The integer 2-domination formula for four-row grids

The conjecture holds for EVERY $n\ge49$, with an exact finite-state induction certificate.

Fix the height $h$, let $\mathcal C=\{0,1,2\}^{h}$, put $\mathbf0=(0,\ldots,0)$, and write $|b|=\sum_i b_i$. For $a,b,c\in\mathcal C$, extend $b$ by $b_0=b_{h+1}=0$ and define

$$Q(a,b,c)\iff a_i+b_{i-1}+b_i+b_{i+1}+c_i\ge2\quad(1\le i\le h).$$

This is exactly the closed-neighborhood constraint for the middle column. Let $V_n(a,b)$ be the minimum weight of an $n$-column labeling with last columns $a,b$, all columns before $b$ already valid, and the fictitious left boundary zero. Then

$$V_1(a,b)=\begin{cases}|b|&a=\mathbf0,\\+\infty&a\ne\mathbf0,\end{cases} \qquad V_{n+1}(b,c)=|c|+\min_{a:Q(a,b,c)}V_n(a,b).$$

The exact domination number is obtained by adjoining the zero right boundary:

$$\gamma_{\{2\}}(G_{h,n}) =\min_{a,b:Q(a,b,\mathbf0)}V_n(a,b).$$

The time-independent transfer operator $T$ satisfies $T(V+d)=T(V)+d$. Therefore an identity of COMPLETE state vectors $V_{N}=V_{N-p}+d$, including every state, implies by induction

$$V_{n+p}=V_n+d,\qquad \gamma_{\{2\}}(G_{h,n+p})=\gamma_{\{2\}}(G_{h,n})+d \quad(n\ge N-p).$$

For $h=4$, direct exact evaluation gives the COMPLETE 6561-coordinate identity

$$V_{78}(a,b)=V_{68}(a,b)+19 \qquad(a,b\in\{0,1,2\}^{4}).$$

The exact terminal optimum agrees with

$$f(n)=2n-\left\lfloor\frac{n-9}{10}\right\rfloor$$

for every base width $49\le n\le77$. The full state-vector identity propagates

$$\gamma_{\{2\}}(G_{4,n+10})=\gamma_{\{2\}}(G_{4,n})+19 \qquad(n\ge68).$$

Since $f(n+10)=f(n)+19$, these base cases and induction prove

$$\boxed{\gamma_{\{2\}}(G_{4,n}) =2n-\left\lfloor\frac{n-9}{10}\right\rfloor \quad\text{for every }n\ge49.}$$

The following complete, standard-library-only exact certificate checks every required base width and EVERY coordinate of BOTH 6561-state vectors. Missing dictionary entries represent $+\infty$; the final cardinality checks exclude unreachable-state ambiguity.

from itertools import product

h, N, period, increment = 4, 78, 10, 19
C = tuple(product(range(3), repeat=h))
Z = (0,) * h
weight = {c: sum(c) for c in C}

def valid(a, b, c):
    return all(a[i] + b[i] + c[i]
               + (b[i-1] if i else 0)
               + (b[i+1] if i+1 < h else 0) >= 2
               for i in range(h))

successors = {(a,b): tuple(c for c in C if valid(a,b,c))
              for a in C for b in C}
V = {(Z,b): weight[b] for b in C}
history = {}
for n in range(1, N+1):
    history[n] = V
    optimum = min(v for (a,b),v in V.items() if valid(a,b,Z))
    if n >= 49:
        assert optimum == 2*n - (n-9)//10
    W = {}
    for (a,b),v in V.items():
        for c in successors[a,b]:
            state = (b,c)
            W[state] = min(W.get(state, 10**9), v+weight[c])
    V = W
A, B = history[N-period], history[N]
assert len(A) == len(B) == len(C)**2
assert A.keys() == B.keys()
assert all(B[state] == A[state]+increment for state in A)
print("PASS:", h, len(B), N, N-period, increment)

The checked equality is between complete transfer-state vectors, so time-independent min-plus homogeneity proves the formula for all remaining widths, not merely for the finitely checked base interval.