4
$\begingroup$

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 $$\frac1{\pi(1)+\pi(2)}+\frac1{\pi(2)+\pi(3)}+\ldots+\frac1{\pi(n-1)+\pi(n)}+\frac1{\pi(n)+\pi(1)}=1.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.