\section{Denominators after differentiation}\label{sec:arithmetic}

\subsection{A progression estimate}

Put $d_n=\lcm(1,\ldots,n)$ and $\delta_n=4d_n$. We first
control the Taylor coefficients of the numerator blocks. Taking
a coefficient of degree $j$ removes at most $j$ divisible
constants at each prime-power level; $d_n^j$ pays for that loss.

\begin{lemma}\label{lem:progression}
For $\theta=v/d\in\Theta$ in lowest terms, define
\[
 B_\theta(t)=\frac{A_d^n}{(n!)^L}
             \prod_{j=0}^{Ln-1}(t-rn+j+v/d).
\]
For every integer $v_0$ and every $j\ge0$,
\begin{equation}\label{eq:block-integrality}
 d_n^j[t^j]B_\theta(t+v_0)\in\Z.
\end{equation}
\end{lemma}
\begin{proof}
Clear $d$ from each factor using the factor $d^{Ln}$ in $A_d^n$.
Fix a prime $q\nmid d$. At each level $q^e\le n$, split the
$Ln$ progression terms into $L$ blocks of length $n$. Each block
has at least $\lfloor n/q^e\rfloor$ divisible constants.
A term in the coefficient of $t^j$ replaces $j$ constants by
variable terms and loses at most $j$ contributions at that level.
Its valuation is therefore at least
\[
 L v_q(n!)-j v_q(d_n).
\]
A term retaining a zero constant is zero and needs no estimate.
If $q\mid d$, the cleared constants are units and the extra
factor in $A_d^n$ has valuation
\[
 n\left\lceil\frac{L}{q-1}\right\rceil
       \ge\frac{Ln}{q-1}\ge L v_q(n!).
\]
These bounds prove \eqref{eq:block-integrality}. Products and
integer powers preserve the same coefficient rule by convolution.
\end{proof}

\subsection{Partial-fraction coefficients}

\begin{proposition}\label{prop:denominators}
For every positive odd $n$,
\begin{align}
 \delta_n^{S-i}a_{i,h}(n)&\in\Z,\label{eq:a-integral}\\
 \delta_n^{S+k-\nu}A_{\nu,h}(n)&\in\Z,
 &\delta_n^{S+k-\nu}\rho_\nu(n)&\in\Z,
 &\delta_n^{S+k}q_b(n)&\in\Z.\label{eq:integral}
\end{align}
\end{proposition}
\begin{proof}
Write $B_n(t)=K^{mn}P_n(t)^m/(n!)^{mDL}$ and suppress the
index $n$ on the coefficients. Expanding at $-4h$ gives
\begin{align}
 a_{i,h}={}&(n+1)^S(-1)^{Sh}\binom nh^S\notag\\[-2pt]
 &\quad\times[t^{S-i}]
 \left\{W_n(t-4h)B_n(t-4h)
       \prod_{\ell\ne h}\left(1+\frac{t}{4(\ell-h)}\right)^{-S}
 \right\}.\label{eq:local-coefficient}
\end{align}
The constants from the other poles have product
$4^{Sn}(-1)^{Sh}(h!(n-h)!)^S$. Their cancellation with $C_n$
leaves the displayed integer prefactor. Each $4(\ell-h)$ divides
$\delta_n$, and $W_n$ has integer coefficients.
Lemma~\ref{lem:progression} and the binomial series prove
\eqref{eq:a-integral}. Formula~\eqref{eq:derivative-coefficients}
then proves the first two bounds in \eqref{eq:integral}.

The finite sums in \eqref{eq:q} can introduce prime powers not
covered by $\delta_n$, so the coefficient bounds alone do not
prove integrality term by term. We use the zeros that survive
differentiation to cancel these denominators after grouping,
as in~\cite[Section~4]{Tcount}.
Fix an odd $j\le4bh$ and put $t_0=j/b-4h$.
This is a grid point in $(-4n,0)$ and is
not a pole, so $V_n(t_0)=0$. With
$x_\ell=j/b+4(\ell-h)$, this says
\begin{equation}\label{eq:grouping}
 \sum_\nu\frac{A_{\nu,h}}{x_h^\nu}
   =-\sum_{\ell\ne h}\sum_\nu\frac{A_{\nu,\ell}}{x_\ell^\nu}.
\end{equation}
At a prime $q$ put $v=v_q(\delta_n)$. At most one $x_\ell$ has
valuation greater than $v$: otherwise the difference of two such
terms, a nonzero $4(\ell-\ell')$ dividing $\delta_n$, would too.
The coefficient bound gives
\begin{equation}\label{eq:group-bound}
 v_q\!\left(\frac{\delta_n^{S+k}A_{\nu,\ell}}{x_\ell^\nu}\right)
                    \ge\nu\bigl(v-v_q(x_\ell)\bigr).
\end{equation}
If $v_q(x_h)\le v$, the left group in \eqref{eq:grouping} is
integral after scaling. Otherwise every group on the right is
integral after scaling, so the same conclusion follows from
\eqref{eq:grouping}. This holds at every prime. Summing the
integral groups with the signs in \eqref{eq:q} proves the last
bound in \eqref{eq:integral}.
\end{proof}

The exponent in the common multiplier is $S+k$. Multiplicity
$m$ in the numerator preserves the zeros needed for the grouped
argument, while differentiation shifts a coefficient of order $i$
to order $i+k$. Its denominator bound remains $\delta_n^{S-i}$,
which is $\delta_n^{S+k-(i+k)}$. This explains the additional
arithmetic cost of differentiation.