Extending $(0,1)$ by one even target $t\equiv2\pmod4$ with $\varepsilon_0=0$ forces $t=2\cdot3^n$ and odd letters $=\{3^i\}$

Fix $\varepsilon:\mathbb{N}\to\{0,1\}$ and the $2$-adic total order on $\mathbb{Z}$ given by $x\prec_\varepsilon y$ iff $x\neq y$ and the $k$-th two's-complement digit of $x$ equals $\varepsilon_k$, where $k=\nu_2(y-x)$. A finite word $Q$ of integers is $\varepsilon$-safe if the sequence "the letters of $Q$ in order, then all remaining integers in $\prec_\varepsilon$-order" contains no monotone $4$-term arithmetic progression. Let $t>0$ with $t\equiv 2\pmod 4$, and let $Q$ be an $\varepsilon$-safe word that begins with $0,1$, contains $t$, and whose only even letters are $0$ and $t$. If $\varepsilon_0=0$, then there is $n\ge0$ with

Thus, outside the very thin family $t=2\cdot3^n$, extending the safe word $(0,1)$ by one even target $t\equiv2\pmod4$ forces $\varepsilon_0=1$, and even in that family the odd letters are completely determined. This complements the case $t\equiv0\pmod4$, where $\varepsilon_0=1$ is always forced (Ho-style constructions of $4$-AP-free permutations of $\mathbb{Z}$, Erdős Problem 195).

Formalization note: Safe, epsLt are from Definitions.Def_Erdos195_defs; "begins with $0,1$" is [0, 1] <+: Q; $\varepsilon_0=0$ is ε 0 = false; the set equality is stated as membership of all $3^i$ ($i\le n$) plus every odd letter being some $3^i$ with $i\le n$.