Primary minimal dom-path length conjecture for linear Jaco graphs

Consider $J_{33}$, and write $\varphi=(1+\sqrt5)/2$. For $i<j$, the Jaco adjacency rule is

$$v_i v_j\in E(J_{33}) \quad\Longleftrightarrow\quad j\le U(i), \qquad U(i)=2i-\left\lfloor\frac{i+1}{\varphi^2}\right\rfloor.$$

The four closed neighborhoods

$$\begin{aligned} N[v_1]&=\{v_1,v_2\},\\ N[v_4]&=\{v_3,\ldots,v_7\},\\ N[v_{12}]&=\{v_8,\ldots,v_{20}\},\\ N[v_{33}]&=\{v_{21},\ldots,v_{33}\} \end{aligned}$$

are pairwise disjoint. Therefore $\gamma(J_{33})\ge4$. Conversely,

$$D=\{v_2,v_7,v_{20},v_{33}\}$$

dominates $J_{33}$, so $\gamma(J_{33})=4$.

Since $U$ is increasing, the successive maximal forward reaches from $v_1$ have indices

$$1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ 34.$$

Thus $d(v_1,v_{33})=7$. Starting from any larger vertex reaches at least as far, so every vertex pair has distance at most $7$, and hence

$$\operatorname{diam}(J_{33})=7.$$

Every primary minimal dom-path $P$ must satisfy

$$\gamma(P)=\gamma(J_{33})=4.$$

Because $\gamma(P)=\lceil |V(P)|/3\rceil$, such a path has at least ten vertices and therefore at least nine edges. Equality is attained by

$$P=(v_1,v_2,v_3,v_4,v_7,v_{11},v_{12},v_{20},v_{32},v_{33}).$$

Its consecutive vertices are adjacent, and $D$ occupies path positions $2,5,8,10$. Thus $D$ is a minimum dominating set of both $P$ and $J_{33}$, so $P$ is a primary minimal dom-path of length $9$.

Consequently,

$$|P_d|-\operatorname{diam}(J_{33})=9-7=2>1,$$

contradicting the conjectured bound.