Which rectangles can be tiled with triangles $T_n$, when only two orientations are allowed? This question is a generalization of another question asked here: Which rectangles can be tiled with L-trominos, when only two orientations are allowed?
A triangle $T_n$ is a polyomino with columns on the same base with lengths $1,2,3,\cdots,n$ with area $n(n+1)/2$. If we only allow two orientations (copies of the tile and a 180-degree rotated copies), what $a \times b$ rectangles can be tiled for $n \geq 2$?
Below is the tile set for $T_3$.

Since two tiles from the set can form $n \times (n + 1)$ and $(n = 1) \times n$ rectangles, we can tile all rectangles tileable by these, that include rectangles of the form $an \times b(n + 1)$ and $a(n + 1) \times bn$ (for integers $a$ and $b$), and a lot of others, for example, $n(n+1) \times (2n + 1)$. In fact, all rectangles with area divisible by $n(n + 1)$ if they are large enough... see https://mathoverflow.net/questions/285018/what-rectangles-can-a-set-of-rectangles-tile).

(This is an example of a tiling of $n(n+1) \times (2n + 1)$ for $n = 3$.)
But can we tile any others?

*

*Can we tile rectangles that are not tileable by copies of $n \times (n + 1)$ and $(n + 1) \times n$ rectangles?

*Can we tile rectangles with area divisible by $n(n+1)/2$ but not divisible by $n(n+1)$?

In the special case linked above (for $T_2$), the answer is "no".
The technique used there uses group theory and is quite sophisticated; I do not understand it well enough to know whether it applies. If it does it would be a very nice example that could help illuminate that technique. If not, it would be interesting to know of another way to tackle this problem.
(This question about triangular tilings is related but possibly much more difficult: Are the only polyominoes that can tile triangles right trominoes?)
 A: 
Theorem: For $n\ge 2$, if an $h\times w$ rectangle can be tiled by right-handed copies of $T_n$, then either $h$ or $w$ is a multiple of $n$, and either $h$ or $w$ is a multiple of $n+1$.
Provided $\min(h,w)\ge n^2-n$, the converse is also true.

Therefore, if a rectangle can be tiled by $T_n$, then its area is a multiple of $n(n+1)$. Furthermore, letting $R_n$ be the $n\times (n+1)$ rectangle, both $T_n$ and and $R_n$ can tile the same sufficiently large rectangles.
In what follows, I will assume $n\ge 3$. My answers to the previous question gave a sketch of the proof for $n=2$, which turns out to be much harder.
We will use the tiling group invented by Conway and Lagarius [1]. The two tiles are described by the boundary words $x^ny^n(x^{-1}y^{-1})^n$ and $y^nx^n(y^{-1}x^{-1})^n$, so the tiling group is
$$
G:= \langle x,y \mid  x^ny^n(x^{-1}y^{-1})^n=y^nx^n(y^{-1}x^{-1})^n=1\rangle
$$
The boundary word of a $h\times w$ rectangle is $x^wy^hw^{-w}y^{-h}$. If the rectangle can be tiled, then its boundary is the identity in $G$, so we will show that $x^wy^hx^{-w}y^{-h}=1$ implies the divisibility conditions.
It suffices to consider the image of $G$ in a certain permutation group.  Both $x$ and $y$ will be realized as permutations of $(\Bbb Z/n\Bbb Z)^2$. For $x$, we use the permutation $(i,j)\mapsto (i+1,j)$, while the image of $y$ is the permutation defined by
$$
\begin{align}
y(0,j) &= (0,j)
\\
y(1,j) &= (1,j+2),
\\
\text{For } i\ge 2,\;\; y(i,j)&=(i,j+1)
\end{align}
$$
It should be clear that $x^n=y^n=1$. With some thought, you can show $(x^{-1}y^{-1})^n=(y^{-1}x^{-1})^n=1$ as well. This further implies that both of the relations defining $G$ are satisfied, so this defines a valid homomorphism. However, $x^w$ does not commute with $y^h$, except in the trivial cases where $n\mid h$ or $n\mid w$. To see this, it suffices to consider the images of $(0,0)$ and $(-1,0)$.
$$
\begin{align}
w\neq 0,1\implies (y^h\circ x^w)(0,0)&=(w,h),&\qquad (x^w\circ y^h)(0,0)&=(w,0)
\\
w=1\implies (y^h\circ x^w)(-1,0)&=(0,0),&\qquad (x^w\circ y^h)(-1,0)&=(0,h)
\end{align}
$$
We have shown that being able to tile the $h\times w$ rectangle implies that $n\mid h$ or $n\mid w$, so we are half way done. For the other half, it helps to use the following alternate presentation of $G$, which attained by conjugating the two relations in $G$:
$$
G = \langle x,y \mid  x^{n+1}y^{n+1}(y^{-1}x^{-1})^{n+1}=y^{n+1}x^{n+1}(x^{-1}y^{-1})^{n+1}=1\rangle
$$
This shows shows that we can represent $x$ and $y$ as permutations of $(\mathbb Z/(n+1)\mathbb Z)^2$, in a way exactly analogous to before, which proves that $x^w$ and $y^h$ do not commute unless $(n+1)\mid h$ or $(n+1)\mid w$.

A harder related question to yours is this; for which integers $n>m$ can $T_n$ be tiled by right-handed copies of $T_m$? The case $m=2$ was analyzed by Conway and Lagarius; they showed that $T_n$ can be tiled by $T_2$ if and only if
$$
n\equiv 0,2,9,\text{ or }11\pmod{12}
$$
It is a fun puzzle to try to tile $T_{12}$ with these triangles. Later, Thurston [2] showed that $n> m\ge 3$ implies that $T_n$ cannot be tiled by $T_m$.

The permutation representation looks like it came out of nowhere, so let me add some motivation. When presented with a complicated group like $G$, a natural first step is to look for a large normal subgroup, $N$, and see if the quotient $G/N$ is sufficient. This is exactly what Thurston did in $[2]$, for (approximately) this same group. He found that $N=\langle x^{n(n+1)},y^{n(n+1)},(xy)^{n(n+1)}\rangle $ is a central subgroup, and that the quotient splits as a direct product of two von Dyck groups. For integers $\ell,m,n\ge 2$, the von Dyck group is defined to be
$$
D(\ell,m,n)=\langle x,y\mid x^\ell=y^m=(xy)^n=1\rangle,
$$
and Thurston proved
$$
G/N\cong D(n,n,n)\oplus D(n+1,n+1,n+1)
$$
The von Dyck group has a geometric realization; given a triangle whose angles are $\pi/\ell,\pi/m,$ and $\pi/n$, the von Dyck group consists of symmetries of the plane generated by rotations by $2\pi/\ell,2\pi/m$, and $2\pi/n$ (resp.) through the vertices of this triangle. The plane has either spherical, Euclidean, or hyperbolic geometry, according to the whether the sum of the triangle angles is greater than, equal to, or less than $2\pi$. Therefore, we reduce the problem to two questions; how can you prove that $x^w$ and $y^h$ do not commute in $D(n,n,n)=\langle x,y\mid x^n=y^n=(xy)^n\rangle $, and the same question when $n$ is replaced with $n+1$?
When $n=3$, $D(3,3,3)$ consists of  Euclidean symmetries, and it is easy to visualize them and prove $x^w$ and $y^h$ do not commute. (The Cayley graph of $D(3,3,3)$ is the tessellation of hexagons and triangles mentioned in the other answer). For larger $n$, the plane is hyperbolic; undoubtedly someone familiar with hyperbolic geometry could easily prove these symmetries did not commute, but I needed some sort of representation. After googling for representations of triangle groups, I came across this paper, which inspired me to look for permutations of a discrete $n\times n$ grid.

Sources:

*

*J.H Conway, J.C Lagarias, Tiling with polyominoes and combinatorial group theory, Journal of Combinatorial Theory, Series A, Volume 53, Issue 2, 1990, Pages 183-208, https://doi.org/10.1016/0097-3165(90)90057-4.


*Thurston, William P. Conway’s Tiling Groups. The American Mathematical Monthly, vol. 97, no. 8, 1990, pp. 757–73. JSTOR, https://doi.org/10.2307/2324578.
A: I believe the shear transform argument in the (original) answer to the linked question provides enough for an informal proof that the answer is No.
Transform the L to a triangle of hexagons, as shown there. But those tile as if they were just equilateral triangles. We have transformed the question into: "Can we tile an equilateral $60^{\circ}$ rhombus with congruent equilateral triangles other than in the trivial/obvious way?"
But consider placing two equilateral triangles in any manner that doesn't have one side from each completely coincident. So imagine an up-pointing triangle and a down-pointing triangle with their horizontal sides along the same line, but overlapping only halfway.
You have just a couple of options for further triangles. You can keep going along the defined line, but the pairs will never line up. Or you can place more triangles with similar overlaps... but those will leave gaps with side length half that of the triangles, and which therefore can't be filled by the same triangles. If they could, the gaps would have the same side length as the triangle... which would mean the original two have no overlap.
Obviously this is not sufficient as a formal proof but I can't see how the informal proof can be broken.
