==== TABLE OF CONTENTS ====
Catalans-constant-is-irrational//certificate.tex
Catalans-constant-is-irrational//conclusion.tex
Catalans-constant-is-irrational//energy.tex
Catalans-constant-is-irrational//foundations.tex
Catalans-constant-is-irrational//introduction.tex
Catalans-constant-is-irrational//main.tex
Catalans-constant-is-irrational//nonvanishing.tex
Catalans-constant-is-irrational//odd-primes.tex
Catalans-constant-is-irrational//real-place.tex
Catalans-constant-is-irrational//references.bib
===========================

==== BEGIN Catalans-constant-is-irrational//certificate.tex ====
\section{An exact certificate for the two barriers}\label{sec:certificate}
All finite decimals in this Section denote exact rational numbers.  We specify
trial sequences for Equation~\eqref{eq:energy-dual}, isolate every stationary
point of its two one-variable functions, and bound their values using rational
arithmetic.

\begin{proposition}\label{prop:real-upper}
For the determinants of Equation~\eqref{eq:determinant},
\begin{equation}\label{eq:certified-real-upper}
 \limsup_{N\to\infty}\left(\frac{\log|\Delta_N|}{n^2}
                    -\frac12\log2\right)
 \le -2.290939875<-2.2909.
\end{equation}
The separate bounds for the majorants in cases $\kappa=2$ and $\kappa=1$
are $-2.290939875$ and $-2.296789875$, respectively.
\end{proposition}

\subsection{Exact trial sequences}
For $\kappa=2$ take $d=10$ and $\lambda_2=0$.  For $\kappa=1$ take
$d=8$ and $\lambda_1=2.47405979$.  Write each sequence as
\begin{equation}\label{eq:certificate-sequences}
 u_k=l_k+\sum_{z}r_z z^k,\qquad l_k=0\quad(k>d).
\end{equation}
The following tables give all coefficients.  Every integer coefficient in
them is to be multiplied by $10^{-8}$.  The finite parts are listed in
increasing order of $k$:
\begin{center}
\begin{tabular}{ccr r r r r}
\toprule
$\kappa$&$u$&\multicolumn{5}{c}{finite coefficients}\\
\midrule
2&$p$&45559127&-50750856&-6578767&13970217&4786184\\
 &&-4292433&-576704&1311615&671564&-346453\\
2&$v$&-23910158&21152432&-2885110&-11558199&6485289\\
 &&1456821&-1912176&-2263524&2742210&-1162454\\
\bottomrule
\end{tabular}

\medskip
\begin{tabular}{ccr r r r}
\toprule
$\kappa$&$u$&\multicolumn{4}{c}{finite coefficients}\\
\midrule
1&$p$&11913521&-79993701&-21956443&37903579\\
 &&10744022&-2073193&570246&-24939103\\
1&$v$&-89913025&52874280&45168341&-30708629\\
 &&-19269841&33922112&9819141&-16992389\\
\bottomrule
\end{tabular}
\end{center}
Case $\kappa=1$ has no exponential terms.  The exponential terms for
$\kappa=2$ are as follows.  For a real base $z$ the displayed $a$ is the
coefficient of $z^k$.  For a nonreal base the two entries $a,b$ are the
coefficients of $\Re(z^k),\Im(z^k)$, in that order.  In
Equation~\eqref{eq:certificate-sequences} such a pair is represented by
$r_z=(a-ib)/2$ and $r_{\bar z}=\overline{r_z}$, with the factor $10^{-8}$
applied to $a,b$.  Each real base appears once and each nonreal base together
with its conjugate.
\begin{center}
\begin{tabular}{ccrr}
\toprule
$u$&base $z$&$a$&$b$\\
\midrule
$p$&$.85$&-9338452&---\\
&$.94$&-2141509&---\\
&$.7i$&-66277922&-31907569\\
&$.85i$&-1231651&6002645\\
&$.092+.92i$&-3225918&8928234\\
&$-.092+.92i$&-2105536&-9091287\\
\midrule
$v$&$-.8$&15199211&---\\
&$-.96$&4451662&---\\
&$.88$&2545398&---\\
&$.95$&-4932634&---\\
&$.984$&11618157&---\\
&$.78i$&27238714&-38447936\\
&$.9i$&-31341084&-30188786\\
&$.955i$&-6693542&11912254\\
&$.984i$&2055213&-21715849\\
\bottomrule
\end{tabular}
\end{center}
In particular these are real exponentially decaying sequences.  There are
ten $p$ tail terms and thirteen $v$ tail terms when conjugates are counted,
and their base moduli are at most $.94$ and $.984$, respectively.
All the $|l_k|$ and $|r_z|$ are less than $1$.

The convergent power series for the logarithm give the finite formulas
\begin{align}
 \|u\|_*^2
 &=\sum_{k=1}^d\frac{l_k^2+2l_k\sum_zr_zz^k}{k}
             -\sum_{z,z'}r_zr_{z'}\operatorname{Log}(1-zz'),\notag\\
 T(u,x)&=\sum_{k=1}^d\frac{l_kT_k(x)}{k}
             -\frac12\sum_zr_z\operatorname{Log}(1-2xz+z^2),\notag\\
 S(u,x)&=\sum_{k=1}^d\frac{l_kx^k}{k}
             -\sum_zr_z\operatorname{Log}(1-xz).
 \label{eq:certificate-log-formulas}
\end{align}
Here $\operatorname{Log}$ is the principal complex logarithm; all sums are
real by conjugation.  There is no branch ambiguity in these formulas.
For the norm and power terms the log arguments have positive real part.
For the cosine term, writing $x=\cos\theta$ gives
\[
 1-2xz+z^2=(1-ze^{i\theta})(1-ze^{-i\theta}).
\]
Both factors have positive real part, and their principal arguments sum
strictly inside $(-\pi,\pi)$.  The sum of their logarithms is therefore
exactly the principal logarithm of the product, including when the product
has negative real part.

For these choices put
\begin{align}
 X_\kappa(x)&=W_\kappa(x)+\lambda_\kappa D(x)
                         -2\kappa T(p,x)-S(v,x),\notag\\
 Y_\kappa(x)&=\beta\log x+\gamma\log(1-x)+2T(v,x).
 \label{eq:certificate-barriers}
\end{align}
The domain of $X_\kappa$ is $[-1,1)\setminus\{0\}$, and that of
$Y_\kappa$ is $(0,1)$.

\subsection{Derivative numerators and exhaustive root brackets}
Define the rational functions
\begin{align*}
 t_u(x)&=\sum_{k=1}^dl_kU_{k-1}(x)
               +\sum_z\frac{r_zz}{1-2xz+z^2},\\
 h_u(x)&=\sum_{k=1}^dl_kx^{k-1}
               +\sum_z\frac{r_zz}{1-xz}.
\end{align*}
Differentiation of Equation~\eqref{eq:certificate-log-formulas} yields
\begin{align}
 X_\kappa'(x)
 &=\frac{\alpha+2\gamma}{x}-\frac{2\eta}{1-x}
 -(\kappa+2\alpha+2\eta+2\gamma)\frac{x}{1+x^2}
 \notag\\
 &\quad-\frac{4\lambda_\kappa x}{(1+x^2)^2}
 -2\kappa t_p(x)-h_v(x),\notag\\
 Y_\kappa'(x)&=\frac\beta x-\frac\gamma{1-x}+2t_v(x).
 \label{eq:certificate-derivatives}
\end{align}
For a completely specified rational numerator, use
\begin{align}
 Q_X(x)&=x(1-x)(1+x^2)^{3-\kappa}
       \prod_{z\text{ in }p}(1-2xz+z^2)
       \prod_{z\text{ in }v}(1-xz),\notag\\
 Q_Y(x)&=x(1-x)\prod_{z\text{ in }v}(1-2xz+z^2),\notag\\
 A_X(x)&=Q_X(x)X_\kappa'(x),\notag\\
 A_Y(x)&=Q_Y(x)Y_\kappa'(x).
 \label{eq:certificate-numerators}
\end{align}
Empty products equal $1$.  These equations, the coefficient tables, and
$T_0=1,T_1=x$, $U_0=1,U_1=2x$, with recurrence
$V_{k+1}=2xV_k-V_{k-1}$, specify the four polynomials over $\mathbb Q$
without any numerical root calculation.
The denominators do not vanish on the indicated open domains:
$|1-xz|\ge1-|z|>0$, and
$|1-2xz+z^2|\ge(1-|z|)^2>0$ for real $|x|\le1$.
Conjugate factors in $Q_X,Q_Y$ have positive products, and real-base
factors are positive.  Hence $Q_X$ has the sign of $x$ on its domain and
$Q_Y>0$ on $(0,1)$.

Here is an exact root-count certificate.  If
$A(x)=\sum_{j=0}^{d_0}A_jx^j$ is the specified numerator, on a subinterval
$(b,c)$ form
\begin{equation}\label{eq:certificate-descartes-transform}
 (1+t)^{d_0}A\left(\frac{b+ct}{1+t}\right)
 =\sum_{h=0}^{d_0}a_ht^h,\qquad
 a_h=\sum_{j=0}^{d_0}\sum_{k=0}^j
 A_j\binom jk b^{j-k}c^k\binom{d_0-j}{h-k},
\end{equation}
where out-of-range binomial coefficients are zero.  Delete zero
coefficients and count consecutive sign changes.  The degrees and the
resulting counts, in the order of the consecutive intervals, are
\begin{center}
\begin{tabular}{cccll}
\toprule
$\kappa$&function&$d_0$&division points&sign variations\\
\midrule
2&$X$&36&$-1,0,1$&$9,9$\\
2&$Y$&24&$0,.25,.5,.75,1$&$5,2,2,6$\\
1&$X$&13&$-1,-.5,0,.5,1$&$1,1,2,1$\\
1&$Y$&9&$0,1$&$5$\\
\bottomrule
\end{tabular}
\end{center}
The sign-variation bound is Descartes' rule of signs
\cite[Book~III, p.~373]{Descartes1637}.
For completeness, the required bound follows by factoring
out positive real roots: multiplication of a real polynomial by $t-r$,
$r>0$, increases the number of variations by at least one.  To see this,
positive rescaling of the variable reduces to $r=1$, and zero initial
coefficients can be removed.  If $a_0,\ldots,a_d$ are the old coefficients
and $b_0,\ldots,b_{d+1}$ the new ones, then
\[
 \sum_{j=0}^h b_j=-a_h\quad(0\le h\le d),\qquad b_{d+1}=a_d.
\]
Each sign change between nonzero partial sums forces an intervening $b_j$
with the sign of the later partial sum.  Starting with $b_0=-a_0$,
these choices give an ordered subsequence with all the variations of the
$a_h$; the final coefficient $b_{d+1}$ has the opposite sign to the last
partial sum and supplies one further variation.  Factoring successively
therefore proves that the variation count bounds
the number of positive roots with multiplicity.  The substitution in
Equation~\eqref{eq:certificate-descartes-transform} sends $t>0$ bijectively
to $(b,c)$ and does not change multiplicities.

The following integers $m$ specify disjoint open brackets
$(m,m+2)/10^{10}$ for roots of the corresponding derivative:
\begin{center}
\begin{tabular}{ccrrr}
\toprule
$\kappa$&function&\multicolumn{3}{c}{$m$}\\
\midrule
2&$X$&-9601109148&-8942317572&-7608305633\\
&&-6503394794&-5185864065&-4015634158\\
&&-3108806646&-2067921826&-1589849496\\
&&1531948062&2072448179&3208186484\\
&&4381119427&5851354199&7269030693\\
&&8390277402&9332614564&9709786219\\
\midrule
2&$Y$&176402802&330649406&764952882\\
&&1227250753&2149465998&3048189112\\
&&4322699096&5564757994&6801929373\\
&&8031988371&8877037851&9577761832\\
&&9838463999&9972727815&9992037196\\
\midrule
1&$X$&-9917299785&-2259572153&2543808026\\
&&4437270259&6348298970&\\
\midrule
1&$Y$&532669786&2504239325&5701738806\\
&&7966939383&9454490138&\\
\bottomrule
\end{tabular}
\end{center}
For each integer in this table the exact sign check is
\begin{equation}\label{eq:certificate-bracket-sign}
 N_A(m)N_A(m+2)<0,\qquad
 N_A(s)=\sum_{j=0}^{d_0}A_j(10^{10})^{d_0-j}s^j.
\end{equation}
Equations~\eqref{eq:certificate-numerators},
\eqref{eq:certificate-descartes-transform}, and
\eqref{eq:certificate-bracket-sign} give rational addition and
multiplication recipes for every entry of the root certificate.
The numbers of brackets in each consecutive interval equal the displayed
variation counts.  The intermediate value theorem and the variation
bound therefore give exactly one simple root per bracket and no other
roots in those open intervals.  Possible roots at the division points
are harmless because all finite division points are evaluated separately
below.

\subsection{Rational logarithms and all candidate values}
Here is the rational procedure used for the value bounds.  For a positive
rational $s$, write $s=2^m y$ with $1\le y\le2$, and put
\begin{equation}\label{eq:certificate-H}
 H(y)=2\sum_{j=0}^{17}\frac1{2j+1}
               \left(\frac{y-1}{y+1}\right)^{2j+1}.
\end{equation}
The substitution for $\log s$ is $mH(2)+H(y)$.  If
$q=(y-1)/(y+1)$, its unscaled positive remainder is bounded by
\[
 0\le\log y-H(y)\le\frac{2q^{37}}{37(1-q^2)}.
\]
For a nonzero rational complex number $a+ib$, its real log part is half
the real log of $a^2+b^2$.  For its principal argument choose an octant:
rotate by $-k\pi/4$, $-4\le k\le4$, so the rotated real part is positive
and the imaginary-to-real ratio $t$ satisfies $|t|\le1/2$.  Select the
rotation for which $k\pi/4+\arctan t$ is the principal argument; at the
negative real axis use the value $\pi$.  The ratio $t$ is rational, since
the rotation can be performed, up to a positive scale, by repeated maps
$(a,b)\mapsto(a+b,b-a)$ or $(a-b,a+b)$.
With
\begin{equation}\label{eq:certificate-J}
 J(t)=\sum_{j=0}^{23}\frac{(-1)^jt^{2j+1}}{2j+1},
\end{equation}
substitute $k(J(1/2)+J(1/3))+J(t)$ for the argument.  The identity
\[
 \arctan(1/2)+\arctan(1/3)=\pi/4
\]
follows from the tangent addition
formula and the fact that both summands are positive and their sum is
less than $\pi/2$.  The alternating-series estimate gives
$|\arctan t-J(t)|\le |t|^{49}/49$.

All positive inputs used here, including squared moduli, lie between
$2^{-100}$ and $2^{100}$.  Indeed, the real arguments at the evaluation
points are between $.0007$ and $2$, and the squared moduli of all complex
arguments lie between $(.016)^4$ and $(1+.984)^4$, by the factor bounds
following Equation~\eqref{eq:certificate-numerators}.
Thus at most $100$ scaling steps are needed.  To make the error allowance
explicit, set
\[
 h_0=\frac{2(1/3)^{37}}{37(1-1/9)},\qquad
 j_0=\frac{(1/2)^{49}}{49}.
\]
A real log substitution costs at most $101h_0$, and an argument costs
at most $9j_0$.  In particular
$2\cdot101h_0+9j_0<7\cdot10^{-16}$, so $10^{-15}$ is an upper
bound for the absolute error of a complex log.  This also proves the
coarser allowance $10^{-11}$ per log.

All rational polynomial terms in Equation~\eqref{eq:certificate-log-formulas}
are evaluated exactly.  The total absolute log coefficient mass in
$\kappa\|p\|_*^2+\|v\|_*^2/2$ is at most
$2\cdot10^2+13^2/2=284.5$, in $X_\kappa$ it is less than
$2\cdot10+13+2=35$, and in $Y_\kappa$ it is less than $13+1=14$.
Their resulting substitution errors are respectively less than
$2.85\cdot10^{-13}$, $3.5\cdot10^{-14}$, and $1.4\cdot10^{-14}$,
all less than $10^{-8}$.
One may enclose each rational series term between consecutive multiples
of $10^{-40}$, add the displayed remainder bounds with outward signs,
and then propagate intervals linearly.  This additional rounding changes
the preceding allowances by less than $10^{-32}$.
For a complex weight $c$, use
$\Re(c\operatorname{Log}z)=\Re(c)\Re(\operatorname{Log}z)
-\Im(c)\Im(\operatorname{Log}z)$, retaining the sign of each weight in
interval multiplication.  This fully specifies a rational interval
calculation; no numerical logarithms are needed.

For transparency, Table~\ref{tab:certificate-values} gives all fifty
point-value upper bounds.  Its argument column is $10^{10}x$.  A row
marked $B$ is the left endpoint of the corresponding root bracket above;
a row marked $P$ is a finite division point.  Each displayed bound is a
rational upper bound rounded upwards to twelve decimal places after the
series remainder has been included.  The largest unrounded interval
width in this table is less than $2.16146856249281\cdot10^{-16}$.
\begin{longtable}{ccrrr}
\caption{Certified rational upper bounds at every candidate evaluation point.}\label{tab:certificate-values}\\
\toprule
$\kappa$&function&$10^{10}x$&type&upper bound\\
\midrule
\endfirsthead
\multicolumn{5}{c}{\tablename~\thetable\ (continued)}\\
\toprule
$\kappa$&function&$10^{10}x$&type&upper bound\\
\midrule
\endhead
\bottomrule
\endfoot
2&$X$&-9601109148&B&-0.984034048775\\
2&$X$&-8942317572&B&-0.984375363807\\
2&$X$&-7608305633&B&-0.984034052640\\
2&$X$&-6503394794&B&-0.984105522615\\
2&$X$&-5185864065&B&-0.984034053414\\
2&$X$&-4015634158&B&-0.984092061375\\
2&$X$&-3108806646&B&-0.984034037901\\
2&$X$&-2067921826&B&-0.984364031075\\
2&$X$&-1589849496&B&-0.984034038450\\
2&$X$&1531948062&B&-0.984033385315\\
2&$X$&2072448179&B&-0.984776982913\\
2&$X$&3208186484&B&-0.984034026898\\
2&$X$&4381119427&B&-0.984264010107\\
2&$X$&5851354199&B&-0.984034029391\\
2&$X$&7269030693&B&-0.984245044505\\
2&$X$&8390277402&B&-0.984034016294\\
2&$X$&9332614564&B&-0.984567630027\\
2&$X$&9709786219&B&-0.984033926884\\
2&$X$&-10000000000&P&-0.986727371546\\
2&$Y$&176402802&B&-1.608946411646\\
2&$Y$&330649406&B&-1.609949505120\\
2&$Y$&764952882&B&-1.608960295828\\
2&$Y$&1227250753&B&-1.609094597854\\
2&$Y$&2149465998&B&-1.608960426500\\
2&$Y$&3048189112&B&-1.608992193672\\
2&$Y$&4322699096&B&-1.608960428486\\
2&$Y$&5564757994&B&-1.608979509517\\
2&$Y$&6801929373&B&-1.608960430478\\
2&$Y$&8031988371&B&-1.608990548812\\
2&$Y$&8877037851&B&-1.608960429811\\
2&$Y$&9577761832&B&-1.609055802895\\
2&$Y$&9838463999&B&-1.608960412835\\
2&$Y$&9972727815&B&-1.609694899071\\
2&$Y$&9992037196&B&-1.608958123669\\
2&$Y$&2500000000&P&-1.608973617030\\
2&$Y$&5000000000&P&-1.608971701898\\
2&$Y$&7500000000&P&-1.608976908052\\
1&$X$&-9917299785&B&-2.778491531574\\
1&$X$&-2259572153&B&-1.324666731948\\
1&$X$&2543808026&B&-1.324655807046\\
1&$X$&4437270259&B&-1.352236629957\\
1&$X$&6348298970&B&-1.324666329425\\
1&$X$&-10000000000&P&-2.775818077526\\
1&$X$&-5000000000&P&-1.544057819905\\
1&$X$&5000000000&P&-1.347557680876\\
1&$Y$&532669786&B&-1.428286151250\\
1&$Y$&2504239325&B&-1.515602647362\\
1&$Y$&5701738806&B&-1.428335732372\\
1&$Y$&7966939383&B&-1.465686164672\\
1&$Y$&9454490138&B&-1.428335358167\\
\end{longtable}
The same calculation gives the following upper bounds for
$\kappa\|p\|_*^2+\tfrac12\|v\|_*^2$:
\[
 .778415976284\quad(\kappa=2),\qquad
 .931985203901\quad(\kappa=1).
\]
In particular the
rational substituted expressions, before their error allowances are
added, satisfy the following convenient stronger cutoffs:
\begin{center}
\begin{tabular}{cccc}
\toprule
$\kappa$&$\kappa\|p\|_*^2+\|v\|_*^2/2$&$X$ at listed points&$Y$ at listed points\\
\midrule
2&$.77843$&$-.98400$&$-1.60891$\\
1&$.93205$&$-1.32442$&$-1.42805$\\
\bottomrule
\end{tabular}
\end{center}
Each entry means a strict upper bound.  With the $10^{-8}$ allowance,
the weaker cutoffs we shall actually use are
\begin{equation}\label{eq:certificate-coarse-cutoffs}
\begin{array}{c|ccc}
 \kappa&\kappa\|p\|_*^2+\tfrac12\|v\|_*^2&X&Y\\\hline
 2&.77844&-.98399&-1.60890\\
 1&.9321&-1.3244&-1.4280.
\end{array}
\end{equation}
The logarithm procedure also gives $.693146<\log2<.693149$.

\subsection{From bracket values to global suprema}
A single derivative bound suffices on every whole closed bracket in the
root table.  All its points have distance greater than $.0007$ from both
$0$ and $1$.  In particular the bracket nearest $s=1$ has right endpoint
$9992037198/10^{10}$, whose distance from $1$ is exactly $.0007962802$.
For $|x|\le1$, the identity
$U_{k-1}(\cos\theta)=\sin(k\theta)/\sin\theta$ gives
$|U_{k-1}(x)|\le k$, including its endpoint limits.
Using the coefficient and base bounds above, the polynomial part of
$X_\kappa'$ is at most
$4\sum_{k=1}^{10}k+10=230$ in absolute value; its endpoint terms are
bounded by $(19/48+1/12)/.0007<685$; its $x/(1+x^2)$ term is less than
$3$; and its multiplier term is less than $4\lambda_1<10$.
Its two log-tail derivatives are bounded by
\[
 \frac{4\cdot10}{(.06)^2}+\frac{13}{.016}<11924.
\]
Consequently $|X_\kappa'|<12852<120000$ throughout every listed bracket.
For $Y_\kappa'$ the polynomial part is at most $110$, the endpoint
terms are bounded by $(7/48+1/12)/.0007<328$, and the log tails are
bounded by $26/(.016)^2=101562.5$.  Hence
$|Y_\kappa'|<102001<120000$ on every whole bracket as well.
These estimates hold for both cases, with empty tails in case $1$.

It follows from the mean value theorem that moving from a left bracket
endpoint to its stationary point increases either value by less than
\[
 120000\cdot\frac2{10^{10}}=.000024.
\]
Every interior stationary point is in one of these brackets or is a
division point.  All finite division points have been included in
Table~\ref{tab:certificate-values}: $-1$ for $X_2$;
$1/4,1/2,3/4$ for $Y_2$; and $-1,-1/2,1/2$ for $X_1$.
The remaining endpoints are singular and have limit $-\infty$:
$x\to0^\pm$ and $x\to1^-$ for $X_\kappa$, and $x\to0^+$ and
$x\to1^-$ for $Y_\kappa$.  The bounded tangent and multiplier terms do
not alter these limits.  Thus the root exhaustion, the finite endpoint
values, and the derivative estimate together certify the global suprema,
not merely the sampled values.

Finally $-1+\alpha+\gamma=-11/16$.  Apply
Proposition~\ref{prop:energy-reduction}, use
Equation~\eqref{eq:certificate-coarse-cutoffs}, and add $.000024$ for
each supremum.  Since $\log2>.693146$, the right side is strictly less
than, respectively,
\begin{align*}
 -\frac{11}{16}(.693146)+.77844-.98399-1.60890+.000048
    &=-2.290939875,\\
 -\frac{11}{16}(.693146)+.9321-1.3244-1.4280+.000048
    &=-2.296789875.
\end{align*}
Taking the larger of these constants proves
Proposition~\ref{prop:real-upper}.
==== END Catalans-constant-is-irrational//certificate.tex ====

==== BEGIN Catalans-constant-is-irrational//conclusion.tex ====
\section{Conclusion}\label{sec:conclusion}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Suppose that $G$ is rational. Proposition~\ref{prop:rationality} then
makes $\Delta_N$ rational for every $N$.
By Proposition~\ref{prop:nonvanishing}, $\Delta_p\ne0$ for every
sufficiently large prime $p$. Apply the product-formula lower bound of
Proposition~\ref{prop:finite-lower} along this sequence:
\[
\liminf_{\substack{p\to\infty\\p\ {\rm prime}}}
\left(\frac{\log|\Delta_p|}{(48p)^2}-\frac12\log2\right)>-2.29084.
\]
Proposition~\ref{prop:real-upper} gives, along the same sequence, an
upper limit strictly smaller than $-2.2909$. Since
$-2.29084>-2.2909$, this is a contradiction. Therefore $G$ is irrational.
\end{proof}

\paragraph{Hyperbolic volumes.}
Normalize sectional curvature to $-1$. Agol's theorem identifies $4G$ as
the minimum volume of an orientable complete finite-volume hyperbolic
three-manifold with exactly two cusps, attained by the Whitehead-link
and $(-2,3,8)$-pretzel-link complements
\cite[Introduction and Theorem~3.6]{Agol2010TwoCusps}.
Theorem~\ref{thm:main} therefore makes this minimum and both link volumes
irrational. More generally, every orientable arithmetic hyperbolic
three-orbifold defined over $\mathbb Q(i)$ has irrational volume.
Here its lattice $\Gamma\le\operatorname{PSL}_2(\mathbb C)$ is,
up to conjugacy, commensurable with the projective norm-one group
$\Gamma^1(\mathcal O)=\mathcal O^1/\{\pm1\}$ of a maximal order
$\mathcal O$ in some quaternion algebra $B/\mathbb Q(i)$
\cite[Definition~38.3.4]{Voight2021Quaternion}.
For each such algebra, with finite reduced discriminant $\mathfrak D$,
the volume formula gives
\[
\operatorname{vol}\bigl(\Gamma^1(\mathcal O)\backslash\mathbb H^3\bigr)
 =\frac{G}{3}\prod_{\mathfrak p\mid\mathfrak D}(N\mathfrak p-1),
\]
where $N\mathfrak p=|\mathbb Z[i]/\mathfrak p|$; this follows by inserting
$\zeta_{\mathbb Q(i)}(2)=\zeta(2)L(2,\chi_{-4})=(\pi^2/6)G$
into \cite[Theorem~39.1.13]{Voight2021Quaternion}.
After conjugating, a common finite-index subgroup $H$ gives
\[
\operatorname{vol}(\Gamma\backslash\mathbb H^3)
 =\frac{[\Gamma^1(\mathcal O):H]}{[\Gamma:H]}
   \operatorname{vol}\bigl(\Gamma^1(\mathcal O)\backslash\mathbb H^3\bigr),
\]
so this volume is a positive rational multiple of $G$ and is irrational.
In the split case $B=M_2(\mathbb Q(i))$, $\mathcal O=M_2(\mathbb Z[i])$,
the empty product yields
$\operatorname{vol}(\operatorname{PSL}_2(\mathbb Z[i])\backslash\mathbb H^3)=G/3$
\cite[Example~39.1.16]{Voight2021Quaternion}.
==== END Catalans-constant-is-irrational//conclusion.tex ====

==== BEGIN Catalans-constant-is-irrational//energy.tex ====
\section{A uniform energy bound}\label{sec:energy}
The purpose of this Section is to replace the two many-variable
majorants by one-variable suprema and explicit quadratic terms. The
bound must be uniform before taking those suprema. Matched damping
and retained endpoint corrections provide this uniformity.

We retain the constants $\alpha,\beta,\gamma,\eta$, the two cases
$\kappa=2,1$, and the nonnegative majorants $\mathcal I_{\kappa,n}$ of
Equation~\eqref{eq:real-integrand-majorants}.  Put
\begin{align}
 W_\kappa(x)&=(\alpha+2\gamma)\log|x|+2\eta\log(1-x)
 -(\kappa/2+\alpha+\eta+\gamma)\log(1+x^2),\notag\\
 W_s(s)&=\beta\log s+\gamma\log(1-s),\notag\\
 D(x)&=2\gamma-\frac{2x^2}{1+x^2}.
 \label{eq:energy-fields}
\end{align}
Here $D(x)$ is a scalar function, distinct from the row polynomials $D_r(t)$.
For a real sequence $u=(u_k)_{k\ge1}$ with $|u_k|\le K r^k$ for some
$K<\infty$ and $0<r<1$, define
\begin{equation}\label{eq:energy-trial-notation}
 \|u\|_*^2=\sum_{k\ge1}\frac{u_k^2}{k},\qquad
 T(u,x)=\sum_{k\ge1}\frac{u_kT_k(x)}{k},\qquad
 S(u,x)=\sum_{k\ge1}\frac{u_kx^k}{k}.
\end{equation}
These series converge uniformly for $-1\le x\le1$.

The sequences $p,v$ below are freely chosen trial coefficients. The
inequality $-a^2\le b^2-2ab$ lets them replace two negative quadratic
sums in the logarithmic interactions by affine upper bounds, leaving
the two one-variable suprema. Each admissible choice gives a valid
bound; the certificate will supply one successful choice for each case.

\begin{proposition}\label{prop:energy-reduction}
Let $p,v$ be two such sequences, and take $\lambda=0$ in case $\kappa=2$
or any fixed $\lambda\ge0$ in case $\kappa=1$.  Uniformly over configurations
in the indicated case,
\begin{align}
 &\limsup_{N\to\infty}\ \sup_{\text{case }\kappa}
 \left(\frac{\log\mathcal I_{\kappa,n}}{n^2}
                  -\frac12\log2\right)\notag\\
 &\qquad\le (-1+\alpha+\gamma)\log2
       +\kappa\|p\|_*^2+\frac12\|v\|_*^2\notag\\
 &\qquad\quad+\sup_{\substack{-1\le x<1\\x\ne0}}
       \{W_\kappa(x)+\lambda D(x)-2\kappa T(p,x)-S(v,x)\}\notag\\
 &\qquad\quad+\sup_{0<s<1}\{W_s(s)+2T(v,s)\}.
 \label{eq:energy-dual}
\end{align}
Consequently the maximum of the right sides for the two cases bounds
\[
 \limsup_{N\to\infty}\left(n^{-2}\log|\Delta_N|-\tfrac12\log2\right).
\]
We use $\log0=-\infty$.
\end{proposition}

\begin{proof}
For a one-variable function $F$, write
$\langle F(x)\rangle=n^{-1}\sum_iF(x_i)$, and similarly for $s$.
In a double average $\langle K(x,x')\rangle_{\ne}$ we sum over ordered
pairs $i\ne j$ and divide by $n^2$; a double average without the subscript
includes every pair.  The substitution $t=2x/(1+x^2)$ gives
\begin{equation}\label{eq:energy-transform-identities}
 |t-t'|=\frac{2|x-x'|(1-xx')}{(1+x^2)(1+x'^2)},\qquad
 1-ts=\frac{1-2xs+x^2}{1+x^2},\qquad
 1-t=\frac{(1-x)^2}{1+x^2}.
\end{equation}
In particular, the exponent of $|x_i|$ in either majorant is exactly
\[
 (C-1)+g-(n-1)=C+g-n=a+2g;
\]
there is no singular finite-size correction at $x=0$.
Keeping the other powers as well gives the following exact identity:
\begin{align}
 \frac{\log\mathcal I_{\kappa,n}}{n^2}-\frac12\log2
 &=c_{\kappa,n}\log2
   +\left\langle W_\kappa(x)+\frac{\kappa+2}{2n}\log(1+x^2)\right\rangle
   +\langle W_s(s)\rangle\notag\\
 &\quad+\frac\kappa2\langle\log|x-x'|\rangle_{\ne}
       +\frac12\langle\log(1-xx')\rangle_{\ne}
       +\langle\log|s-s'|\rangle_{\ne}\notag\\
 &\quad-\langle\log(1-2xs+x^2)\rangle,
 \label{eq:energy-exact-finite}\\
 c_{\kappa,n}&=\frac Cn+\frac{\kappa-2}{2}
                       -\frac{\kappa+1}{2n}.\notag
\end{align}
The displayed correction involving $\log(1+x^2)$ is bounded by
$2\log2/n$ on the whole domain.

We use the classical Chebyshev expansion of the logarithmic kernel,
in the form of Haagerup's identity presented in
\cite[Lemma~3.1 and its proof]{GaroufalidisPopescu2013}.
We first record its damped form on $[-1,1]$. If
$u=\cos\theta$, $u'=\cos\theta'$, and $0\le r<1$, set
\begin{align}
 L_r(u,u')
 &=-\log2+\log|1-re^{i(\theta+\theta')}|
                 +\log|1-re^{i(\theta-\theta')}|\notag\\
 &=-\log2-2\sum_{k\ge1}\frac{r^kT_k(u)T_k(u')}{k}.
 \label{eq:energy-cosine-kernel}
\end{align}
This follows from the power series for $\log(1-z)$ and the addition
formula for cosines.  Letting $r\uparrow1$ at distinct $u,u'$ gives
$L_1(u,u')=\log|u-u'|$, since the product of the two chord lengths
is $2|u-u'|$.  The other two identities, for $|z|,|z'|<1$, are
\begin{equation}\label{eq:energy-power-cross-kernels}
 \log(1-zz')=-\sum_{k\ge1}\frac{z^kz'^k}{k},\qquad
 -\log(1-2zs+z^2)=2\sum_{k\ge1}\frac{z^kT_k(s)}{k}.
\end{equation}
For the last identity the logarithm is the logarithm of a positive real
number, equal to $\log|1-ze^{i\arccos s}|^2$.
The singular kernels cannot be summed at empirical diagonals. We
regularize all appearances of the same moment by the same factor so
that the pure and cross interactions still complete a square. Near
$(x,s)=(1,1)$, the identity
$|1-xe^{i\arccos s}|^2=(1-x)^2+2x(1-s)$ identifies the scales
$1-x$ and $\sqrt{1-s}$ used in the following damping factors.

Fix $0<\varepsilon<1/8$ and define
\begin{equation}\label{eq:energy-damping}
 \tau=1-\varepsilon,\qquad
 \rho(x)=1-\varepsilon(1-x),\qquad
 \sigma(s)=1-\varepsilon\sqrt{1-s}.
\end{equation}
Replace the $x$ cosine kernel in Equation~\eqref{eq:energy-exact-finite}
by $L_{\tau^2}(x,x')$ and the $s$ cosine kernel by
$L_{\sigma(s)\sigma(s')}(s,s')$.  Replace the power kernel by
\[
 \log(1-\rho(x)\rho(x')xx'),
\]
and the cross kernel by
\[
 -\log|1-\rho(x)\sigma(s)xe^{i\arccos s}|^2.
\]
Every original kernel is bounded above by its replacement plus
$O(\varepsilon)$, with an absolute constant independent of the configuration.
We give the global estimates, including the signs in the power kernel.

For $|\zeta|=1$ and $0<r\le1$,
\[
 |1-r\zeta|^2-r|1-\zeta|^2=(1-r)^2\ge0.
\]
Thus each of the cosine kernels costs at most $-\log r$ in the
comparison, and here $r\ge(1-\varepsilon)^2$.  For the power kernel put
$u=xx'$ and $r=\rho(x)\rho(x')$.  If $u\ge0$, monotonicity gives
$\log(1-u)\le\log(1-ru)$.  If $u<0$, write $q=-u\le1$; then
\begin{equation}\label{eq:energy-opposite-sign}
 \log(1+q)-\log(1+rq)
 \le\frac{(1-r)q}{1+rq}\le1-r\le4\varepsilon.
\end{equation}
For the cross kernel write
$a=1-x$, $b=\sqrt{1-s}$, $\theta=\arccos s$,
$z=1-xe^{i\theta}$, and $z_\varepsilon=1-\rho(x)\sigma(s)xe^{i\theta}$.
Then
$|z_\varepsilon-z|\le\varepsilon|x|(a+b)$.
On $x\le0$ we have $|z|\ge1$ and $|x|(a+b)\le3$.
On $0\le x\le1/2$ we have $|z|\ge1/2$ and
$|x|(a+b)\le1$.  On $1/2\le x<1$ we have
\[
 |z|^2=a^2+2xb^2\ge a^2+b^2,
 \qquad |x|(a+b)\le\sqrt2\,|z|.
\]
Consequently $|z_\varepsilon-z|\le3\varepsilon|z|$ on all three domains,
and in the direction required for an upper bound,
\begin{equation}\label{eq:energy-cross-comparison}
 -\log|z|^2\le-\log|z_\varepsilon|^2
                +2\log(1+3\varepsilon)
 \le-\log|z_\varepsilon|^2+6\varepsilon.
\end{equation}

For each fixed finite interior configuration all the damped series are
absolutely convergent.  Put
\[
 A_k=\langle\tau^kT_k(x)\rangle,\qquad
 B_k=\langle\rho(x)^kx^k\rangle,\qquad
 C_k=\langle\sigma(s)^kT_k(s)\rangle.
\]
Inserting the diagonals gives the two exact negative-square identities
\begin{align}
 \frac\kappa2\langle L_{\tau^2}(x,x')\rangle
 &=-\frac\kappa2\log2-\kappa\sum_{k\ge1}\frac{A_k^2}{k},
 \label{eq:energy-first-square}
\end{align}
\begin{align}
 &\frac12\langle\log(1-\rho(x)\rho(x')xx')\rangle
 +\langle L_{\sigma(s)\sigma(s')}(s,s')\rangle\notag\\
 &\qquad-\langle\log|1-\rho(x)\sigma(s)xe^{i\arccos s}|^2\rangle\notag\\
 &\quad=-\log2-\frac12\sum_{k\ge1}\frac{(B_k-2C_k)^2}{k}.
 \label{eq:energy-second-square}
\end{align}
In particular the $B_k$ and $C_k$ in the cross term are exactly the same
ones as in the pure power and cosine terms.  Adding the constants in
these identities to $c_{\kappa,n}\log2$ gives
\[
 \left(-1+\alpha+\gamma-\frac{\kappa+1}{2n}\right)\log2.
\]
The omitted-diagonal constants are part of the next $O_\varepsilon(1/n)$
correction.

We control this correction before taking any supremum.  A damped cosine
diagonal is bounded below by $-\log2+2\log(1-r)$, so the $x$ cosine
diagonals cost $O_\varepsilon(1/n)$.  For the power diagonal,
\[
 1-\rho(x)^2x^2\ge1-x\quad(x\ge0),\qquad
 1-\rho(x)^2x^2\ge1-(1-\varepsilon)^2\ge\varepsilon\quad(x\le0).
\]
Since $1\le1-x\le2$ when $x\le0$, their upward correction is at most
\[
 O_\varepsilon(1/n)-\frac1{2n}\langle\log(1-x)\rangle.
\]
Finally,
\[
 1-\sigma(s)^2
 =\varepsilon\sqrt{1-s}\,(2-\varepsilon\sqrt{1-s})
 \ge\varepsilon\sqrt{1-s},
\]
so the $s$ cosine diagonals cost at most
$O_\varepsilon(1/n)-n^{-1}\langle\log(1-s)\rangle$.
The cross term already contains all pairs.  There is no loss involving
$\log|x|$ or $\log(1+x)$.

For real numbers $a,b$ we have $-a^2\le b^2-2ab$.  Apply this to the
first square with $b=p_k$, and to the second with $b=v_k$; after dividing
by $k$ and summing, the two inequalities are
\begin{align*}
 -\kappa\sum_k A_k^2/k
 &\le\kappa\|p\|_*^2-2\kappa\sum_kp_kA_k/k,\\
 -\tfrac12\sum_k(B_k-2C_k)^2/k
 &\le\tfrac12\|v\|_*^2-\sum_kv_kB_k/k+2\sum_kv_kC_k/k.
\end{align*}
For $\kappa=1$, failure of Equation~\eqref{eq:case-condition} says
\[
 1-2\left\langle\frac{x^2}{1+x^2}\right\rangle
 >1-\frac1n-2\gamma,
 \qquad\text{hence}\qquad \langle D(x)\rangle>-1/n.
\]
It follows that adding $\lambda\langle D(x)\rangle$ costs at most
$\lambda/n$ for an upper bound; this explains both $\lambda\ge0$ and its
sign in Equation~\eqref{eq:energy-dual}.

Let $T_\varepsilon(p,x)$, $S_\varepsilon(v,x)$, and
$T_\varepsilon(v,s)$ denote the three series in
Equation~\eqref{eq:energy-trial-notation} with factors
$\tau^k$, $\rho(x)^k$, and $\sigma(s)^k$, respectively.
The preceding pointwise bounds reduce the upper estimate to the sum of
two one-variable suprema, with functions
\begin{align*}
 X_{n,\varepsilon}(x)
 &=\frac{19}{48}\log|x|
   +\left(\frac1{12}-\frac1{2n}\right)\log(1-x)
   \\
 &\quad-(\kappa/2+\alpha+\eta+\gamma)\log(1+x^2)
   +\lambda D(x)\\
 &\quad-2\kappa T_\varepsilon(p,x)-S_\varepsilon(v,x),\\
 Y_{n,\varepsilon}(s)
 &=\frac7{48}\log s
   +\left(\frac1{12}-\frac1n\right)\log(1-s)
   +2T_\varepsilon(v,s),
\end{align*}
plus the constant in Equation~\eqref{eq:energy-dual} and
$O(\varepsilon)+O_\varepsilon(1/n)$.  All bounded finite-size terms from
Equation~\eqref{eq:energy-exact-finite} are included in this last error.
For $n\ge24$ both weakened endpoint coefficients are at least $1/24$.

Set $P_1=\sum_k|p_k|$ and $V_1=\sum_k|v_k|$, which are finite.
The inequality $1-(1-d)^k\le kd$ shows uniformly that
\begin{align*}
 |T_\varepsilon(p,x)-T(p,x)|&\le\varepsilon P_1,\\
 |S_\varepsilon(v,x)-S(v,x)|&\le2\varepsilon V_1,\\
 |T_\varepsilon(v,s)-T(v,s)|&\le\varepsilon V_1.
\end{align*}
Thus the changes in the two tangent functions are at most
$2\kappa\varepsilon P_1+2\varepsilon V_1$ and
$2\varepsilon V_1$, respectively.  These are common summable bounds;
no uniform convergence of the raw square series is required.
All remaining nonsingular terms are bounded uniformly in $n\ge24$ and
$0<\varepsilon<1/8$.  The unchanged positive coefficients $19/48,7/48$
and the weakened coefficients at least $1/24$ force uniform decay to
$-\infty$ near $x=0$, $x=1$, $s=0$, and $s=1$.
The values at the fixed comparison points $x=-1/2$, $s=1/2$ have a
common finite lower bound. For the first supremum we include the
regular endpoint $x=-1$ by continuous extension.
There are therefore compact sets, independent
of $n\ge24$ and small $\varepsilon$, containing maximizers for both
suprema. On these sets the functions converge uniformly as $n\to\infty$
with $\varepsilon$ fixed, and subsequently as $\varepsilon\downarrow0$.
Taking the limits in precisely this order removes
$O_\varepsilon(1/n)$ first and proves Equation~\eqref{eq:energy-dual}.
Finally apply Equation~\eqref{eq:real-supremum-reduction}.
\end{proof}
==== END Catalans-constant-is-irrational//energy.tex ====

==== BEGIN Catalans-constant-is-irrational//foundations.tex ====
\section{Polynomial rows and mixed moments}\label{sec:foundations}

We construct the determinant and prove that its entries are rational
under $G\in\mathbb Q$. The row design serves two purposes: Taylor
contact cancels $\zeta(2)$, while integral Chebyshev coefficients permit
the finite-prime estimates. After the moment calculations, we establish
the estimate at $2$; odd primes are treated in the next Section.

For each positive integer $N$, set
\begin{equation}\label{eq:parameters}
\begin{gathered}
n=48N,\qquad a=11N,\qquad b=7N,\qquad q=g=4N,\qquad h=2N,\\
L=n+a=n+b+q=59N,\qquad C=L+g=63N,\\
H=C+h=65N,\qquad A=a+2g=19N.
\end{gathered}
\end{equation}
Write $f(t)=\sqrt{1-t^2}$, with the positive branch on $(-1,1)$,
and $w=t/(1+f)$. Then
\[
t=\frac{2w}{1+w^2},\qquad f=\frac{1-w^2}{1+w^2}.
\]
The involution denoted by a star sends $w$ to $w^{-1}$, fixes $t$,
and sends $f$ to $-f$. For $0\le r<n$, define
\begin{equation}\label{eq:rows}
R_r=(1-t)^h t^{C-1}w^{r-g},\qquad
P_r=\frac{R_r+R_r^*}{2},\qquad
D_r=\frac{t(R_r^*-R_r)}{2f}.
\end{equation}
Here and below identities involving $f$ near zero use its Taylor branch
with $f(0)=1$.

Let $T_d,U_d$ be the Chebyshev polynomials, normalized by
$T_0=1,T_1=x,U_{-1}=0,U_0=1$, and the recurrence
$S_{d+1}=2xS_d-S_{d-1}$. The associated Laurent expressions are the
standard formulas \cite[Equations~(18.5.1)--(18.5.2)]{DLMFChebyshev}.
Since
$(w+w^{-1})/2=1/t$ and $(w^{-1}-w)/2=f/t$, Equation~\eqref{eq:rows}
becomes, with $d=|r-g|$,
\[
P_r=(1-t)^h t^{C-1}T_d(1/t),\qquad
D_r=\operatorname{sgn}(r-g)(1-t)^h t^{C-1}U_{d-1}(1/t).
\]
In particular these are integer polynomials; the second is zero when $r=g$.
The inequality $d\le n-1-g$ gives
\begin{equation}\label{eq:row-support}
\operatorname{supp}P_r\subseteq\{A,\ldots,H-1\},\qquad
\operatorname{supp}D_r\subseteq\{A+1,\ldots,H-1\}.
\end{equation}
Moreover $w=t/2+O(t^3)$, so that
\begin{equation}\label{eq:contact}
\frac{tP_r}{f}-D_r=\frac{tR_r}{f}=O(t^{C+r-g})=O(t^L).
\end{equation}

For nonnegative integers $i,j$, define
\begin{equation}\label{eq:mixed-moments}
\begin{split}
M(i,j)&=\int_{-1}^1\int_0^1
 \frac{|t|}{f(t)}\frac{t^i s^j}{1-ts}\,ds\,dt,\\
Z(i,j)&=\int_0^1\int_0^1\frac{t^i s^j}{1-ts}\,ds\,dt.
\end{split}
\end{equation}
Extend these expressions bilinearly to polynomial arguments.
The integrals converge absolutely. Indeed, for $0<u<1$,
\[
\int_0^1\frac{ds}{1-us}=\frac{-\log(1-u)}{u},
\]
and $(-\log(1-u))/\sqrt{1-u^2}$ is integrable on $(0,1)$.
This dominates the absolute $M$ integrand after integration in $s$;
the corresponding assertion for $Z$ is weaker. Bounded polynomial factors
preserve this domination. It also justifies termwise integration of the
geometric series used below.

For each raw column index $b\le j<L$, let
\[
F_j(r)=M(P_r,j)-\frac32 Z(D_r,j)\qquad(0\le r<n).
\]
The determinant used throughout the proof is
\begin{equation}\label{eq:determinant}
\Delta_N=\det_{0\le r,k<n}
\left(M\bigl(P_r,s^{b+k}(1-s)^q\bigr)
      -\frac32Z\bigl(D_r,s^{b+k}(1-s)^q\bigr)\right).
\end{equation}
Thus, if $\mathcal F$ is the $n$ by $n+q$ matrix with columns $F_j$,
the matrix in Equation~\eqref{eq:determinant} is $\mathcal F\mathcal T$,
where the integer matrix $\mathcal T$ has entries
\begin{equation}\label{eq:integer-filter}
\mathcal T_{j,k}=(-1)^{j-b-k}\binom{q}{j-b-k},
\qquad b\le j<L,\quad 0\le k<n.
\end{equation}
A binomial coefficient outside its usual range is zero. In particular,
all raw columns required by the filter lie in the contact range $j<L$.

\subsection{Moment evaluation and cancellation}

Set
\begin{equation}\label{eq:moment-values}
c_l=4^{-l}\binom{2l}{l}\quad(l\ge0),\qquad
m_i=\int_{-1}^1t^i\frac{|t|}{f(t)}\,dt
=\begin{cases}\displaystyle\frac{2}{(i+1)c_{i/2}},&i\ge0\text{ even},\\
0,&i\ge0\text{ odd},\end{cases}
\end{equation}
and use the convention $m_{-1}=0$.
The moment formula follows from symmetry, $m_0=2$, and integration by
parts, which gives $(i+1)m_i=i m_{i-2}$ for $i\ge1$.
Throughout, $c_z$ means zero unless $z$ is a nonnegative integer.
Termwise integration gives
\begin{equation}\label{eq:moment-series}
M(i,j)=\sum_{u\ge0}\frac{m_{i+u}}{j+u+1},\qquad
M(i,j)-M(i+1,j+1)=\frac{m_i}{j+1}.
\end{equation}

Put $K^-_d=M(d,0)$ and $K^+_d=M(0,d)$. Their boundary recurrences are
\begin{equation}\label{eq:boundary-recurrences}
\begin{aligned}
dK^-_d&=(d-1)K^-_{d-2}+m_{d-2}+m_{d-1} &&(d\ge2),\\
K^-_1&=2,\\
(d-1)K^+_d&=(d-2)K^+_{d-2}+\frac{2}{d-1} &&(d\ge2).
\end{aligned}
\end{equation}
For the first recurrence, the moment recurrence gives the termwise identity
\[
\frac{d m_{d+u}-(d-1)m_{d+u-2}}{u+1}
=m_{d+u-2}-m_{d+u}.
\]
The right side telescopes, since $m_v\to0$ as $v\to\infty$.
The same identity for $d=1$, with the term multiplied by $d-1$ omitted,
gives $K^-_1=m_{-1}+m_0=2$. For the last recurrence, shifting the
first series by two leaves, for $u\ge2$, the terms
\[
\frac{(d-1)m_{u-2}-(d-2)m_u}{d+u-1}=m_{u-2}-m_u.
\]
Their sum is $2$; the remaining boundary term is
$-2(d-2)/(d-1)$, giving the claimed result.

The two independent starting values are
\begin{equation}\label{eq:boundary-starts}
K^-_0=K^+_0=4G,\qquad K^+_1=\frac{\pi^2}{4}=\frac32\zeta(2).
\end{equation}
To verify the first, integrate in $s$ and put $t=\sin\theta$ on the
positive half interval. This gives
\[
K^-_0=\int_0^{\pi/2}\log\frac{1+\sin\theta}{1-\sin\theta}\,d\theta
=-4\int_0^{\pi/4}\log\tan v\,dv=4G.
\]
The last equality follows on setting $x=\tan v$ and integrating the
geometric series for $(1+x^2)^{-1}$ against $-\log x$.
For the other start, Equation~\eqref{eq:moment-series} and
Equation~\eqref{eq:moment-values} yield
\[
K^+_1=\sum_{k\ge1}\frac{1}{2k^2c_k}.
\]
The differential equation
$(1-x^2)y''-xy'=2$, with $y(0)=y'(0)=0$, determines the Taylor
series of $y=(\arcsin x)^2$ as
$\sum_{k\ge1}x^{2k}/(2k^2c_k)$. Its nonnegative coefficients allow
passage to $x=1$ by monotone convergence, giving $\pi^2/4$.
For completeness, integrating
\[
\sum_{k\ge1}\frac{\rho^k\sin(kx)}{k}
=\operatorname{Im}\bigl(-\log(1-\rho e^{ix})\bigr),\qquad0<\rho<1,
\]
over $0<x<\pi$ and letting $\rho\uparrow1$ gives
$2\sum_{k\text{ odd}}k^{-2}=\pi^2/4$.
Here the imaginary part is uniformly bounded and tends to $(\pi-x)/2$,
so dominated convergence applies. The odd sum is $3\zeta(2)/4$,
establishing the final equality in Equation~\eqref{eq:boundary-starts}.

Define $M^0(i,j)$ by the rational recurrences
\eqref{eq:moment-series}--\eqref{eq:boundary-recurrences}, replacing
only the starts $K^-_0=K^+_0$ and $K^+_1$ by zero.
More explicitly, let $k^-_d,k^+_d$ satisfy
Equation~\eqref{eq:boundary-recurrences} with
\begin{equation}\label{eq:rational-starts}
k^-_0=k^+_0=k^+_1=0,\qquad k^-_1=2.
\end{equation}
Then the following formulas specify the whole array without ambiguity:
\begin{equation}\label{eq:diagonal-reduction}
M^0(i,j)=
\begin{cases}
\displaystyle k^-_{i-j}-\sum_{k=1}^{j}\frac{m_{i-j+k-1}}{k},&i\ge j,\\[6pt]
\displaystyle k^+_{j-i}-\sum_{k=j-i+1}^{j}
                  \frac{m_{k-(j-i)-1}}{k},&i<j.
\end{cases}
\end{equation}
Empty sums are zero. We shall also use the algebraic convention
\begin{equation}\label{eq:negative-boundary}
M^0(-1,j)=k^+_{j+1}\qquad(j\ge0);
\end{equation}
this does not define an additional integral.
Let $B_i^{(d)}=\sum_{k=1}^i k^{-d}$, with $B_0^{(d)}=0$, and put
\begin{equation}\label{eq:rational-z}
Z^0(i,j)=\begin{cases}
-B_i^{(2)},&i=j,\\[2pt]
\displaystyle\frac{B_i^{(1)}-B_j^{(1)}}{i-j},&i\ne j.
\end{cases}
\end{equation}
The evaluations of the moments are
\begin{equation}\label{eq:rational-decomposition}
\begin{split}
M(i,j)&=M^0(i,j)+4G c_{(i-j)/2}
                      +\frac32\zeta(2)c_{(j-i-1)/2},\\
Z(i,j)&=Z^0(i,j)+[i=j]\zeta(2).
\end{split}
\end{equation}
Indeed, the homogeneous parts of the minus and plus boundary recurrences
propagate the starts by precisely these two $c$-kernels. Diagonal reduction
preserves both index differences. The formula for $Z$ follows from
\[
Z(i,j)=\sum_{u\ge0}\frac{1}{(i+u+1)(j+u+1)}
\]
by partial fractions when $i\ne j$, and directly when $i=j$.

For later valuation calculations, it is useful to solve the rational
boundary recurrences explicitly. Define
\begin{equation}\label{eq:h-star}
H_z^*=\begin{cases}
c_{z/2},&z\ge0\text{ even},\\
\displaystyle\frac{1}{z c_{(z-1)/2}},&z\ge1\text{ odd}.
\end{cases}
\qquad\frac{H_{z+2}^*}{H_z^*}=\frac{z+1}{z+2}.
\end{equation}
Then
\begin{equation}\label{eq:explicit-boundary-arrays}
\begin{split}
k^-_u&=\begin{cases}
\displaystyle H_u^*\sum_{\substack{1\le z\le u\\z\ \text{even}}}
            \frac{2}{z^2(H_z^*)^2},&u\text{ even},\\[6pt]
\displaystyle H_u^*\sum_{\substack{1\le z\le u\\z\ \text{odd}}}
            \frac{2}{z},&u\text{ odd},
\end{cases}\\
k^+_{u+1}&=H_u^*\sum_{\substack{1\le z\le u\\z\equiv u\ (2)}}
                 \frac{2}{z^2H_z^*}\qquad(u\ge0).
\end{split}
\end{equation}
To see this, divide each recurrence by the appropriate $H^*$ and use
the ratio in Equation~\eqref{eq:h-star}. For the minus recurrence the
inhomogeneous increment after division is $2/(z^2(H_z^*)^2)$ when
$z$ is even and $2/z$ when $z$ is odd. The odd case starts with
$k^-_1/H_1^*=2$. For the plus recurrence the increment is
$2/(z^2H_z^*)$, with initial values $k^+_1=0$ and $k^+_2=2$.
This proves all the formulas, including their empty-sum cases.

\begin{proposition}\label{prop:rationality}
Each raw entry $F_j(r)$ lies in $\mathbb Q+\mathbb QG$.
Consequently, if $G\in\mathbb Q$, then $\Delta_N\in\mathbb Q$ for
every positive integer $N$.
\end{proposition}
\begin{proof}
Since $f(t)^{-1}=\sum_{l\ge0}c_l t^{2l}$, the coefficient of
$\zeta(2)$ in $F_j(r)$ is
\[
\frac32\left(\sum_i[t^i]P_r\,c_{(j-i-1)/2}-[t^j]D_r\right)
=\frac32[t^j]\left(\frac{tP_r}{f}-D_r\right)=0
\]
by Equation~\eqref{eq:contact} and $j<L$. More explicitly,
\begin{equation}\label{eq:rational-raw-entry}
F_j(r)=\sum_i[t^i]P_r
       \left(M^0(i,j)+4G c_{(i-j)/2}\right)
       -\frac32\sum_i[t^i]D_r Z^0(i,j).
\end{equation}
The assertion follows from the rational arrays and the integer filter.
\end{proof}

\subsection{A uniform estimate at the prime two}

We normalize $v_p(p)=1$ and set $v_p(0)=+\infty$.
In the rest of this section assume $G\in\mathbb Q$, viewed also in
$\mathbb Q_2$. Constants allowed to depend on this fixed rational number
will carry a subscript $G$.

\begin{proposition}\label{prop:two-adic}
With $\delta=(q+g+h)/n=5/24$, one has, for every positive integer $N$,
\begin{equation}\label{eq:two-adic-bound}
v_2(\Delta_N)\ge
-\left(\frac{\delta}{2}+\frac{\delta^2}{8}\right)n^2
-O_G\bigl(n\log(n+2)\bigr)
=-\frac{505}{4608}n^2-O_G\bigl(n\log(n+2)\bigr).
\end{equation}
The same bound holds for every full minor of the raw column matrix.
\end{proposition}
\begin{proof}
We first construct a $2$-adic version of the geometric moment series:
\begin{equation}\label{eq:two-adic-smoothing}
M_{\rm s}(i,j)=\sum_{k\ge0}\frac{m_{i+k}}{j+k+1}\quad\text{in }\mathbb Q_2.
\end{equation}
For even $u$, the factorial formula implies
\[
v_2(m_u)=1+u-v_2\binom{u}{u/2}
\ge u+1-\log_2(u+1).
\]
The last bound follows, for example, by subtracting the factorial-floor
formulas: each binary place contributes at most one to the binomial
valuation. Odd moments are zero. If $0\le i,j<H$, every nonzero term
of Equation~\eqref{eq:two-adic-smoothing} therefore has valuation at least
\begin{equation}\label{eq:smoothing-tail-bound}
i+k+1-2\log_2(H+k)
\ge i+1-2\log_2 H+k-2\log_2(k+1)
\ge i-2\log_2 H-1.
\end{equation}
The middle expression tends to infinity with $k$. Thus the infinite
tail converges, and the final bound is uniform over all its terms and
over $0\le i,j<H$. In particular
$v_2(M_{\rm s}(i,j))\ge i-2\log_2H-1$.

The diagonal recurrence and both boundary recurrences hold for
$M_{\rm s}$ as well. The same finite telescoping calculations prove
this because their tails tend to zero $2$-adically. In particular its
minus start at one is forced to be $2$. There are therefore exactly
two homogeneous discrepancies between $M_{\rm s}$ and the rational
part $M^0(i,j)+4G c_{(i-j)/2}$. Write
\begin{equation}\label{eq:two-adic-discrepancies}
e_1=4G-M_{\rm s}(0,0),\qquad e_2=-M_{\rm s}(0,1).
\end{equation}
These are fixed elements of $\mathbb Q_2$, independent of $N$ and of
all degree indices, and
\[
M^0(i,j)+4G c_{(i-j)/2}
=M_{\rm s}(i,j)+e_1c_{(i-j)/2}+e_2c_{(j-i-1)/2}.
\]
Contracting the last kernel with $P_r$ and applying
Equation~\eqref{eq:contact} expresses each raw column as the sum of
three column vectors:
\begin{equation}\label{eq:two-adic-column-types}
\begin{split}
F_j(r)&=S_j(r)+E_j(r)+B_j(r),\\
S_j(r)&=M_{\rm s}(P_r,j),\\
E_j(r)&=e_1\sum_i[t^i]P_r\,c_{(i-j)/2},\\
B_j(r)&=\sum_i[t^i]D_r\left(e_2[i=j]-\frac32Z^0(i,j)\right).
\end{split}
\end{equation}
If either discrepancy is zero, the corresponding summands vanish;
there is no need to assign a finite valuation to a zero constant.

Here are the coefficient and denominator bounds needed for these columns.
Remove the factor $(1-t)^h$ and write the resulting polynomials as
\begin{equation}\label{eq:unconvolved-coefficients}
\frac{P_r(t)}{(1-t)^h}=\sum_{u\ge0}p_{r,u}t^{C-1-u},\qquad
\frac{D_r(t)}{(1-t)^h}=\sum_{u\ge0}d_{r,u}t^{C-1-u}.
\end{equation}
All sums here are finite. Induction in the Chebyshev recurrence shows
that the coefficient of $x^u$ in $T_d$ has valuation at least $u-1$
and that in $U_d$ has valuation at least $u$; the assertion at $u=0$
uses only integrality. Hence
\begin{equation}\label{eq:coefficient-valuation}
v_2(p_{r,u})\ge u-1,\qquad v_2(d_{r,u})\ge u-1.
\end{equation}
Convolution with $(1-t)^h$ uses integer coefficients. Combining
Equations~\eqref{eq:smoothing-tail-bound} and
\eqref{eq:coefficient-valuation}, every entry of $S_j$ has valuation
at least $C-2\log_2H-3$.

Set $L_2=\lfloor\log_2H\rfloor$. Equation~\eqref{eq:rational-z} gives
\begin{equation}\label{eq:harmonic-two-adic}
v_2(Z^0(i,j))\ge-2L_2\qquad(0\le i,j<H).
\end{equation}
Indeed, the harmonic sum of order $d$ has valuation at least $-dL_2$;
when $i\ne j$, division by $i-j$ loses at most another $L_2$.
The ultrametric inequality introduces no loss for the number of terms
in these sums or in polynomial convolutions.

Choose a fixed nonnegative integer $K_G$ with
$v_2(e_1),v_2(e_2)\ge-K_G$. Let $\mathbf p_u=(p_{r,u})_r$ and
$\mathbf d_u=(d_{r,u})_r$. The exceptional columns in
Equation~\eqref{eq:two-adic-column-types} have expansions
\[
E_j=\sum_u\mathbf p_u\,a_{u,j},\qquad
B_j=\sum_u\mathbf d_u\,b_{u,j},
\]
where
\begin{align*}
a_{u,j}&=e_1\sum_{v=0}^h(-1)^v\binom hv
                    c_{(C-1-u+v-j)/2},\\
b_{u,j}&=\sum_{v=0}^h(-1)^v\binom hv
 \left(e_2[C-1-u+v=j]-\frac32Z^0(C-1-u+v,j)\right).
\end{align*}
Only $u$ with a nonzero coefficient vector need be included, so all
moment indices in these expressions are between $0$ and $H-1$.
Since $v_2(c_l)\ge-2l$, every contributing summand yields
\begin{equation}\label{eq:exceptional-scalar-bounds}
v_2(a_{u,j})\ge u+j-H+1-K_G,\qquad
v_2(b_{u,j})\ge-2L_2-1-K_G.
\end{equation}

We now apply the bounds in the order needed to preserve alternation.
First expand $\det(\mathcal F\mathcal T)$ by Cauchy--Binet.
Each term is an integer times a full raw minor, whose column indices
$j_1,\ldots,j_n$ are distinct. Fix such a minor, expand each column
by Equation~\eqref{eq:two-adic-column-types}, and consider a term
containing $m$ columns of type $E$, $l$ of type $B$, and $n-m-l$
columns of type $S$.

Expand the $E$ columns through the fixed vectors $\mathbf p_u$ and
the $B$ columns through the fixed vectors $\mathbf d_u$.
Repeated indices within the first group give equal coefficient vectors
and hence zero determinants; the same holds within the second group.
No distinctness between the two groups is asserted or needed.
For nonzero terms we therefore have
\[
\sum_{E\text{ columns}}u\ge\frac{m(m-1)}2,\qquad
\sum_{B\text{ columns}}u\ge\frac{l(l-1)}2.
\]
The already fixed, distinct raw column indices also give
\[
\sum_{E\text{ columns}}j\ge mb+\frac{m(m-1)}2.
\]
Equations~\eqref{eq:coefficient-valuation} and
\eqref{eq:exceptional-scalar-bounds} show that the first group contributes
at least $2\sum u+\sum j-Hm-K_Gm$.
The second contributes at least $\sum u-O_G(l\log(H+2))$;
the smoothed columns contribute at least
$C(n-m-l)-O((n-m-l)\log(H+2))$.
Thus every term, every raw minor, and finally the filtered determinant
satisfies
\begin{equation}\label{eq:two-adic-quadratic}
\begin{split}
v_2(\Delta_N)\ge{}&-O_G(n\log(n+2))\\
&+\min_{\substack{m,l\ge0\\m+l\le n}}
\left\{-(H-b)m+\frac32m^2+\frac12l^2+C(n-m-l)\right\}.
\end{split}
\end{equation}
Finite summation causes no further loss in valuation.

For completeness minimize over real $m,l$; this can only lower the
minimum over integer choices. Since $C\ge n$, the expression decreases
as $l$ increases up to $n-m$, so its minimum occurs at $l=n-m$.
Using $H-b=n(1+\delta)$, the remaining expression is
\[
\frac{n^2}{2}-(2+\delta)nm+2m^2
=2\left(m-\frac{(2+\delta)n}{4}\right)^2
 -\left(\frac\delta2+\frac{\delta^2}{8}\right)n^2.
\]
This proves Equation~\eqref{eq:two-adic-bound} and the stated
bound for each raw minor.
\end{proof}
==== END Catalans-constant-is-irrational//foundations.tex ====

==== BEGIN Catalans-constant-is-irrational//introduction.tex ====
\section{Introduction}\label{sec:introduction}

Catalan's constant is the real number
\[
G=\beta(2)=\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)^2},
\qquad
\beta(s)=\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)^s}\quad(s>0).
\]
It is the simplest even value of the Dirichlet beta function, or
equivalently $L(2,\chi_{-4})$ for the odd quadratic character of
conductor $4$. Its arithmetic illustrates a basic difficulty in the
study of special values: rapidly converging rational approximations
need not be sufficiently accurate after their denominators are cleared.
We prove the following statement about this individual value.

\begin{theorem}\label{thm:main}
Catalan's constant is irrational.
\end{theorem}

\subsection{Earlier results and the approximation problem}

For odd positive integers, the beta values are rational multiples of the
corresponding powers of $\pi$. The even values present a different
arithmetic problem \cite[Introduction]{RivoalZudilin2003}.
Rivoal and Zudilin proved that infinitely many even beta values are
irrational, and that at least one of
$\beta(2),\beta(4),\ldots,\beta(14)$ is irrational
\cite[Theorems~1 and~2]{RivoalZudilin2003}.
Zudilin reduced this finite collection to the six values through
$\beta(12)$ \cite[Theorem~1]{Zudilin2019}.
Lai and Zhou subsequently reduced it to the five values
$\beta(2),\beta(4),\beta(6),\beta(8),\beta(10)$
\cite[Theorem~6.1]{LaiZhou2022}.
Fischler obtained stronger quantitative results for irrationality and
linear independence in families of Dirichlet $L$-values
\cite[Theorems~1 and~2]{Fischler2020}.
Such family results guarantee irrational members without identifying
the particular value $\beta(2)$.

For an individual value, the basic approximation criterion requires
nonzero integer linear forms $A_mG-B_m$ that tend to zero.
Convergence of $B_m/A_m$ to $G$ alone does not suffice: multiplication
by $A_m$ may remove the decay.
Ap\'ery's recurrence constructions for $\zeta(2)$ and $\zeta(3)$
\cite{Apery1979}, and Beukers's integral proofs of their irrationality
\cite{Beukers1979} succeed because the analytic decay survives the
necessary denominator clearing.
For Catalan's constant, Zudilin constructed Ap\'ery-like recurrences,
a continued fraction and double-integral representations
\cite[Theorems~1--3]{Zudilin2003}.
His discussion following Theorem~1 makes the obstruction explicit:
the displayed integer linear forms do not tend to zero, despite the
rapid convergence of their rational quotients.

The denominators themselves became an important part of the problem.
Rivoal connected Pad\'e approximation with the hypergeometric
construction and proved its conjectured denominator bounds
\cite[Theorems~1 and~2]{Rivoal2006Catalan}.
Krattenthaler and Rivoal subsequently gave a more direct hypergeometric
proof of these bounds \cite[Theorems~1 and~2]{KrattenthalerRivoal2008Catalan}.
These results establish arithmetic cancellation that is invisible in
a crude estimate of the individual summands. They still leave a
denominator cost too large for the small-linear-form criterion.
Krattenthaler and Zudilin later identified two apparently different
hypergeometric constructions of the same approximants
\cite[Section~2 and Theorem~2]{KrattenthalerZudilin2019}.
Such identities matter arithmetically because different expressions can
make different denominator factors visible.

Nesterenko developed effective approximations using half-integer
hypergeometric series and double Euler integrals
\cite{Nesterenko2016Catalan}.
Viola announced joint work with Marcovecchio improving an approximation
exponent through the Rhin--Viola permutation-group method
\cite{Viola2022Catalan}; their report gives the exponent $0.6293\ldots$
for its explicit approximants. A recent preprint of Eskandari constructs
further explicit rational approximations satisfying
$0<|G-p_m/q_m|\le q_m^{-0.62}$ for all sufficiently large $m$
\cite[Theorem~1.1]{Eskandari2026Approximations}.
These are bounds for particular constructions, not irrationality
measures. An error bound with exponent below $1$ does not by itself
make the integer forms $q_mG-p_m$ tend to zero.

Results at other characters and other places give useful comparisons.
Calegari, Dimitrov and Tang proved the $\mathbb Q$-linear independence
of $1$, $\pi^2$ and $L(2,\chi_{-3})$
\cite[Theorem~A]{CalegariDimitrovTang2024}.
Their character has conductor $3$, rather than the conductor $4$ of $G$.
Their discussion of two Catalan approximation families again identifies
the denominator cost as an obstruction for those constructions
\cite[Remark~11.1.17]{CalegariDimitrovTang2024}.
Calegari proved irrationality of a $2$-adic analogue
$G_2\in\mathbb Q_2$ \cite[Theorem~4.2]{Calegari2005}.
As explained by Calegari, Dimitrov and Tang
\cite[Section~5.2, footnote~2]{CalegariDimitrovTang2025}, the related
real identity contains an additional $\pi^2$ term that is absent in
the $2$-adic identity. This illustrates why the $2$-adic result does
not settle the real problem. In our construction, cancellation of an
additional $\zeta(2)$ term is likewise essential, although the moment
identities and the cancellation argument are proved directly below.

Sun~\cite[Theorem~1.1]{Sun2026} has also announced a proof of the same
qualitative irrationality statement. No result from that preprint is
used here.

\subsection{The determinant strategy}

We construct determinants $\Delta_N$ of size $n=48N$ whose entries
combine moments of two elementary kernels. Each moment is a rational
linear combination of $1$, $G$ and $\zeta(2)$. The polynomial rows are
chosen to have high Taylor contact: two prescribed expressions in the
rows have identical initial Taylor coefficients. This agreement cancels the $\zeta(2)$ term
in every entry. Consequently, the single hypothesis $G\in\mathbb Q$
makes all the determinants rational.

For a nonzero rational number, its ordinary absolute value is determined
by its prime valuations. Bounds on the denominators of $\Delta_N$
therefore give a lower bound on $\log|\Delta_N|$. An integral formula
for the same determinant gives an upper bound. The contradiction comes
from making these bounds incompatible as the size grows.
Zudilin's determinantal criterion gives a useful precedent for this
comparison \cite[Proposition~2 and Section~2]{Zudilin2017Determinantal}.
There, a positive moment representation produces Hankel determinants
with squared-Vandermonde integrals. Positivity supplies nonvanishing,
and the denominator and integral estimates are compared on the scale
of the square of the matrix size. His discussion of Catalan's constant
explains why the denominator growth of the approximation family
considered there still prevents application of the criterion
\cite[Section~6]{Zudilin2017Determinantal}.
Our determinant is mixed and signed; its nonvanishing and its
real-place estimate require separate arguments.

Three features of the construction make this comparison possible.
First, integral Chebyshev rows provide both the Taylor contact and
useful divisibility. At the large odd primes relevant to the leading
bound, denominators of order $p^2$ and $p$ must both be controlled.
The arithmetic estimate follows these two layers separately; cancellation
at the first layer alone would not give the required denominator bound.

Second, nonvanishing is arithmetic. Under the hypothesis $G\in\mathbb Q$,
when $N=p$ is itself a sufficiently large prime, Frobenius and a pair of
polynomial bases reduce the growing matrix to three fixed rational
matrices. An exact finite certificate proves their nonvanishing.
The reduction then gives nonzero determinants along an unbounded
prime sequence, without asserting positivity of the mixed determinant.

Third, the real estimate respects both branches of the rational
parametrization used to construct the rows. A bounded holomorphic
interpolant controls their mixed evaluations in one range of
configurations; a Hadamard bound treats the complementary range.
The interpolation estimate is uniform even when nodes approach one
another. Its proof uses a finite-dimensional Hardy-space operator,
so the evaluation determinant cancels algebraically rather than
introducing an inverse-Vandermonde estimate.
Andr\'eief's integration identity \cite{Forrester2018Andreief},
Cauchy's double alternant \cite{Krattenthaler1999}, and the
kernel/compression viewpoint of analytic interpolation
\cite{Sarason1967} supply the classical context; the
specific estimates are proved below.

The resulting Vandermonde products have logarithmic interactions,
as in logarithmic potential theory \cite{Saff2010}.
A Chebyshev expansion of the logarithmic kernel
\cite[Lemma~3.1]{GaroufalidisPopescu2013} reduces the estimate to
quadratic sums and two one-variable functions in each case. We regularize
their interactions together, control the omitted diagonals and endpoint
terms, and only then take a supremum and pass to the limit.
Explicit rational trial coefficients finish the argument. Their
certificate exhausts all stationary points and bounds entire root
brackets, rather than relying on a numerical search for the maxima.

For orientation, the two estimates are stated for the same normalized
logarithm
\begin{equation}\label{eq:normalized-logarithm}
\mathcal L_N=\frac{\log|\Delta_N|}{(48N)^2}-\frac12\log 2.
\end{equation}
Under the rationality hypothesis, Proposition~\ref{prop:finite-lower}
gives $\liminf\mathcal L_N>-2.29084$ along every sequence of nonzero
determinants, and Proposition~\ref{prop:nonvanishing} supplies such a
sequence with $N$ prime. Independently,
Proposition~\ref{prop:real-upper} gives
$\limsup\mathcal L_N\le-2.290939875<-2.2909$.
These inequalities are incompatible. The argument proves qualitative
irrationality; a bound for an irrationality measure would require
additional control of rational approximations.
Section~\ref{sec:conclusion} also deduces irrationality of the minimum
volume of an orientable complete finite-volume hyperbolic three-manifold
with exactly two cusps, and of certain arithmetic hyperbolic volumes.

\subsection{Organization and conventions}

Section~\ref{sec:foundations} constructs the rows and moments, proves
cancellation and rationality, and gives the estimate at the prime $2$.
Section~\ref{sec:odd-primes} treats odd primes and derives the
finite-place lower bound. Its only prime-distribution input is the
ordinary prime number theorem; the weighted consequence needed here
is proved locally. Section~\ref{sec:nonvanishing} establishes
nonvanishing at prime scales. Sections~\ref{sec:real-place}
and~\ref{sec:energy} give the uniform real-place and energy estimates.
Section~\ref{sec:certificate} supplies the rational data certifying the real-place bound,
and Section~\ref{sec:conclusion} assembles the contradiction.

All logarithms are natural. We use $v_p(p)=1$ and $v_p(0)=+\infty$;
for real estimates we put $\log 0=-\infty$.
Constants in asymptotic estimates may depend on the fixed rational value
hypothetically assigned to $G$, but never on $N$ or on the integration
points. The finite certificates are specified by rational data and
arithmetic instructions in the text; supplementary programs reproduce them.
==== END Catalans-constant-is-irrational//introduction.tex ====

==== BEGIN Catalans-constant-is-irrational//main.tex ====
\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amssymb,amsthm,mathtools}
\usepackage{booktabs,longtable,array}
\usepackage{microtype}
\usepackage[numbers,sort&compress]{natbib}
\usepackage[colorlinks=true,linkcolor=blue,citecolor=blue,urlcolor=blue]{hyperref}
\hypersetup{pdftitle={Catalan's constant is irrational},pdfauthor={OpenAI}}
\newtheorem{theorem}{Theorem}[section]
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\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
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\newtheorem{remark}[theorem]{Remark}
\title{Catalan's constant is irrational}
\author{OpenAI}
\date{September 24, 2026}

\begin{document}
\maketitle
\begin{abstract}
We prove that Catalan's constant
$G=\sum_{j\geq0}(-1)^j/(2j+1)^2$ is irrational.
\end{abstract}

\input{sections/introduction}
\input{sections/foundations}
\input{sections/odd-primes}
\input{sections/nonvanishing}
\input{sections/real-place}
\input{sections/energy}
\input{sections/certificate}
\input{sections/conclusion}

\bibliographystyle{plainnat}
\begingroup
\raggedright
\bibliography{references}
\endgroup
\end{document}
==== END Catalans-constant-is-irrational//main.tex ====

==== BEGIN Catalans-constant-is-irrational//nonvanishing.tex ====
\section{Nonvanishing along the prime sequence}
\label{sec:nonvanishing}

The next proposition supplies the nonzero determinants needed in
Proposition~\ref{prop:finite-lower}.

\begin{proposition}\label{prop:nonvanishing}
Assume that $G\in\mathbb Q$. For every sufficiently large prime $p$, the
determinant with scale parameter $N=p$ satisfies
\begin{equation}\label{eq:nonvanishing-exact-valuation}
 v_p(\Delta_p)=-96p.
\end{equation}
In particular, $\Delta_p\ne0$ for every sufficiently large prime $p$.
\end{proposition}

We prove the proposition by reducing a matrix of size $48p$ to three fixed
rational matrices of size $48$. Throughout this section, an underlined row
polynomial is formed with $N=1$, so that
\[
 n_0=48,\quad b_0=7,\quad q_0=g_0=4,\quad h_0=2,\quad
 L_0=59,\quad C_0=63,\quad H_0=65.
\]
In addition to the usual rows $0,\ldots,47$, define the auxiliary row $48$
by exactly the formulas in Equation~\eqref{eq:rows}. For $0\le r\le48$ and
$0\le k<48$, put
\begin{equation}\label{eq:nonvanishing-base-matrix}
 \mathcal B(r,k)=\sum_{j=0}^{4}(-1)^j\binom4j
 \left\{M^0(\underline P_r,7+k+j)
              -\frac32 Z^0(\underline D_r,7+k+j)\right\}.
\end{equation}
Here $M^0$ and $Z^0$ act linearly on the first polynomial argument. Thus
$\mathcal B$ is a fixed rational $49$ by $48$ matrix. Define
\[
 \mathcal B_0=(\mathcal B(r,k))_{0\le r,k<48},\qquad
 \mathcal B_1=(\mathcal B(r+1,k))_{0\le r,k<48}.
\]
The arithmetic certificate below proves
\begin{equation}\label{eq:nonvanishing-fixed-determinants}
 \det\mathcal B_0\ne0,\qquad
 \det(\mathcal B_0+\mathcal B_1)\ne0,\qquad
 \det(\mathcal B_0-\mathcal B_1)\ne0.
\end{equation}
First we show why these three fixed assertions imply the proposition.

\subsection{Two palindromic bases}

Let $p$ be an odd prime and work over $\mathbb F_p$. The vector space
\[
 \mathcal P_p=\{Q\in\mathbb F_p[w]:\deg Q\le2p-2,
                \ w^{2p-2}Q(1/w)=Q(w)\}
\]
has dimension $p$: its coefficients of $w^0,\ldots,w^{p-1}$ determine
all its coefficients. For $0\le\ell,i<p$, consider
\[
 U_\ell(w)=(2w)^\ell(1+w^2)^{p-1-\ell},\qquad
 Q_i(w)=w^i\sum_{j=0}^{p-i-1}w^{2j}.
\]
Both families belong to $\mathcal P_p$. The lowest terms of $U_\ell$
and $Q_i$ are respectively $2^\ell w^\ell$ and $w^i$. Triangularity
of their coefficients in degrees $0,\ldots,p-1$ proves that each family
is a basis. Define $a_{\ell i}\in\mathbb F_p$ by
\[
 Q_i=\sum_{\ell=0}^{p-1}a_{\ell i}U_\ell,
 \qquad \mathbf a=(a_{\ell i})_{0\le\ell,i<p}.
\]
The same lowest-term comparison gives
\begin{equation}\label{eq:nonvanishing-triangular-unit}
 a_{\ell i}=0\quad(\ell<i),\qquad a_{ii}=2^{-i},\qquad
 \det\mathbf a=2^{-p(p-1)/2}\ne0.
\end{equation}
In particular, invertibility of this varying-size matrix introduces no
exceptional odd primes.

Define $Q'_0=0$ and $Q'_i=Q_{p-i}$ for $1\le i<p$, and write
$Q'_i=\sum_\ell b_{\ell i}U_\ell$. Equivalently,
\begin{equation}\label{eq:nonvanishing-transition}
 \mathbf b=\mathbf a\Pi,\qquad
 \Pi e_0=0,\qquad \Pi e_i=e_{p-i}\quad(1\le i<p),
\end{equation}
where $e_i$ are the standard coordinate vectors. The explicit forms are
\[
 Q_i=w^i\frac{1-w^{2(p-i)}}{1-w^2},\qquad
 Q'_i=w^{p-i}\frac{1-w^{2i}}{1-w^2};
\]
the latter formula gives zero also when $i=0$. Their sum satisfies
\[
 Q_i+w^pQ'_i
 =w^i\sum_{j=0}^{p-1}w^{2j}
 =w^i(1-w^2)^{p-1},
\]
because $\binom{p-1}{j}=(-1)^j$ in $\mathbb F_p$. With
\[
 t=\frac{2w}{1+w^2},\qquad f=\frac{1-w^2}{1+w^2},\qquad
 E(t)=(1-t^2)^{(p-1)/2}=f^{p-1},
\]
division by $(1+w^2)^{p-1}$ therefore proves the rational-function identity
\begin{equation}\label{eq:nonvanishing-palindromic-expansion}
 E(t)w^i=\sum_{\ell=0}^{p-1}t^\ell
              (a_{\ell i}+b_{\ell i}w^p).
\end{equation}

\subsection{Frobenius and the two extraction shifts}

Set $N=p$, and write every row index uniquely as
$r=pr_0+i$, with $0\le r_0<48$ and $0\le i<p$. Put
$t'=t^p$, $w'=w^p$, and $f'=f^p$. Frobenius gives
\[
 t'=\frac{2w'}{1+(w')^2},\qquad
 f'=\frac{1-(w')^2}{1+(w')^2},\qquad (f')^2=1-(t')^2.
\]
Using $C=63p$, $h=2p$, and $g=4p$ in the row definition gives
\begin{align}
 tR_rE(t)
 &= (1-t')^2(t')^{63}(w')^{r_0-4}E(t)w^i\notag\\
 &=\sum_{\ell=0}^{p-1}t^\ell t'
       \bigl(a_{\ell i}\underline R_{r_0}(t',w')
            +b_{\ell i}\underline R_{r_0+1}(t',w')\bigr).
 \label{eq:nonvanishing-frobenius-row}
\end{align}
Here the single factor $w'$ in the second term creates exactly one successor
row; its index is at most $48$.

To separate the polynomial parts, star fixes $t,t'$ and negates $f,f'$.
Moreover $fE=f'$, so the row identities give
\[
 tR_rE=tP_rE-f'D_r,
 \qquad t'\underline R_j=t'\underline P_j-f'\underline D_j.
\]
Taking the star-invariant part of
Equation~\eqref{eq:nonvanishing-frobenius-row} and then its anti-invariant
part, and cancelling the nonzero rational function $f'$ in the latter,
yields two separate polynomial identities:
\begin{align}
 tP_r(t)E(t)
 &=\sum_{\ell=0}^{p-1}t^\ell t'
       \bigl(a_{\ell i}\underline P_{r_0}(t')
              +b_{\ell i}\underline P_{r_0+1}(t')\bigr),
       \label{eq:nonvanishing-P-identity}\\
 D_r(t)
 &=\sum_{\ell=0}^{p-1}t^\ell
       \bigl(a_{\ell i}\underline D_{r_0}(t')
              +b_{\ell i}\underline D_{r_0+1}(t')\bigr).
       \label{eq:nonvanishing-D-identity}
\end{align}
These are polynomial identities since the row polynomials on both sides
are polynomials, and substitution $t=2w/(1+w^2)$ is injective on
$\mathbb F_p[t]$.

In the notation of Equation~\eqref{eq:column-layer}, the extractions for a
fixed residue $\ell$ are
\[
 P'_u=[t^{(u+1)p+\ell}]tP_r(t)E(t),\qquad
 D'_u=[t^{up+\ell}]D_r(t).
\]
Every polynomial has a unique expression
$\sum_{\ell=0}^{p-1}t^\ell A_\ell(t^p)$. Consequently
Equations~\eqref{eq:nonvanishing-P-identity} and
\eqref{eq:nonvanishing-D-identity} give
\begin{equation}\label{eq:nonvanishing-extracted-rows}
 (P'_u,D'_u)=[(t')^u]
 \left\{a_{\ell i}(\underline P_{r_0},\underline D_{r_0})(t')
        +b_{\ell i}(\underline P_{r_0+1},\underline D_{r_0+1})(t')\right\}.
\end{equation}
Thus the $t'$ in the first identity is exactly consumed by the $u+1$
in the $P$ extraction. The $D$ extraction has no such shift. Negative
coefficient indices are zero; in particular $P'_{-1}=0$ here.

There is a degree issue at the last successor row which is useful to make
explicit. The auxiliary polynomials have
\begin{equation}\label{eq:nonvanishing-auxiliary-degrees}
 \min\deg\underline P_{48}=62-44=18,\qquad
 \min\deg\underline D_{48}=63-44=19.
\end{equation}
These coefficients must be retained, even though the first $48$ base rows
have minimum degrees at least $19$ and $20$. For the last block
$r=47p+i$, $i>0$, Equation~\eqref{eq:nonvanishing-transition} gives
$b_{\ell i}=a_{\ell,p-i}=0$ for $\ell<p-i$, and
$b_{p-i,i}=2^{-(p-i)}\ne0$. The successor contribution in either
$tP_rE$ or $D_r$ therefore starts at
\[
 19p+(p-i)=20p-i.
\]
The first-row contribution starts no earlier than $20p+i$.
Directly from the original rows, $|r-g|=43p+i$, so $tP_r$ and $D_r$
also start at $C-|r-g|=20p-i$; since $E(0)=1$, this agrees with the
extraction. The leading Chebyshev coefficients are powers of $2$, hence
are nonzero in odd characteristic. When $i=0$, the successor term is
zero and the first term starts at $20p$. The lower degree of the
auxiliary row therefore introduces no forbidden coefficient into the
growing matrix. Its contact identity is also valid:
$t\underline R_{48}/f=O(t^{107})$, since $63+48-4=107>59$.

\subsection{The residue blocks and their determinant}

For this paragraph take $p>260$ and exclude primes dividing the denominator
of the hypothesized rational number $G$. Then
\[
 p>2\sqrt{65p}=2\sqrt H,
\]
so Equations~\eqref{eq:digit-layers} and \eqref{eq:column-layer} apply at
the same prime $p$ used as the scale parameter. To spell out the role of
$G$, the rational raw entry is
\[
 F_j(r)=M^0(P_r,j)-\frac32 Z^0(D_r,j)
          +4G\sum_v[t^v]P_r\,c_{(v-j)/2}.
\]
The last sum is $p$-integral: its polynomial coefficients are integers,
and every $c_d=4^{-d}\binom{2d}{d}$ is integral at an odd prime.
The $G$ term therefore reduces to zero after multiplication by $p^2$.
All entries of the scaled matrix are $p$-integral by the cited layer
formulas. Substitution of Equation~\eqref{eq:nonvanishing-extracted-rows}
into Equation~\eqref{eq:column-layer} gives the reduced raw entry
\begin{align}
 p^2F_{kp+\ell}(pr_0+i)
 \equiv{}&a_{\ell i}\left\{M^0(\underline P_{r_0},k)
                         -\frac32 Z^0(\underline D_{r_0},k)\right\}
                         \notag\\
 &+b_{\ell i}\left\{M^0(\underline P_{r_0+1},k)
                         -\frac32 Z^0(\underline D_{r_0+1},k)\right\}
                         \pmod p.
 \label{eq:nonvanishing-raw-bridge}
\end{align}
This applies to every raw column in the specified range. The base
entries here have denominators with prime factors at most $65$, as
the rational recipes below also show, so their reductions exist for
$p>260$.

The integer filter also respects residues. Write the column index
uniquely as $k=\ell+pk_0$, where
$0\le\ell<p$ and $0\le k_0<48$. In $\mathbb F_p[s]$,
\begin{equation}\label{eq:nonvanishing-residue-filter}
 s^{7p+k}(1-s)^{4p}
   =s^\ell(s^p)^{7+k_0}(1-s^p)^4.
\end{equation}
Multiplying each raw entry by $p^2$ before reduction makes this polynomial
congruence applicable to the column operation: coefficient differences
divisible by $p$ multiply integral scaled entries. The fourth finite
difference on the right uses exactly the five base raw indices
$7+k_0,\ldots,11+k_0$,
all in $7,\ldots,58$. Each of the $p$ residue groups thus contains
exactly $48$ base filtered columns, and no operation changes $\ell$.

Order rows by $(i,r_0)$ and columns by $(\ell,k_0)$. The reduction of
the scaled $48p$ by $48p$ matrix has $(i,\ell)$ block
\[
 a_{\ell i}\mathcal B_0+b_{\ell i}\mathcal B_1.
\]
Writing this matrix as $\mathcal L_p$ makes its orientation explicit:
\begin{equation}\label{eq:nonvanishing-block-matrix}
 \begin{split}
 \mathcal L_p&=\mathbf a^T\otimes\mathcal B_0
                       +\mathbf b^T\otimes\mathcal B_1,\\
 \mathcal L_p\bigl((\mathbf a^T)^{-1}\otimes I_{48}\bigr)
      &=I_p\otimes\mathcal B_0+\Pi^T\otimes\mathcal B_1.
 \end{split}
\end{equation}
Indeed $\mathbf b^T=\Pi^T\mathbf a^T$, which explains both the
transpose and the side of multiplication.

The nonzero indices $1,\ldots,p-1$ fall into $(p-1)/2$ disjoint pairs
$\{i,p-i\}$. On each pair the vectors $e_i+e_{p-i}$ and
$e_i-e_{p-i}$ are eigenvectors of $\Pi$ with eigenvalues $1$ and
$-1$. Together with $e_0$, of eigenvalue zero, they form a basis since
$2$ is invertible. Thus the eigenvalue multiplicities of $\Pi^T$ are
$1,(p-1)/2,(p-1)/2$ for $0,1,-1$, respectively. A similarity on the
$p$-dimensional factor in Equation~\eqref{eq:nonvanishing-block-matrix}
now proves the exact identity in this block ordering:
\begin{align}
 \det\mathcal L_p
  ={}&(\det\mathbf a)^{48}\det\mathcal B_0\notag\\
    &\quad\cdot\det(\mathcal B_0+\mathcal B_1)^{(p-1)/2}
          \det(\mathcal B_0-\mathcal B_1)^{(p-1)/2}.
 \label{eq:nonvanishing-factorization}
\end{align}
There is no determinant factor from the similarity. Returning to the
original row and column orders changes at most the sign.

\subsection{An exact certificate for the fixed matrices}

All entries in Equation~\eqref{eq:nonvanishing-base-matrix} can be
constructed by the following rational arithmetic. This also specifies
their reduction modulo $101$ without any numerical approximation.
First form $m_0=2$, $m_i=0$ for odd $i$, and
$m_i=i m_{i-2}/(i+1)$ for even $2\le i\le64$. Set
\[
 k^-_0=0,\quad k^-_1=2,\qquad k^+_0=k^+_1=0,
\]
and use
\begin{align*}
 k^-_d&=\frac{(d-1)k^-_{d-2}+m_{d-2}+m_{d-1}}{d}
                  &&(2\le d\le64),\\
 k^+_d&=\frac{(d-2)k^+_{d-2}+2/(d-1)}{d-1}
                  &&(2\le d\le58).
\end{align*}
On $0\le i\le64$, $0\le j\le58$, form
\[
 Y_{i0}=k^-_i,\qquad Y_{0j}=k^+_j,\qquad
 Y_{ij}=Y_{i-1,j-1}-\frac{m_{i-1}}j\quad(i,j\ge1).
\]
Thus $Y_{ij}=M^0(i,j)$. With $B_i^{(d)}=\sum_{u=1}^i u^{-d}$ and
$B_0^{(d)}=0$, set
\[
 J_{ij}=\frac32\begin{cases}
       -B_i^{(2)},&i=j,\\[2pt]
       (B_i^{(1)}-B_j^{(1)})/(i-j),&i\ne j.
      \end{cases}
\]
This gives $J_{ij}=\tfrac32Z^0(i,j)$, with the sign on its diagonal
included.

For the polynomial construction define
\[
 E_0=t^{62},\quad E_1=t^{61},\qquad I_0=0,\quad I_1=t^{62},
\]
and, for either family $S=E,I$, use
\[
 S_m=2S_{m-1}/t-S_{m-2}\qquad(2\le m\le44).
\]
These are ordinary integer polynomials: they are
$E_m=t^{62}T_m(1/t)$ and $I_m=t^{62}U_{m-1}(1/t)$.
For $0\le r\le48$ let $u=|r-4|$ and form the coefficient vectors,
in degrees $0,\ldots,64$, of
\[
 y_r(t)=(1-t)^2E_u(t),\qquad
 z_r(t)=\operatorname{sign}(r-4)(1-t)^2I_u(t).
\]
They are exactly $\underline P_r$ and $\underline D_r$.
Contract them with the arrays to obtain the length-$52$ vector
\[
 v_r(j)=\sum_{i=0}^{64}\bigl([t^i]y_r\,Y_{ij}
                                  -[t^i]z_r\,J_{ij}\bigr),
               \qquad j=7,\ldots,58.
\]
Replace a vector $v$ four times successively by its vector of consecutive
differences $(v(j)-v(j+1))_j$. The resulting length-$48$ vector is
row $r$ of $\mathcal B$. In particular, the construction includes all
coefficients of the auxiliary row, including its degree-$18$ term.

Every scalar denominator in these recipes is a product of nonzero integers
of absolute value at most $65$. Every rational array entry therefore has
denominator prime factors at most $65$, so each denominator is a unit
modulo $101$. Polynomial division by $t$ above shifts exponents of
polynomials divisible by $t$ and introduces no scalar denominator.
Consequently the entire construction can be performed over
$\mathbb F_{101}$ and agrees with reduction of the rational matrices.

For completeness, the elimination rule producing the certificate is as
follows. For $\sigma\in\{0,1,-1\}$, start with row vectors
$R_i=\mathcal B(i,{\cdot})+\sigma\mathcal B(i+1,{\cdot})$,
$0\le i<48$, in increasing order. At step $i$, if $R_i(i)=0$, swap
row $i$ with the first later row having a nonzero entry in column $i$.
Record $d_i=R_i(i)$ and replace every row $j>i$ by
\[
 R_j-\frac{R_j(i)}{d_i}R_i.
\]
There are no swaps for $\sigma=0,-1$. For $\sigma=1$ the only swaps,
using indices starting at zero, are $(30,31)$ and $(45,46)$ at their
respective steps. The pivots are given in
Table~\ref{tab:modular-pivots}; its three lines for each $\sigma$
list the pivots consecutively, with $16$ entries per line.

\begin{table}[htbp]
\centering
\small
\begin{tabular}{cl}
\toprule
$\sigma$ & Successive pivots in $\mathbb F_{101}$\\
\midrule
$0$ & $60,68,62,79,47,32,69,57,30,72,35,66,43,20,85,48$\\
& $88,4,77,54,60,79,26,68,83,39,40,65,1,68,78,24$\\
& $15,98,32,22,94,9,99,10,15,75,4,2,25,53,90,79$\\[2pt]
$1$ & $38,51,90,70,5,21,88,55,45,20,35,41,77,10,18,25$\\
& $76,14,38,72,6,66,56,35,83,58,56,11,17,20,30,24$\\
& $28,11,25,70,79,99,66,38,4,41,63,91,63,17,90,98$\\[2pt]
$-1$ & $82,92,26,87,21,74,87,88,88,3,14,23,38,58,36,20$\\
& $26,33,94,74,78,45,93,86,73,66,45,30,61,3,88,27$\\
& $20,58,69,48,78,39,48,1,66,18,86,93,52,92,39,77$\\
\bottomrule
\end{tabular}
\caption{Exact Gaussian pivots for $\mathcal B_0+\sigma\mathcal B_1$
modulo $101$.}\label{tab:modular-pivots}
\end{table}

Every listed pivot is nonzero. Since the swaps and row additions are
invertible, all three matrices are invertible modulo $101$ and hence
have nonzero determinants over $\mathbb Q$. This proves
Equation~\eqref{eq:nonvanishing-fixed-determinants}. The modulus $101$
is used only for this fixed finite certificate; the growing determinant
uses arbitrary sufficiently large primes as follows.

\begin{proof}[Completion of the proof of Proposition~\ref{prop:nonvanishing}]
Let $\mathcal E$ consist of $2$, the primes dividing the denominator of
$G$, the primes dividing any denominator of an entry of $\mathcal B$,
and the primes dividing the numerator or denominator, in lowest terms,
of any of the three nonzero rational determinants in
Equation~\eqref{eq:nonvanishing-fixed-determinants}. This is a fixed
finite set. For $p>260$ outside $\mathcal E$, all three fixed matrices
reduce to invertible matrices over $\mathbb F_p$. Equation~
\eqref{eq:nonvanishing-triangular-unit} and the factorization in
Equation~\eqref{eq:nonvanishing-factorization} then show that
$\det\mathcal L_p\ne0$.

Let $\mathcal F_p$ be the original filtered matrix whose determinant is
$\Delta_p$. Its size is $n=48p$, and $p^2\mathcal F_p$ is integral
over the local ring $\mathbb Z_{(p)}$. Its reduction, after the stated
row and column permutations, is $\mathcal L_p$. Therefore
\[
 p^{2n}\Delta_p=\det(p^2\mathcal F_p)
\]
is a unit in $\mathbb Z_{(p)}$. This proves
$v_p(\Delta_p)=-2n=-96p$. Every sufficiently large prime is outside
$\mathcal E$ and exceeds $260$, as required.
\end{proof}
==== END Catalans-constant-is-irrational//nonvanishing.tex ====

==== BEGIN Catalans-constant-is-irrational//odd-primes.tex ====
\section{Odd primes and the finite-place lower bound}
\label{sec:odd-primes}

Throughout this section assume that $G\in\mathbb Q$, so that the columns
$F_j$ of Section~\ref{sec:foundations} are rational.  We regard each $F_j$
as a column in $\mathbb Q^n$.  For an odd prime $p$, reduction modulo $p$
always means reduction of a member of $\mathbb Z_p$; in particular, every
congruence below includes the assertion that its two sides are locally
integral.

The raw matrix has $n+q$ columns, whereas the filtered determinant has
size $n$.  At the large primes treated first, each raw column has at most
two powers of $p$ in its denominator.  We shall make invertible column
changes over $\mathbb Z_p$ and bound how many columns can still have
denominator $p^2$ or $p$.  Such bounds control every full minor of the
raw matrix, and Cauchy--Binet then transfers them to the integer column
filter.  The first reduction identifies residues after multiplication
by $p^2$; for $p>H/2$ a second reduction identifies the remaining
residues after multiplication by $p$.

\subsection{Digit reduction of the rational moments}

We first prove the reductions needed at primes
\[
  2\sqrt H<p\le H.
\]
We may suppose that $p$ does not divide the denominator of $G$: as $N$
tends to infinity, every fixed denominator prime eventually lies below
$2\sqrt H$.  Put
\[
 E(t)=(1-t^2)^{(p-1)/2}=\sum_{d=0}^{p-1}E_dt^d
 \quad\text{in }\mathbb F_p[t],\qquad
 \epsilon=(-1)^{(p-1)/2}.
\]
We set $E_d=0$ outside $0\le d<p$.  Within this range,
\[
 E_d=\begin{cases}c_{d/2}\pmod p,&d\text{ even},\\0,&d\text{ odd},\end{cases}
 \qquad E_{p-1-d}=\epsilon E_d.
\]
The first identity follows from
$(-1)^k\binom{(p-1)/2}{k}\equiv4^{-k}\binom{2k}{k}$; the second follows
by reversing the coefficients of $E$.

\begin{lemma}[Digit reductions]\label{lem:odd-digit-reductions}
For $0\le i,j<H$, let
\[
 \ell=j\bmod p,\quad d=(j-i-1)\bmod p,
 \qquad
 i'=\frac{i+1+d-\ell}{p}-1,\quad j'=\left\lfloor\frac jp\right\rfloor,
 \qquad 0\le\ell,d<p.
\]
Then $i'\ge-1$, and, with the convention
$M^0(-1,j')=k^+_{j'+1}$,
\begin{equation}\label{eq:digit-layers}
 \begin{split}
 p^2M^0(i,j)&\equiv E_dM^0(i',j')\pmod p,\\
 p^2Z^0(i,j)&\equiv
 \begin{cases}
 Z^0(\lfloor i/p\rfloor,j'),&i\equiv j\pmod p,\\
 0,&i\not\equiv j\pmod p.
 \end{cases}
 \end{split}
\end{equation}
\end{lemma}

\begin{proof}
We first give the parity and carry calculation for the factors $H_z^*$
in the explicit formulas of Section~\ref{sec:foundations}.  Write
$z=Pp+r$, $0\le r<p$, $0\le z<H$.  Since $H<p^2/4$, all factors at
the reduced indices $P$ are $p$-adic units.  If $z$ is even, then
\begin{equation}\label{eq:odd-even-H}
 \begin{cases}
 H_z^*\equiv H_P^*c_{r/2}\pmod p,&r\text{ even},\\
 v_p(H_z^*)=1,&r\text{ odd}.
 \end{cases}
\end{equation}
Indeed, when $r$ is even, $P$ is even and the base-$p$ digits of $z/2$
are $P/2,r/2$.  Expanding $(1+x)^z$ modulo $p$ by
$(1+x)^p=1+x^p$ gives the first assertion, including the factor $4^{-z/2}$.
When $r$ is odd, $P$ is odd and the low digit of $z/2$ is $(p+r)/2$.
The factorial formula for $\binom z{z/2}$ has exactly one carry:
\[
 \left\lfloor\frac zp\right\rfloor
       -2\left\lfloor\frac{z/2}{p}\right\rfloor=1,
\]
and there is no contribution from $p^2$.  Thus, for positive even $z$,
$v_p(zH_z^*)$ is either zero or one.  It is one precisely when $r=0$
or $r$ is odd.

For odd $z$ one has
\begin{equation}\label{eq:odd-odd-H}
 \frac1{H_z^*}\in\mathbb Z_p,\qquad
 pH_z^*\equiv
 \begin{cases}
 \epsilon c_{r/2}H_P^*,&r\text{ even},\\
 0,&r\text{ odd}.
 \end{cases}
\end{equation}
If $r$ is odd, both $z$ and $c_{(z-1)/2}$ are units by the same
digit calculation.  If $r$ is even, $P$ is odd.  At $z=Pp$, digit
expansion gives
\[
 c_{(Pp-1)/2}\equiv c_{(P-1)/2}c_{(p-1)/2}
       =\epsilon c_{(P-1)/2},
 \qquad pH_{Pp}^*\equiv\epsilon H_P^*.
\]
The recurrence $H_{z+2}^*/H_z^*=(z+1)/(z+2)$, applied through
$r=0,2,\ldots,p-1$, now proves the nonzero case of
Equation~\eqref{eq:odd-odd-H}.  Its denominators on this interval are
units.  Since $m_{z-1}=2H_z^*$ for odd $z$ and is zero for even $z$,
these formulas imply
\begin{equation}\label{eq:odd-moment-digit}
 pm_{z-1}\equiv E_{p-1-r}m_{P-1}\pmod p.
\end{equation}
This includes $z=0$, using $m_{-1}=0$.

We next reduce the two starting arrays.  For $u=Pp+r$, $0\le u<H$,
we claim
\begin{equation}\label{eq:odd-starting-digits}
 p^2k^-_u\equiv E_{p-1-r}k^-_P,
 \qquad
 p^2k^+_{u+1}\equiv E_r k^+_{P+1}\pmod p.
\end{equation}
For odd $u$, write the first formula as
\[
 p^2 k^-_u=(pH_u^*)
           \sum_{\substack{1\le z\le u\\z\text{ odd}}}\frac{2p}{z}.
\]
Only $z=mp$, with $m$ odd, survives in the sum.  If $r$ is odd the
first factor vanishes modulo $p$.  If $r$ is even, $P$ is odd and
Equation~\eqref{eq:odd-odd-H} gives
$\epsilon E_rH_P^*\sum_{m\le P,\ m\text{ odd}}2/m
 =E_{p-1-r}k^-_P$, as required.

For even $u$ the quantities $p/(zH_z^*)$, for positive even $z\le u$,
are integral.  Thus
\[
 p^2k^-_u=H_u^*
     \sum_{\substack{1\le z\le u\\z\text{ even}}}
                  2\left(\frac{p}{zH_z^*}\right)^2
\]
is integral, and is zero modulo $p$ when $r$ is odd.  If $r$ is even,
then $P$ is even.  The nonzero summands form precisely the complete
blocks
\[
 z=mp-\lambda,\qquad
 m=2,4,\ldots,P,\qquad \lambda=0,2,\ldots,p-1.
\]
To see completeness, the indices with $z\bmod p$ odd lie immediately
below an even multiple $mp$, while that multiple supplies $\lambda=0$.
The upper limit $u=Pp+r$ contains the whole block ending at $Pp$,
and the next such block begins at $(P+1)p+1>u$.  No partial block is
present.  The recurrence, started at $mp$ and followed downwards, gives
\[
 \frac{p}{(mp-\lambda)H_{mp-\lambda}^*}
       \equiv\frac{c_{\lambda/2}}{mH_m^*}\pmod p.
\]
For example, each downward step from $z$ to $z-2$ multiplies this
quantity by $(z-1)/(z-2)$; these multipliers give successively
$(2j-1)/(2j)$ modulo $p$.  Finally, with $s=(p-1)/2$,
\[
 \sum_{j=0}^{s}c_j^2
  \equiv\sum_{j=0}^{s}\binom{s}{j}^2
  =\binom{2s}{s}\equiv(-1)^s=\epsilon\pmod p.
\]
The equality is the coefficient identity obtained from
$(1+x)^s(1+x)^s$.  Multiplying the block sums by
$H_u^*\equiv E_rH_P^*$ proves the first congruence of
Equation~\eqref{eq:odd-starting-digits} also for even $u$.

For the second congruence, suppose first that $u$ is even.  In
\[
 p^2k^+_{u+1}
   =H_u^*\sum_{\substack{1\le z\le u\\z\equiv u\ (2)}}
                     \frac{2p^2}{z^2H_z^*},
\]
every term with $p\nmid z$ vanishes modulo $p$, because
$v_p(H_z^*)\le1$.  At $z=mp$, necessarily $m$ is even and
$H_{mp}^*\equiv H_m^*$.  The result is $E_r k^+_{P+1}$ when $r$ is
even, and zero when $r$ is odd.  For odd $u$ instead write the expression
as
\[
 (pH_u^*)\sum_{\substack{1\le z\le u\\z\text{ odd}}}
                         \frac{2p}{z^2H_z^*}.
\]
The reciprocal $1/H_z^*$ is integral.  Again only multiples $z=mp$
can survive; there
$pH_{mp}^*\equiv\epsilon H_m^*$, and the two factors of $\epsilon$
cancel.  If $r$ is odd the prefactor is zero.  This proves the second
congruence in every case and also proves the local integrality asserted
in Equation~\eqref{eq:odd-starting-digits}.

If $i\ge j$, put $u=i-j=Pp+r$.  Then $d=p-1-r$ and $i'=j'+P$.
The diagonal reduction is
\[
 M^0(i,j)=k^-_u-\sum_{k=1}^{j}\frac{m_{u+k-1}}k.
\]
After multiplication by $p^2$, terms with $p\nmid k$ vanish by
Equation~\eqref{eq:odd-moment-digit}.  At $k=mp$ the surviving term is
$E_{p-1-r}m_{P+m-1}/m$.  The first starting congruence therefore gives
\[
 p^2M^0(i,j)\equiv E_{p-1-r}
       \left(k^-_P-\sum_{m=1}^{j'}\frac{m_{P+m-1}}m\right)
       =E_dM^0(i',j').
\]
If $i<j$, put $u=j-i-1=Pp+r$, so $d=r$ and $i'=j'-P-1$.  Now
\[
 M^0(i,j)=k^+_{u+1}-\sum_{k=u+2}^{j}\frac{m_{k-u-2}}k.
\]
For a multiple $k=mp$ in the sum,
\[
 k-u-1=(m-P-1)p+(p-1-r).
\]
Equation~\eqref{eq:odd-moment-digit} gives the reduced summand
$E_r m_{m-P-2}/m$.  A possible first multiple with $m=P+1$ contributes
zero, since its reduced moment is $m_{-1}$.  Thus the reduced sum is
over $m=P+2,\ldots,j'$, as in the formula for $M^0(i',j')$.
The expression for $i'$ is also
$i'=\lfloor(i-\ell)/p\rfloor$, so $i'\ge-1$.  When $i'=-1$,
$j'=P$, the sum is empty and the starting term is exactly
$k^+_{j'+1}=M^0(-1,j')$.  This proves the first assertion of
Equation~\eqref{eq:digit-layers}, including its boundary convention.
The reduced array indices are less than $H/p<p/4$; even the boundary
index $j'+1$ is less than $p$.  Thus the explicit rational formulas
show that all quantities on its right side are integral at $p$.

For $Z^0$, write $I=\lfloor i/p\rfloor$ and $J=\lfloor j/p\rfloor$.
Because $i,j<p^2$, the harmonic sums satisfy
\[
 pB_i^{(1)}\equiv B_I^{(1)},\qquad
 p^2B_i^{(2)}\equiv B_I^{(2)}\pmod p.
\]
The diagonal formula follows at once.  Off the diagonal, if
$i\not\equiv j\pmod p$, the denominator $i-j$ is a unit and
$p^2Z^0(i,j)$ is zero modulo $p$.  If the residues agree,
$i-j=p(I-J)$ with $I-J$ a unit; substituting the first harmonic
congruence gives $Z^0(I,J)$.  This also proves integrality of
$p^2Z^0(i,j)$.
\end{proof}

\subsection{The leading layer and paired raw columns}

We now gather the moment reductions into column vectors.  This will
identify pairs of raw columns whose sum has at most one power of $p$ in
its denominator.  For a fixed residue $0\le\ell<p$, define column vectors over
$\mathbb F_p$ by
\[
 P'_u=[t^{(u+1)p+\ell}]\,tP(t)E(t)\quad(u\ge-1),
 \qquad
 D'_u=[t^{up+\ell}]\,D(t)\quad(u\ge0).
\]
Here $P,D$ denote the vectors of all row polynomials; the dependence
of these extracted vectors on $\ell$ is understood.  Their supports
are finite.  Lemma~\ref{lem:odd-digit-reductions}, gathered by coefficient
residue, gives
\begin{equation}\label{eq:column-layer}
 X_{kp+\ell}:=p^2F_{kp+\ell}\bmod p
   =\sum_{u\ge-1}P'_uM^0(u,k)
          -\frac32\sum_{u\ge0}D'_uZ^0(u,k).
\end{equation}
For a term $[t^i]P$, its unique residue $d$ satisfies
$i+1+d=(i'+1)p+\ell$, so that it contributes exactly to $P'_{i'}$,
including $i'=-1$.  The $Z^0$ reduction survives only when
$i=up+\ell$, which is precisely the extraction defining $D'_u$.
The coefficient of $G$ in $F_j$ is a sum of integral multiples
of $c_{(i-j)/2}$.  These coefficients are integral at every odd prime,
so this term disappears after multiplication by $p^2$.  Every raw
column therefore lies in $p^{-2}\mathbb Z_p^n$.

If $\ell\ge H-2p$, the degree bounds
$\deg(tPE)\le H+p-1$ and $\deg D<H$ show that $P'_u,D'_u$ vanish
for $u>1$.  Put
\[
 V_\ell=2P'_1-\frac32D'_1,
 \qquad U_\ell=2P'_{-1}-\frac32D'_0.
\]
The needed small values of the reduced arrays are
\[
 \begin{array}{c|rrr}
 u&-1&0&1\\\hline
 M^0(u,0)&0&0&2\\
 M^0(u,1)&2&0&-2
 \end{array}
 \qquad
 \begin{array}{c|rr}
 u&0&1\\\hline
 Z^0(u,0)&0&1\\
 Z^0(u,1)&1&-1
 \end{array}.
\]
Consequently, whenever the indicated columns exist,
\begin{equation}\label{eq:odd-paired-layer}
 X_\ell=V_\ell,\qquad X_{p+\ell}=U_\ell-V_\ell.
\end{equation}
Since $tP$ and $D$ have no terms below degree $A+1$,
$U_\ell=0$ for $\ell\le A$.

We record explicitly how changes in the full column pool will be used.
Let $T$ be the $n$ by $(n+q)$ matrix of all raw columns.  If
$Q\in\operatorname{GL}_{n+q}(\mathbb Z_p)$ and every full minor of
$TQ$ has valuation at least $-D$, the same holds for every full minor
of $T=(TQ)Q^{-1}$ by Cauchy--Binet.  The coefficients in this expansion
are minors of the integral matrix $Q^{-1}$.  Applying Cauchy--Binet
once more to the integer matrix defining the prescribed filter gives
$v_p(\Delta_N)\ge-D$.  This transfer applies even when the columns or
their leading residues are linearly dependent.

For $2\sqrt H<p\le H/2$, use each pair of columns with
\[
 \max(b,H-2p)\le\ell<\min(p,L-p,A).
\]
Their number is exactly
\[
 d_0=\bigl(\min(p,L-p,A)-\max(b,H-2p)\bigr)_+,
 \qquad y_+=\max(y,0).
\]
Replacing its high column by the sum of the pair is an integral
elementary column operation with integral inverse.  The pair sum lies
in $p^{-1}\mathbb Z_p^n$ by Equation~\eqref{eq:odd-paired-layer}; all
other columns still lie in $p^{-2}\mathbb Z_p^n$.  These pairs are
disjoint, since $\ell<p$.  Any selection of $n$ columns from the pool
of $n+q$ contains at least $(d_0-q)_+$ of the improved columns.
The transfer just explained gives
\begin{equation}\label{eq:odd-lower-pair-range}
 v_p(\Delta_N)\ge-2n+(d_0-q)_+.
\end{equation}

\subsection{The central layer and integral column elimination}

For $p>H/2$, the next lemma identifies the residues of central columns
after multiplication by $p$.  It expresses them in terms of the same
vectors $V_\ell$ that occur in the leading residues of noncentral
columns.  The elimination proof will retain noncentral columns and use
$p$ times those columns to cancel the corresponding $V_\ell$ terms.
\begin{lemma}[Central layer]\label{lem:odd-central-layer}
Suppose $H/2<p\le H$, and write
$K=L-p$ and $J=H-p$.  For every central column, by which we mean
\[
 b\le j<\min(p,L),\qquad j\ge J,
\]
one has
\begin{equation}\label{eq:central-layer}
 pF_j\in\mathbb Z_p^n,\qquad
 pF_j\equiv-\sum_{0\le\ell<J}\frac{V_\ell}{j-\ell}\pmod p.
\end{equation}
All denominators $j-\ell$ occurring here are units at $p$.
\end{lemma}

\begin{proof}
For every contributing row degree $0\le i<H$,
$i-j\le H-1-J=p-1$ and $j-i<p$.  Hence $|i-j|<p$, and the
starting value $k^-_{i-j}$ or $k^+_{j-i}$ in the diagonal reduction
is integral at $p$.  In either case that reduction can be written as
the starting value minus
\[
 \sum_{0\le\ell<\min(i,j)}\frac{m_{i-\ell-1}}{j-\ell}.
\]
Here $1\le j-\ell<p$.  By Equation~\eqref{eq:odd-moment-digit}, a
nonzero reduction after multiplication by $p$ requires
$p\le i-\ell<2p$, and then
\[
 pm_{i-\ell-1}\equiv2E_{2p+\ell-1-i}.
\]
Such an index necessarily has $0\le\ell<J$.  Conversely its
coefficient can be collected over all $i$ with no truncation, since
$j\ge J>\ell$ and $i\ge p+\ell>\ell$.  Thus
\[
 pM^0(P,j)\equiv-\sum_{0\le\ell<J}\frac{2P'_1}{j-\ell}.
\]
The extracted vector $P'_1$ in each summand is the one belonging to
that residue $\ell$.
For $Z^0(i,j)$, the harmonic sums have a pole only if
$i=p+\ell$ with $0\le\ell<J$.  The denominator $i-j$ is then a
unit: equality modulo $p$ would require $j=\ell<J$.  Therefore
\[
 pZ^0(p+\ell,j)\equiv\frac1{\ell-j}.
\]
All other $Z^0(i,j)$ are integral.  Combining these two calculations
with the factor $-3/2$ proves Equation~\eqref{eq:central-layer}; the
$G$ term is integral here as before.
\end{proof}

We now carry out these cancellations in the full column pool.  In the
resulting pool, $R$ will count the retained columns with possible
denominator $p^2$, and $S$ will bound the number of other columns with
possible denominator $p$; every remaining column will be integral.
The proof first uses the $V_\ell$ supplied by retained noncentral
columns to modify the central columns.  We then use the rank of their
remaining scaled residues to bound the number of central columns that
can still have denominator $p$ after an invertible column change.

\begin{lemma}[Elimination over the local integers]
\label{lem:odd-local-elimination}
For $H/2<p\le H$, define
\begin{align*}
 R&=K_++(K-A)_++(J-\max(b,K))_+,\\
 S&=(\min(A,K)-b)_+
        +(\min(b,J)-\max(0,K))_+.
\end{align*}
Then every full raw minor, and hence the filtered determinant, satisfies
\begin{equation}\label{eq:odd-two-layer-bound}
 v_p(\Delta_N)\ge-\min(2n,n+R,2R+S).
\end{equation}
\end{lemma}

\begin{proof}
The noncentral columns are precisely the high columns
$F_{p+\ell}$ for $0\le\ell<K$ and the low columns $F_\ell$ for
$b\le\ell<J$.  Empty ranges are allowed.  Indeed $p>b$, $J<p$,
and $K<J$.  For
\[
 b\le\ell<\min(A,K)
\]
retain the low column and replace the high column by
$F_{p+\ell}+F_\ell$.  Let
\[
 B=(\min(A,K)-b)_+
\]
be the number of these pair sums.  Each lies in $p^{-1}\mathbb Z_p^n$.
Retain every other noncentral column as a column in
$p^{-2}\mathbb Z_p^n$, whether or not its leading residue depends on
the others.  The number of columns so retained is
\begin{equation}\label{eq:odd-retained-count}
 R_0=K_++(J-b)_+-B.
\end{equation}
It equals the stated $R$.  To verify the identity, split $K$ into
$K\le0$, $0<K\le b$, $b<K\le A$, and $K>A$; use $K<J$ in the
last two cases.  The resulting formulas are respectively
$(J-b)_+$, $K+(J-b)_+$, $J$, and $K+J-A$.

These retained columns furnish a representative of every vector
$V_\ell$ with $0\le\ell<J$ except possibly those in
\[
 \mathcal E=\{\ell:\max(0,K)\le\ell<\min(b,J)\}.
\]
For $b\le\ell<J$, the retained low column has leading residue
$p^2F_\ell\equiv V_\ell$.  For $0\le\ell<\min(b,K)$, the
unpaired high column has leading residue $-V_\ell$, because
$\ell<b<A$ makes $U_\ell=0$.  These two ranges exhaust the complement
of $\mathcal E$.  Denote the size of $\mathcal E$ by
\[
 e=|\mathcal E|=(\min(b,J)-\max(0,K))_+.
\]

For each represented residue choose a retained column $Y_\ell$ and
a sign $\sigma_\ell\in\{1,-1\}$ such that
$p^2Y_\ell\equiv\sigma_\ell V_\ell$.  For every central $j$, choose
$a_{\ell j}\in\mathbb Z_p$ reducing to
$\sigma_\ell/(j-\ell)$ and replace that column by
\[
 C_j=F_j+p\sum_{\substack{0\le\ell<J\\\ell\notin\mathcal E}}
                                      a_{\ell j}Y_\ell.
\]
This is a shear of the whole column pool with integral coefficients
and inverse obtained by changing the signs of the added coefficients.
It preserves every retained column and every pair sum.  The resulting
central columns lie in $p^{-1}\mathbb Z_p^n$ and satisfy
\[
 pC_j\equiv-\sum_{\ell\in\mathcal E}\frac{V_\ell}{j-\ell}
       \pmod p.
\]
This step does not require any information about the next coefficient
of $Y_\ell$.  Explicitly, if
$Y_\ell=p^{-2}y_0+p^{-1}y_1+y_2$ with integral lifts $y_0,y_1,y_2$,
then $pY_\ell=p^{-1}y_0+y_1+py_2$; its unknown second coefficient is
already integral.

Let $c$ be the number of central columns.  Their scaled residues
define a linear map
\[
 \varphi:\mathbb F_p^c\longrightarrow\mathbb F_p^n,
 \qquad (a_j)_j\longmapsto\sum_j a_j(pC_j\bmod p).
\]
Its image is contained in the span of the $e$ exceptional vectors,
so $d:=\operatorname{rank}\varphi\le\min(c,e)$.  Choose a basis of
$\mathbb F_p^c$ whose last $c-d$ vectors form a basis of
$\ker\varphi$.  Lift its basis matrix arbitrarily to a matrix
$W\in\operatorname{Mat}_{c}(\mathbb Z_p)$.  Its determinant is a
unit, so $W\in\operatorname{GL}_c(\mathbb Z_p)$ and the adjugate
formula gives $W^{-1}\in\operatorname{Mat}_{c}(\mathbb Z_p)$.
Applying this change to the central columns leaves at most $d$
columns in $p^{-1}\mathbb Z_p^n$; the other $c-d$ are integral,
because their scaled residues vanish and they already belonged to
$p^{-1}\mathbb Z_p^n$.  If $c=0$ this step is the empty identity.

The resulting full pool consequently contains $R$ columns with
possible denominator $p^2$, at most $B+d\le B+e=S$ other columns
with possible denominator $p$, and integral remaining columns.
No noncentral column was discarded, nor was its denominator reduced
on the strength of a leading linear dependence.  In a full minor,
if $r$ columns come from the first group and $s$ from the second,
the loss is at most $2r+s$.  The inequalities
\[
 2r+s\le2n,\qquad 2r+s\le n+R,\qquad 2r+s\le2R+S
\]
hold since $r+s\le n$, $r\le R$, and $s\le S$.  All changes used
above and their inverses are integral, so the Cauchy--Binet transfer
proves Equation~\eqref{eq:odd-two-layer-bound} for all original full
minors and for $\Delta_N$.
\end{proof}

The omission of the optional pair at $\ell=A$ is harmless: its
physical column is included in $R$.  The preceding proof also covers
$K\le0$ and $J\le b$.  In particular, if $L\le p\le H$, there
are no high or low noncentral columns, $R=0$, and $S=H-p$; all raw
columns are central.  At the formal endpoint $p=H$, all of them are
integral.

\subsection{Summing the prime losses}

For a nonzero rational determinant, the valuations weighted by
$\log p$ sum to its ordinary logarithmic absolute value.  We now
combine the odd-prime estimates with the bound at $2$ to obtain the
lower bound used in the final comparison.

\begin{proposition}[Finite-place bound]\label{prop:finite-lower}
Assume $G\in\mathbb Q$.  Along any sequence of positive integers
$N\to\infty$ for which $\Delta_N\ne0$, one has
\begin{equation}\label{eq:finite-lower-final}
 \liminf_{N\to\infty}
 \left(\frac{\log|\Delta_N|}{n^2}-\frac12\log2\right)
 \ge -\frac{8609}{4608}
       -\left(\frac12+\frac{505}{4608}\right)\log2
 >-2.29084.
\end{equation}
\end{proposition}

\begin{proof}
We first compute the complete loss function.  Put $x=p/N$.  For
$x\le65/2$, Equation~\eqref{eq:odd-lower-pair-range} has
\[
 \frac{d_0}{N}
  =\bigl(\min(x,59-x,19)-\max(7,65-2x)\bigr)_+.
\]
For $0\le x\le23$ this is zero; for $23\le x\le29$ it is
$2x-46$; and for $29\le x\le65/2$ it is $12$.  Only the part
exceeding $q/N=4$ improves a full minor.  For $x>65/2$, the formulas
of Lemma~\ref{lem:odd-local-elimination} give the following values;
the table retains the intermediate breakpoint $52$ at which the
counting formula changes:
\[
\begin{array}{c|cc}
 x\text{ range}&R/N&S/N\\\hline
 {[65/2,40]}&105-2x&12\\
 {[40,52]}&65-x&52-x\\
 {[52,58]}&117-2x&x-52\\
 {[58,59]}&59-x&6\\
 {[59,65]}&0&65-x
\end{array}.
\]
The entries agree at their common endpoints.  Taking
$\min(96,48+R/N,2R/N+S/N)$ yields the additional breakpoint $69/2$.
Thus for every relevant prime $p>2\sqrt H$ the valuation is bounded
below by $-Nd(p/N)$, where $d$ is continuous, is zero for $x\ge65$,
and has the following exact pieces:
\begin{equation}\label{eq:odd-loss-table}
\begin{array}{c|c|c|c}
 x\text{ range}&d(x)&
       \text{endpoint values}&\displaystyle\int_{\text{range}}d(x)\,dx\\\hline
 {[0,25]}&96&96,96&2400\\
 {[25,29]}&146-2x&96,88&368\\
 {[29,65/2]}&88&88,88&308\\
 {[65/2,69/2]}&153-2x&88,84&172\\
 {[69/2,40]}&222-4x&84,62&803/2\\
 {[40,58]}&182-3x&62,8&630\\
 {[58,59]}&124-2x&8,6&7\\
 {[59,65]}&65-x&6,0&18
\end{array}.
\end{equation}
In particular,
\begin{equation}\label{eq:odd-loss-area}
 \int_0^{65}d(x)\,dx=\frac{8609}{2}.
\end{equation}

For completeness, the omitted small odd primes cost only $o(n^2)$
in the logarithm of the determinant.  At any odd prime, the factorial
formula gives
\[
 0\le v_p\binom{2l}{l}\le1+\frac{\log H}{\log p}
 \qquad(2l\le H).
\]
Each factorial-floor difference is zero or one.  The displayed
formulas for $H_z^*,m_i,k_d^\pm,Z^0$, and diagonal reduction then
give, with an absolute constant $C_0$ and with
$\operatorname{den}G$ denoting the positive reduced denominator of $G$,
\[
 v_p(F_j(r))\ge
   -C_0\left(1+\frac{\log H}{\log p}\right)-v_p(\operatorname{den}G).
\]
For instance, each summand in $k^-_u$ uses at most two powers of
$z$ and two powers of $H_z^*$ in its denominator; no additional
loss arises from summing.  All row and filter coefficients are
integers.  Expanding a determinant therefore multiplies this entry
bound by at most $n$, and summing it over $p\le2\sqrt H$ with
weights $\log p$ gives
\[
 O\bigl(n\sqrt H\log H+n\log(\operatorname{den}G)\bigr)
       =o(n^2).
\]
This estimate uses only that the number of such primes is at most
$2\sqrt H$.  For sufficiently large $N$, every prime $p>H$ is
integral throughout the calculation: factorials and degree
denominators introduce only prime factors at most $H$, and the fixed
denominator of $G$ has no prime factor above $H$.  Thus these primes
give no negative contribution.

We use the classical prime number theorem in the form
\[
 \theta(y):=\sum_{p\le y}\log p\sim y
       \qquad(y\to\infty);
\]
see \cite[Step~VI, p.~707]{Zagier1997}, which presents Newman's analytic
proof. The weighted consequence needed here follows directly:
\begin{equation}\label{eq:odd-weighted-primes}
 \frac1N\sum_{p\le65N}d(p/N)\log p
       \longrightarrow\int_0^{65}d(x)\,dx.
\end{equation}
Here is a direct justification of the weight.  For a fixed partition
$0=x_0<\cdots<x_m=65$, the prime number theorem gives
\[
 \frac{\theta(Nx_j)-\theta(Nx_{j-1})}{N}
       \longrightarrow x_j-x_{j-1}.
\]
Upper and lower step functions formed from the supremum and infimum
of $d$ on each interval bound the weighted sum.  Their limiting
difference tends to zero as the mesh decreases, since $d$ is
uniformly continuous.  Individual partition endpoints change the
sum by at most $O(\log N/N)$.  This proves
Equation~\eqref{eq:odd-weighted-primes}.  The prime $2$ and the
primes below $2\sqrt H$ may be removed from that sum at cost $o(1)$,
using $\|d\|_\infty=96$ and the elementary bound
$\theta(2\sqrt H)\le2\sqrt H\log(2\sqrt H)$.

Combining these estimates with Equation~\eqref{eq:odd-loss-area}
and $n^2=2304N^2$ gives
\[
 \sum_{p\text{ odd}}v_p(\Delta_N)\log p
       \ge-\frac{8609}{4608}n^2-o(n^2).
\]
Proposition~\ref{prop:two-adic}, with $\delta=10/48$, gives
\[
 v_2(\Delta_N)\ge
  -\left(\frac\delta2+\frac{\delta^2}{8}\right)n^2-o(n^2)
  =-\frac{505}{4608}n^2-o(n^2).
\]
For a nonzero rational number its numerator and denominator
factorizations give the exact identity
\[
 \log|\Delta_N|=\sum_pv_p(\Delta_N)\log p.
\]
Subtracting $\tfrac12\log2$ after dividing by $n^2$ proves the
non-strict inequality in Equation~\eqref{eq:finite-lower-final}.

The final numerical comparison also has a short rational check.
The identity
\[
 \log2=2\sum_{j=0}^{\infty}\frac{3^{-(2j+1)}}{2j+1}
\]
and its positive geometric tail give
\[
 \log2\le
 2\sum_{j=0}^{5}\frac{3^{-(2j+1)}}{2j+1}
       +\frac{2\,3^{-13}}{13(1-3^{-2})}
 <\frac{693149}{10^6}.
\]
Consequently the lower constant is strictly larger than
\[
 -\frac{8609}{4608}-\frac{2809}{4608}\frac{693149}{10^6}
 =-\frac{10556055541}{4608000000}>-2.29084,
\]
which completes the proof.
\end{proof}
==== END Catalans-constant-is-irrational//odd-primes.tex ====

==== BEGIN Catalans-constant-is-irrational//real-place.tex ====
\section{The real-place determinant estimates}\label{sec:real-place}

We estimate the determinant defined in Equation~\eqref{eq:determinant}
without a rationality assumption.  Write
\begin{equation}\label{eq:real-normalized-parameters}
 \begin{gathered}
 \alpha=\frac an=\frac{11}{48},\quad
 \beta=\frac bn=\frac7{48},\quad
 \gamma=\frac gn=\frac qn=\frac4{48},\quad
 \eta=\frac hn=\frac2{48},\\
 D_*=n-1-2g=40N-1.
 \end{gathered}
\end{equation}
In particular, $0<D_*<n$ and $n\ge48$.  For a real list $y$ of
length $n$, let
\[
 V(y)=\prod_{i<j}(y_j-y_i),\qquad \mathcal V(y)=|V(y)|.
\]
All constants in the estimates of this Section are independent of the
locations and separations of the integration variables.

\subsection{The integral and its two sheets}

Put $d\mu(t)=|t|\,dt/f(t)$ on $(-1,1)$, and set
\[
 A_r(t)=P_r(t)-\frac32\boldsymbol 1_{\{t>0\}}\frac{f(t)}tD_r(t),
 \qquad \psi_k(s)=s^{b+k}(1-s)^q.
\]
The value at $t=0$ is immaterial; the displayed expression has a finite
limit there because $D_r(t)/t$ is a polynomial.  The determinant entries
are exactly
\[
 \int_{-1}^1\int_0^1 A_r(t)\frac{\psi_k(s)}{1-ts}\,ds\,d\mu(t).
\]
The measure has mass $\mu((-1,1))=2$.  The scalar majorant
$\int\!\int(1-|t|s)^{-1}\,ds\,d\mu(t)$ is finite: integration in $s$
gives at worst a logarithmic singularity at $|t|=1$, which is integrable
against $(1-t^2)^{-1/2}dt$.  The functions $A_r$ and $\psi_k$ are
bounded for each fixed $n$.  Thus all permutation expansions below are
absolutely integrable.

We apply Andr\'eief's determinant integration identity twice
\cite[Equation~(1.7) and Section~2.2]{Forrester2018Andreief}.
Expanding determinants and relabeling integration variables gives
the identity
\begin{equation}\label{eq:real-double-integral}
 \Delta_N=\frac1{(n!)^2}\int_{(-1,1)^n}\int_{(0,1)^n}
 \det[A_r(t_i)]_{r,i}\,
 \det\left[\frac1{1-t_i s_j}\right]_{i,j}\,
 \det[\psi_k(s_j)]_{j,k}\,ds\,d\mu^n(t).
\end{equation}
Here $0\le r,k<n$ and $1\le i,j\le n$.  To see the factor $1/n!$
at each application directly, expand the two determinants sharing an
integration list: every one of the $n!$ permutations of that list gives
the same determinant of single integrals.  Fubini's theorem applies by
the preceding majorant, also after each permutation expansion.

Cauchy's double alternant in multiplicative variables
\cite[Section~2.1, Equation~(2.7)]{Krattenthaler1999} is
\begin{equation}\label{eq:real-cauchy-determinant}
 \det\left[\frac1{1-t_i s_j}\right]_{i,j}
 =\frac{V(t)V(s)}{\prod_{i,j}(1-t_i s_j)}.
\end{equation}
For completeness, multiplying by the denominator gives a polynomial
alternating separately in $t$ and $s$, of degree at most $n-1$ in each
individual variable.  Dividing by $V(t)V(s)$ therefore leaves a constant.
Expanding $(1-t_i s_j)^{-1}$ at $s=0$, the first nonzero homogeneous
part of the determinant is $V(t)V(s)$, so that constant is one.  Also
$\det[\psi_k(s_j)]_{j,k}=V(s)\prod_j s_j^b(1-s_j)^q$.

Use the real coordinate
\begin{equation}\label{eq:real-coordinate}
 x_i=\frac{t_i}{1+\sqrt{1-t_i^2}},\qquad
 t_i=\frac{2x_i}{1+x_i^2},\qquad
 d\mu(t_i)=\frac{4|x_i|}{(1+x_i^2)^2}\,dx_i.
\end{equation}
The coordinate maps $(-1,1)$ bijectively to $(-1,1)$.  Except on a
set of measure zero we may assume that all $x_i$ are distinct and
nonzero.  Define
\[
 (\xi_{\mathrm f}(x),\xi_{\mathrm n}(x))=
 \begin{cases}(1/2,1/2),&x<0,\\(-1/4,5/4),&x>0,\end{cases}
 \qquad
 \mathcal R_n(x)=
 \det\left[\xi_{\mathrm f}(x_i)x_i^{g-r}
             +\xi_{\mathrm n}(x_i)x_i^{r-g}\right]_{r,i}.
\]
The subscripts $\mathrm n$ and $\mathrm f$ refer to the near and far
evaluations at $x_i$ and $1/x_i$, respectively, inside and outside the
unit circle.
Indeed, $fD_r/t=(R_r^*-R_r)/2$, so the negative half-interval
has the row $(R_r+R_r^*)/2$, whereas the positive half-interval has
$(5R_r-R_r^*)/4$.  Factoring $(1-t_i)^ht_i^{C-1}$ out of the
$i$th column therefore gives the exact identity
\begin{equation}\label{eq:real-sheet-integral}
 \Delta_N=\frac1{(n!)^2}\int\!\int
 \mathcal R_n(x)\,
 \frac{V(t)V(s)^2}{\prod_{i,j}(1-t_i s_j)}
 \prod_j s_j^b(1-s_j)^q
 \prod_i t_i^{C-1}(1-t_i)^h\,ds\,d\mu^n(t).
\end{equation}
In particular, there is one $t$ Vandermonde and two $s$
Vandermondes.  All the denominators in this formula are positive.

\subsection{A uniform interpolating function}

The interpolation problem comes from factoring out the far-sheet weight.
Put $c_i=\xi_{\mathrm n}(x_i)/\xi_{\mathrm f}(x_i)$; the denominator
is nonzero on both half-intervals. For $0\le r<n$,
\begin{equation}\label{eq:real-sheet-factorization}
 \xi_{\mathrm f}(x_i)x_i^{g-r}
   +\xi_{\mathrm n}(x_i)x_i^{r-g}
 =\xi_{\mathrm f}(x_i)x_i^{g-(n-1)}
   \left(x_i^{n-1-r}+c_i x_i^{D_*}x_i^r\right).
\end{equation}
Thus a holomorphic function with values $h_*(x_i)=c_i x_i^{D_*}$
turns the expression in parentheses into
$x_i^{n-1-r}+h_*(x_i)x_i^r$. We shall bound the resulting evaluation
determinant by controlling the norm of this function. The first estimate
applies when
\begin{equation}\label{eq:case-condition}
 \sum_{i=1}^n\frac{1-x_i^2}{1+x_i^2}\le D_*.
\end{equation}
The following construction is uniform even when the nodes cluster.

\begin{lemma}\label{lem:real-uniform-interpolation}
Let $x_1,\ldots,x_n$ be distinct nonzero real numbers in $(-1,1)$
satisfying Equation~\eqref{eq:case-condition}.  There is a function
$h_*$ holomorphic on a neighborhood of the closed unit disk such that
\[
 h_*(x_i)=c_i x_i^{D_*},\qquad
 c_i=\begin{cases}1,&x_i<0,\\-5,&x_i>0,\end{cases}
 \qquad
 \sup_{|z|\le1}|h_*(z)|\le K_0 n,
 \quad K_0=10e^{12}.
\]
The neighborhood may depend on the node list; the displayed bound does not.
\end{lemma}

\begin{proof}
Set
\[
 B(z)=\prod_i\frac{z-x_i}{1-x_i z},\qquad
 F(z)=\frac{z^{D_*}}{B(z)}.
\]
For $0<u\le1$ and a real $x$ with $0<|x|<1$, the logarithmic
derivative of one factor on the imaginary diameter is
\[
 \frac{d}{d\log u}\frac12\log\frac{u^2+x^2}{1+x^2u^2}
 =\frac{(1-x^4)u^2}{(u^2+x^2)(1+x^2u^2)}
 \le\frac{1-x^2}{1+x^2}.
\]
The inequality follows on dividing the denominator by $u^2$ and
using $u^2+u^{-2}\ge2$.  Integration from $u$ to $1$ gives
\[
 |B(iu)|\ge u^{\sum_i(1-x_i^2)/(1+x_i^2)}\ge u^{D_*}.
\]
The same holds at $-iu$, and $F(0)=0$.  For $1\le u\le1+2/n$,
each factor has modulus at least one because
$(u^2+x^2)-(1+x^2u^2)=(u^2-1)(1-x^2)\ge0$.
Consequently
\begin{equation}\label{eq:real-diameter-bound}
 |F(iu)|\le e^2\qquad (|u|\le1+2/n).
\end{equation}

Orient the fixed segment $\Gamma=[-i(1+2/n),i(1+2/n)]$ upwards,
and, off that segment, define the fixed Cauchy integral
\[
 \mathcal C(z)=\frac1{2\pi i}\int_\Gamma\frac{6F(w)}{w-z}\,dw.
\]
Write $c_-=1$, $c_+=-5$, with the sign specifying the left or right
half-plane, and put
\begin{equation}\label{eq:real-glued-interpolant}
 h_*(z)=c_\pm z^{D_*}-B(z)\mathcal C(z)
 \quad\text{when }\ \pm\operatorname{Re}z>0.
\end{equation}
The density is holomorphic in a neighborhood of every point of
$\Gamma$, since its poles $x_i$ are nonzero and real.  Local contour
deformation and the Cauchy integral formula show that the left lateral
value of $\mathcal C$ minus its right lateral value is $6F$.
More explicitly, move a short upward piece of $\Gamma$ to its left;
the closed contour formed by the original piece followed by the reversed
new piece is positively oriented, and its residue at $w=z$ is $6F(z)$.
Thus the difference of the two local analytic continuations in
Equation~\eqref{eq:real-glued-interpolant} is
$(c_--c_+)z^{D_*}-6B(z)F(z)=0$.  This proves holomorphic gluing
across the segment, including its two crossings of the unit circle.

For each fixed node list the poles $1/x_i$ of $B$ and the endpoints
of $\Gamma$ lie strictly outside the closed unit disk.  A sufficiently
small exterior neighborhood avoids all these points, and the same local
gluing works there.  It follows that $h_*$ is holomorphic on that
neighborhood.  Since $B(x_i)=0$ and $\mathcal C$ is regular at the
nonzero real point $x_i$, the required interpolation follows.

It remains to bound this one fixed function.  In either radius-$1/4$
disk about $i$ or $-i$, the quantities $|w|$, $|w-x_i|$, and
$|1-x_iw|$ are at least $3/4$.  Hence $F$ is nonzero there and
\begin{equation}\label{eq:real-logarithmic-derivative}
 \left|\frac{F'(w)}{F(w)}\right|
 \le\frac43D_*+\frac83n<4n.
\end{equation}
As $|F(i)|=|F(-i)|=1$, integration along a segment in either disk
gives $|F(w)|\le e^{12}$ whenever $|w-i|\le3/n$ or
$|w+i|\le3/n$.

If $|z|=1$ and $|\operatorname{Re}z|\ge1/(10n)$, the original
segment has distance at least $1/(10n)$ from $z$; its length is at
most $3$.  Equation~\eqref{eq:real-diameter-bound} gives
$|\mathcal C(z)|\le30e^2n$.
If instead $0<|\operatorname{Re}z|<1/(10n)$, $z$ is near one
of $\pm i$.  Near $i$, replace the portion of $\Gamma$ from
$i(1-1/n)$ to $i(1+2/n)$ by the three sides of the rectangle whose
other vertical side has real part
$-\operatorname{sign}(\operatorname{Re}z)/n$.  Near $-i$, make
the identical replacement of the bottom portion with imaginary parts
from $-1-2/n$ to $-1+1/n$.  The swept rectangle is in the opposite
half-plane from $z$, and it is contained in the radius-$1/4$ disk
about the relevant endpoint.  It contains neither $z$ nor a pole of
$F$, so this deformation leaves the value of the fixed integral
$\mathcal C(z)$ unchanged.

Each new side is within $3/n$ of $i$ or $-i$.  The new contour has
length at most $3$ and distance at least $1/(2n)$ from $z$.
For the latter assertion the vertical side has horizontal separation
at least $1/n$, the outer horizontal side has vertical separation
at least $2/n$, and the inner horizontal side and remaining diameter
have separation at least
\[
 \frac1n-\bigl(1-\sqrt{1-(10n)^{-2}}\bigr)>\frac1{2n}.
\]
Thus $|\mathcal C(z)|\le6e^{12}n$ in this case.  Only the two
endpoint neighborhoods were deformed; the middle diameter uses its
absolute bound \eqref{eq:real-diameter-bound}, with no derivative
estimate near zero.  Finally $|B(z)|=1$ on $|z|=1$, so
Equation~\eqref{eq:real-glued-interpolant} gives
$|h_*(z)|\le K_0n$ there.  At $z=\pm i$ use continuity of the
glued function.  The maximum modulus principle proves the bound in
the disk.  No uniform lower bound for the exterior neighborhood was used.
\end{proof}

\subsection{Evaluation in a finite-dimensional Hilbert space}

The following argument uses the kernel and compression viewpoint of
bounded analytic interpolation; see \cite{Sarason1967}.
We prove the needed finite-dimensional facts directly.

\begin{proposition}[Interpolation estimate]\label{prop:first-sheet-bound}
Under Equation~\eqref{eq:case-condition},
\begin{equation}\label{eq:real-first-sheet-bound}
 |\mathcal R_n(x)|\le(1+K_0n)^n
       \mathcal V(1/x)\prod_i|x_i|^g.
\end{equation}
In particular, the prefactor is $\exp(O(n\log n))$ uniformly in the nodes.
\end{proposition}

\begin{proof}
Let $Q(z)=\prod_i(1-x_i z)$, and give
\[
 \mathcal H_Q=\{v/Q:\ v\in\mathbb C[z],\ \deg v<n\}
\]
the norm inherited from the Hilbert space $H^2$ of analytic functions
with square-summable Taylor coefficients:
\[
 \|v/Q\|^2=\frac1{2\pi}\int_0^{2\pi}
          \frac{|v(e^{i\theta})|^2}{|Q(e^{i\theta})|^2}\,d\theta.
\]
For each fixed list the zeros of $Q$ lie outside the closed disk, so
this is a well-defined norm.  Define
$\widetilde v(z)=z^{n-1}v(1/z)$ and
$R(v/Q)=\widetilde v/Q$.  This is a complex-linear involution and
an isometry: on the circle
$|\widetilde v(e^{i\theta})|=|v(e^{-i\theta})|$, while the real
coefficients of $Q$ imply
$|Q(e^{i\theta})|=|Q(e^{-i\theta})|$.

The functions $k_i(z)=(1-x_i z)^{-1}$ belong to $\mathcal H_Q$.
They are linearly independent, as their distinct poles $1/x_i$ show,
and hence form a basis.  For $u\in H^2$, the Taylor coefficient
inner product gives $\langle u,k_i\rangle=u(x_i)$.
If $\Pi_Q$ is orthogonal projection from $H^2$ to $\mathcal H_Q$,
it follows that
\[
 (\Pi_Q u)(x_i)=u(x_i)\quad(1\le i\le n).
\]
Multiplication by the function $h_*$ of
Lemma~\ref{lem:real-uniform-interpolation} has $H^2$ operator norm
at most $\|h_*\|_\infty$.  Consequently
\[
 J=\Pi_Q M_{h_*}|_{\mathcal H_Q},\qquad L=I+JR
 \quad\text{satisfy}\quad \|L\|\le1+K_0n.
\]

Let $E:\mathcal H_Q\to\mathbb C^n$ send $v/Q$ to
$(v(x_i))_i$.  It is invertible because a polynomial of degree less
than $n$ cannot vanish at all the distinct nodes.
For $u\in\mathcal H_Q$, $E(u)_i=Q(x_i)u(x_i)$.
Since $LR=R+J$ and projection preserves function values, applying
this identity to $LR(v/Q)$ cancels the factors $Q(x_i)$
algebraically and gives
\begin{equation}\label{eq:real-evaluation-cancellation}
 E L R(v/Q)=\bigl(\widetilde v(x_i)+h_*(x_i)v(x_i)\bigr)_i.
\end{equation}
In the basis $1/Q,z/Q,\ldots,z^{n-1}/Q$, $\det E=V(x)$,
and $R$ has determinant of absolute value one.  Hence the determinant
of the mixed evaluations divided by $V(x)$ has absolute value
$|\det L|\le\|L\|^n\le(1+K_0n)^n$.  This is an algebraic
cancellation of the evaluation determinant.  In particular no bound
for the inverse evaluation matrix, which might be poorly conditioned,
enters the argument.

Apply this bound to the mixed evaluations in
Equation~\eqref{eq:real-sheet-factorization}. Since
$|\xi_{\mathrm f}(x_i)|\le1/2\le1$ and
$\mathcal V(1/x)=\mathcal V(x)\prod_i|x_i|^{-(n-1)}$,
Equation~\eqref{eq:real-first-sheet-bound} follows.
\end{proof}

\begin{proposition}[Hadamard estimate]\label{prop:second-sheet-bound}
For every distinct nonzero real list in $(-1,1)$, and in particular
when Equation~\eqref{eq:case-condition} fails,
\begin{equation}\label{eq:real-second-sheet-bound}
 |\mathcal R_n(x)|\le
 (3/2)^n n^{n/2}2^{-\binom n2}
 \prod_i(1+x_i^2)^{(n-1)/2}|x_i|^{g-(n-1)}.
\end{equation}
The factor $(3/2)^n n^{n/2}$ is $\exp(O(n\log n))$; the power
$2^{-\binom n2}$ is retained in the principal integrand below.
\end{proposition}

\begin{proof}
Expand the determinant by choosing one sheet in each column.  For
$z_i\in\{x_i,1/x_i\}$ the absolute monomial determinant is
$\mathcal V(z)\prod_i|z_i|^{-g}$.  Write
$z_i=\tan\phi_i$ with $-\pi/2<\phi_i<\pi/2$.  Then
\[
 \mathcal V(z)=\prod_i(1+z_i^2)^{(n-1)/2}
              \prod_{i<j}|\sin(\phi_j-\phi_i)|.
\]
The last product is $2^{-\binom n2}$ times the Vandermonde on
the unit-circle points $e^{2i\phi_i}$.  Its monomial matrix has
column norms $\sqrt n$, so Hadamard's inequality gives the bound
$2^{-\binom n2}n^{n/2}$.  The ratio of the remaining weight on
the far sheet to that on the near sheet is
\[
 \frac{|1/x|^{-g}(1+x^{-2})^{(n-1)/2}}
      {|x|^{-g}(1+x^2)^{(n-1)/2}}
 =|x|^{-D_*}\ge1.
\]
Thus all the weights can be bounded by their far-sheet values.
Finally the sum of absolute sheet coefficients is at most
$\prod_i(|\xi_{\mathrm f}(x_i)|+|\xi_{\mathrm n}(x_i)|)
\le(3/2)^n$, proving the result.
\end{proof}

\subsection{The two principal integrands}

To state precisely the quantities needed for the real-place energy
estimate, put $t_i=2x_i/(1+x_i^2)$ and
\[
 \mathcal J_n(x,s)=
 \frac{\mathcal V(t)\mathcal V(s)^2}{\prod_{i,j}(1-t_i s_j)}
 \prod_j s_j^b(1-s_j)^q
 \prod_i |t_i|^{C-1}(1-t_i)^h.
\]
Define
\begin{equation}\label{eq:real-integrand-majorants}
 \begin{split}
 \mathcal I_{2,n}(x,s)
   &=\mathcal J_n(x,s)\mathcal V(1/x)\prod_i|x_i|^g,\\
 \mathcal I_{1,n}(x,s)
   &=\mathcal J_n(x,s)2^{-\binom n2}
       \prod_i(1+x_i^2)^{(n-1)/2}|x_i|^{g-(n-1)}.
 \end{split}
\end{equation}
Let $\Omega_{2,n}$ be the set of interior configurations with distinct
nonzero $x_i$ satisfying Equation~\eqref{eq:case-condition}, and let
$\Omega_{1,n}$ be its complementary case among such configurations.
We use the interpolation estimate on $\Omega_{2,n}$ and the
Hadamard estimate on $\Omega_{1,n}$; the latter estimate is valid
in both cases. After substituting $t=2x/(1+x^2)$, $\kappa$ is the
exponent of the absolute $x$-Vandermonde in the corresponding
majorant: $\kappa=2$ for the interpolation estimate and
$\kappa=1$ for the Hadamard estimate.
The $s_j$ always range over $(0,1)$.

The preceding propositions and Equation~\eqref{eq:real-sheet-integral}
give
\[
 |\Delta_N|\le\frac{2^n K_n}{(n!)^2}
 \max_{\kappa\in\{1,2\}}\sup_{(x,s)\in\Omega_{\kappa,n}}
      \mathcal I_{\kappa,n}(x,s),
 \quad
 K_n=\max\{(1+K_0n)^n,(3/2)^n n^{n/2}\}.
\]
Indeed $d\mu^n(t)\,ds$ has total mass $2^n$, and the two case
sets partition its domain up to a null set.  In logarithmic form,
with $\log0=-\infty$, this yields
\begin{equation}\label{eq:real-supremum-reduction}
 \frac{\log|\Delta_N|}{n^2}-\frac12\log2
 \le\max_{\kappa\in\{1,2\}}
 \sup_{(x,s)\in\Omega_{\kappa,n}}
 \left(\frac{\log\mathcal I_{\kappa,n}(x,s)}{n^2}
                -\frac12\log2\right)
 +O\left(\frac{\log(n+2)}n\right).
\end{equation}
The error is independent of both integration lists.  Factorials,
measure masses, and sheet sums have all been accounted for explicitly.

Repeated nonzero $x_i$ give a zero original row determinant and are
also handled by continuity in the first estimate; repeated $s_j$
give a zero $s$ Vandermonde.  The sets with a zero node or a boundary
node have measure zero for the absolutely continuous measures above.
At such points the original expression in
Equation~\eqref{eq:real-double-integral}, rather than its factored
Laurent expression, supplies the integrable definition.  Approaches
to these exceptional sets require no separation condition in any
bound proved here.  In particular all configurations arbitrarily
close to them remain included in the suprema in
Equation~\eqref{eq:real-supremum-reduction}.
==== END Catalans-constant-is-irrational//real-place.tex ====

==== BEGIN Catalans-constant-is-irrational//references.bib ====

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@misc{Sun2026,
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  note          = {Submitted 3 September 2026},
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@book{Voight2021Quaternion,
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}
==== END Catalans-constant-is-irrational//references.bib ====