# Making bigger penny triangles from smaller

I thought up a problem a long time ago, and googling doesn't even close to turn up the answer. It is this:

Given unlimited 3-penny triangles (e.g. triangles with 3 pennies touching each other,) for which $N$ is it possible to create penny triangles with $N$ on each side?

N must be either $3x+2$ or $3x+3$, otherwise the total number of pennies is not divisible by 3.

We can also show that $N=3, 5, 6$ and $8$ doesn't work by brute force.

N can also always be $2+12x$, $9+12x$, $11+12x$ or $12x$. We can show this with brute force. We can also prove that ($x$ works) means ($x+12$ works) because we could cut an $X$-triangle off the top and a 12-triangle off one of the sides, leaving an $X$ by $12$ parallelogram which can be filled in easily.

But I'm wondering about the others. When I try to fill in, say, $n=8$, I'm always left with 3 pennies in a row. That suggests a parity argument, but the problem is, when I try to use a sum of coordinates, it doesn't quite work, e.g. we assume they're not quite equilateral and

$(2x, 4y) (2x+2, 4y) (2x+1, 2y +/- 2)$

$(2x+1, 4y +/- 2) (2x+2, 4y) (2x, 4y)$ (yes, this isn't quite equilateral, but I wanted to deal with integers)

and compare the coordinate sum to the sum of coordinates of the triangle, but you can always make things add up. In addition, trying to label possible vertices

-4-5-6
1-2-3


turns up nothing.

I hope this question is appropriate for this forum. Maybe it is better somewhere else? I'd like to know, because this problem has interested me off and on, and I suspect someone has solved it in some form or another and I don't know how/where to ask.

Edited to say thanks to the person adding formatting, and to add 12x+2. And for the quick answer! I appreciate it so quickly, and it definitely makes me feel welcome asking again.

One can visualize this problem in an hexagonal grid (where each penny is an hexagon). A 3-penny triangle becomes an 'hexagonal triangle' of side $2$ (that is, two hexagons on each side), also called $T_2$. Similarly, an 'hexagonal triangle' of side $n$ is denoted by $T_n$.
The question then becomes for which $n$ can $T_n$ be tiled by $T_2$'s. This has been answered in a famous paper by William Thurston, Conway's Tiling Groups: such a tiling exists if and only if $n$ is congruent to $0,2,9$ or $11$ modulo $12$.