Back-and-forth harmonic progression -- do we get all of $\mathbb{Q}?$ I was attempting a problem earlier and this side question came about.
Set up:
We create a set of rationals $A$ by hopping along $\mathbb{Q}$. It's awkward to write out, but the idea is very simple, I attached a picture of the first couple steps to illustrate.

*

*Start at $0\in A$

*Jump to the right by $1$, add that value to set (in other words, $1\in A$)

*Jump to left by $\frac12$, add value to set (i.e. $\frac{1}{2}\in A$), jump again to the left by $\frac13$, add value to set (i.e. $\frac16\in A$), jump to the left by $\frac14$ and so on.

*... till we reach just before $-1$ (this happens after left-jumping $\frac{1}{10}$). Now continue to the right, following the progression i.e. to the right by $\frac{1}{11}$, then $\frac{1}{12}$ etc.

*... till we reach just before $+2$. Now continue to the left, following the progression etc. till we reach just over $-2$, then back to right till just under $+3$ etc.


In other words, we are going back and forth over progressively larger intervals where the jumps are getting smaller, following a harmonic progression. Since $\sum 1/n$ diverges this never converges.
Now the resulting set $A$ does not contain all rationals, for instance it misses all integers other than $0, 1$. It seems messy and probably not very interesting to look at $A$ alone, but it led me to think about the following:
Question
If we label the above $A$ as $A_0$, and consider sets $A_k$ obtained in the same way but starting at the integer $k$, rather than $0$, does:
$$\bigcup_{k\in\mathbb{Z}} A_k = \mathbb{Q}?$$
And moreover, is $\left|A_i\cap A_j\right|\leqslant 1$ for $i\neq j$?
This appears to get messy rather quickly if one tries a hands-on approach, but I am having trouble getting a feel of what is happening. I wrote a program to get an idea but the numerators/denominators get large very quickly
 A: Whatever is meant by "in the same way but starting at the integer k" (I can think of at least 3 different ways to do "in the same way" and they don't give the same $A_k$s)
I'll give a proof of why I said it doesn't hit $\frac13$ or any integral translates of it.
Claim: The procedure does not hit any rational number whose lowest form has $3$ in the denominator.
Proof: Whatever "in the same way" may be, the procedure only hits rationals that are of the form
$$
k\pm\frac11\pm\frac12\pm\frac13\pm\dots\pm\frac1n
$$
We check that none of the signs, for $n\leq 4$, hits a rational with 3 in the denominator unless it is an integer.  For $n>4$, note that Bertrand's postulate gives a prime $p$ between $\lceil(n+1)/2\rceil$ and $n$ inclusive (in particular, $>3$ starting from $p=n=5$) and so its appearance in the denominator is guaranteed.  QED.
As to whether $\lvert A_i\cap A_j\rvert\leq 1$, that really depends on how you define $A_k$ "in the same way".  If you just mean translation by $k$ then it is true.  If you mean some other ways, e.g., starting at $k$ move $\pm 1$ to near $0$ (+ if at 0), move $\pm\frac12$ ... reversing only when you travelled more than say $(\lvert k\rvert+2)$ times the previous distance then I just defeated this. by having $1,0,\frac12\in A_0\cap A_1$.
