Is it true that $\sum_{k=m}^n\frac{\sigma(k)}k\not\in\mathbb Z$ for all derangements $\sigma\in S_n$ and $1\le m\le n$?

Let $$S_n$$ be the symmetric group of all the permutations of $$\{1,\ldots,n\}$$. Recall that a permutation $$\sigma\in S_n$$ is called a derangemnt if $$\sigma(k)\not=k$$ for all $$k=1,\ldots,n$$.

Motivated by the well-known result that $$\sum_{k=m}^n\frac1k\not\in\mathbb Z$$ whenever $$n\ge m>1$$ (Kurschak, 1918), here I ask the following question.

QUESTION: Is it true that whenever $$n\ge m\ge1$$ we have $$\sum_{k=m}^n\frac{\sigma(k)}k\not\in\mathbb Z$$ for all derangements $$\sigma\in S_n$$?

If $$n$$ is a prime number $$p$$ and $$\sum_{k=m}^n\frac{\sigma(k)}k\in\mathbb Z$$ with $$\sigma\in S_n$$, then $$\sigma$$ is not a derangement since $$\sigma(p)=p$$. Thus the question has a positive answer if $$n$$ is prime. I conjecture that the question always has a positive answer, and I have verified this for every $$n=1,\ldots,11$$. For $$n=4$$, note that $$\frac41+\frac12+\frac33+\frac24\in\mathbb Z$$ but the permutation $$(4,1,3,2)$$ of $$\{1,2,3,4\}$$ is not a derangement since it fixes the number $$3$$.

• If $m$ is smaller than the last prime before $n$, then this prime appears in the denominator. – WhatsUp Nov 27 '18 at 13:09
• I think that this approach based on Bertrand's postulate shows that your conjecture is true as long as the interval contains a prime (then $\sigma$ does not fix the largest prime $p$ in the interval so you can split the sum into one with $p$ in the denominator and one without). – Nathaniel Johnston Nov 27 '18 at 13:15

I extended the search to $$n \leq 111111$$ and find this: $$\frac{10090}{110990} + \frac{36997}{110991} + \frac{15856}{110992} + \frac{6529}{110993} + \frac{8538}{110994} + \frac{22199}{110995} + \frac{55498}{110996} + \frac{36999}{110997} + \frac{55499}{110998} + \frac{95142}{110999} + \frac{55500}{111000} + \frac{100910}{111001} + \frac{55501}{111002} + \frac{74002}{111003} + \frac{27751}{111004} + \frac{88804}{111005} + \frac{55503}{111006} + \frac{102468}{111007} + \frac{27752}{111008} + \frac{74006}{111009} + \frac{104480}{111010} = 10.$$

In case you want to verify this more efficiently, here is the "code version":

10090/110990 + 36997/110991 + 15856/110992 + 6529/110993 + 8538/110994 + 22199/110995 + 55498/110996 + 36999/110997 + 55499/110998 + 95142/110999 + 55500/111000 + 100910/111001 + 55501/111002 + 74002/111003 + 27751/111004 + 88804/111005 + 55503/111006 + 102468/111007 + 27752/111008 + 74006/111009 + 104480/111010

EDIT:

Now that almost one year passed, I would like to confess that the "extended the search to $$n \leq 111111$$" part was a joke...

Sadly nobody seems to find it funny ...

Or perhaps nobody ever cares (including the OP, who is probably busy conjecturing all kinds of things) ...