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Suppose $a_i\in \mathbb{N} $ such that $a_i\ne a_j$ and $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ for $i \ne j$ $($take $a_{n+1}=a_1)$ and $\displaystyle A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$.
Then for what values of $n\ge 2$ can $A$ be an integer?

I found that if $a_i=(n-1)^i$ then $A$ is a integer but I realised that this violates the condition $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$. It was easy to prove that $A$ is not an integer for $n=2$, but I was stuck for the cases where for $n>2$. I then I plugged in some values for $a_i$ but for $n>2$ but I found no solution.

Intuitively I think there is no solution but I am unable to prove it. Any help would be appreciated.

Edit-1: As pointed out by @WillJagy it is possible for $n=3,4,5$. So can we actually prove it is true for $n>6$ or are there some $n>6$ for which it isn't true?

Edit-2: I was able to find a solution for $n=5$ from a solution for $n=4$. $$~~~~~~~~~~\frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6$$ $$\implies \frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 5$$ $$\implies \frac{6}{2} +\frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 8$$
But I could not find a solution for $n=6$ from a solution for $n=5$ because when I rewrite $\frac{4}{3} =1+\frac{1}{3}$, the $1$ in the numerator of $\frac{1}{3}$ is used by another $a_i$. I don't know whether we can prove by induction. Any suggestions is welcome.

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  • $\begingroup$ artofproblemsolving.com/community/c6h114708p651437 $\endgroup$
    – Integrand
    Commented Jun 1, 2021 at 1:12
  • $\begingroup$ @FearfulSymmetry This question has an additional constraint of $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ $\endgroup$
    – Asher2211
    Commented Jun 1, 2021 at 1:19
  • $\begingroup$ Sure; try and start there and see if you can eliminate some of the general solutions. $\endgroup$
    – Integrand
    Commented Jun 1, 2021 at 1:33

3 Answers 3

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$$ \frac{81}{2} + \frac{36}{81} + \frac{2}{36} = 41 $$

$$ \frac{98}{12} + \frac{63}{98} + \frac{12}{63} = 9 $$

$$ \frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6 $$

$$ \frac{6}{5} + \frac{5}{3} + \frac{3}{10} + \frac{10}{12} + \frac{12}{6} = 6 $$

$$ \frac{3}{12} + \frac{12}{9} + \frac{9}{10} + \frac{10}{8} + \frac{8}{5} + \frac{5}{3} = 7 $$

$$ \frac{2}{8} + \frac{8}{10} + \frac{10}{6} + \frac{6}{5} + \frac{5}{4} + \frac{4}{3} + \frac{3}{2}= 8 $$

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  • $\begingroup$ What was your approach for finding these numbers? $\endgroup$
    – Asher2211
    Commented May 31, 2021 at 20:15
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    $\begingroup$ @Asher2211 I just wrote a program. It's worth checking something before trying to prove it; with integers, pretty firm indications can come from a computer. There is probably some sort of order to these, the only integers I've gotten ( with 3 variables) are 9, 41, 66 $\endgroup$
    – Will Jagy
    Commented May 31, 2021 at 20:21
  • $\begingroup$ Have you assumed $a_i$ to be ordered in the program? I got $6$ with $a = (2, 12, 9)$ and it seems to follow the restrictions. $\endgroup$
    – AnilCh
    Commented May 31, 2021 at 20:31
  • $\begingroup$ @AnilCh good point, I did have them ordered, apparently that was not a valid restriction. $\endgroup$
    – Will Jagy
    Commented May 31, 2021 at 20:32
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    $\begingroup$ @Asher2211 alright. I forgot that condition. I suspect it will force the $a_i$ to be a bit larger, but the job seems to be getting easier with more variables. As far as your wish for a proof, it is a bit early to be asking about that, given that you began in the opposite direction. If it is important, you will write your own programs and look for patterns. $\endgroup$
    – Will Jagy
    Commented May 31, 2021 at 23:28
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I guess you do not want any ratio $a_i/a_{i+1}$ to be an integer, otherwise trivial solutions can be produced.

For $n=3$, a solution can be produced as already seen in the answers.

We will use Egyptian fractions to prove that there always exists a sequence of such $a_i$'s for $n\gt 3$

$$1=\frac 12+\frac 13+\frac 16$$

We can break the last unit fraction $\dfrac 16$ using the fact that $\dfrac 1m=\dfrac 1{m+1}+\dfrac 1{m(m+1)}$

In this way, you can create an Egyptian fraction representation (with distinct unit fractions) for $1$ of arbitrary length, ie, for any $n\gt 3$, there exist a sequence of distinct $p_i (\neq 1)$'s ($i=1...(n-1)$) such that $1=\displaystyle\sum_1^{n-1}\frac 1{p_i}$

All that is left is to define our $a_i$'s in a suitable way. We define $a_1=1$ and $a_{i+1}=a_ip_i$ for all $i=1...(n-1)$ so that all the $a_i$'s are distinct and $\dfrac{a_i}{a_{i+1}}=\dfrac 1{p_i}$ are distinct for all $i=1...(n-1)$ and sum up to $1$ and $\dfrac{a_n}{a_1}=a_n$ is an integer, so the resulting sum $A=a_n+1$

Examples:

$$\frac 12+\frac 26+\frac 6{36}+\frac{36}1=37\\ \frac 12+\frac 26+\frac 6{42}+\frac{42}{1764}+\frac{1764}1=1765$$

Here's a Tio.run link if you want to generate a sequence of $a_i$'s for any $n$, but beware that the $a_i$'s grow larger quite quickly due to the construction.


Hmm, then again, perhaps this answer also generates trivial solutions because $a_n/a_1$ is an integer. I'm unsure if I can fix that. I'll update if I can.

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$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{15}{6}+\frac{24}{15}+\frac{10}{24}+\frac{2}{10}=10$$

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{30}{7}+\frac{14}{30}+\frac{24}{14}+\frac{2}{24}=13$$

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{36}{9}+\frac{15}{36}+\frac{42}{15}+\frac{2}{42}=15$$

$$\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{9}{7}+\frac{10}{ 9}+\frac {28}{ 10}+\frac {27}{ 28}+\frac {36}{ 27}+\frac {2}{ 36}=14$$

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