\section{Denominators of the local solution}\label{sec:arithmetic}

Our aim is to turn the rational linear forms in \eqref{eq:value-row}
into integer linear forms without losing their smallness. Put $u=z+1$
and let $U(u)$ be the fundamental matrix of \eqref{eq:base-system},
normalized by $U(0)=I$. Thus the solution with initial column $x$ at $-1$
is $U(u)x$. The Taylor coefficients of $U$ record how the derivatives of
a solution depend on its initial values. Since the polynomial row
$C^{[p]}$ has integer coefficients, denominators for these Taylor
coefficients will suffice for our forms.

An integrating factor will turn the system into successive integrations.
The essential denominator saving is already visible in two integrations:
for nonnegative integers $m_1,m_2$,
\[
 \int_0^u x^{m_2}\left(\int_0^x t^{m_1}\dd t\right)\dd x
   =\frac{u^{m_1+m_2+2}}{(m_1+1)(m_1+m_2+2)}.
\]
The denominator factors are the degrees reached at the two integration
steps. The second is strictly larger than the first. Further integrations
give a strictly increasing list of degrees, all at most the final degree.
This motivates the distinct-product least common multiple from
\cite[Lemma~2]{F26}; here positive indices suffice.

For integers $d\geq1$ and $k\geq0$, define
\begin{equation}\label{eq:Delta-def}
 \Delta_{d,k}
   =\lcm\left\{\prod_{t\in I}t:
        I\subseteq\{1,\ldots,k\},\ |I|\leq d\right\},
 \qquad \Dt_{d,k}=4^k\Delta_{d,k}.
\end{equation}
Empty products and $\lcm(1,\ldots,0)$ are taken to be $1$.

\begin{lemma}\label{lem:lcm}
One has
\begin{equation}\label{eq:lcm-product}
 \Delta_{d,k}=\prod_{j=1}^{d}\lcm(1,\ldots,\lfloor k/j\rfloor).
\end{equation}
In particular, $\Dt_{d,k}$ divides $\Dt_{d,k+1}$, and for fixed $d$,
\begin{equation}\label{eq:D-rate}
 \log\Dt_{d,k}=(\log4+\Ht_d)k+o(k).
\end{equation}
\end{lemma}

\begin{proof}
For a prime $p$, the maximum valuation of a product of at most $d$
distinct integers in $\{1,\ldots,k\}$ is the sum of the $\min\{d,k\}$ largest
valuations in that set. It equals
\[
 \sum_{e\geq1}\min\{d,\lfloor k/p^e\rfloor\}.
\]
The product on the right of \eqref{eq:lcm-product} has the same valuation:
for each $e$, there are exactly $\min\{d,\lfloor k/p^e\rfloor\}$ indices
$j$ with $p^e\leq\lfloor k/j\rfloor$.
Divisibility follows either from the definition or from this formula.
Finally, the prime number theorem gives
$\log\lcm(1,\ldots,K)=K+o(K)$.
Sum this estimate over the fixed set $1\leq j\leq d$ to obtain
\eqref{eq:D-rate}.
\end{proof}

The dependence on $d$ is important. Allowing $d$ arbitrary factors from
$\{1,\ldots,k\}$ would give the larger multiplier
$\lcm(1,\ldots,k)^d$, with logarithm asymptotic to $dk$.
Distinct factors give $\Ht_d k$ instead. In our application $d$ is
proportional to $a$ and $k$ is of order $n$. First let $n$ grow with
$a$ fixed, then compare the resulting exponential rates as $a$ increases.
The naive multiplier would cost an exponent of order $an$, overwhelming
the intended decay on the scale $n\sqrt{a\log a}$. The distinct factors
reduce this cost to order $n\log a$, which leaves that decay available
for the dimension bound.

\begin{lemma}\label{lem:denominators}
For the normalized fundamental matrix $U$ defined above,
\[
 \Dt_{d,k}[u^k]U(u)\in M_{d+1}(\Z)
 \qquad(k\geq0).
\]
\end{lemma}

\begin{proof}
The common diagonal term in the equations is removed by an integrating
factor. Put $\eta(u)=(1-u)^{-1/2}$, set $y_i=\eta(u)v_i$ for $i\geq1$,
and leave $y_0=v_0$ unchanged. Since $\eta(0)=1$, this does not change
initial values.
The transformed equations are
\begin{equation}\label{eq:local-gauge}
 v_0'=0,\qquad
 v_1'=-\frac{\eta(u)}{1-u/2}v_0,\qquad
 v_i'=\frac{v_{i-1}}{1-u}\quad(i\geq2).
\end{equation}
Both $\eta(u)$ and $-\eta(u)/(1-u/2)$ have their degree-$j$ coefficient
in $4^{-j}\Z$. For $\eta$, this follows from
\[
 \eta(u)=\sum_{j\geq0}\binom{2j}{j}\frac{u^j}{4^j};
\]
multiplication by $(1-u/2)^{-1}$ preserves the asserted denominator bound.
The coefficients of $(1-u)^{-1}$ are integers.

Each entry of the normalized solution of \eqref{eq:local-gauge} is a
finite sum of iterated integrals of depth at most $d$, and multiplication
by $\eta$ restores $U$. Expand one contribution to a coefficient of degree
$k$. At each integration, the degree increases strictly. Its integration
denominators are therefore distinct integers
\[
 1\leq t_1<\cdots<t_e\leq k,\qquad e\leq d.
\]
Their product divides $\Delta_{d,k}$.
The total degrees supplied by coefficients of the multiplied series,
including the final factor $\eta$, are at most $k$; their powers of $4$
divide $4^k$. This controls each contribution and hence their sum.
The argument includes the column with $v_0(0)=1$, whose first integration
uses the inhomogeneous term displayed in \eqref{eq:local-gauge}.
\end{proof}

\begin{corollary}\label{cor:integer}
For every $p,k,i$ in the construction,
\[
 \Dt_{d,k-1}\frac{Q_{k,i}^{[p]}(-1)}{(k-1)!}\in\Z.
\]
For $k\geq2M+2$ this gives an integer linear form in
$1,\beta(2),\ldots,\beta(2\lfloor d/2\rfloor)$ whose value is
\begin{equation}\label{eq:integer-form}
 \Lambda_{n,p,k}
       =\Dt_{d,k-1}\frac{S_{n,p}^{(k-1)}(-1)}{(k-1)!}.
\end{equation}
\end{corollary}

\begin{proof}
The row $C^{[p]}(-1+u)$ has integral Taylor coefficients.
The coefficient of $u^{k-1}$ in $C^{[p]}(-1+u)U(u)$ is the row
$Q_k^{[p]}(-1)/(k-1)!$, by \eqref{eq:universal-row}.
Use Lemma~\ref{lem:denominators} and
$\Dt_{d,j}\mid\Dt_{d,k-1}$ for $j\leq k-1$.
Substitution of \eqref{eq:H-values} multiplies even columns by integers
and sets odd contributions to zero.
\end{proof}