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

QUESTION: Is it true that for each $n=8,9,\ldots$ we have $$\sum_{0<k<n}\frac1{\pi(k)^2-\pi(k+1)^2}=0\tag{$*$}$$ for some $\pi\in S_n$?

Let $s(n)$ denote the number of permutations $\pi\in S_n$ with $\pi(1)<\pi(n)$ satisfying $(*)$. Via Mathematica, I find that $$s(1)=s(2)=\ldots=s(7)=0,\ s(8)=1,\ s(9)=s(10)=4,\ s(11)=55.$$ When $n=8$ we can only take $$(\pi(1),\ldots,\pi(8))=(4,5,2,7,3,1,6,8)$$ and indeed $$\frac1{4^2-5^2}+\frac1{5^2-2^2}+\frac1{2^2-7^2}+\frac1{7^2-3^2}+\frac1{3^2-1^2}+\frac1{1^2-6^2}+\frac1{6^2-8^2}$$ is zero. For $n=9$ we may choose $\pi\in S_9$ with $$(\pi(1),\ldots,\pi(9))=(4,1,9,3,5,2,7,8,6).$$ When $n=10$ we may take $$(\pi(1),\ldots,\pi(10))=(5,4,7,8,3,9,1,10,2,6).$$ For $n=11$, we may take $$(\pi(1),\ldots,\pi(11))=(1,3,5,4,6,2,10,8,7,11,9).$$

I conjecture that the question has a positive answer. In my opinion this question is quite challenging.

I also have some other similar conjectures (cf. http://oeis.org/A322069 and http://oeis.org/A322070). For example, I conjecture that for any integer $n>5$ there is a permutation $\pi\in S_n$ with $$\sum_{0<k<n}\frac1{\pi(k)\pi(k+1)}=1,$$ and that for each integer $n>7$ there is a permutation $\pi\in S_n$ such that $$\frac1{\pi(1)+\pi(2)}+\frac1{\pi(2)+\pi(3)}+\ldots+\frac1{\pi(n-1)+\pi(n)}+\frac1{\pi(n)+\pi(1)}=1.$$

  • $\begingroup$ The question has a positive answer for $n=12$ since we may take $$(\pi(1),\ldots,\pi(12))=(1,3,7,5,4,8,6,2,10,11,9,12).$$ $\endgroup$ – Zhi-Wei Sun Nov 27 '18 at 16:46
  • $\begingroup$ The question also has a positive answer for $n=13$ since we may take $$(\pi(1),\ldots,\pi(13))=(1, 6, 2, 9, 11, 5, 3, 7, 13, 8, 4, 10, 12).$$ See also oeis.org/A322099. $\endgroup$ – Zhi-Wei Sun Nov 28 '18 at 0:09

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