# Does this sequence of product of two primes always exist

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 $$n=6$$.

• There is a problem in your example for $n=6$: $bd$ and $db$ are equal. – ajotatxe Jan 8 '19 at 15:13
• @ajotatxe sorry I overlooked this one! – jeea Jan 8 '19 at 15:23

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.