17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 32, 33, 36, 40, 44, 45, 48...

16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 32, 33, 36, 40, 44, 45, 48, 72, 80, 84, 90, and 120.Assume towards a contradiction that χ( ˙G

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) = 3. So, let us colour the vertices of ˙G

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with three colours x, y, and z. Without loss of generality, let us colour 14 with colourx. Since, the four numbers 13, 14, 22, and 40 form a pair of triangles sharing the edge(13,40), 22 must get the same colour as 14, so we have to colour 22 also with x. Now,since 22, 23, and 120 form a triangle, 23 and 120 must get the colours y and z. Thus,without loss of generality, let us colour 23 with z and 120 withy. Finally, because of theedges (24,23) and (14,24), and since 23 has colour z and 14 has colour x, 24 must getcolour y. So far, we have the following colouring: [14, x],[22, x],[23, z],[24, y],[120, y].Consequently we get the following: