\section{The series for \texorpdfstring{$q_4$}{q4}}
\label{app:series}

We write $q_4^{(a)}(y)=\sum_na_ny^n$ and $q_4^{(b)}(y)=\sum_nb_ny^n$ with $y=\sqrt x$. Tables~\ref{tab:q4a} and~\ref{tab:q4b} give the coefficients, the size of each term at $y_H=2-\sqrt3$ and the partial sums.

\begin{table}[ht]
\centering
\begin{tabular}{rrrr}
\toprule
$n$ & $a_n$ & $a_ny_H^n$ & partial sum \\
\midrule
6 & $-450$ & $-0.16654$ & $-0.1665$ \\
7 & $-2184$ & $-0.21658$ & $-0.3831$ \\
8 & $-6246$ & $-0.16597$ & $-0.5491$ \\
9 & $-13632$ & $-0.09706$ & $-0.6461$ \\
10 & $-231931/9$ & $-0.04916$ & $-0.6953$ \\
11 & $-132776/3$ & $-0.02262$ & $-0.7179$ \\
12 & $-5104045/72$ & $-0.00971$ & $-0.7276$ \\
13 & $-107405.298$ & $-0.00394$ & $-0.7316$ \\
14 & $-155890.649$ & $-0.00153$ & $-0.7331$ \\
15 & $-218506.809$ & $-0.00058$ & $-0.7337$ \\
16 & $-297723.117$ & $-0.00021$ & $-0.7339$ \\
\bottomrule
\end{tabular}
\caption{Piece (a), the tree kernel. The coefficients through $y^{12}$ are exact rational numbers. The rest were obtained in floating-point arithmetic and are rational too. Successive terms fall by a factor that tends to $0.37$, and the remainder after $y^{16}$ is $-0.0001$.}
\label{tab:q4a}
\end{table}

\begin{table}[ht]
\centering
\begin{tabular}{rrrr}
\toprule
$n$ & $b_n$ & $b_ny_H^n$ & partial sum \\
\midrule
4 & $-42$ & $-0.21650$ & $-0.2165$ \\
5 & $-384$ & $-0.53039$ & $-0.7469$ \\
6 & $-2181/2$ & $-0.40359$ & $-1.1505$ \\
7 & $-2100$ & $-0.20825$ & $-1.3587$ \\
8 & $-118841/36$ & $-0.08772$ & $-1.4464$ \\
9 & $-128452/27$ & $-0.03387$ & $-1.4803$ \\
10 & $-306863/48$ & $-0.01220$ & $-1.4925$ \\
\bottomrule
\end{tabular}
\caption{Piece (b), the loop kernel. Each coefficient is the limit $J\to\infty$ of a sequence computed at finite cut-off, identified as a rational number. The coefficients at $n=4,5,6$ were obtained by two independent codes. A geometric estimate of the remainder gives $-0.007$, and Pad\'e approximants give totals between $-1.4987$ and $-1.5044$.}
\label{tab:q4b}
\end{table}

The leading powers have a simple origin. In piece (a) the first term needs three letters, and the lightest are three scalars of energy one, which gives $x^3=y^6$. In piece (b) two letters are enough, and two scalars give $x^2=y^4$.

The one-loop functions $c_3(x)$ and $Q_4(x)$ were obtained as power series in $y$ in the same way, to high enough order that the truncation error is below the precision quoted in \eqref{eq:schemes}. In the $su(2)$ sector they reduce to $c_3=2x^3$ and $Q_4=4x^3+8x^4$, which can be checked by hand.