We will show that a distance-regular graph with intersection array $\{(2r+1)(4r+1)(4t-1), 8r(4rt-r+2t), (r+t)(4r+1); 1, (r+t)(4r+1), 4r(2r+1)(4t-1)\}$, where $r, t \ge 1$, does not exist. The intersection array arises for graphs of diameter $3$ with $b_2 = c_2$ and $p^3_{33}$ even which are $Q$-polynomial with respect to the natural ordering of eigenvalues and contain a maximal $1$-code that is both locally regular and last subconstituent perfect. See Extremal $1$-codes in distance-regular graphs of diameter $3$ by A. Jurišić and J. Vidali, where the theory for dealing with such intersection arrays has been developed.
%display latex
import drg
This family is not entirely feasible, however, we will find two infinite feasible subfamilies.
r, t = var("r t")
p = drg.DRGParameters([(2*r+1)*(4*r+1)*(4*t-1), 8*r*(4*r*t-r+2*t), (r+t)*(4*r+1)],
[1, (r+t)*(4*r+1), 4*r*(2*r+1)*(4*t-1)])
p.order(expand=True, factor=True)
The two feasible subfamilies can be obtained by setting $t = 4r^2$ and $t = 2r(2r+1)$, respectively.
pA = p.subs(t == 4*r^2)
pA.intersectionArray(expand=True, factor=True)
show(pA)
show(pA.order(expand=True, factor=True))
pB = p.subs(t == 2*r*(2*r+1))
pB.intersectionArray(expand=True, factor=True)
show(pB)
show(pB.order(expand=True, factor=True))
Let us check that the first members of each family are indeed feasible. We skip the family nonexistence check since the intersection array of the entire family is already included.
pA1 = pA.subs(r == 1)
show(pA1)
show(pA1.order())
pA1.check_feasible(skip=["family"])
pB1 = pB.subs(r == 1)
show(pB1)
show(pB1.order())
pB1.check_feasible(skip=["family"])
We now compute the Krein parameters. We have $q^1_{13} = q^1_{31} = q^3_{11} = 0$, so the graph would be $Q$-polynomial with respect to the natural ordering of the eigenvalues.
p.set_vars([t, r])
[p.q[1, 1, 3], p.q[1, 3, 1], p.q[3, 1, 1]]
We now compute the triple intersection numbers with respect to three vertices $u, v, w$ at mutual distances $3$. Let us first check that $p^3_{33}$ is positive.
p.p[3, 3, 3].factor().simplify_full()
The parameter $\alpha$ will denote the number of vertices at distance $3$ from all of $u, v, w$. Let us count the number of vertices at distance $1$ or $2$ from one of $u, v, w$ and $3$ from the other two vertices.
alpha = var("alpha")
S333 = p.tripleEquations(3, 3, 3, params={alpha: (3, 3, 3)})
[S333[s].expand().factor() for s in [(1, 3, 3), (3, 1, 3), (3, 3, 1), (2, 3, 3), (3, 2, 3), (3, 3, 2)]]
Note that for the above expressions to be nonnegative, we must have $a = 4r - 1$, and then they are all equal to zero. Consequently, all of the $a_3$ vertices adjacent to one of $u, v, w$ which are at distance $3$ from another of these vertices are at distance $2$ from the remaining vertex in the triple.
S333a = S333.subs(alpha == 4*r - 1)
show(p.a[3].expand().factor())
[S333a[s].expand().factor() for s in [(1, 2, 3), (3, 1, 2), (2, 3, 1), (2, 1, 3), (3, 2, 1), (1, 3, 2)]]
The above results mean that any two vertices $v, w$ at distance $3$ uniquely define a set $C$ of $4r + 2$ vertices mutually at distance $3$ containing $v, w$ - i.e., a $1$-code in the graph. Furthermore, since $a_3$ is nonzero, for each $u$ in $C \setminus \{v, w\}$, there are vertices at distances $3, 1, 2$ from $u, v, w$. We now check that $c_3 = a_3 p^3_{33}$.
show(p.c[3].expand().factor())
show((p.a[3] * p.p[3, 3, 3]).expand().factor())
Let $u'$ be a neighbour of $v$. Since $u'$ is not in $C$, it may be at distance $3$ from at most one vertex of $C$. As there are $c_3$ and $a_3$ neighbours of $v$ that are at distances $2$ and $3$ from $w$, respectively, the above equality implies that each neighbour of $v$ is at distance $3$ from precisely one vertex of $C$.
Suppose now that $u'$ is at distance 2 from $w$. Let us count the number of vertices at distances $1, 1, 3$ from $u', v, w$.
beta = var("beta")
S123 = p.tripleEquations(1, 2, 3, params={beta: (3, 3, 3)}).subs(beta == 1)
S123[1, 1, 3]
As this value is nonintegral, we conclude that a graph with intersection array $\{(2r+1)(4r+1)(4t-1), 8r(4rt-r+2t), (r+t)(4r+1); 1, (r+t)(4r+1), 4r(2r+1)(4t-1)\}$ and $r, t \ge 1$ does not exist.