Easy. Just ensure that any 2 number whose position differ by 2 have different parities. Their average will not be integer. In general, (1,n-k,2,n-k+1,3....)

It works for $1,2,3,4$.Let we assume that it works for $n$.By induction we have to prove that it works also for $n+1$. I

Olympus_mountaineer wrote: Prove that for all positive integers $n$ we can find a permutation of {$1,2,...,n$} such that the average of two numbers doesn't appear in-between them.For example {$1,3,2,4$}works,but {$1,4,3,2$} doesn't because $2$ is between $1$ and $3$. Shouldn't it be {$1,4,2,3$} instead of {$1,4,3,2$} in the last line?

meet18 wrote: Easy. Just ensure that any 2 number whose position differ by 2 have different parities. Their average will not be integer. In general, (1,n-k,2,n-k+1,3....) It won' work. As you can see in your own example {$1,n-k,2,n-k+1,3...$}, $2$ is between $1$ and $3$

DeZade2002 wrote: Olympus_mountaineer wrote: Prove that for all positive integers $n$ we can find a permutation of {$1,2,...,n$} such that the average of two numbers doesn't appear in-between them.For example {$1,3,2,4$}works,but {$1,4,3,2$} doesn't because $2$ is between $1$ and $3$. Shouldn't it be {$1,4,2,3$} instead of {$1,4,3,2$} in the last line? Yes.I fixed the problem.

meet18 wrote: Easy. Just ensure that any 2 number whose position differ by 2 have different parities. Their average will not be integer. In general, (1,n-k,2,n-k+1,3....) DeZade2002 wrote: Olympus_mountaineer wrote: Prove that for all positive integers $n$ we can find a permutation of {$1,2,...,n$} such that the average of two numbers doesn't appear in-between them.For example {$1,3,2,4$}works,but {$1,4,3,2$} doesn't because $2$ is between $1$ and $3$. Shouldn't it be {$1,4,2,3$} instead of {$1,4,3,2$} in the last line? Yes, I misunderstood the question

Sorry for my mistake.I think now you are understanding the problem.

2019 BDMO NATIONAL SECONARY PROBLEM 8 Prove that for all positive integers n we can find a permutation of {1,2...n} such that the average of two numbers doesn't appear in-between them.For example {1324}works,but {1,4,2,3} doesn't because 2 is between 1 and 3. SOLUTION We will prove it by strong Induction. Call a permutation a1,a2...an "GOOD" if the average of any two numbers doesn't appear in between them and call it "VERY GOOD" if it is "GOOD" and for every integer x belongs to1,2...n, there exists a i such that ai=x Lemma: If a1,a2...an is a "GOOD" permutation then so is 2a1−1,2a2−1...2an−1 and 2a1,2a2...2an. Proof: Let bi=2ai−1 and ci=2ai for i element of 1,2....n We will prove that if for integers k,l,m; ak=(al+am)/2 then also bk=(bl+bm)/2 and ck=(cl+cm)/2 holds. (bl+bm)/2=(2al−1+2am−1)/2 =al+am−1 =2ak−1 =bk Similiarly,ck=(cl+cm)/2 So, If ak is not between al and am, then bk is also not between bl and bm and ck is not between cl and cm.So, we have proved our lemma. The base case for n=1 is true.Now, we will consider two cases. Case 1: When n=2k for any positive integer k. By strong Induction hypothesis,there exists a permutation a1,a2...ak which is "VERY GOOD" and by our lemma, the permutations 2a1−1,2a2−1....2ak−1 and 2a1,2a2...2ak are also "GOOD". So, 2a1−1,2a2−1...2ak−12a12a22ak is a "VERY GOOD" permutation for n=2k. Case 2: When n=2k+1 By strong Induction hypothesis,there exists permutations a1a2ak and d1d2dk+1 which are "VERY GOOD" permutations for n=k and n=k+1 respectively. By our lemma, the permutations 2d1−1,2d2−1....2dk+1 and 2a1,2a2...2ak are also "GOOD" So, 2d1−1,2d2−1....2(dk+1) and 2a1,2a2.....2ak is a "VERY GOOD" permutation for n=2k+1 And we are done.

That's seongbin's problem.