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Nov 26, 2018 at 4:04 comment added Brian Hopkins @ Zhi-Wei Your first permutation, $\pi = (2,3,\ldots, n, 1)$, is the ``not hard'' $n$-cycle solution from the math.stackexchange problem. When $n$ is even, $\pi$ is odd and vice versa. Your second permutation is a different $n$-cycle for $n$ odd, giving another even permutation. The remaining parts of the question ask for an even permutation of $S_{2k}$ (with $k \ge 4$) and an odd permutation of $S_{2k+1}$ (also with $k \ge 4)$ that satisfy the sum condition.
Nov 26, 2018 at 2:39 history edited Zhi-Wei Sun CC BY-SA 4.0
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Nov 26, 2018 at 2:32 history answered Zhi-Wei Sun CC BY-SA 4.0