\section{Rational functions for a shifted interval}\label{sec:construction}

\subsection{Repeated zeros and differentiation}

Fix a positive odd integer $m$ and an integer $B\ge1$, and put
$k=m-1$. The $b$th form will be an alternating sum over the
points $j/b$ with $j$ odd. We need zeros on these grids both to
remove initial summands and to control the denominators of the
rational constants. The signs on odd summation indices repeat
after a shift by $4$, so poles at $0,-4,\ldots,-4n$ allow the
partial-fraction sums to be shifted back to the beta series
without changing their signs. A polynomial multiplier will
provide freedom to choose the seed constants.

The fraction classes needed for all $1\le b\le B$ are
\[
 \Theta=\left\{\frac vd:1\le d\le B,\ 1\le v\le d,\ (v,d)=1\right\},
 \qquad D=|\Theta|=D_B.
\]
The integer class is represented by $1$.

Choose an integer $r\ge3$ and put $L=2r+4$. At index $n$,
the rightmost zero will be $rn$; reflection about $-2n$ puts the
leftmost zero at $-(r+4)n$. The length of this interval is $Ln$.
Let $S$ be an odd integer satisfying $S>mDL$, and put $s=S-1$.
Define
\begin{equation}\label{eq:K}
 A_d=d^L\prod_{p\mid d}p^{\lceil L/(p-1)\rceil},\qquad
 K=\prod_{\theta\in\Theta}A_{\operatorname{den}(\theta)}.
\end{equation}
The first product is over primes; the progression estimate in
Section~\ref{sec:arithmetic} explains the normalization.

For every positive odd integer $n$, define
\begin{align}
 P_n(t)&=(t-rn)\prod_{\theta\in\Theta}
                  \prod_{j=0}^{Ln-1}(t-rn+j+\theta),\label{eq:P}\\
 C_n&=4^{Sn}K^{mn}\frac{((n+1)!)^S}{(n!)^{mDL}},
 &R_n(t)&=\frac{C_nP_n(t)^m}{\prod_{h=0}^n(t+4h)^S}.
 \label{eq:R}
\end{align}
At $m=1$, these are the rational functions of~\cite[Section~2]{Tcount}.
All roots of $P_n$ are simple. They include the required grid
points in $[-(r+4)n,rn]$ and every denominator point
$0,-4,\ldots,-4n$. Thus the poles of $R_n$ have order at most
$S-m$. The root set is reflection invariant and $P_n$ has odd
degree. Since $m,n,S$ are odd,
\[
 P_n(-4n-t)=-P_n(t),\qquad R_n(-4n-t)=-R_n(t).
\]

We will choose an integer polynomial $W_n(t)$, odd about $-2n$
and homogeneous of a fixed total degree $E$ in $n,t$. Set
\begin{equation}\label{eq:T-V}
 T_n(t)=W_n(t)R_n(t),\qquad V_n(t)=\frac1{k!}T_n^{(k)}(t).
\end{equation}
The degree will satisfy $S-m-E>0$. Then $T_n$ is proper, with
degree at infinity at most $-(S-mDL)n-(S-m-E)$.
Both $T_n$ and $V_n$ are reflection invariant, because $k$ is even.
At every non-pole grid point in the zero interval, $T_n$ has a
zero of order at least $m=k+1$, so $V_n$ vanishes there. Its poles
have order at most $S-m+k=S-1$.

\subsection{The linear forms}

Write the partial fractions before and after differentiation as
\begin{align}
 T_n(t)&=\sum_{h=0}^n\sum_{i=1}^{S-m}
                       \frac{a_{i,h}(n)}{(t+4h)^i},\notag\\
 V_n(t)&=\sum_{h=0}^n\sum_{\nu=1}^{S-1}
                       \frac{A_{\nu,h}(n)}{(t+4h)^\nu}.
 \label{eq:partials}
\end{align}
The only possibly nonzero differentiated coefficients are
\begin{equation}\label{eq:derivative-coefficients}
 A_{i+k,h}(n)=\binom{i+k-1}{k}a_{i,h}(n),
                    \qquad 1\le i\le S-m.
\end{equation}
Reflection gives $A_{\nu,n-h}(n)=(-1)^\nu A_{\nu,h}(n)$.
Consequently
\[
 \rho_\nu(n)=\sum_{h=0}^n A_{\nu,h}(n)=0
                         \quad\hbox{for odd }\nu.
\]

To write the alternating signs compactly, let $\ch(j)$ be $0$
for even $j$, $1$ for $j\equiv1\pmod4$, and $-1$ for
$j\equiv3\pmod4$. For $1\le b\le B$, put
\begin{equation}\label{eq:F}
 F_{n,b}=\sum_{j\ge1}\ch(j)V_n(j/b).
\end{equation}
In the term belonging to the pole $-4h$, shift the summation index
by $4bh$. The character is unchanged, and the omitted initial terms
give a rational constant. Thus
\begin{align}
 F_{n,b}&=q_b(n)+
   \sum_{\substack{m+1\le\nu\le S-1\\\nu\text{ even}}}
          b^\nu\rho_\nu(n)\beta(\nu),\label{eq:forms}\\
 q_b(n)&=-\sum_{h=0}^n\sum_{\nu=1}^{S-1}A_{\nu,h}(n)b^\nu
                     \sum_{j=1}^{4bh}\frac{\ch(j)}{j^\nu}.
 \label{eq:q}
\end{align}
For $k\ge2$ the individual series are absolutely convergent.
When $k=0$, the individual exponent-one series converge
conditionally and their total beta coefficient is zero. There
are only finitely many partial-fraction series, so the decomposition
is valid in this case too.

\subsection{A rational criterion for the endpoints}

The criterion below states the two estimates we will need. The
positive rational $u$ bounds the location of the largest terms,
measured beyond $rn$ in units of $n$. The quantity $U$ bounds the
fourth power of the exponential rate after clearing denominators.
Put $e_0=68/25$, a rational upper bound for $\ep$.

\begin{proposition}\label{prop:criterion}
With the parameters above, suppose that a positive rational $u$
satisfies
\begin{align}
 f(u)&=\left(\frac{u+L}{u}\right)^{4mD}
       \left(\frac{u+r}{u+r+4}\right)^S<1,\label{eq:f}\\
 U&=e_0^{4(S+k)}K^{4m}(L+u)^{4mDL}4^{4S}
       \left(\frac{r^r}{(r+4)^{r+4}}\right)^S<1.
 \label{eq:U}
\end{align}
Then at least $B$ of
$\beta(m+1),\beta(m+3),\ldots,\beta(S-1)$ are irrational.
\end{proposition}

The extra factor $e_0^{4k}$ in \eqref{eq:U} pays for the
denominators introduced by differentiation. We still need a seed
constant that survives the elimination of the possible irrational
values.

For example, when $B=2$, the combination
\[
 2^\nu F_{n,1}-F_{n,2}
\]
eliminates $\beta(\nu)$ at every $n$, by \eqref{eq:forms}.
If $q_1(1)=0$ and $q_2(1)=\eta\ne0$, its rational constant at
$n=1$ is $-\eta$, whichever index $\nu$ must be eliminated.

For general $B$, the elimination in Section~\ref{sec:contradiction}
can be performed with a nonzero coefficient on the last form.
We therefore choose $W_n$ so that
\begin{equation}\label{eq:seed-target}
 (q_1(1),\ldots,q_B(1))=\eta e_B,\qquad
                        \eta\in\Q\setminus\{0\},
\end{equation}
where $e_B=(0,\ldots,0,1)$. The rational constant of the combined
form at $n=1$ is then $\eta$ times its nonzero last coefficient.
This is the seed constant used by the prime-dilation argument
in Section~\ref{sec:dilation}.