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Consider a sequence $A: a_1, a_2, a_3 .. a_n$ such that each element of this sequence is product of two primes from $p_1, p_2 .. p_m$ so that:

  1. Each $a_i$ is distinct.

  2. $m$ is the least integer so that $\frac{m(m-1)}{2} \ge n$

  3. Any three consecutive elements of sequence are coprime, but any two are not.

For example,

  1. for $n = 3$, we have $m = 3$ and so taking the 3 primes as $a,b,c$, we can say the sequence $A$ is $ab, bc, ac$

  2. For $n = 4$, $m = 4$ and $A: ab, bc, cd, da$

  3. For $n = 6$, $m = 4$ and $A: ab, bd, da, ac, cd, db$

The question is to show that such a sequence exists for any $n$.

If this sequence exists for all $n$, is there a simple pattern for $A$ ?\

Edit

There was an error in example 3, no such sequence exists for $n=6$.

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    $\begingroup$ There is a problem in your example for $n=6$: $bd$ and $db$ are equal. $\endgroup$ – ajotatxe Jan 8 '19 at 15:13
  • $\begingroup$ @ajotatxe sorry I overlooked this one! $\endgroup$ – jeea Jan 8 '19 at 15:23
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Your example for $n=6$ doesn't work, because the terms $bd$ and $db$ are the same number, and in fact no valid sequence exists in this case.

This is really a graph theory question: terms of the sequence correspond to edges of the complete graph $K_m$ and the conditions require that two consecutive edges share a vertex but three consecutive edges don't. Equivalently, we are looking for a trail of length $n$. If $m$ is odd there will always be such a trail (because $K_m$ is Eulerian), but if $m$ is even and $n$ is close to the maximum value it won't exist.

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