# Is there a permutation $\pi\in S_n$ with $\sum\limits_{0<k<n}\frac1{\pi(k)^2-\pi(k+1)^2}=0$ for each $n>7$?

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 $$\sum_{0 for some $$\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 $$\sum_{0 and that for each integer $$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.$$

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