[Note: OP has revised the question with a slightly different barycentric indexing scheme, which makes the following no longer quite right. I'll fix it up in a minute.]

Barycentric coordinates describe points rather than triangles. So let's identify each triangle by its "top centre" point. That's its top vertex for an upward-pointing triangle, and the centre of its top edge for a downward-pointing one.

Number the rows of triangles from 0. If the big triangle has $n$ little triangles on each side, then row $k$ contains the triangles whose reference points have barycentric coords $(n-k,b,c)$ for some $b,c$. And it begins with triangle number $k^2$. (Rows have 1,3,5,... triangles; sums of odd numbers are squares.)

Then the reference points are spaced evenly with distance 1/2 along each row.

Therefore, the triangle whose reference point has barycentric coords $(a,b,c)$ is in row $n-a$ which begins with triangle $(n-a)^2$; within that row it's in (0-based) position $2c$; so its index number is $(n-a)^2+2c$.

Let's sanity-check this by looking at the first triangle whose number isn't given in the diagram above. Its reference point is at (6,3/2,1/2) and $n=8$ so its index should be $(8-6)^2+2\cdot\frac12=5$, which is correct. Looks like it works!

[Note: OP has revised the question with a slightly different barycentric indexing scheme, which makes the following no longer quite right. I'll fix it up in a minute.]

Number the rows of triangles from 0. If the big triangle has $n$ little triangles on each side, then row $k$ contains the triangles whose reference points have barycentric coords $(n-1-k,b,c)$ for some $b,c$. And it begins with triangle number $k^2$. (Rows have 1,3,5,... triangles; sums of odd numbers are squares.)

Then the (0-based) position within its row of the triangle with barycentric indices $(a,b,c)$ is $2c+\varepsilon$ where $\varepsilon$ is 0 for upward-pointing triangles and 1 for downward-pointing ones; we can distinguish these because the sum of all the barycentric indices is $n-1$ for upward triangles and $n-2$ for downward. In other words, $\varepsilon=n-1-(a+b+c)$.

Therefore, the triangle with barycentric indices $(a,b,c)$ is in row $n-1-a$ which begins with triangle $(n-1-a)^2$; within that row its (0-based) position is $2c+n-1-(a+b+c)=n-1-a-b+c$; so its index number is $(n-1-a)^2+n-1-a-b+c$.

Let's do a spot check. Triangle 7 has $(a,b,c)=(0,0,1)$ and $n-1=2$. So the index number is $(2-0)^2+2-0-0+1$ which is indeed 7. Looks like it works.

(Remark: $n-1$ is probably a nicer parameter to work with than $n$ itself.)

Barycentric coordinates describe points rather than triangles. So let's identify each triangle by its "top centre" point. That's its top vertex for an upward-pointing triangle, and the centre of its top edge for a downward-pointing one.

Number the rows of triangles from 0. If the big triangle has $n$ little triangles on each side, then row $k$ contains the triangles whose reference points have barycentric coords $(n-k,b,c)$ for some $b,c$. And it begins with triangle number $k^2$.

Then the reference points are spaced evenly with distance 1/2 along each row.

Therefore, the triangle whose reference point has barycentric coords $(a,b,c)$ is in row $n-a$ which begins with triangle $(n-a)^2$; within that row it's in (0-based) position $2c$; so its index number is $(n-a)^2+2c$.

Let's sanity-check this by looking at the first triangle whose number isn't given in the diagram above. Its reference point is at (6,3/2,1/2) and $n=8$ so its index should be $(8-6)^2+2\cdot\frac12=5$, which is correct. Looks like it works!

[Note: OP has revised the question with a slightly different barycentric indexing scheme, which makes the following no longer quite right. I'll fix it up in a minute.]

Barycentric coordinates describe points rather than triangles. So let's identify each triangle by its "top centre" point. That's its top vertex for an upward-pointing triangle, and the centre of its top edge for a downward-pointing one.

Number the rows of triangles from 0. If the big triangle has $n$ little triangles on each side, then row $k$ contains the triangles whose reference points have barycentric coords $(n-k,b,c)$ for some $b,c$. And it begins with triangle number $k^2$. (Rows have 1,3,5,... triangles; sums of odd numbers are squares.)

Then the reference points are spaced evenly with distance 1/2 along each row.

Therefore, the triangle whose reference point has barycentric coords $(a,b,c)$ is in row $n-a$ which begins with triangle $(n-a)^2$; within that row it's in (0-based) position $2c$; so its index number is $(n-a)^2+2c$.

Let's sanity-check this by looking at the first triangle whose number isn't given in the diagram above. Its reference point is at (6,3/2,1/2) and $n=8$ so its index should be $(8-6)^2+2\cdot\frac12=5$, which is correct. Looks like it works!