Euler-exponential gamma-set conjecture for the infinite linear Jaco graph

The proposed Euler-exponential set is not even a dominating set. In fact, it leaves infinitely many vertices undominated.

For the infinite linear Jaco graph, the established indegree formula is

$$d^-(v_i) = \left\lfloor\frac{2(i+1)}{3+\sqrt5}\right\rfloor.$$

For $i<j$, its underlying undirected adjacency rule is

$$v_iv_j\in E \quad\Longleftrightarrow\quad j\le U(i), \qquad U(i)=2i-d^-(v_i).$$

Since successive indegrees differ by zero or one, $U(i)$ is strictly increasing.

The proposed set is

$$X=\{v_{\lfloor e^t\rfloor}:t=0,1,2,\ldots\} =\{v_1,v_2,v_7,v_{20},v_{54},v_{148},\ldots\}.$$

Consider $v_{88}$. The last selected index preceding 88 is 54, and

$$d^-(v_{54})=21, \qquad U(54)=2\cdot54-21=87<88.$$

By monotonicity of $U$, no selected vertex of index at most 54 is adjacent to $v_{88}$.

On the other hand,

$$d^-(v_{88})=33, \qquad U(88)=2\cdot88-33=143<148.$$

Thus no selected vertex of index at least 148 is adjacent to $v_{88}$. Since $v_{88}\notin X$,

$$N[v_{88}]\cap X=\varnothing.$$

Therefore $X$ is not dominating and cannot be a minimum dominating set.

Moreover, put $\varphi=(1+\sqrt5)/2$ and $a_t=\lfloor e^t\rfloor$. The indegree formula gives

$$U(i)=\varphi i+O(1),$$

whereas

$$a_{t+1}=e\,a_t+O(1).$$

Since $e>\varphi^2$,

$$U\bigl(U(a_t)+1\bigr) = \varphi^2a_t+O(1) < a_{t+1}$$

for all sufficiently large $t$. Hence each vertex

$$v_{U(a_t)+1}$$

has no neighbor among the proposed indices at most $a_t$ or among those at least $a_{t+1}$. Consequently the conjectured set misses infinitely many vertices.