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One more motivated by recent questions of Zhi-Wei Sun.

Let $S_n$ be the group of permutations of $\{1,2,\ldots, n\}$.

Is it true that, for every $n \ge 8$, there is at least one even permutation $\pi \in S_n$ and at least one odd permutation $\tau \in S_n$ with $$\sum_{k=1}^n \frac{1}{k \, \pi(k)} = \sum_{k=1}^n \frac{1}{k \, \tau(k)} = 1?$$

One case for each $n$ is not hard; I made it a math.stackexchange question that was successfully answered in 12 minutes. Hopefully the other case is more interesting.


Clarification: As per the MSE question referenced above and Zhi-Wei's comment, the $n$-cycle $(1,2, \dots, n) \in S_n$ satisfies the sum condition. An $n$-cycle is an odd permutation for even $n$ and an even permutation for odd $n$.

Here are the remaining parts of the conjecture.

a. For $n$ even and $n \ge 8$, there is an even $\pi \in S_n$ satisfying $\sum_{k=1}^n \frac{1}{k \, \pi(k)} = 1$.

b. For $n$ odd and $n \ge 9$, there is an odd $\tau \in S_n$ satisfying $\sum_{k=1}^n \frac{1}{k \, \tau(k)} = 1$.


Here are the numbers of even and odd permutations satisfying the sum condition for small $n$.

\begin{array}{c|rr} n\backslash \text{sgn} & +1 & -1 \\ \hline 1 & 1 & 0 \\ 2 & 0 & 1 \\ 3 & 2 & 0 \\ 4 & 0 & 2 \\ 5 & 4 & 0 \\ 6 & 0 & 2 \\ 7 & 4 & 0\\ 8 & 6 & 4\\ 9 & 12 & 24\\ 10 & 90 & 88 \end{array}

One of the first ``interesting'' permutations is the even permutation (in cycle notation) $(1,2,5,8,7,6)(3,4) \in S_8$ which gives \begin{align*} \frac{1}{1\cdot2} + \frac{1}{2\cdot5}+ \frac{1}{3\cdot4}+ \frac{1}{4\cdot3}+ \frac{1}{5\cdot8}+ \frac{1}{6\cdot1}+ \frac{1}{7\cdot6}+ \frac{1}{8\cdot7}\\ = \frac{1}{2} + \frac{1}{10}+ \frac{1}{12}+ \frac{1}{12}+ \frac{1}{40}+ \frac{1}{6}+ \frac{1}{42}+ \frac{1}{56}=1. \end{align*} Not coincidentally, $n=8$ is the smallest value for which there are non-$n$-cycle permutations that satisfy the sum condition.

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Clearly, \begin{align}&\frac1{1\cdot2}+\frac1{2\cdot3}+\ldots+\frac1{(n-1)n}+\frac1{n\cdot1} \\=& 1-\frac12+\frac12-\frac13+\ldots+\frac1{n-1}-\frac1n+\frac1n=1. \end{align} If $n=2k+1$, then we also have \begin{align}&\frac1{1\cdot2}+\frac1{2\cdot3}+\ldots+\frac1{k(2k+1)}+\frac1{(2k+1)(k+1)} \\&+\frac1{(k+1)(k+2)}+\cdots+\frac1{(2k-1)2k}+\frac1{2k\cdot1} \\=&1.\end{align} See also my related conjectures available from http://oeis.org/A322069 and http://oeis.org/A322070.

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  • $\begingroup$ @ 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. $\endgroup$ – Brian Hopkins Nov 26 '18 at 4:04

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