\section{A Gram matrix for the seed constants}\label{sec:gram} We construct $B$ candidate multipliers. Each produces a vector of $B$ seed constants, one for each odd grid. These vectors form the columns of a rational matrix $\mathcal M$. If $\mathcal M$ is nonsingular, a rational combination of its columns gives $e_B$; clearing denominators gives the seed vector \eqref{eq:seed-target} with integer coefficients on the multipliers. We adapt the Gram representation of~\cite[Section~3]{Tcount}, first showing the choice with two grids. \subsection{A worked example with two grids}\label{sec:worked-gram} Take $B=2$, $m=1$, $r=2$ and $a=70$, the parameters that will give two irrational values through $\beta(70)$. There is no differentiation, since $k=0$. The grids have $b=1,3$. We want $q_1(1)=0$ and $q_3(1)=\eta\ne0$. Indeed, for any possible irrational value $\beta(\nu)$, the fixed combination $3^\nu F_{n,1}-F_{n,3}$ cancels its coefficient at every $n$. Its seed constant is then $-\eta\ne0$. At $n=1$, put $z=t+1$. The two kernels are \[ H_1(t)=\frac1z,\qquad H_3(t)=\frac1{z-2/3}-\frac1z+\frac1{z+2/3}. \] Expansion at $t=0$ gives $q_b(1)=\Res_{t=0}T_1(t)H_b(t)$. We will choose independent polynomials whose pairwise products, integrated against one positive weight, give the seed matrix up to a common sign. The line $z=iy$ lies midway between the poles $t=0,-2$ and turns every symmetric factor $z^2-\alpha^2$ into $-(y^2+\alpha^2)$, with fixed sign. Write $x=y^2$. On this line, $H_b(-1+iy)=iy\,g_b(x)$, with \[ g_1(x)=-\frac1x,\qquad g_3(x)=\frac1x-\frac2{x+4/9}. \] Multiplying by their common denominator $x(9x+4)$ gives two independent polynomials, \[ h_1(x)=-9x-4,\qquad h_3(x)=4-9x. \] To obtain products $h_bh_j$ in the integral, choose multipliers whose values on this line reproduce the polynomials: $W_{1,j}(-1+iy)=h_j(y^2)$. Substituting $-(t+1)^2$ for $x$ and homogenizing in $n$ gives \[ W_{n,1}(t)=9(t+n)^2-4n^2,\qquad W_{n,3}(t)=9(t+n)^2+4n^2. \] These are integer polynomials, even about $-n$ and homogeneous of degree $2$. Let $\mathcal M$ have rows indexed by the seed constants and columns indexed by these multipliers. Closing the midpoint contour clockwise gives $\mathcal M=-\mathcal N$, where \[ \mathcal N_{b,j}=\int_{-\infty}^{\infty} w(y)h_b(y^2)h_j(y^2)\,dy,\qquad b,j\in\{1,3\}, \] and the positive weight is \[ w(y)=\frac{2^{85}3^{16}}{18\pi} \frac{\displaystyle \prod_{\alpha\in\{4/3,\,2,\,8/3,\,10/3,\,4,\,14/3\}} (y^2+\alpha^2)} {(1+y^2)^{70}}. \] The numerator's central zero cancels the kernel pole at $z=0$; its factor $z^2-4/9$ cancels the paired poles at $z=\pm2/3$. On the midpoint line, $P_1$ is $-iy$ times a positive product, since it has seven symmetric root pairs. Multiplying by the kernel's $iy$ factor removes that sign, and the clockwise contour supplies the minus sign in $\mathcal M=-\mathcal N$. Section~\ref{sec:positive-weight} gives the general calculation. Put $\mu_\ell=\int_{-\infty}^{\infty}w(y)y^{2\ell}\,dy$ for $\ell=0,1,2$. The matrix is explicitly \[ \mathcal N= \begin{pmatrix} 16\mu_0+72\mu_1+81\mu_2&-16\mu_0+81\mu_2\\ -16\mu_0+81\mu_2&16\mu_0-72\mu_1+81\mu_2 \end{pmatrix}. \] Its determinant is \[ \det\mathcal N=5184(\mu_0\mu_2-\mu_1^2)>0. \] The inequality is strict Cauchy--Schwarz: $y^2$ is not constant under the positive weight $w$. The rational combination \[ W_n=\mathcal N_{1,3}W_{n,1}-\mathcal N_{1,1}W_{n,3} \] has seed vector $(0,\det\mathcal N)^{\mathsf T}$. Its coefficients are rational because $\mathcal M$ is a matrix of rational residues. Clearing their fixed denominators gives an integer multiplier with seed vector $(0,\eta)$, $\eta\ne0$. We now make the same choice with $B$ independent polynomials. \subsection{Moving the derivatives onto the kernels} At $n=1$, the poles are $0$ and $-2$. For a proper rational function $V$ with these poles and antisymmetric about $-1$, its partial-fraction coefficients satisfy $A_{\nu,1}=(-1)^{\nu+1}A_{\nu,0}$. For odd $b\le M$, define \[ H_b(t)=\sum_{\substack{1\le j<2b\\j\text{ odd}}} \frac{\ch(j)}{t+j/b}. \] Expansion at $0$ and \eqref{eq:q} give $q_b=\Res_{t=0}V(t)H_b(t)$. If $V=T^{(k)}/k!$, repeated integration by parts for residues gives \begin{equation}\label{eq:residue} q_b=\Res_{t=0}T(t)\frac{H_b^{(k)}(t)}{k!}. \end{equation} The sign is positive because $k$ is even. Transferring the derivatives to the kernel leaves the numerator of $T$ intact. Put $c=(m+1)/2$ and $z=t+1$. The central kernel term is $\ch(b)/z$. The terms at distances $\pm\alpha$, $\alpha=2j/b$, have equal weights $\ch(b-2j)=\ch(b+2j)$. On $z=iy$ we therefore have \[ \frac{H_b^{(k)}(-1+iy)}{k!}=iy\,g_b(y^2), \] where \begin{equation}\label{eq:g} \begin{split} g_b(x)={}&\frac{(-1)^c\ch(b)}{x^c}\\ &-2\sum_{j=1}^{(b-1)/2}\ch(b-2j) \frac{\displaystyle\sum_{\ell=0}^{c-1}(-1)^\ell \binom m{2\ell+1}\alpha^{m-2\ell-1}x^\ell} {(x+\alpha^2)^m},\qquad \alpha=\frac{2j}{b}. \end{split} \end{equation} This follows by combining $(iy-\alpha)^{-m}+(iy+\alpha)^{-m}$. For $b=1$ the sum is empty: $g_1=-1/x$ when $m=1$, and $g_1=1/x^2$ when $m=3$. \subsection{Independent integer polynomials} Let $\mathcal A$ be the distinct positive distances in \eqref{eq:g}. In reduced form they are $v/d$ with $d\le M$ odd, $v$ even and $01$, its pole at $x=-((b-1)/b)^2$ has exact order $m$ and is absent from every earlier kernel: it is farther from $0$ than any of their negative poles. The order is exact because the corresponding term of $H_b^{(k)}/k!$ has nonzero principal coefficient, and $x=-z^2$ has nonzero derivative there. If $b=1$, then $g_1$ is nonzero. Thus no relation is possible. Multiplication by the common denominator preserves independence. \end{proof} Homogenize the polynomials to obtain \begin{align} W_{n,b}(t) &=n^{2J}h_b\!\left(-\frac{(t+n)^2}{n^2}\right)\notag\\ &=\sum_{j=0}^J h_{b,j}(-1)^j n^{2J-2j}(t+n)^{2j}. \label{eq:W} \end{align} They have integer coefficients, are even about $-n$, and have total degree $E=2J=m(D-1)$, as required in \eqref{eq:E}. In particular, \begin{equation}\label{eq:homogeneity} W_{p,b}(pt)=p^E W_{1,b}(t). \end{equation} \subsection{The common positive weight}\label{sec:positive-weight} At $n=1$, the symmetric simple roots give \[ P_1(t)=z\prod_{\alpha\in\Lambda}(z^2-\alpha^2), \qquad z=t+1,\qquad |\Lambda|=\frac{DL-1}{2}, \qquad\mathcal A\subseteq\Lambda. \] All $\alpha$ in $\Lambda$ are positive. Let $\mathcal M_{b,j}$ be the $b$th seed constant associated with $W_{1,j}$. Every pole of $H_b^{(k)}/k!$ in $(-2,0)$, including the central pole, has order at most $m$ and is canceled by $P_1^m$. Thus the product in \eqref{eq:residue} has poles only at $0,-2$. Close the upward line $t=-1+iy$ by a large right semicircle, which encloses $0$ clockwise. Since $H_b^{(k)}(t)=O(t^{-m})$, the integrand has degree at most $mDL+E-2a-m<-1$ at infinity. The arc tends to zero. On the midpoint line, $P_1^m$ contributes $i^m(-1)^{|\Lambda|}y^m$ times a positive product. The kernel contributes $iy\,h_b(y^2)/(y^{2c}Q(y^2)^m)$. Since $2c=m+1$, the powers of $y$ cancel. The clockwise orientation supplies the final minus sign, giving \begin{equation}\label{eq:Gram} \mathcal M_{b,j}=\frac{\sigma\kappa_1}{2\pi} \int_{-\infty}^{\infty}\omega(y)h_b(y^2)h_j(y^2)\,dy, \qquad \sigma=(-1)^{c+(DL-1)/2+1}, \end{equation} where \[ \omega(y)=\frac{\displaystyle \prod_{\alpha\in\Lambda}(y^2+\alpha^2)^m} {(y^2+1)^a Q(y^2)^m}. \] This weight is strictly positive for all real $y$. The central singularity has canceled, and the degree estimate proves absolute convergence. Lemma~\ref{lem:independence} therefore makes $\sigma\mathcal M$ positive definite: its quadratic form is the integral of the square of a nonzero polynomial against this weight. The entries of $\mathcal M$ are rational by \eqref{eq:residue}. Solve $\mathcal M\mathbf c=e_B$ over $\Q$ and clear denominators. This gives integer coefficients $c_j$ and a nonzero rational $\eta$ such that $\mathcal M\mathbf c=\eta e_B$. Set \[ W_n=\sum_{\substack{1\le j\le M\\j\text{ odd}}}c_j W_{n,j}. \] This proves \eqref{eq:seed-target}. The coefficients remain fixed as $n$ varies, so their sizes do not affect an exponential rate.