I'm not certain this is what you're looking for, but there is a pattern for building an index set for a regular grid of this sort. (This will be off the top of my head, so there may be errors in my description.)

I'll assume the first index is 0 rather than 1, as that's more typical and makes the math easier.

The grid has a size in cells, and a size in vertices. For each dimension the size in vertices is one greater than the size in cells. In your example the size in cells for each dimension is 5, and the size in vertices is 6.

Each cell in the grid has xy coordinates, starting with 0, 0 in the lower left (with respect to your drawing).

For any cell y coordinate, there are two rows of vertices of interest, starting at 'cellY * vertsPerRow' and '(cellY + 1) * vertsPerRow'. Finding the 4 indices for a cell might look something like this:

index0 = cellY * vertsPerRow + cellX;
index1 = index0 + 1;
index2 = (cellY + 1) * vertsPerRow + cellX;
index3 = index2 + 1;

The index arrangement is for each cell is:

2----3
|    |
|    |
0----1

You can then triangulate that however you want.

3 minutes ago, VanillaSnake21 said:

But don't I need 6 indices per 4 vertex square? Three for each triangle? 

6 indices total, but only 4 unique indices. You can see that in the first example for your diagram:

1 2 8 8 7 1

There are 6 indices there, but only 4 unique indices, that is:

1 2 7 8

It'll be the same for every cell (just with different specific indices).

If that's still not clear, I can probably provide a more concrete example.

You could do something like that, provided the indices are added to the index array in the right order. Or you could just add them to the array directly without creating duplicates, e.g.:

index0 = cellY * vertsPerRow + cellX;
index1 = index0 + 1;
index2 = (cellY + 1) * vertsPerRow + cellX;
index3 = index2 + 1;

indices.push_back(index0);
indices.push_back(index1);
indices.push_back(index3);
indices.push_back(index0);
indices.push_back(index3);
indices.push_back(index2);

Also, just for clarity, the above (or similar) should only appear in your code once. To generate the indices for all the cells in the grid, a nested loop should be used. (That may be obvious, but since your last example appears to 'hard code' the indices for the first and second cells, I thought it might be worth mentioning that a loop should be used instead.)