Let $Y$ be a tight $4$-design in the Hamming scheme $H(n, q)$ (i.e., an orthogonal array of strength $4$ meeting the Rao bound). Noda proved that then one of the following holds:
Cases 1 and 2 uniquely determine a binary code with $4$ data bits and one parity bit (i.e., the dual of the binary repetition code of length $5$), and the dual of ternary Golay code, respectively. No examples are known for Case 3.
Gavrilyuk, Suda and Vidali show that a $Q$-polynomial association scheme with $d = 4$ classes and Krein array $\{9a^2 - 4, 9a^2 - 9, 10, 1; 1, 2, 9a^2 - 9, 9a^2 - 4\}$ would arise from an example of Case 3. Writing $r = 3a$, we may consider the two-parametric family of Krein arrays $\{r^2 - 4, r^2 - 9, {12 (s-1) \over s}, 1; 1, {12 \over s}, r^2 - 9, r^2 - 4\}$ (taking $s = 6$ gives the previous Krein array). Here, we show that no corresponding schemes with $s \ge 4$ exist, and consequently also no example of Case 3 exists. Note however that we consider all positive integral values of $r$.
%display latex
import drg
By dismantlability (see Van Dam, Martin and Muzychuk), it suffices to show that no association scheme with the above Krein array and $s = 4$ exists. Such a scheme would have $r^2 (r^2 - 1)/3$ vertices.
r, t = var("r t")
s = 4
p = drg.QPolyParameters([r^2 - 4, r^2 - 9, 12*(s-1)/s, 1], [1, 12/s, r^2 - 9, r^2 - 4])
p.order(factor=True, simplify=True)
Such a scheme is $Q$-antipodal, so it has $q^h_{ij} = 0$ whenever $h, i, j$ do not satisfy the triangle inequality, or $h+i+j > 2d$ and $d-h, d-i, d-j$ do not satisfy the triangle inequality.
p.kreinParameters(factor=True, simplify=2)
The intersection numbers can be checked to be nonnegative and integral for odd values of $r \ge 5$.
p.pTable(expand=True, factor=True, simplify=True)
Let us check that the first member of the family satisfies the known feasibility conditions. We skip the family nonexistence check since the Krein array of the entire family is already included.
p.subs(r == 5).check_feasible(skip=["family"])
We now compute the triple intersection numbers with respect to three vertices $u, v, w$ mutually in relation $1$. Note that we have $p^1_{11} = (r^2 - 3r + 6)(r^2 - 1)/24 > 0$ for all $r \ge 5$, so such triples must exist. The parameter $\alpha$ will denote the number of vertices in relations $1, 2, 3$ to $u, v, w$, respectively.
S111 = p.tripleEquations(1, 1, 1, params={'alpha': (1, 2, 3)})
S111[1, 1, 1].expand().factor()
The above triple intersection number is integral for odd values of $r$ whenever $r^4 + 2r^2 + 9r - 3 - 9/r$ is divisible by $24$.
(S111[1, 1, 1] + (r^4 + 2*r^2 + 9*r - 3 - 9/r)/24).subs(r == 2*t + 1).factor()
The above expression is integral only when $r$ divides $9$. As we must have $r \ge 5$, this leaves $r = 9$ as the only candidate solution. In this case, however, the above expression is then not divisible by $24$, giving a nonintegral value for the triple intersection number.
S111[1, 1, 1].subs(r == 9)
We thus conclude that a $Q$-polynomial association scheme with Krein array $\{r^2 - 4, r^2 - 9, {12 (s-1) \over s}, 1; 1, {12 \over s}, r^2 - 9, r^2 - 4\}$ and $s \ge 4$ does not exist. Consequently, no tight $4$-design in $H((9a^2+1)/5, 6)$ exists, thus completing the classification of orthogonal arrays of strength $4$ meeting the Rao bound.