Tiling problem

So I was considering triangles of cells (like WYPS-boards), and the possibility of tiling them with two types of tiles: Three cells in a straight line (type A), and three mutually adjacent cells (type B). Of course the total number of cells must be divisible by 3, so that the side length of the original triangle must be == 0 or 2 (mod 3). The curious thing I noticed was that for any fixed side length, the number of A-tiles appears to have the same parity for any tiling. For example, take side length 11:

i / Number of tilings with exactly i A-tiles
--------------------------------------------
0 / 8
1 / 0
2 / 3870
3 / 0
4 / 236,643
5 / 0
6 / 4,532,867
7 / 0
8 / 34,937,847
9 / 0
10 / 123,920,874
11 / 0
12 / 210,843,044
13 / 0
14 / 167,121,108
15 / 0
16 / 55,629,282
17 / 0
18 / 5,879,766
19 / 0
20 / 0
21 / 0
22 / 0

I would think there was some simple reason for this phenomenon, but I can't seem to find one. Any suggestions? It is worth noting that it is not necessarily true for other shapes than triangles: If we take a triangle with side length 4 and remove the central cell, we can tile it with either 3 A-tiles or 3 B-tiles.

From a rapid analysis, here it goes:

Size/ parity
3/ odd
5/ odd
6/ odd
8/ odd
9/ even
11/ even
12/ even
14/ even
. . .
3(4k+1)[+2]/ odd
3(4k+2)[+2]/ odd
3(4k+3)[+2]/ even
3(4k)[+2]/ even

An explanation I see is that starting from the size 3 triangle with one A-tile, you can then tile the band 4+5 with an even number of A-tiles (0 or 2) as neither 4 nor 5 can be divided by 3.
Then adding a 6 raw, you add another even number of A-tiles (2) as 3(4k+2) has an even number of A-tiles.
Then again the band 7+8 has an even number of A-tiles as neither 7 nor 8 is divisible by 3, so the odiness continues for 3(4k+2)+2.
When you add the 9 raw, you add an odd number of A-tiles (hum only if k==0 (mod 3) and so the parity goes back to even, and stays there for triangles of sizes 11,12,14.

I'm aware this is not a proof.

Edit: 3(4k+3) can alway be tiled by an odd number of A-tiles.

I was also wondering if it was not simpler to look at the dual triangular triangle grid where the B-tile is just the small triangle (edges) ?

Conway and Lagarias have studied similar problems, maybe their technics could be applied for your problem: http://www.sciencedirect.com/science/article/pii/0097316590900574 ?

http://faculty.plattsburgh.edu/don.west/tiling/ is the solution to your problem, very neat!

If you want a hint before you read it, define the valence of a tiling to be the difference between the number of B-tiles with point up and point down, it is [(n+1)/3] for a canonical tiling.
A simple operation preserving valence can replace two opposite triangles by two A-tiles, thus preserving the parity. This operation coverts any tiling to the canonical one.

Thanks!! I'm glad I hadn't missed something more trivial (well, as far as we know). ;-)

What was your motivation on this problem?
How did you compute the enumerations?

No particular reason; I just drew 21 dots in a triangular pattern in a notebook while sitting in my office, then started encircling them 3 by 3. The enumeration is from a turbo pascal (!) program I wrote to count them one by one (so it took a little while to go through all 600,000,000+).

I first looked at side length 5 and 6 manually and noticed the parity phenomenon (2 and 3 being trivial).