# Does $\{\tau(1)\tau(2)+\cdots+\tau(n-1)\tau(n)+\tau(n)\tau(1):\ \tau\in S_n\}$ contain a unique multiple of $n^2$ for each $n\ge6$?

Motivated by Question 397575, here I pose a related question.

Question. Does the set $$T_n:=\{\tau(1)\tau(2)+\cdots+\tau(n-1)\tau(n)+\tau(n)\tau(1):\ \tau\in S_n\}$$ contain a unique multiple of $$n^2$$ for each $$n\ge6$$?

I conjecture that the answer is positive. I have verified this for $$n=6,\ldots,10$$. For $$n=6$$, we have \begin{align*}&2\times4+4\times1+1\times3+3\times5+5\times6+6\times2 \\=&3\times5+5\times1+1\times2+2\times4+4\times6+6\times3=2\times6^2. \end{align*} For $$n=7$$, we have $$1\times3+3\times4+4\times5+5\times6+6\times2+2\times7+7\times1=2\times7^2.$$ For $$n=8$$, we have $$1\times5+5\times3+3\times6+6\times4+4\times7+7\times2+2\times8+8\times1=2\times8^2.$$ For $$n=9$$, we have $$1\times2+2\times3+3\times5+5\times4+4\times6+6\times8+8\times7+7\times9+9\times1=3\times9^2.$$ For $$n=10$$, we have $$\begin{gather*}1\times2+2\times3+3\times6+6\times8+8\times4+4\times9+9\times7+7\times5+5\times10+10\times1 \\=3\times10^2.\end{gather*}$$

• Perhaps, $T_n$ contains $\lfloor n/3\rfloor n^2$ for each integer $n\ge6$. Jul 15 at 9:15

No, for $$n = 11$$ this fails:
Running the code I wrote to check this a little more, there is more than one multiple of $$n^2$$ in the set you describe for all $$11\leq n \leq 50$$, see here for two permutations leading to different multiples of $$n^2$$ for each such $$n$$.