Which letter may precede the shield: $(x,0,1,6,3)$ is tree-safe iff $x\equiv1$ (mod 4) or $x\in\{7,16\}$

A tree witness is a function $\theta(k,r)\in\{0,1\}$ on the nodes $(k,r)$, $0\le r<2^k$, of the binary tree of residues. For distinct integers $x,y$ with $k=\nu_2(y-x)$, put $x\prec_\theta y$ iff $\mathrm{bit}_k(x)=\theta(k,\,x\bmod 2^k)$ (two's-complement digits). A finite word $Q$ of integers is $\theta$-safe if the sequence "the letters of $Q$ in order, then all remaining integers in $\prec_\theta$-order" contains no monotone $4$-term arithmetic progression; it is tree-safe if it is $\theta$-safe for some $\theta$. An in-order pair $(a,b)$ of $Q$ ($a$ occurs before $b$) is open if $2b-a\notin Q$ and $3b-2a\notin Q$; an open pair lies at the node $N(a,b)=(k,\,a\bmod 2^k)$, $k=\nu_2(b-a)$, and $\theta$-safety requires $\theta(N(a,b))=\mathrm{bit}_k(b)$ there. Let $x$ be an integer with $x\notin\{0,1,3,6\}$. Then the five-letter word $(x,0,1,6,3)$ is tree-safe if and only if

Here $(0,1,6,3)$ is the shield word $S(0,1)$ (tree-safe, closing the root pair $(0,1)$). The result classifies which single letter may precede the shield. Obstructions: for even $x\neq 16$ the root pair $(1,6)$ (forcing $\mathrm{bit}_0=0$) conflicts at the root with $(x,1)$ or $(x,3)$; for $x\equiv 3\pmod 4$, $x\neq 7$, the pair $(1,3)$ conflicts with $(x,1)$ at level $1$; $x\in\{-1,-4,-6\}$ create an in-order $3$-AP without its fourth term. The exceptions $x=7$, $x=16$ close $(1,3)$ resp. $(1,6)$.

Formalization note: treeLt θ is from Definitions.Def_Erdos195_tree; Safe, bitOne are from Definitions.Def_Erdos195_defs. Positions in a word are List.idxOf; P ++ [c] appends c.