Equality between k-Grundy Z-domination and k-forcing number

The conjecture is false for EVERY $k\ge2$, even for connected graphs satisfying its minimum-degree hypothesis.

Fix $k\ge2$. Let

$$A=\{u,v,a_1,\ldots,a_k\}, \qquad B=\{b_1,\ldots,b_k\},$$

and let $G_k$ be obtained from the complete bipartite graph $K_{k+2,k}$ on $A\sqcup B$ by adding the edge $uv$. Then

$$|V(G_k)|=2k+2,\qquad \delta(G_k)=k.$$

First, no singleton is a $k$-forcing set. An initial vertex $u$, $v$, or $b_j$ has more than $k$ white neighbors and cannot begin forcing. An initial vertex $a_i$ forces all $k$ vertices of $B$, but each newly blue $b_j$ then has $k+1$ remaining white neighbors, so forcing stops.

On the other hand, start with the two blue vertices $\{u,a_1\}$. Vertex $a_1$ forces all of $B$, after which any $b_j$ forces the remaining $k$ vertices of $A$. Consequently

$$F_k(G_k)=2.$$

Now consider the sequence

$$(a_1,a_2,\ldots,a_k,u,v,b_1,b_2,\ldots,b_{k-1}).$$

This is a valid $k$-$Z$-Grundy sequence:

• For $a_1,\ldots,a_k$, use the open-neighborhood witness $b_k$. The respective numbers of previous selected closed neighborhoods containing $b_k$ are $0,1,\ldots,k-1$.
• For $u$, use witness $v$, which lies in zero previous selected closed neighborhoods.
• For $v$, use witness $u$, which lies in one previous selected closed neighborhood.
• For $b_1,\ldots,b_{k-1}$, use witness $a_1$. The respective counts are $1,2,\ldots,k-1$.

All these counts are strictly less than $k$. Therefore

$$\gamma_{\mathrm{gr}}^{Z,k}(G_k)\ge2k+1.$$

A sequence containing all $2k+2$ vertices is impossible. At its final step, for every possible open-neighborhood witness $w$ of the final vertex, the previous selected closed neighborhoods contain $w$ exactly

$$\deg(w)\ge\delta(G_k)=k$$

times, violating the defining strict inequality. Hence

$$\gamma_{\mathrm{gr}}^{Z,k}(G_k)=2k+1.$$

It follows that

$$\boxed{ \gamma_{\mathrm{gr}}^{Z,k}(G_k) =2k+1 > 2k =|V(G_k)|-F_k(G_k) } \qquad(k\ge2).$$

Already for $k=2$, the graph has six vertices and gives $5\ne6-2=4$. Thus the asserted equality fails for every forcing parameter $k\ge2$.