\texttt{p28\_full\_closure\_certificate.wl} and
\texttt{p28\_kernel\_contiguity\_certificate.sage}.
\end{proof}

For \(N=0\), the standard ascension identity 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}
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}.

\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}, 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}
The proof is an exact differential-gauge calculation in
\(\Q(n,z)\).  The nonterminating tail
\[
 \F43\left(
 \begin{matrix}n,n+\frac16,n+\frac12,n+\frac56\\
 2n,2n,2n
 \end{matrix};z\right)
\]
has a \(4\times4\) Euler companion system.  Direct simplification gives
\[
 \mathcal C_n(z)\,[-zM(2n+1,-z/(1-z))]
 -\theta[-zM(2n+1,-z/(1-z))]
 -[-zM(2n+1,-z/(1-z))]\mathcal C_{n+1}(z)=0.
\]
The transformed seed \(-zC(-z/(1-z))\) is a horizontal adjoint row.
Eliminating its other three coordinates from the horizontal equation
produces exactly
\[
 \left[
 \theta(\theta-2n)^3
 -z(\theta-n)(\theta-n-\tfrac16)
       (\theta-n-\tfrac12)(\theta-n-\tfrac56)
 \right]\widehat Q_N=0.
\]
The analytic solution normalized at \(z=0\) is the terminating
\(\F43\) in \eqref{eq:qhat}.

For completeness, the CAS certificate does not rely only on this
differential equation.  It computes the actual one-step scalar operator
and verifies its generic coefficient identity, its \(k=0\) normalization,
and the separate top boundary \(k=n+1\).  Every remainder simplifies
identically to zero.  This proves the statement for all \(n\), not merely
for sampled values.
\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 \(u/(m+1)\le5/2\); summing four entries in
each row gives the deliberately loose uniform bound
\[
 \|\mathcal B_m\|_\infty\le4\cdot10^8,\qquad
 \|M_0\|_\infty<10^7.
\]
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}
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 C(x_0)\,\beta(x_0)^N,
\end{equation}
where \(C(x_0)<\infty\) 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}

For completeness, we recall the exact finite-frame reduction already used
to establish convergence of the recurrence.  The balanced transfer tends
to
\[
 \mathcal S=
\begin{pmatrix}
64R-44&96R-54&48R-17&8R\\
-8&-12&-6&-1\\
R^{-1}&-4R^{-1}&-6R^{-1}&-2R^{-1}\\
2R^{-2}&(17R-8)R^{-2}&4(5R-3)R^{-2}&(6R-4)R^{-2}
\end{pmatrix}.
\]
Its characteristic polynomial is \(Q_R(t)/R^2\), where
\[
\begin{aligned}
Q_R(t)={}&R^2t^4-(64R^3-56R^2-4)t^3\\
&+(48R^2-262R+220)t^2-(12R-8)t+1.
\end{aligned}
\]
The quartic is irreducible.  Its spectral separation is also exact: on
\(|t|=1\), the absolute value of its cubic coefficient exceeds the sum of
the other coefficient magnitudes, because
\[
 (64R^3-56R^2-4)-(49R^2-250R+213)
 =64R^3-105R^2+250R-217>0.
\]
Rouch\'e's theorem therefore places exactly three roots in \(|t|<1\) and
the remaining root \(\rho\) in \(|t|>1\).  Hence \(\rho\) is the unique
root of maximal modulus.

We next remove any possible nonvanishing assumption about the denominator.
The positivity estimate above and \(Q_N=x_0^nq_N\) give
\begin{equation}\label{eq:q-lower}
 q_N(x_0)\ge
 18\cdot29^N(N!)^2
 \left(\frac{1-x_0}{x_0}\right)^{N+1}.
\end{equation}
The scalar recurrence obtained from the first cyclic coordinate is of
Poincar\'e type after the \((N!)^2\) balancing.  The discrete
Birkhoff--Poincar\'e theorem \([4,\text{ Chapters 3 and 5}]\) applies because
the balanced coefficients are rational in \(N\), have full expansions in
\(N^{-1}\), and the limiting spectrum is simple.  If the coefficient of the
\(\rho\)-mode in \(q_N\) were zero, the three-root separation just proved
would give, for some \(\tau<1\),
\[
 |q_N(x_0)|\le K_\tau (N!)^2\tau^N.
\]
This contradicts \eqref{eq:q-lower}.  Thus the dominant denominator
coefficient is nonzero by a wholly exact argument.

It remains to transfer the first-column result to the other columns.  For
\(r\ge1\), put
\[
\begin{aligned}
 F_r&=[\,\e_1,M_r\e_1,M_rM_{r+1}\e_1,
              M_rM_{r+1}M_{r+2}\e_1\,],\\
 \gamma_{r,k}&=\prod_{\ell=1}^{k}(r+\ell)^2,\\
 C_r&=[\,\e_1,\mathcal B_r\e_1,
 \mathcal B_r\mathcal B_{r+1}\e_1,
 \mathcal B_r\mathcal B_{r+1}\mathcal B_{r+2}\e_1\,].
\end{aligned}
\]
The balancing telescopes exactly:
\[
 F_r=D(r)C_r\diag(\gamma_{r,0},\ldots,\gamma_{r,3}),
 \qquad
 C_r\longrightarrow
 C=[\,\e_1,\mathcal S\e_1,\mathcal S^2\e_1,\mathcal S^3\e_1\,].
\]
The limiting cyclic frame is nonsingular:
\[
 \det C
 =-\frac{4(27R-11)(128R^2-149R-43)}{R^6}\ne0.
\]
Thus \(F_r\) is invertible for all sufficiently large \(r\).  If
\(y_r(a)=aG_r\e_1\), exact inversion of this frame gives
\[
 aG_r\e_j=r^{-(j-1)}
 \sum_{k=0}^{3}(C_r^{-1})_{k+1,j}
 \frac{y_{r+k}(a)}{\gamma_{r,k}}.
\]
The same Birkhoff--Poincar\'e theorem supplies a linear dominant functional
\(\Lambda\) and an exponent \(\sigma\) such that, for fixed \(k\),
\[
 \frac{y_{r+k}(a)}
 {(r!)^2\rho^r r^\sigma\gamma_{r,k}}
 \longrightarrow\Lambda(a)\rho^k.
\]
Consequently
\[
 \frac{aG_r\e_j}
 {(r!)^2\rho^r r^{\sigma-(j-1)}}
 \longrightarrow\Lambda(a)\,\widetilde w_j,\qquad
 \widetilde w=[1,\rho,\rho^2,\rho^3]C^{-1}.
\]
An explicit left eigenvector is obtained from the first row of
\(R^2\operatorname{adj}(tI-\mathcal S)\).  Each of its four coordinate
polynomials is coprime to \(Q_R\); hence no coordinate vanishes at \(\rho\).
It is a nonzero multiple of \(\widetilde w\), so
\(\widetilde w_j\ne0\) for every \(j\).  Applying the last limit to
\(a=A_0,A_1\), using the exact denominator nonvanishing above, gives
\[
 \lim_{N\to\infty}\frac{P_{N,j}}{Q_{N,j}}
 =\frac{\Lambda(A_0)}{\Lambda(A_1)}
 \qquad(j=1,2,3,4).
\]
Equation \eqref{eq:first-column} evaluates this common ratio.  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}
& Authoritative differential gauge; nonterminating tail contiguity;
terminating adjoint equation; coefficientwise \(n\)-contiguity;
normalization and top boundary; exact spectral and cyclic-frame closure.\\
\path{p28_kernel_contiguity_certificate.sage}
& Independent coefficient/Ore proof of \(M_Nk_{N+1}=k_N\).\\
\path{p28_lattice_hypotheses_certificate.sage}
& 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}
& Balanced limit, Rouch\'e separation, nonzero eigenvector coordinates,
and invertible cyclic frame.\\
\path{p28_parametric_pade_probe.py}
& Dependency-free finite exact regression of the predicted valuations.\\
\bottomrule
\end{tabular}
\end{center}

The Wolfram certificate performs symbolic identities over
\(\Q(n,z)\); it uses no numerical samples.  The Python constants check uses
only the standard library's \texttt{fractions.Fraction}.  The SageMath
files are independent exact cross-checks.

\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 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 S. Bodine and D. A. Lutz,
  \emph{Asymptotic Integration of Differential and Difference Equations},
  Lecture Notes in Mathematics 2129, Springer, 2015, Chapters 3 and 5.
\item The Ramanujan Machine,
  \href{https://www.ramanujanmachine.com/ramanujan-challenge/}
  {Ramanujan Challenge}, Problem 2.8.
\end{enumerate}

\end{document}
