I found a theorem that states that if $A\subset \mathbb{Z}$ such that the upper Banach density is non-zero, then $A$ contains arbitrarily long arithmetic progressions, this is called Szemerédi's Theorem. The converse however is not true. In fact T. Tao showed that the prime numbers admit arbitrarily long arithmetic progressions and it is an easy consequence of the prime number theorem that the Banach density of the prime numbers is zero.
Inspired by this theory, I tried to get some intuition :
Consider the following sequence:
$0,1,2,4,5,7,10,11,13,16,20,21,23,26,30,35,36,38,41,45,50,56,57,59,\dots$
The scheme is as follows, we start with $0$ and add $1$, then we add $1$ again and then we add $2$. We start again by adding $1$ then $2$ and then $3$, and again $1$,$2$,$3$,$4$ and so on. (I hope I didn't make a silly mistake in writing down the sequence).
It easy to show that this sequence has Banach density zero, so the question is does this sequence still admit arbitrarily long arithmetic progressions?
For example, we can find one of length six, i.e. $1,4,7,10,13,16$, but in a way you can feel that the formula determining the sequence destroys arithmetic progressions.
What would be a reasonable way to tackle this problem?