The answer is yes. The initial values are \[ (\gamma(1),\ldots,\gamma(14)) =(3,4,4,4,4,4,6,8,8,8,8,8,10,11). \] For every \(n\geq9\), \[ \gamma(n+6)=\gamma(n)+4. \] Equivalently, write \(n=6q+r\), where \(0\leq r<6\). Then for \(n\geq9\), \[ \gamma(n)=4q+c_r, \qquad (c_0,c_1,c_2,c_3,c_4,c_5)=(0,2,3,4,4,4). \]

The finite-state proof is recorded in `ksd6-claim-min-plus-certificate`. Its dynamic program computes the optimum over every column-mask sequence, so each displayed value includes both a construction and a lower bound. The scalar recurrence is checked directly for \(9\leq n\leq18\). A componentwise min-plus vector identity at widths 19 and 25 propagates the result through every larger width.

Represent a set of rows in one column by a three-bit mask. For a mask \(m\), let \(L_1(m)\) be the rows reached one column away by a knight move, so rows 0 and 2 are interchanged and row 1 reaches nothing. Let \(L_2(m)\) be the rows reached two columns away, so each selected row reaches its adjacent rows.

After processing a prefix, use the state \[ (a,b,u,v)\in\{0,\ldots,7\}^4. \] Here \(a,b\) are the selected masks in the last two columns and \(u,v\) are their currently dominated masks. Choosing mask \(x\) in the next column is allowed exactly when \[ u\mathbin{\vert}L_2(x)=7, \] because this is the final chance to dominate the older column. The new state is \[ (b,x,\ v\mathbin{\vert}L_1(x),\ x\mathbin{\vert}L_1(b)\mathbin{\vert}L_2(a)), \] with cost \(|x|\). Start at \((0,0,7,7)\), using two already-satisfied padding columns. After \(n\) actual transitions, two forced zero-mask transitions enforce domination in the final two actual columns. Every set of board squares gives one path, and every accepted path gives a dominating set.

Let \(f_n(s)\) be the minimum path cost to state \(s\), with infinity for unreachable states. The state space has \(8^4=4096\) elements. Exactly 351 states are reachable at width 6, and this reachable set remains fixed. Its transfer graph has 2,046 labeled edges. Exact evaluation gives \[ f_{25}(s)=f_{19}(s)+4 \] for all 351 finite entries, with the same states infinite on both sides. If \(T\) is the min-plus transfer, then \(f_{n+1}=Tf_n\), and \(T(z+4)=Tz+4\). Applying \(T^k\) proves \(f_{25+k}=f_{19+k}+4\) for every \(k\geq0\). The two-zero finalization also commutes with an additive constant, so \(\gamma(n+6)=\gamma(n)+4\) for every \(n\geq19\). Direct exact evaluation covers \(9\leq n\leq18\).

Searches for ordinary domination on a three-row rectangular knight graph, knight-cover strips, and \(KN_{n,3}\) did not locate this six-column formula.

David C. Fisher's 2003 paper studies the ordinary knight-cover problem on square \(n\times n\) boards. Todd Fenstermacher, Soumendra Ganguly, and Renu Laskar study perfect domination on rectangular knight graphs. They define the same underlying graph \(KN_{n,m}\), but perfect domination requires every unselected vertex to have exactly one selected neighbor. Their three-row results consequently concern a different invariant and exhibit width classes modulo 8. That paper also points to Sinko and Slater's work on efficient domination, another stricter variant.

This search gives useful context and leaves publication novelty unresolved. The exact ordinary-domination recurrence in this entry rests on the independent finite-state proof.

The program implements the four-mask transition in `ksd6-claim-min-plus-certificate`. It retains the minimum cost for every target state, which is exact because all selections in a new column are among the eight three-bit masks. The two forced zero transitions in `finish` impose the right boundary condition.

The reachable-state counts at widths zero through six are \[ 1,8,64,114,196,279,351. \] The state set then stays at 351 through every vector used in the certificate. The canonical transfer serialization orders states by the 12-bit encoding \(a+(b\ll3)+(u\ll6)+(v\ll9)\), then orders the chosen mask from 0 through 7. Its SHA-256 digest is `5a6b35842eb54931e9d1b0de3ed239567576c98df3f7fa81420150232bcf9567`.

The program checks all 25 initial values, the ten scalar recurrence instances starting at widths 9 through 18, equality of the finite supports of \(f_{19}\) and \(f_{25}\), and all 351 componentwise equations \(f_{25}(s)=f_{19}(s)+4\). The stable nine-line output has SHA-256 digest `115b3c754e1663b8922a609ee2e637e95cab85fb8359886b7f1dbd36e9ffefb4`.