Quite a while back, I did this post. In it, I made a bit of a boo-boo. And I didn't realise it at the time. I figured it out by finding this page here, looking at the same problem. I discovered quite quickly that I'd been missing out all of the downwards-facing triangles that weren't unit triangles. Which wasn't a problem for the first three triangles, because there aren't any. The fourth has only one. But then it goes up.
So, my previous formula of (x^3 - x)/6 doesn't work at all. So I tried to find a new one.
First, I tried doing it while maintaining the x^2 out the front. I discovered that the rest of the sum was the series (0, 1, 4, 11, 23, 42, 69....), which, when typed into the online encyclopaedia of integer sequences (quite handy), said that it was the sequence "A019298 - Number of balls in pyramid with base either a regular hexagon or a hexagon with alternate sides differing by 1 (balls in hexagonal pyramid of height n taken from hexagonal close-packing)." You can have a look here if you like. Unfortunately, it looks like they don't have formulas for the nth term in the series in any simple form, though I'm probably just reading it badly.
So I tried it a different way - separating out the upward-facing (which I knew was just a sum of triangular numbers, and was easy enough), from all the downward-facing. The latter yielded this series: 0, 1, 3, 7, 13, 22, 34..... which also came up, this time as "A002623 - G.f.: 1/((1-x)^4*(1+x))." I've just found out that G.f. stands for Generating function. Again, though, it didn't have any simple formulas.
There were formulas available for both. But all of them included extra expressions. They're all given in the format a(n)=, n being the nth number in that series, and a being that series. But then they'll have a(n-1)*a(n+2) in the formula or something, which isn't helpful! I need it just in terms of n. But, they didn't have that.
So, at present, I have no formula for this. Unfortunately. I knew you were all hanging on for me to pull out something nice, right, but I've got nothing for you. :/ If any of my maths friends can figure it out, let me know!