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Jun 16 at 17:02 comment added codeR In the recurrence, i think it should be $n$ (not $n-1$). Why not verify the final answer with a few examples?
Jun 15 at 21:38 comment added einpoklum So, $|F_{n+1}| = |F_n| + (n-1) \times 2k$ then? And $F_n = 2+k \times n \times (n-1)$?
Jun 14 at 18:40 history edited codeR CC BY-SA 4.0
generalization
Jun 14 at 16:18 comment added codeR Yes thus $|V| = 4 + 4 + 8 = 16$ which is same as $4k$.
Jun 14 at 14:31 history answered codeR CC BY-SA 4.0