“…To summarize: Fix ∈ {2, 5, 8, ..., 3k − 1}.• If k = 2k , then by(1) and Cases 3 and 4, the subgraphs H 2,5 , ..., H 6k −4,6k −1 of I (C 12k +1 ) all have order 24k + 2, andH 2,5 ∼ = • • • ∼ = H 6k −4,6k −1 ∼ = C 24k +2 .• If k = 2k + 1, then by (2) and Cases 3 and 4, the subgraphs H 2,5 , ..., H 6k −4,6k −1 of I (C 12k +7 ) all have order 24k + 14, andH 2,5 ∼ = • • • ∼ = H 6k −4,6k −1 ∼ = C 24k +14 .However, the subgraph H 6k +2,6k +5 of I (C 12k +7 ) has order 12k +7 and H 6k +2,6k +5 ∼ = C 12k +7 . In Figure11, the subgraph H 8,11 of I (C 19 ) is shown with blue (dotted) edges.…”