<s>[INST] Number of possible triangle for the given triangle. An equilateral triangle has n equally spaced dots  on each side. How many triangles can be formed (of any size)?

Analysis
If there are n dots on each side then total no of dots  =n(n+1)/2
Number of combination using 3 at a time =${n(n+1)/2 \choose 3}$
but all set of 3 pts wont be ∆ some will be co linear points
like x1,x2,x3 are coliner on line AB , similarly on CD,EF n so on2
n similarly on other side of ∆ BT, GH & so on & AT ,SJ & so on
combination of 3 dots on AB = ${n \choose 3}$
on CD = ${n-1 \choose 3}$
on EF = ${n-2 \choose 3}$
so total = ${n \choose 3}+{n-1 \choose 3}+{n-2 \choose 3}....+{3 \choose 3}$
as there are 3 side
$\therefore  \ $ total number of such combination = $3\cdot\left[{n \choose 3}+{n-1 \choose 3}+{n-2 \choose 3}....+{3 \choose 3}\right]$
= $\left[3 \cdot \sum\limits_{k = 3}^n \binom{k}{3} \right]$
$\therefore  \ $ total number of triangle 
= ${n(n+1)/2 \choose 3}-\left[3 \cdot \sum\limits_{k = 3}^n \binom{k}{3} \right]$

But this is not the answer
as x1,x4,x6,x7 are also collinear
also as the length of triangle increases ,we find more co-linearity at different 
angles
 [/INST] This question has previously been considered for triangles created by points in a square lattice; see here.
As with the square case, the challenge on the triangular lattice is determining the number of collinear triples. A000938 in OEIS gives the answer for the square lattice as
$$\left(2 \;\sum _{m=2}^n\; \sum _{k=2}^n (n-k+1) (n-m+1) \gcd (k-1,m-1)\right)-\frac{1}{6} n^2 (n^2-1).$$
This can be rearranged as 
$$\left(2 \;\sum _{m=2}^n\; \sum _{k=2}^n (n-k+1) (n-m+1) (\gcd (k-1,m-1)-1)\right)+\frac{1}{3} n^2 (n-1)(n-2)$$
where each summand is the number of collinear triples with end-points offset by $(k-1,m-1)$, and the final term is the number of horizontal and vertical triples.
For the triangular lattice, a similar analysis gives the number of collinear triples as 
$$
3\; \sum _{k=2}^n \;\sum _{m=2}^k \frac{1}{2} (n-k+1) (n-k+2) (\gcd (k-1,m-1)-1).
$$
After rearranging, we get the total number of triangles in the triangular lattice as
$$
\frac{1}{2} \left(n^2+n+2\right) \binom{n+2}{4}
-\frac{3}{2}\; \sum _{k=2}^n\; \sum _{m=2}^k (n-k+1) (n-k+2) \gcd (k-1,m-1).
$$
The first few values are $0, 1, 17, 105, 407, 1216, 3036, 6696, 13428, 25005, 43861, 73277$. As commented below, this is A194131 in OEIS.
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