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For each natural number $n$, find a sequence of $n^2$ real numbers which contains no monotonic subsequence of more than $n$ terms.

I've been stuck on this for a while. Can somebody please point me in the right direction? Many Thanks.

Kola B.
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Richard
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  • Hint: The $n^2$ bound is strict; if the sequence consists of $(n^2+1)$ numbers, it would certainly contain a monotonic subsequence of more than $(n+1)$ terms. Thus, the $n^2$ case is quite likely to contain many increasing sequences of exactly $n$ elements. Can you find such a sequence of $n=2$ or $n=3$? – Peter Košinár Nov 27 '14 at 23:09

1 Answers1

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HINT: $n=4$:

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Brian M. Scott
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  • @sciona: Thanks! – Brian M. Scott Nov 27 '14 at 23:15
  • Nice! So we just extend same structure to n^2 terms? And how would you write the general term? – Richard Nov 27 '14 at 23:48
  • @Richard: You can do it for any $n$, not just $4$. Write down a permutation of $[n^2]={1,\ldots,n^2}$: for $n=4$ the diagram yields the permutation $$4,3,2,1,8,7,6,5,12,11,10,9,16,15,14,13;.$$ In general you get $n$ blocks of $n$; the $k$-th block starts with $kn$, so it’s not too hard to describe the $i$-th term of the $k$-th block. – Brian M. Scott Nov 27 '14 at 23:52