Integer Triples and Reflection tiling $1,2,\ldots,n$ $\forall a,b\in\mathbb Z,\ \exists n\in \mathbb N$ such that the numbers $1,2,\ldots,n$ can be tiled using translates of $\{0,\ a,\ a+b\}$ and $\{0,\ -a,\ -(a+b)\}$ ?
In other words for every integer triple $T$ does there exist a finite contiguous range of $n$ integers that can be tiled with $T+x$ and the reflection $T^{-1}+x$.
Can the minimal $n$ be given in terms of $a$ and $b$?

 A: Well, it is certainly required that $n$ is a multiple of $3$. In fact, a multiple of $3\cdot\gcd(a,a+b),$ if you think about it. So in the following I assume $\gcd(a,b)=1.$
I strongly believe the situation to be this, though I have no proof:

*

*If $a \equiv b \equiv 1\pmod 3$ (equivalently, $0,a,a+b \equiv 0,1,2 \pmod 3$) there is an $N$ so that there is a tiling of $0,1,2,\cdots,n-1$ for any $n=3m \ge N.$


*Otherwise there must be an equal number of each orientation so $n \equiv 0 \pmod 6$ is necessary. Then there is an $N$ so that there is a tiling of $0,1,2,\cdots,n-1$ for any $n=6m \ge N.$
Below are two diagrams from a paper which show why the "all large enough $n$" claims are plausible. To understand the  decorations, read the paper, it's great! (imho)
There are  $2^{a+b-1}$ situations which can arise in a tiling (or attempted tiling) of $\mathbb{Z}$ or $\mathbb{N}.$ By situation I mean pattern of full and empty starting from the first empty and ending at the last full as we work left to right. This includes "dead ends" (no way to leave)  and "Garden of Eden" situations (no way to arrive). If we look at how one can get from one to another we have a finite directed graph where each node has indegree and outdegree $2$ or less (each edge involving three more filled spaces). Your questions are about the possible length of closed walks starting and ending at the state I call $\emptyset.$
Here is the full digraph for the tile $0,a,a+b=0,1,5.$ Looking only at the big region, one sees two ways to get an interval of length $9$ which I might call $ABA$ and $BAB.$ One can also get an iterval of length $12$ from $AABB.$ So in this case the possible lengths are $9,12,18,21,24,27,30\cdots.$

And here is the full digraph for $0,a,a+b=0,2,5.$ One observes that there are interval tilings must be $AAB(BA)^kABB$ for any $k \ge 0.$ So intervals of lengths $12,18,24,30,\cdots$ are tilable.

For larger $a,b$ the digraph will have (many) more cycles, some relatively short. And having two of  lengths $\ell_1,\ell_2$ (first case) or $2\ell_1,2\ell_2$ (second case) with $\gcd(\ell_1,\ell_2)=1$ seems to always happen (as one might expect from a random digraph where a healthy number of nodes have indegree = outdegree = $2$). This is sufficient for the "all large enough $n$" claims, although having a larger set of cycle lengths with $\gcd(\ell_1,\ell_2,\cdots,\ell_j)=1$ would also be enough.
In the first case there are several connected components but in the second it seems that any "reasonable" situation can reach any other. By reasonable I meant that there are arbitrarily long paths arriving at and leaving that node. A possible counterexample would be two closed cycles with a one way bridge from the first to the second.
As far as the shortest tilable interval, it is no longer than $6(a+b+2)$. I quote theorem 4 from Some Results On One-Dimensional Tilings:

THEOREM 4. Let $a = q - 1$, $b = \lambda q + (q - 1) + m$, where $\lambda$ and $m$ are non-negative integers and $m < q$. Then:

*

*If $\lambda \equiv 1 (\operatorname{mod} 3)$, then the shortest interval that can be tiled by $1(a)1(b)1$ is of length $(2\lambda + 4)q = 2(a + b + 2 - m)$;

*If $\lambda \equiv 0$ or $2 (\operatorname{mod} 3)$, then $1(a)1(b)1$ will tile an interval of length $6(a + b + 2)$.

